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# zbMATH — the first resource for mathematics

Numerical methods and asymptotic error expansions for the Emden-Fowler equations. (English) Zbl 0854.65067
The author solves the boundary value problem $y''(x) + cx^p[y(x)]^q = 0,\quad x \in (0,1), \quad y(0) = 1,\quad y(1) = 0.$ The approximate solution is obtained in two stages. In the first stage, the Picard method is applied for linearization. In the second stage, for finding in each iteration an approximate solution of the appropriate linear boundary value problem the method of grids is applied. On the basis of asymptotic expansions for the discretization error, the author uses the E-algorithm of Brezinski to accelerate the convergence of the numerical solutions obtained by the finite difference method, for the three cases $$p = -{1\over 2}$$, $$p = -1$$, $$p = -{5\over 4}$$. In the end some numerical results are presented.
Reviewer: V.Makarov (Kiev)

##### MSC:
 65L10 Numerical solution of boundary value problems involving ordinary differential equations 34B15 Nonlinear boundary value problems for ordinary differential equations 65L12 Finite difference and finite volume methods for ordinary differential equations 65L70 Error bounds for numerical methods for ordinary differential equations
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##### References:
 [1] Brezinski, C., A general extrapolation algorithm, Numer. math., 35, 175-187, (1980) · Zbl 0444.65001 [2] Brezinski, C.; Redivo-Zaglia, M., Extrapolation methods — theory and practice, (1991), North-Holland Amsterdam · Zbl 0744.65004 [3] Fedoriuk, M.V., Asymptoticheskie metody delia lineinykh obyknovennykh differentsialnykh uravnenii, (1983), Nauka Moscow, (in Russian) [4] Håvie, T., Generalized neville type extrapolation schemes, Bit, 19, 204-213, (1979) · Zbl 0404.65001 [5] Kwong, M.K., On the kolodner-coffman method for the uniqueness problem of Emden-Fowler BVP, J. appl. math. phys. (ZAMP), 41, 79-104, (1990) · Zbl 0708.34026 [6] Lima, P.M.; Graça, M.M., Convergence acceleration for boundary value problems with singularities using the E-algorithm, J. comput. appl. math., 61, 139-164, (1995) · Zbl 0840.65083 [7] Marchuk, G.I.; Shaidurov, V.V., Difference methods and their extrapolations, (1983), Springer New York · Zbl 0511.65076 [8] Mayers, D.F., The deferred approach to the limit in ordinary differential equations, Comput. J., 7, 54-57, (1964) · Zbl 0127.34003 [9] Mooney, J.W., Solution of a Thomas-Fermi problem using linear approximants, Comput. phys. comm., 76, 51-57, (1993) · Zbl 0854.65068 [10] Mooney, J.W., Numerical schemes for degenerate boundary-value problems, J. phys. A, 26, L413-L421, (1993) · Zbl 0776.65056 [11] Nayfeh, A., Introduction to perturbation techniques, (1981), Wiley New York · Zbl 0449.34001 [12] Richardson, L.; Gaunt, J., The deferred approach to the limit, Trans. roy. soc. (London), A226, 299-361, (1927) [13] Stetter, H.J., Asymptotic expansions for the error of discretization algorithms for nonlinear functional equations, Numer. math., 7, 18-31, (1965) · Zbl 0148.39003 [14] Wong, J.W., On the generalized Emden-Fowler equation, SIAM rev., 17, 339-360, (1975) · Zbl 0295.34026
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


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# Fulgencio Lopez: Banach spaces from Construction Schemes

Place: Fields Institute (Room 210)

Date: May 27th, 2016 (13:30-15:00)

Speaker: Fulgencio Lopez

Title: Banach spaces from Construction Schemes

Abstract: In 2011 J. Lopez-Abad and S. Todorcevic used forcing to construct a
Banach space with an $\varepsilon$-biorthogonal system that didn’t have $\delta$-biorthogonal systems for every $1\leq\delta<\varepsilon$. We show that there is a Banach space with the same property provided there is a capturing Construction Scheme.

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# Railways NTPC (Technical Ability) Metrology and Inspection

Metrology and Inspection

Category : Railways

Metrology and Inspection

• Metrology is the science of measurement. Metrology includes all theoretical and practical aspects of measurement.
• Metrology is defined by the International Bureau of Weights and Measures (BIPM) as "the science of measurement, embracing both experimental and theoretical determinations at any level of uncertainty in any field of science and technology.
• Metrology is a very broad field and may be divided into three basic activities, though there is considerable overlap between the activities:
• Realization of these units of measurement in practice
• Application of chains of traceability linking measurements made in practice to reference standards.
• Metrology also has three basic subfields, all of which make use of the three basic activities, though in varying proportions:

$-$Scientific or fundamental metrology

$-$Applied, technical or industrial metrology

$-$Legal metrology

• Scientific or fundamental metrology concerns the establishment of quantity systems, unit systems, units of measurement, the development of new measurement methods, realisation of measurement standards and the transfer of traceability from these standards to users in society.
• The BIPM maintains a database of the metrological calibration and measurement capabilities of various institutes around the world. These institutes, whose activities are peer-reviewed, provide the top-level reference points for metrological traceability.
• In the area of measurement the BIPM has identified nine metrology areas including length, mass and time.
• Applied, technical or industrial metrology concerns the application of measurement science to manufacturing and other processes and their use in society, ensuring the suitability of measurement instruments, their calibration and quality control of measurements.
• Although the emphasis in this area of metrology is on the measurements themselves, traceability of the calibration of the measurement devices is necessary to ensure confidence in the measurements.
• Legal metrology "concerns activities which result from statutory requirements and concern measurement, units of measurement, measuring instruments and methods of measurement and which are performed by competent
• Such statutory requirements might arise from, amongst others, the needs for protection of health, public safety, the environment, enabling taxation, protection of consumers and fair trade.
• The OIML was set up to assist in harmonising such regulations across national boundaries to ensure that legal requirements do not inhibit trade. In Europe WELMEC was established to promote cooperation on the field of legal metrology.
• A core concept in metrology is metrological traceability defined by the Joint Committee for Guides in Metrology as "property of a measurement result whereby the result can be related to a reference through a documented unbroken chain of calibrations, each contributing to the measurement uncertainty".
• Metrological traceability permits comparison of measurements, whether the result is compared to the previous result in the same laboratory, a measurement result a year ago, or to the result of a measurement performed anywhere else in the world.
• Traceability is most often obtained by calibration, establishing the relation between the indication of a measuring instrument and the value of a measurement standard. These standards are usually coordinated by national metrological institutes.
• Traceability is used to extend measurement from a method that works in one regime to a different method that works in a different regime, by calibrating the two using an overlapping range where both work.
• Traceability, accuracy, precision, systematic bias, evaluation of measurement uncertainty are critical parts of a quality management system.
• Mistakes can make measurements and counts incorrect. Even if there are no mistakes, nearly all measurements are still inexact. The term 'error' is reserved for that inexactness, also called measurement uncertainty. Among the few exact measurements are:
• The absence of the quantity being measured, such as a voltmeter with its leads shorted together: the meter should read zero exactly.
• Measurement of an accepted constant under qualifying conditions, such as the triple point of pure water: the thermometer should read 273.16 Kelvin (0.01 degrees Celsius, 32.018 degrees Fahrenheit) when qualified equipment is used correctly.
• Self-checking ratio metric measurements, such as a potentiometer: the ratio in between steps is independently adjusted and verified to be beyond influential inexactness.
• All other measurements either have to be checked to be sufficiently correct or left to chance. Metrology is the science that establishes the correctness of specific measurement situations.
• This is done by anticipating and allowing for both mistakes and error. The precise distinction between measurement error and mistakes is not settled and varies by country.
• Calibration is the process where metrology is applied to measurement equipment and processes to ensure conformity with a known standard of measurement, usually traceable to a national standards board.
• Metrology laboratories are places where both metrology and calibration work are performed. Calibration laboratories generally specialize in calibration work only.
• Both metrology and calibration laboratories must isolate the work performed from influences that might affect the work. Temperature, humidity, vibration, electrical power supply, radiated energy and other influences are often controlled.
• Generally, it is the rate of change or instability that is more detrimental than whatever value prevails.
• Calibration technicians execute calibration work. In large organizations, the work is further divided into three groups.
• Metrology technicians perform investigation work in addition to calibrations. They also apply proven principles to known situations and evaluate unexpected or contradictory results.
• Specific education in metrology was formerly limited to sub-professional work. Most of the branches of the US Military train 'enlisted-grade' technicians to meet their specific needs.
• Large industrial organizations also develop people who demonstrate aptitude in testing functions.
• Mythologists are people who perform metrology work at and above the technician levels.
• The metrology and calibration work described above is always accompanied by documentation. The documentation can be divided into two types: one related to the task, and the other related the administrative program.
• Task documentation includes calibration procedures and the collected data. Administrative program documentation includes equipment identification data, 'calibration certificates', calibration time interval information and 'as-found' or 'out-of-tolerance' notifications.
• Standards are objects or ideas that are designated as being authoritative for some accepted reason. Whatever value they possess is useful for comparison to unknowns for the purpose of establishing or confirming an assigned value based on the standard.
• The design of this comparison process for measurements is metrology. The execution of measurement comparisons for the purpose of establishing the relationship between a standard and some other measuring device is calibration.
• The ideal standard is independently reproducible without uncertainty. This is what the creators of the "metre" length standard were attempting to do in the 19th century when they defined a metre as one ten-millionth of the distance from the equator to one of the Earth’s poles.
• It was later learned that the Earth's surface is an unreliable basis for a standard, as the Earth is not spherical and it is constantly changing in shape. But the special alloy metre bars that were created and accepted in that time period standardized international length measurement until the 1950s.
• Careful calibrations allowed tolerances as small as 10 parts per million to be distributed and reproduced in metrology laboratories worldwide, regardless of whether the rest of the metric system was implemented and in spite of the shortfalls of the metre's original basis.
• Currently, five independent units of measure are internationally recognized:temperature interval, linear distance, electrical current, frequency and mass. Any measurement can be based on one or more of these measurement units.
• To supplement these five, two units of angle measurement that are also independent are recognized. For example, Ohm's law is a widely known concept in electrical study. Of the three units of measure involved, only current (ampere) is an independent unit. Voltage and resistance units are dependent on current units, as defined by ohm’s law.
• In industrial metrology, several issues beyond accuracy constrain the usability of metrology methods. These include:
• The speed with which measurements can be accomplished on parts or surfaces in the process of manufacturing, which must match the takt time of the production line.
• The completeness with which the manufactured part can be measured such as described in high-definition metrology.
• The ability of the measurement mechanism to operate reliably in a manufacturing plant environment considering temperature, vibration, dust, and a host of other potential hostile factors.
• The ability of the measurement results, as they are presented, to be assimilated by the manufacturing operators or automation in time to effectively the manufacturing process variables.

#### Other Topics

##### Notes - Metrology and Inspection

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+1 vote
208 views

Consider the following C++ function f():

unsigned int f(unsigned int n) {
unsigned int b=0;
while (n) {
b+=n & 1;
n>>1;
}
return b;
}

The function f() returns the int that represents the ____P____ in the binary representation of positive integer n, where P is

1. number of $0$’s
2. number of bits
3. number of consecutive $1$’s
4. number of $1$’s

edited | 208 views

+1 vote
Ans $4)$

Program counting number of $1.$

It is given in question, that $n$ represent, binary representation for $n.$

Now, if we check 'while' loop, $b$ is increments it's value only  when last digit of $n$ is $1.$ and then it is right sifting the value.

So, $b$ is returning nothing but number of $1's$ in $n.$
by Veteran (119k points)


text:
# The hazard rate function, an introduction

The goal of this post is to introduce the concept of hazard rate function by modifying one of the postulates of the approximate Poisson process. The rate of changes in the modified process is the hazard rate function. When a “change” in the modified Poisson process means a termination of a system (be it manufactured or biological), the notion of the hazard rate function leads to the concept of survival models. We then discuss several important examples of survival probability models that are defined by the hazard rate function. These examples include the Weibull distribution, the Gompertz distribution and the model based on the Makeham’s law.

We consider an experiment in which the occurrences of a certain type of events are counted during a given time interval or on a given physical object. Suppose that we count the occurrences of events on the interval $(0,t)$. We call the occurrence of the type of events in question a change. We assume the following three conditions:

1. The numbers of changes occurring in nonoverlapping intervals are independent.
2. The probability of two or more changes taking place in a sufficiently small interval is essentially zero.
3. The probability of exactly one change in the short interval $(t,t+\delta)$ is approximately $\lambda(t) \delta$ where $\delta$ is sufficiently small and $\lambda(t)$ is a nonnegative function of $t$.

For the lack of a better name, throughout this post, we call the above process the counting process (*). The approximate Poisson process is defined by conditions 1 and 2 and the condition that the $\lambda(t)$ in condition 3 is a constant function. Thus the process we describe here is a more general process than the Poisson process.

Though the counting process indicated here can model the number of changes occurred in a physical object or a physical interval, we focus on the time aspect by considering the counting process as models for the number of changes occurred in a time interval where a change means “termination” or ‘failure” of a system under consideration. In many applications (e.g. in actuarial science and reliability engineering), the interest is on the time until termination or failure. Thus, the distribution for the time until failure is called a survival model. The rate of change function $\lambda(t)$ indicated in condition 3 is called the hazard rate function. It is also called the failure rate function in reliability engineering. In actuarial science, the hazard rate function is known as the force of mortality.

Two random variables naturally arise from the counting process (*). One is the discrete variable $N_t$, defined as the number of changes in the time interval $(0,t)$. The other is the continuous random variable $T$, defined as the time until the occurrence of the first (or next) change.

Claim 1. Let $\displaystyle \Lambda(t)=\int_{0}^{t} \lambda(y) dy$. Then $e^{-\Lambda(t)}$ is the probability that there is no change in the interval $(0,t)$. That is, $\displaystyle P[N_t=0]=e^{-\Lambda(t)}$.

We are interested in finding the probability of zero changes in the interval $(0,y+\delta)$. By condition 1, the numbers of changes in the nonoverlapping intervals $(0,y)$ and $(y,y+\delta)$ are independent. Thus we have:

$\displaystyle P[N_{y+\delta}=0] \approx P[N_y=0] \times [1-\lambda(y) \delta] \ \ \ \ \ \ \ \ (a)$

Note that by condition 3, the probability of exactly one change in the small interval $(y,y+\delta)$ is $\lambda(y) \delta$. Thus $[1-\lambda(y) \delta]$ is the probability of no change in the interval $(y,y+\delta)$. Continuing with equation $(a)$, we have the following derivation:

$\displaystyle \frac{P[N_{y+\delta}=0] - P[N_y=0]}{\delta} \approx -\lambda(y) P[N_y=0]$

$\displaystyle \frac{d}{dy} P[N_y=0]=-\lambda(y) P[N_y=0]$

$\displaystyle \frac{\frac{d}{dy} P[N_y=0]}{P[N_y=0]}=-\lambda(y)$

$\displaystyle \int_0^{t} \frac{\frac{d}{dy} P[N_y=0]}{P[N_y=0]} dy=-\int_0^{t} \lambda(y)dy$

Integrating the left hand side and using the boundary condition of $P[N_0=0]=1$, we have:

$\displaystyle ln P[N_t=0]=-\int_0^{t} \lambda(y)dy$

$\displaystyle P[N_t=0]=e^{-\int_0^{t} \lambda(y)dy}$

Claim 2
As discussed above, let $T$ be the length of the interval that is required to observe the first change in the counting process (*). Then the following are the distribution function, survival function and pdf of $T$:

• $\displaystyle F_T(t)=\displaystyle 1-e^{-\int_0^t \lambda(y) dy}$
• $\displaystyle S_T(t)=\displaystyle e^{-\int_0^t \lambda(y) dy}$
• $\displaystyle f_T(t)=\displaystyle \lambda(t) e^{-\int_0^t \lambda(y) dy}$

In Claim 1, we derive the probability $P[N_y=0]$ for the discrete variable $N_y$ derived from the counting process (*). We now consider the continuous random variable $T$. Note that $P[T > t]$ is the probability that the first change occurs after time $t$. This means there is no change within the interval $(0,t)$. Thus $S_T(t)=P[T > t]=P[N_t=0]=e^{-\int_0^t \lambda(y) dy}$. The distribution function and density function can be derived accordingly.

Claim 3
The hazard rate function $\lambda(t)$ is equivalent to each of the following:

• $\displaystyle \lambda(t)=\frac{f_T(t)}{1-F_T(t)}$
• $\displaystyle \lambda(t)=\frac{-S_T^{'}(t)}{S_T(t)}$

Remark
Based on the condition 3 in the counting process (*), the $\lambda(t)$ is the rate of change in the counting process. Note that $\lambda(t) \delta$ is the probability of a change (e.g. a failure or a termination) in a small time interval of length $\delta$. Thus the hazard rate function can be interpreted as the failure rate at time $t$ given that the life in question has survived to time $t$. Claim 3 shows that the hazard rate function is the ratio of the density function and the survival function of the time until failure variable $T$. Thus the hazard rate function $\lambda(t)$ is the conditional density of failure at time $t$. It is the rate of failure at the next instant given that the life or system being studied has survived up to time $t$.

It is interesting to note that the function $\Lambda(t)=\int_0^t \lambda(y) dy$ defined in claim 1 is called the cumulative hazard rate function. Thus the cumulative hazard rate function is an alternative way of representing the hazard rate function (see the discussion on Weibull distribution below).

Examples of Survival Models

Exponential Distribution
In many applications, especially those for biological organisms and mechanical systems that wear out over time, the hazard rate $\lambda(t)$ is an increasing function of $t$. In other words, the older the life in question (the larger the $t$), the higher chance of failure at the next instant. For humans, the probability of a 85 years old dying in the next year is clearly higher than for a 20 years old. In a Poisson process, the rate of change $\lambda(t)=\lambda$ indicated in condition 3 is a constant. As a result, the time $T$ until the first change derived in claim 2 has an exponential distribution with parameter $\lambda$. In terms of mortality study or reliability study of machines that wear out over time, this is not a realistic model. However, if the mortality or failure is caused by random external events, this could be an appropriate model.

Weibull Distribution
This distribution is an excellent model choice for describing the life of manufactured objects. It is defined by the following cumulative hazard rate function:

$\displaystyle \Lambda(t)=\biggl(\frac{t}{\beta}\biggr)^{\alpha}$ where $\alpha > 0$ and $\beta>0$

As a result, the hazard rate function, the density function and the survival function for the lifetime distribution are:

$\displaystyle \lambda(t)=\frac{\alpha}{\beta} \biggl(\frac{t}{\beta}\biggr)^{\alpha-1}$

$\displaystyle f_T(t)=\frac{\alpha}{\beta} \biggl(\frac{t}{\beta}\biggr)^{\alpha-1} \displaystyle e^{\displaystyle -\biggl[\frac{t}{\beta}\biggr]^{\alpha}}$

$\displaystyle S_T(t)=\displaystyle e^{\displaystyle -\biggl[\frac{t}{\beta}\biggr]^{\alpha}}$

The parameter $\alpha$ is the shape parameter and $\beta$ is the scale parameter. When $\alpha=1$, the hazard rate becomes a constant and the Weibull distribution becomes an exponential distribution.

When the parameter $\alpha<1$, the failure rate decreases over time. One interpretation is that most of the defective items fail early on in the life cycle. Once they they are removed from the population, failure rate decreases over time.

When the parameter $1<\alpha$, the failure rate increases with time. This is a good candidate for a model to describe the lifetime of machines or systems that wear out over time.

The Gompertz Distribution
The Gompertz law states that the force of mortality or failure rate increases exponentially over time. It describe human mortality quite accurately. The following is the hazard rate function:

$\displaystyle \lambda(t)=\alpha e^{\beta t}$ where $\alpha>0$ and $\beta>0$.

The following are the cumulative hazard rate function as well as the survival function, distribution function and the pdf of the lifetime distribution $T$.

$\displaystyle \Lambda(t)=\int_0^t \alpha e^{\beta y} dy=\frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}$

$\displaystyle S_T(t)=\displaystyle e^{\displaystyle \frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}}$

$\displaystyle F_T(t)=\displaystyle 1-e^{\displaystyle \frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}}$

$\displaystyle f_T(t)=\displaystyle \alpha e^{\beta t} \thinspace e^{\displaystyle \frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}}$

Makeham’s Law
The Makeham’s Law states that the force of mortality is the Gompertz failure rate plus an age-indpendent component that accounts for external causes of mortality. The following is the hazard rate function:

$\displaystyle \lambda(t)=\alpha e^{\beta t}+\mu$ where $\alpha>0$, $\beta>0$ and $\mu>0$.

The following are the cumulative hazard rate function as well as the survival function, distribution function and the pdf of the lifetime distribution $T$.

$\displaystyle \Lambda(t)=\int_0^t (\alpha e^{\beta y}+\mu) dy=\frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}+\mu t$

$\displaystyle S_T(t)=\displaystyle e^{\displaystyle \frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}+\mu t}$

$\displaystyle F_T(t)=\displaystyle 1-e^{\displaystyle \frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}+\mu t}$

$\displaystyle f_T(t)=\biggl( \alpha e^{\beta t}+\mu t \biggr) \thinspace e^{\displaystyle \frac{\alpha}{\beta} e^{\beta t}-\frac{\alpha}{\beta}+\mu t}$

# Introduction to Buhlmann credibility

In this post, we continue our discussion in credibility theory. Suppose that for a particular insured (either an individual entity or a group of insureds), we have observed data $X_1,X_2, \cdots, X_n$ (the numbers of claims or loss amounts). We are interested in setting a rate to cover the claim experience $X_{n+1}$ from the next period. In two previous posts (Examples of Bayesian prediction in insurance, Examples of Bayesian prediction in insurance-continued), we discussed this estimation problem from a Bayesian perspective and presented two examples. In this post, we discuss the Buhlmann credibility model and work the same two examples using the Buhlmann method.

First, let’s further describe the setting of the problem. For a particular insured, the experience data corresponding to various exposure periods are assumed to be independent. Statistically speaking, conditional on a risk parameter $\Theta$, the claim numbers or loss amounts $X_1, \cdots, X_n,X_{n+1}$ are independent and identically distributed. Furthermore, the distribution of the risk characteristics in the population of insureds and potential insureds is represented by $\pi_{\Theta}(\theta)$. The experience (either claim numbers or loss amounts) of a particular insured with risk parameter $\Theta=\theta$ is modeled by the conditional distribution $f_{X \lvert \Theta}(x \lvert \theta)$ given $\Theta=\theta$.

The Buhlmann Credibility Estimator
Given the observations $X_1, \cdots, X_n$ in the prior exposure periods, the Buhlmann credibility estimate $C$ of the claim experience $X_{n+1}$ is

$\displaystyle C=Z \overline{X}+(1-Z)\mu$

where $Z$ is the credibility factor assigned to the observed experience data and $\mu$ is the unconditional mean $E[X]$ (the mean taken over all members of the risk parameter $\Theta$). The credibility factor $Z$ is of the form $\displaystyle Z=\frac{n}{n+K}$ where $n$ is a measure of the exposure size (it is the number of observation periods in our examples) and $\displaystyle K=\frac{E[Var[X \lvert \Theta]]}{Var[E[X \lvert \Theta]]}$. The parameter $K$ will be further explained below.

The Buhlmann credibility estimator $C$ is a linear function of the past data. Note that it is of the form:

$\displaystyle C=Z \overline{X}+(1-Z)\mu=w_0+\sum \limits_{i=1}^{n} w_i X_i$

where $w_0=(1-Z)\mu$ and $\displaystyle w_i=\frac{Z}{n}$ for $i=1, \cdots, n$.

Not only is the Buhlmann credibility estimator a linear estimator, it is the best linear estimator to the Bayesian predictive mean $E[X_{n+1} \lvert X_1, \cdots, X_n]$ and the hypothetical mean $E[X_{n+1} \lvert \Theta]$ in terms of minimizing squared error loss. In other words, the coefficients $w_i$ are obtained in such a way that the following expectations (loss functions) are minimized where the expectations are taken over all observations and/or $\Theta$ (see [1]):

$\displaystyle L_1=E\biggl( \biggl[E[X_{n+1} \lvert \Theta]-w_0-\sum \limits_{i=1}^{n} w_i X_i \biggr]^2 \biggr)$

$\displaystyle L_2=E\biggl( \biggl[E[X_{n+1} \lvert X_1, \cdots, X_n]-w_0-\sum \limits_{i=1}^{n} w_i X_i \biggr]^2 \biggr)$

The Buhlmann Method
As discussed above, the Buhlmann credibility factor $Z=\frac{n}{n+K}$ is chosen such that $C=Z \overline{X}+(1-Z) \mu$ is the best linear approximation to the Bayesian estimate of the next period’s claim experience. Now we focus on the calculation of the parameter $K$.

Conditional on the risk parameter $\Theta$, $E[X \lvert \Theta]$ is called the hypothetical mean and $Var[X \lvert \Theta]$ is called the process variance. Then $\mu=E[X]=E[E[X \lvert \Theta]]$ is the expected value of hypothetical means (the unconditional mean). The total variance of this random process is:

$\displaystyle Var[X]=E[Var[X \lvert \Theta]]+Var[E[X \lvert \Theta]]$

The first part of the total variance $E[Var[X \lvert \Theta]]$ is called the expected value of process variance (EPV) and the second part $Var[E[X \lvert \Theta]]$ is called the variance of the hypothetical means (VHM). The parameter $K$ in the Buhlmann method is simply the ratio $K=\frac{EPV}{VHM}$.

We can get an intuitive feel of this formula by considering the variability of the hypothetical means $E[X \lvert \Theta]$ across many values of the risk parameter $\Theta$. If the entire population of insureds (and potential insureds) is fairly homogeneous with respect to the risk parameter $\Theta$, then $VHM=Var[E[X \lvert \Theta]]$ does not vary a great deal and is relatively small in relation to $EPV=E[Var[X \lvert \Theta]]$. As a result, $K$ is large and $Z$ is closer to 0. This agrees with the notion that in a homogeneous population, the unconditional mean (the overall mean) is of more value as a predictor of the next period’s claim experience. On the other hand, if the population of insureds is heterogeneous with respect to the risk parameter $\Theta$, then the overall mean is of less value as a predictor of future experience and we should reply more on the experience of the particular insured. Again, the Buhlmann formula agrees with this notion. If $VHM=Var[E[X \lvert \Theta]]$ is large relative to $EPV=E[Var[X \lvert \Theta]]$, then $K$ is small and $Z$ is closer to 1.

Another attractive feature of the Buhlmann formula is that as more experience data accumulate (as $n \rightarrow \infty$), the credibility factor $Z$ approaches 1 (the experience data become more and more credible).

Example 1
In this random experiment, there are a big bowl (called B) and two boxes (Box 1 and Box 2). Bowl B consists of a large quantity of balls, 80% of which are white and 20% of which are red. In Box 1, 60% of the balls are labeled 0, 30% are labeled 1 and 10% are labeled 2. In Box 2, 15% of the balls are labeled 0, 35% are labeled 1 and 50% are labeled 2. In the experiment, a ball is selected at random from bowl B. The color of the selected ball from bowl B determines which box to use (if the ball is white, then use Box 1, if red, use Box 2). Then balls are drawn at random from the selected box (Box $i$) repeatedly with replacement and the values of the series of selected balls are recorded. The value of first selected ball is $X_1$, the value of the second selected ball is $X_2$ and so on.

Suppose that your friend performs this random experiment (you do not know whether he uses Box 1 or Box 2) and that his first selected ball is a 1 ($X_1=1$) and his second selected ball is a 2 ($X_2=2$). What is the predicted value $X_3$ of the third selected ball?

This example was solved in (Examples of Bayesian prediction in insurance) using the Bayesian approach. We now work this example in the Buhlmann approach.

The following restates the prior distribution of $\Theta$ and the conditional distribution of $X \lvert \Theta$. We denote “white ball from bowl B” by $\Theta=1$ and “red ball from bowl B” by $\Theta=2$.

$\pi_{\Theta}(1)=0.8$
$\pi_{\Theta}(2)=0.2$

$\displaystyle f_{X \lvert \Theta}(0 \lvert \Theta=1)=0.60$
$\displaystyle f_{X \lvert \Theta}(1 \lvert \Theta=1)=0.30$
$\displaystyle f_{X \lvert \Theta}(2 \lvert \Theta=1)=0.10$

$\displaystyle f_{X \lvert \Theta}(0 \lvert \Theta=2)=0.15$
$\displaystyle f_{X \lvert \Theta}(1 \lvert \Theta=2)=0.35$
$\displaystyle f_{X \lvert \Theta}(2 \lvert \Theta=2)=0.50$

The following computes the conditional means (hypothetical means) and conditional variances (process variances) and the other parameters of the Buhlmann method.

Hypothetical Means
$\displaystyle E[X \lvert \Theta=1]=0.60(0)+0.30(1)+0.10(2)=0.50$
$\displaystyle E[X \lvert \Theta=2]=0.15(0)+0.35(1)+0.50(2)=1.35$

$\displaystyle E[X^2 \lvert \Theta=1]=0.60(0)+0.30(1)+0.10(4)=0.70$
$\displaystyle E[X^2 \lvert \Theta=2]=0.15(0)+0.35(1)+0.50(4)=2.35$

Process Variances
$\displaystyle Var[X \lvert \Theta=1]=0.70-0.50^2=0.45$
$\displaystyle Var[X \lvert \Theta=2]=2.35-1.35^2=0.5275$

Expected Value of the Hypothetical Means
$\displaystyle \mu=E[X]=E[E[X \lvert \Theta]]=0.80(0.50)+0.20(1.35)=0.67$

Expected Value of the Process Variance
$\displaystyle EPV=E[Var[X \lvert \Theta]]=0.8(0.45)+0.20(0.5275)=0.4655$

Variance of the Hypothetical Means
$\displaystyle VHM=Var[E[X \lvert \Theta]]=0.80(0.50)^2+0.20(1.35)^2-0.67^2=0.1156$

Buhlmann Credibility Factor
$\displaystyle K=\frac{4655}{1156}$

$\displaystyle Z=\frac{2}{2+\frac{4655}{1156}}=\frac{2312}{6967}=0.33185$

Buhlmann Credibility Estimate
$\displaystyle C=\frac{2312}{6967} \frac{3}{2}+\frac{4655}{6967} (0.67)=\frac{6586.85}{6967}=0.9454356$

Note that the Bayesian estimate obtained in Examples of Bayesian prediction in insurance is 1.004237288. Under the Buhlmann model, the past claim experience of the insured in this example is assigned 33% weight in projecting the claim frequency in the next period.

Example 2
The number of claims $X$ generated by an insured in a potfolio of independent insurance policies has a Poisson distribution with parameter $\Theta$. In the portfolio of policies, the parameter $\Theta$ varies according to a gamma distribution with parameters $\alpha$ and $\beta$. We have the following conditional distributions of $X$ and distribution of the risk parameter $\Theta$.

$\displaystyle f_{X \lvert \Theta}(x \lvert \theta)=\frac{\theta^x e^{-\theta}}{x!}$ where $x=0,1,2, \cdots$

$\displaystyle \pi_{\Theta}(\theta)=\frac{\beta^{\alpha}}{\Gamma(\alpha)} \theta^{\alpha-1} e^{-\beta \theta}$ where $\Gamma(\cdot)$ is the gamma function.

Suppose that a particular insured in this portfolio has generated 0 and 3 claims in the first 2 policy periods. What is the Buhlmann estimate of the number of claims for this insured in period 3?

Since the conditional distribution of $X$ is Poisson, we have $E[X \lvert \Theta]=\Theta$ and $Var[X \lvert \Theta]=\Theta$. As a result, the $EPV$, $VHM$ and $K$ are:

$\displaystyle EPV=E[\Theta]=\frac{\alpha}{\beta}$

$\displaystyle VHM=Var[\Theta]=\frac{\alpha}{\beta^2}$

$\displaystyle K=\frac{EPV}{VHM}=\beta$

As a result, the credibility factor for a 2-period experience period is $Z=\frac{2}{2+\beta}$ and the Buhlmann estimate of the claim frequency in the next period is:

$\displaystyle C=\frac{2}{2+\beta} \thinspace \biggl(\frac{3}{2}\biggr)+\frac{\beta}{2+\beta} \thinspace \biggl(\frac{\alpha}{\beta}\biggr)$

To generalize the above results, suppose that we have observed $X_1=x_1, \cdots, X_n=x_n$ for this insured in the prior periods. Then the Buhlmann estimate for the claim frequency in the next period is:

$\displaystyle C=\frac{n}{n+\beta} \thinspace \biggl(\frac{\sum \limits_{i=1}^{n}x_i}{n}\biggr)+\frac{\beta}{n+\beta} \thinspace \biggl(\frac{\alpha}{\beta}\biggr)$

In this example, the Buhlmann estimate is exactly the same as the Bayesian estimate (Examples of Bayesian prediction in insurance-continued).

Reference

1. Klugman S. A., Panjer H. H., Willmot G. E., Loss Models, From Data To Decisions, Second Edition, 2004, John Wiley & Sons, Inc.

# Examples of Bayesian prediction in insurance-continued

This post is a continuation of the previous post Examples of Bayesian prediction in insurance. We present another example as an illustration of the methodology of Bayesian estimation. The example in this post, along with the example in the previous post, serve to motivate the concept of Bayeisan credibility and Buhlmann credibility theory. So these two posts are part of an introduction to credibility theory.

Suppose $X_1, \cdots, X_n,X_{n+1}$ are independent and identically distributed conditional on $\Theta=\theta$. We denote the density function of the common distribution of $X_j$ by $f_{X \lvert \Theta}(x \lvert \theta)$. We denote the prior distribution of the risk parameter $\Theta$ by $\pi_{\Theta}(\theta)$. The following shows the steps of the Bayesian estimate of the next observation $X_{n+1}$ given $X_1, \cdots, X_n$.

The Marginal Distribution
$\displaystyle f_{X_1, \cdots, X_n}(x_1, \cdots, x_n)=\int \limits_{\theta} \biggl[\prod \limits_{i=1}^{n} f_{X \lvert \Theta}(x_i \lvert \theta)\biggr] \pi_{\Theta}(\theta) d \theta$

The Posterior Distribution
$\displaystyle \pi_{\Theta \lvert X_1, \cdots, X_n}(\theta \lvert x_1, \cdots, x_n)$

$\displaystyle =\ \ \ \ \ \ \ \ \ \ \frac{1}{f_{X_1, \cdots, X_n}(x_1, \cdots, x_n)} \biggl[\prod \limits_{i=1}^{n} f_{X \lvert \Theta}(x_i \lvert \theta)\biggr] \pi_{\Theta}(\theta)$

The Predictive Distribution
$\displaystyle f_{X_{n+1} \lvert X_1, \cdots, X_n}(x \vert x_1, \cdots, x_n)$

$\displaystyle =\ \ \ \ \ \ \ \ \ \ \int \limits_{\theta} \biggl(f_{X \lvert \Theta}(x \lvert \theta)\biggr) \thinspace \pi_{\Theta \lvert X_1, \cdots, X_n}(\theta \lvert x_1, \cdots, x_n) d \theta$

The Bayesian Predictive Mean of the Next Period
$\displaystyle E[X_{n+1} \lvert X_1=x_1, \cdots, X_n=x_n]$

$\displaystyle =\ \ \ \ \ \ \ \ \ \ \int \limits_{x} x \thinspace f_{X_{n+1} \lvert X_1, \cdots, X_n}(x \vert x_1, \cdots, x_n) dx$

$\displaystyle E[X_{n+1} \lvert X_1=x_1, \cdots, X_n=x_n]$

$\displaystyle =\ \ \ \ \ \ \ \ \ \ \int \limits_{\theta} E[X \lvert \theta] \thinspace \pi_{\Theta \lvert X_1, \cdots, X_n}(\theta \lvert x_1, \cdots, x_n) d \theta$

Example 2
The number of claims $X$ generated by an insured in a potfolio of independent insurance policies has a Poisson distribution with parameter $\Theta$. In the portfolio of policies, the parameter $\Theta$ varies according to a gamma distribution with parameters $\alpha$ and $\beta$. We have the following conditional distributions of $X$ and prior distribution of $\theta$.

$\displaystyle f_{X \lvert \Theta}(x \lvert \theta)=\frac{\theta^x e^{-\theta}}{x!}$ where $x=0,1,2, \cdots$

$\displaystyle \pi_{\Theta}(\theta)=\frac{\beta^{\alpha}}{\Gamma(\alpha)} \theta^{\alpha-1} e^{-\beta \theta}$ where $\Gamma(\cdot)$ is the gamma function.

Suppose that a particular insured in this portfolio has generated 0 and 3 claims in the first 2 policy periods. What is the Bayesian estimate of the number of claims for this insured in period 3?

Note that the conditional mean $E[X \lvert \Theta]=\Theta$. Thus the unconditional mean $E[X]=E[\Theta]=\frac{\alpha}{\beta}$.

Comment
Note that the unconditional distribution of $X$ is a negative binomial distribution. In a previous post (Compound negative binomial distribution), it was shown that if $N \sim Poisson(\Lambda)$ and $\Lambda \sim Gamma(\alpha,\beta)$, then the unconditional distribution of $X$ has the following probability function. We make use of this result in the Bayesian estimation problem in this post.

$\displaystyle P[N=n]=\frac{\Gamma(\alpha+n}{\Gamma(\alpha) \Gamma(n)} \biggl[\frac{\beta}{\beta+1}\biggr]^{\alpha} \biggl[\frac{1}{\beta+1}\biggr]^n$

The Marginal Distribution
$\displaystyle f_{X_1,X_2}(0,3)=\int_{0}^{\infty} e^{-\theta} \frac{\theta^3 e^{-\theta}}{3!} \frac{\beta^{\alpha}}{\Gamma(\alpha)} \theta^{\alpha-1} e^{-\beta \theta} d \theta$

$\displaystyle =\int_{0}^{\infty} \frac{\beta^{\alpha}}{3! \Gamma(\alpha)} \theta^{\alpha+3-1} e^{\beta+2} d \theta=\frac{\Gamma(\alpha+3)}{6 \Gamma(\alpha)} \frac{\beta^{\alpha}}{(\beta+2)^{\alpha+3}}$

The Posterior Distribution
$\displaystyle \pi_{\Theta \lvert X_1,X_2}(\theta \lvert 0,3)=\frac{1}{f_{X_1,X_2}(0,3)} e^{-\theta} \frac{\theta^3 e^{-\theta}}{3!} \frac{\beta^{\alpha}}{\Gamma(\alpha)} \theta^{\alpha-1} e^{-\beta \theta}$

$\displaystyle =K \thinspace \theta^{\alpha+3-1} e^{-(\beta+2) \theta}$

In the above expression $K$ is a constant making $\pi_{\Theta \lvert X_1,X_2}(\theta \lvert 0,3)$ a density function. Note that it has the form of a gamma distribution. Thus the posterior distribution must be:

$\displaystyle \pi_{\Theta \lvert X_1,X_2}(\theta \lvert 0,3)=\frac{(\beta+2)^{\alpha+1}}{\Gamma(\alpha+3)} \thinspace \theta^{\alpha+3-1} e^{-(\beta+2) \theta}$

Thus the posterior distribution of $\Theta$ is a gamma distribution with parameter $\alpha+3$ and $\beta+2$.

The Predictive Distribution
Note that the predictive distribution is simply the mixture of $Poisson(\Theta)$ with $Gamma(\alpha+3,\beta+2)$ as mixing weights. By the comment above, the predictive distribution is a negative binomial distribution with the following probability function:

$\displaystyle f_{X_3 \lvert X_1,X_2}(x \lvert 0,3)=\frac{\Gamma(\alpha+5)}{\Gamma(\alpha+3) \Gamma(2)} \biggl[\frac{\beta+2}{\beta+3}\biggr]^{\alpha+3} \biggl[\frac{1}{\beta+3}\biggr]^{2}$

The Bayesian Predictive Mean
$\displaystyle E[X_3 \lvert 0,3]=\frac{\alpha+3}{\beta+2}=\frac{2}{\beta+2} \biggl(\frac{3}{2}\biggr)+\frac{\beta}{\beta+2} \biggl(\frac{\alpha}{\beta}\biggr) \ \ \ \ \ \ \ \ \ \ (1)$

Note that $E[X \lvert \Theta]=\Theta$. Thus the Bayesian predictive mean in this example is simply the mean of the posterior distribution of $\Theta$, which is $E[\Theta \vert 0,3]=\frac{\alpha+3}{\beta+2}$.

Comment
Generalizing the example, suppose that in the first $n$ periods, the claim counts for the insured are $X_1=x_1, \cdots, X_n=x_n$. Then the posterior distribution of the parameter $\Theta$ is a gamma distribution.

$\biggl[\Theta \lvert X_1=x_1, \cdots, X_n=x_n\biggr] \sim Gamma(\alpha+\sum_{i=1}^{n} x_i,\beta+n)$

Then the predictive distribution of $X_{n+1}$ given the observations has a negative binomial distribution. More importantly, the Bayesian predictive mean is:

$\displaystyle E[X_{n+1} \lvert X_1=x_1, \cdots, X_n=x_n]=\frac{\alpha+\sum_{i=1}^{n} x_i}{\beta+n}$

$\displaystyle =\frac{n}{\beta+n} \biggl(\frac{\sum \limits_{i=0}^{n} x_i}{n}\biggr)+\frac{\beta}{\beta+n} \biggl(\frac{\alpha}{\beta}\biggr)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

It is interesting that the Bayesian predictive mean of the $(n+1)^{th}$ period is a weighted average of the mean of the observed data ($\overline{X}$) and the unconditional mean $E[X]=\frac{\alpha}{\beta}$. Consequently, the above Bayesian estimate is a credibility estimate. The weight given to the observed data $Z=\frac{n}{\beta+n}$ is called the credibility factor. The estimate and the factor are called Bayesian credibility estimate and Bayesian credibility factor, respectively.

In general, the credibility estimate is an estimator of the following form:

$\displaystyle E=Z \thinspace \overline{X}+ (1-Z) \thinspace \mu_0$

where $\overline{X}$ is the mean of the observed data and $\mu_0$ is the mean based on other information. In our example here, $\mu_0$ is the unconditional mean. In practice, $\mu_0$ could be the mean based on the entire book of business, or a mean based on a different block of similar insurance policies. Another interpretation is that $\overline{X}$ is the mean of the recent experience data and $\mu_0$ is the mean of prior periods.

One more comment about the credibility factor $Z=\frac{n}{\beta+n}$ derived in this example. As $n \rightarrow \infty$, $Z \rightarrow 1$. This makes intuitive sense since this gives more weight to $\overline{X}$ as more and more data are accumulated.

# Examples of Bayesian prediction in insurance

We present two examples to illustrate the notion of Bayesian predictive distributions. The general insurance problem we aim to illustrate is that of using past claim experience data from an individual insured or a group of insureds to predict the future claim experience. Suppose we have $X_1,X_2, \cdots, X_n$ with each $X_i$ being the number of claims or an aggregate amount of claims in a prior period of observation. Given such results, what will be the number of claims during the next period, or what will be the aggregate claim amount in the next period? These two examples will motivate the notion of credibility, both Bayesian credibility theory and Buhlmann credibility theory. We present Example 1 in this post. Example 2 is presented in the next post (Examples of Bayesian prediction in insurance-continued).

Example 1
In this random experiment, there are a big bowl (called B) and two boxes (Box 1 and Box 2). Bowl B consists of a large quantity of balls, 80% of which are white and 20% of which are red. In Box 1, 60% of the balls are labeled 0, 30% are labeled 1 and 10% are labeled 2. In Box 2, 15% of the balls are labeled 0, 35% are labeled 1 and 50% are labeled 2. In the experiment, a ball is selected at random from bowl B. The color of the selected ball from bowl B determines which box to use (if the ball is white, then use Box 1, if red, use Box 2). Then balls are drawn at random from the selected box (Box $i$) repeatedly with replacement and the values of the series of selected balls are recorded. The value of first selected ball is $X_1$, the value of the second selected ball is $X_2$ and so on.

Suppose that your friend performs this random experiment (you do not know whether he uses Box 1 or Box 2) and that his first ball is a 1 ($X_1=1$) and his second ball is a 2 ($X_2=2$). What is the predicted value $X_3$ of the third selected ball?

Though it is straightforward to apply the Bayes’ theorem to this problem (the solution can be seen easily using a tree diagram) to obtain a numerical answer, we use this example to draw out the principle of Bayesian prediction. So it may appear that we are making a simple problem overly complicated. We are merely using this example to motivate the method of Bayesian estimation.

For convenience, we denote “draw a white ball from bowl B” by $\theta=1$ and “draw a red ball from bowl B” by $\theta=2$. Box 1 and Box 2 are conditional distributions. The Bowl B is a distribution for the parameter $\theta$. The distribution given in Bowl B is a probability distribution over the space of all parameter values (called a prior distribution). The prior distribution of $\theta$ and the conditional distributions of $X$ given $\theta$ are restated as follows:

$\pi_{\theta}(1)=0.8$
$\pi_{\theta}(2)=0.2$

$\displaystyle f_{X \lvert \Theta}(0 \lvert \theta=1)=0.60$
$\displaystyle f_{X \lvert \Theta}(1 \lvert \theta=1)=0.30$
$\displaystyle f_{X \lvert \Theta}(2 \lvert \theta=1)=0.10$

$\displaystyle f_{X \lvert \Theta}(0 \lvert \theta=2)=0.15$
$\displaystyle f_{X \lvert \Theta}(1 \lvert \theta=2)=0.35$
$\displaystyle f_{X \lvert \Theta}(2 \lvert \theta=2)=0.50$

The following shows the conditional means $E[X \lvert \theta]$ and the unconditional mean $E[X]$.

$\displaystyle E[X \lvert \theta=1]=0.6(0)+0.3(1)+0.1(2)=0.50$
$\displaystyle E[X \lvert \theta=2]=0.15(0)+0.35(1)+0.5(2)=1.35$
$\displaystyle E[X]=0.8(0.50)+0.2(1.35)=0.67$

If you know which particular box your friend is using ($\theta=1$ or $\theta=2$), then the estimate of the next ball should be $E[X \lvert \theta]$. But the value of $\theta$ is unkown to you. Another alternative for a predicted value is the unconditional mean $E[X]=0.67$. While the estimate $E[X]=0.67$ is easy to calculate, this estimate does not take the observed data ($X_1=1$ and $X_2=2$) into account and it certainly does not take the parameter $\theta$ into account. A third alternative is to incorporate the observed data into the estimate of the next ball. We now continue with the calculation of the Bayesian estimation.

Unconditional Distribution
$\displaystyle f_X(0)=0.6(0.8)+0.15(0.2)=0.51$
$\displaystyle f_X(1)=0.3(0.8)+0.35(0.2)=0.31$
$\displaystyle f_X(2)=0.1(0.8)+0.50(0.2)=0.18$

Marginal Probability
$\displaystyle f_{X_1,X_2}(1,2)=0.1(0.3)(0.8)+0.5(0.35)(0.2)=0.059$

Posterior Distribution of $\theta$
$\displaystyle \pi_{\Theta \lvert X_1,X_2}(1 \lvert 1,2)=\frac{0.1(0.3)(0.8)}{0.059}=\frac{24}{59}$

$\displaystyle \pi_{\Theta \lvert X_1,X_2}(2 \lvert 1,2)=\frac{0.5(0.35)(0.2)}{0.059}=\frac{35}{59}$

Predictive Distribution of $X$
$\displaystyle f_{X_3 \lvert X_1,X_2}(0 \lvert 1,2)=0.6 \frac{24}{59} + 0.15 \frac{35}{59}=\frac{19.65}{59}$

$\displaystyle f_{X_3 \lvert X_1,X_2}(1 \lvert 1,2)=0.3 \frac{24}{59} + 0.35 \frac{35}{59}=\frac{19.45}{59}$

$\displaystyle f_{X_3 \lvert X_1,X_2}(2 \lvert 1,2)=0.1 \frac{24}{59} + 0.50 \frac{35}{59}=\frac{19.90}{59}$

Here is another formulation of the predictive distribution of $X_3$. See the general methodology section below.
$\displaystyle f_{X_3 \lvert X_1,X_2}(0 \lvert 1,2)=\frac{0.6(0.1)(0.3)(0.8)+0.15(0.5)(0.35)(0.2)}{0.059}=\frac{19.65}{59}$

$\displaystyle f_{X_3 \lvert X_1,X_2}(1 \lvert 1,2)=\frac{0.3(0.1)(0.3)(0.8)+0.35(0.5)(0.35)(0.2)}{0.059}=\frac{19.45}{59}$

$\displaystyle f_{X_3 \lvert X_1,X_2}(2 \lvert 1,2)=\frac{0.1(0.1)(0.3)(0.8)+0.5(0.5)(0.35)(0.2)}{0.059}=\frac{19.90}{59}$

The posterior distribution $\pi_{\theta}(\cdot \lvert 1,2)$ is the conditional probability distribution of the parameter $\theta$ given the observed data $X_1=1$ and $X_2=2$. This is a result of applying the Bayes’ theorem. The predictive distribution $f_{X_3 \lvert X_1,X_2}(\cdot \lvert 1,2)$ is the conditional probability distribution of a new observation given the past observed data of $X_1=1$ and $X_2=2$. Since both of these distributions incorporate the past observations, the Bayesian estimate of the next observation is the mean of the predictive distribution.

$\displaystyle E[X_3 \lvert X_1=1,X_2=2]$

$\displaystyle =0 \thinspace f_{X_3 \lvert X_1,X_2}(0 \lvert 1,2)+1 \thinspace f_{X_3 \lvert X_1,X_2}(1 \lvert 1,2)+2 \thinspace f_{X_3 \lvert X_1,X_2}(2 \lvert 1,2)$

$\displaystyle =0 \frac{19.65}{59}+1 \frac{19.45}{59}+ 2 \frac{19.90}{59}$

$\displaystyle =\frac{59.25}{59}=1.0042372$

$\displaystyle E[X_3 \lvert X_1=1,X_2=2]$

$\displaystyle =E[X \lvert \theta=1] \medspace \pi_{\Theta \lvert X_1,X_2}(1 \lvert 1,2)+E[X \lvert \theta=2] \medspace \pi_{\Theta \lvert X_1,X_2}(2 \lvert 1,2)$

$\displaystyle =0.5 \frac{24}{59}+1.35 \frac{35}{59}=\frac{59.25}{59}$

Note that we compute the Bayesian estimate $E[X_3 \vert X_1,X_2]$ in two ways, one using the predictive distribution and the other using the posterior distribution of the parameter $\theta$. The Bayesian estimate is the mean of the hypothetical means $E[X \lvert \theta]$ with expectation taken over the entire posterior distribution $\pi_{\theta}(\cdot \lvert 1,2)$.

Discussion of General Methodology
We now use Example 1 to draw out general methodology. We first describe the discrete case and have the continuous case as a generalization.

Suppose we have a family of conditional density functions $f_{X \lvert \Theta}(x \lvert \theta)$. In Example 1, the bowl B is the distribution of the parameter $\theta$. Box 1 and Box 2 are the conditional distributions with density $f_{X \lvert \Theta}(x \lvert \theta)$. In an insurance application, the $\theta$ is a risk parameter and the conditional distribution $f_{X \lvert \Theta}(x \lvert \theta)$ is the claim experience in a given fixed period (conditional on $\Theta=\theta$).

Suppose that $X_1,X_2, \cdots, X_n,X_{n+1}$ (conditional on $\Theta=\theta$) are independent and identically distributed where the common density function is $f_{X \lvert \Theta}(x \lvert \theta)$. In our Example 1, once a box is selected (e.g. Box 1), then the repeated drawing of the balls are independent and identically distributed. In an insurance application, the $X_k$ are the claim experience from an insured (or a group of insureds) where the insured belongs to the risk class with parameter $\theta$.

We are interested in the conditional distribution of $X_{n+1}$ given $\Theta=\theta$ to predict $X_{n+1}$. In our example, $X_{n+1}$ is the value of the ball in the $(n+1)^{st}$ draw. In an insurance application, $X_{n+1}$ may be the claim experience of an insured (or a group of insureds) in the next policy period. We can use the unconditional mean $E[X]=E[E(X \lvert \Theta)]$ (the mean of the hypothetical means). This approach does not take the risk parameter of the insured into the equation. On the other hand, if we know the value of $\theta$, then we can use $f_{X \lvert \Theta}(x \lvert \theta)$. But the risk parameter is usually unknown. The natural alternative is to condition on the observed experience in the $n$ prior periods $X_1, \cdots, X_n$ rather than conditioning on the risk parameter $\theta$. Thus we derive the predictive distribution of $X_{n+1}$ given the observation $X_1, \cdots, X_n$. Given the observed experience data $X_1=x_1,X_2=x_2, \cdots, X_n=x_n$, the following is the derivation of the Bayesian predictive distribution. Note that the prior distribution of the parameter $\theta$ is $\pi_{\Theta}(\theta)$.

The Unconditional Distribution
$\displaystyle f_X(x)=\sum \limits_{\theta} f_{X \lvert \Theta}(x \lvert \theta) \ \pi_{\Theta}(\theta)$

The Marginal Distribution
$\displaystyle f_{X_1, \cdots, X_n}(x_1, \cdots, x_n)=\sum \limits_{\theta} \biggl[\prod \limits_{i=1}^{n} f_{X_i \lvert \Theta}(x_i \lvert \theta)\biggr] \pi_{\Theta}(\theta)$

The Posterior Distribution
$\displaystyle \pi_{\Theta \lvert X_1, \cdots, X_n}(\theta \lvert x_1, \cdots, x_n)$

$\displaystyle = \ \ \ \ \ \ \ \ \ \ \frac{1}{f_{X_1, \cdots, X_n}(x_1, \cdots, x_n)} \biggl[\prod \limits_{i=1}^{n} f_{X_i \lvert \Theta}(x_i \lvert \theta)\biggr] \pi_{\Theta}(\theta)$

The Predictive Distribution
$\displaystyle f_{X_{n+1} \lvert X_1, \cdots, X_n}(x \vert x_1, \cdots, x_n)$

$\displaystyle =\ \ \ \ \ \ \ \ \ \ \sum \limits_{\theta} f_{X \lvert \Theta}(x \lvert \theta) \thinspace \pi_{\Theta \lvert X_1, \cdots, X_n}(\theta \lvert x_1, \cdots, x_n)$

Another formulation is:
$\displaystyle f_{X_{n+1} \lvert X_1, \cdots, X_n}(x \vert x_1, \cdots, x_n)$

$\displaystyle =\ \ \ \ \ \ \ \ \ \ \frac{1}{f_{X_1, \cdots, X_n}(x_1, \cdots, x_n)} \sum \limits_{\theta} f_{X_{n+1} \lvert \Theta}(x \lvert \theta) \biggl[ \prod \limits_{j=1}^{n}f_{X_j \lvert \Theta}(x_j \lvert \theta)\biggr] \thinspace \pi_{\Theta}(\theta)$

The Bayesian Predictive Mean of the Next Period
$\displaystyle E[X_{n+1} \lvert X_1=x_1, \cdots, X_n=x_n]$

$\displaystyle =\ \ \ \ \ \ \ \ \ \ \sum \limits_{x} x \thinspace f_{X_{n+1} \lvert X_1, \cdots, X_n}(x \vert x_1, \cdots, x_n)$

$\displaystyle E[X_{n+1} \lvert X_1=x_1, \cdots, X_n=x_n]$

$\displaystyle =\ \ \ \ \ \ \ \ \ \ \sum \limits_{\theta} E[X \lvert \theta] \thinspace \pi_{\Theta \lvert X_1, \cdots, X_n}(\theta \lvert x_1, \cdots, x_n)$

We state the same results for the case that the claim experience $X$ is continuous.

The Unconditional Distribution
$\displaystyle f_{X}(x) = \int_{\theta} f_{X \lvert \Theta} (x \lvert \theta) \ \pi_{\Theta}(\theta) \ d \theta$

The Marginal Distribution
$\displaystyle f_{X_1, \cdots, X_n}(x_1, \cdots, x_n)=\int \limits_{\theta} \biggl[\prod \limits_{i=1}^{n} f_{X \lvert \Theta}(x_i \lvert \theta)\biggr] \pi_{\Theta}(\theta) d \theta$

The Posterior Distribution
$\displaystyle \pi_{\Theta \lvert X_1, \cdots, X_n}(\theta \lvert x_1, \cdots, x_n)$

$\displaystyle =\ \ \ \ \ \ \ \ \ \ \frac{1}{f_{X_1, \cdots, X_n}(x_1, \cdots, x_n)} \biggl[\prod \limits_{i=1}^{n} f_{X \lvert \Theta}(x_i \lvert \theta)\biggr] \pi_{\Theta}(\theta)$

The Predictive Distribution
$\displaystyle f_{X_{n+1} \lvert X_1, \cdots, X_n}(x \vert x_1, \cdots, x_n)$

$\displaystyle =\ \ \ \ \ \ \ \ \ \ \int \limits_{\theta} f_{X \lvert \Theta}(x \lvert \theta) \thinspace \pi_{\Theta \lvert X_1, \cdots, X_n}(\theta \lvert x_1, \cdots, x_n) \ d \theta$

Another formulation is:
$\displaystyle f_{X_{n+1} \lvert X_1, \cdots, X_n}(x \vert x_1, \cdots, x_n)$

$\displaystyle =\ \ \ \ \ \ \ \ \ \ \frac{1}{f_{X_1, \cdots, X_n}(x_1, \cdots, x_n)} \int \limits_{\theta} f_{X_{n+1} \lvert \Theta}(x \lvert \theta) \biggl[ \prod \limits_{j=1}^{n}f_{X_j \lvert \Theta}(x_j \lvert \theta)\biggr] \thinspace \pi_{\Theta}(\theta) \ d \theta$

The Bayesian Predictive Mean of the Next Period
$\displaystyle E[X_{n+1} \lvert X_1=x_1, \cdots, X_n=x_n]$

$\displaystyle =\ \ \ \ \ \ \ \ \ \ \int \limits_{x} x \thinspace f_{X_{n+1} \lvert X_1, \cdots, X_n}(x \vert x_1, \cdots, x_n) dx$

$\displaystyle E[X_{n+1} \lvert X_1=x_1, \cdots, X_n=x_n]$

$\displaystyle =\ \ \ \ \ \ \ \ \ \ \int \limits_{\theta} E[X \lvert \theta] \thinspace \pi_{\Theta \lvert X_1, \cdots, X_n}(\theta \lvert x_1, \cdots, x_n) d \theta$

See the next post (Examples of Bayesian prediction in insurance-continued) for Example 2.

# Compound mixed Poisson distribution

Let the random sum $Y=X_1+X_2+ \cdots +Y_N$ be the aggregate claims generated in a fixed period by an independent group of insureds. When the number of claims $N$ follows a Poisson distribution, the sum $Y$ is said to have a compound Poisson distribution. When the number of claims $N$ has a mixed Poisson distribution, the sum $Y$ is said to have a compound mixed Poisson distribution. A mixed Poisson distribution is a Poisson random variable $N$ such that the Poisson parameter $\Lambda$ is uncertain. In other words, $N$ is a mixture of a family of Poisson distributions $N(\Lambda)$ and the random variable $\Lambda$ specifies the mixing weights. In this post, we present several basic properties of compound mixed Poisson distributions. In a previous post (Compound negative binomial distribution), we showed that the compound negative binomial distribution is an example of a compound mixed Poisson distribution (with gamma mixing weights).

In terms of notation, we have:

• $Y=X_1+X_2+ \cdots +Y_N$,
• $N \sim$ Poisson$(\Lambda)$,
• $\Lambda \sim$ some unspecified distribution.

The following presents basic proeprties of the compound mixed Poisson $Y$ in terms of the mixing weights $\Lambda$ and the claim amount random variable $X$.

Mean and Variance

$\displaystyle E[Y]=E[\Lambda] E[X]$

$\displaystyle Var[Y]=E[\Lambda] E[X^2]+Var[\Lambda] E[X]^2$

Moment Generating Function

$\displaystyle M_Y(t)=M_{\Lambda}[M_X(t)-1]$

Cumulant Generating Function

$\displaystyle \Psi_Y(t)=ln M_{\Lambda}[M_X(t)-1]=\Psi_{\Lambda}[M_X(t)-1]$

Measure of Skewness
$\displaystyle E[(Y-\mu_Y)^3]=\Psi_Y^{(3)}(0)$

$\displaystyle =\Psi_{\Lambda}^{(3)}(0) E[X]^3 + 3 \Psi_{\Lambda}^{(2)}(0) E[X] E[X^2]+\Psi_{\Lambda}^{(1)}(0) E[X^3]$

$\displaystyle =\gamma_{\Lambda} Var[\Lambda]^{\frac{3}{2}} E[X]^3 + 3 Var[\Lambda] E[X] E[X^2]+E[\Lambda] E[X^3]$

Measure of skewness: $\displaystyle \gamma_Y=\frac{E[(Y-\mu_Y)^3]}{(Var[Y])^{\frac{3}{2}}}$

Previous Posts on Compound Distributions

# Compound Poisson distribution-discrete example

We present a discrete example of a compound Poisson distribution. A random variable $Y$ has a compound distribution if $Y=X_1+ \cdots +X_N$ where the number of terms $N$ is a discrete random variable whose support is the set of all nonnegative integers (or some appropriate subset) and the random variables $X_i$ are identically distributed (let $X$ be the common distribution). We further assume that the random variables $X_i$ are independent and each $X_i$ is independent of $N$. When $N$ follows the Poisson distribution, $Y$ is said to have a compound Poisson distribution. When the common distribution for the $X_i$ is continuous, $Y$ is a mixed distribution if $P[N=0]$ is nonzero. When the common distribution for the $X_i$ is discrete, $Y$ is a discrete distribution. In this post we present an example of a compound Poisson distribution where the common distribution $X$ is discrete. The compound distribution has a natural insurance interpretation (see the following links).

General Discussion
In general, the distribution function of a compound Poisson random variable $Y$ is the weighted average of all the $n^{th}$ convolutions of the common distribution function of the individual claim amount $X$. The following shows the form of such a distribution function:

$\displaystyle F_Y(y)=\sum \limits_{n=0}^{\infty} F^{*n}(y) P[N=n]$

where $\displaystyle F$ is the common distribution of the $X_n$ and $F^{*n}$ is the $n^{th}$ convolution of $F$.

If the distribution of the individual claim $X$ is discrete, we can obtain the probability mass function of $Y$ by convolutions as follows:

$\displaystyle f_Y(y)=P[Y=y]=\sum \limits_{n=0}^{\infty} p^{*n}(y) P[N=n]$

where $\displaystyle p^{*1}(y)=P[X=y]$
and $\displaystyle p^{*n}=p^* \cdots p^{*}(x)=P[X_1+X_2+ \cdots +X_n=y]$
and $\displaystyle p^{*0}(y)=\left\{\begin{matrix}0&\thinspace y \ne 0\\{1}&\thinspace x=0\end{matrix}\right.$

Example
Suppose the number of claims generated by a portfolio of insurance policies over a fixed time period has a Poisson distribution with parameter $\lambda$. Individual claim amounts will be 1 or 2 with probabilities 0.6 and 0.4, respectively. For the compound Poisson aggregate claims $Y=X_1+ \cdots +X_N$, find $P[Y=k]$ for $k=0,1,2,3,4$.

The probability mass function of $N$ is: $\displaystyle f_N(n)=\frac{\lambda^n e^{-\lambda}}{n!}$ where $n=0,1,2, \cdots$. The individual claim amounnt $X$ has a Bernoulli distribution since it is a two-valued discrete random variable. For convenience, we let $p=0.4$ (i.e. we consider $X=2$ is a success). Then the sum $X_1+ \cdots + X_n$ has a Binomial distribution. Consequently, the $n^{th}$ convolution $p^{*n}$ is simply the distribution function of Binomial(n,p). The following shows $p^{*n}$ for $n=1,2,3,4$.

$\displaystyle p^{*1}(1)=0.6, \thinspace p^{*1}(2)=0.4$

$\displaystyle p^{*2}(2)=\binom{2}{0} (0.4)^0 (0.6)^2=0.36$
$\displaystyle p^{*2}(3)=\binom{2}{1} (0.4)^1 (0.6)^1=0.48$
$\displaystyle p^{*2}(4)=\binom{2}{2} (0.4)^2 (0.6)^0=0.16$

$\displaystyle p^{*3}(3)=\binom{3}{0} (0.4)^0 (0.6)^3=0.216$
$\displaystyle p^{*3}(4)=\binom{3}{1} (0.4)^1 (0.6)^2=0.432$
$\displaystyle p^{*3}(5)=\binom{3}{2} (0.4)^2 (0.6)^1=0.288$
$\displaystyle p^{*3}(6)=\binom{3}{3} (0.4)^3 (0.6)^0=0.064$

$\displaystyle p^{*4}(4)=\binom{4}{0} (0.4)^0 (0.6)^4=0.1296$
$\displaystyle p^{*4}(5)=\binom{4}{1} (0.4)^1 (0.6)^3=0.3456$
$\displaystyle p^{*4}(6)=\binom{4}{2} (0.4)^2 (0.6)^2=0.3456$
$\displaystyle p^{*4}(7)=\binom{4}{3} (0.4)^3 (0.6)^1=0.1536$
$\displaystyle p^{*4}(8)=\binom{4}{4} (0.4)^4 (0.6)^0=0.0256$

Since we are interested in finding $P[Y=y]$ for $y=0,1,2,3,4$, we only need to consider $N=0,1,2,3,4$. The following matrix shows the relevant values of $p^{*n}$. The rows are for $y=0,1,2,3,4$. The columns are $p^{*0}$, $p^{*1}$, $p^{*2}$, $p^{*3}$, $p^{*4}$.

$\displaystyle \begin{pmatrix} 1&0&0&0&0 \\{0}&0.6&0&0&0 \\{0}&0.4&0.36&0&0 \\{0}&0&0.48&0.216&0 \\{0}&0&0.16&0.432&0.1296\end{pmatrix}$

To obtain the probability mass function of $Y$, we simply multiply each row by $P[N=n]$ where $n=0,1,2,3,4$.

$\displaystyle P[Y=0]=e^{-\lambda}$
$\displaystyle P[Y=1]=0.6 \lambda e^{-\lambda}$
$\displaystyle P[Y=2]=0.4 \lambda e^{-\lambda}+0.36 \frac{\lambda^2 e^{-\lambda}}{2}$
$\displaystyle P[Y=3]=0.48 \frac{\lambda^2 e^{-\lambda}}{2}+0.216 \frac{\lambda^3 e^{-\lambda}}{6}$
$\displaystyle P[Y=4]=0.16 \frac{\lambda^2 e^{-\lambda}}{2}+0.432 \frac{\lambda^3 e^{-\lambda}}{6}+0.1296 \frac{\lambda^4 e^{-\lambda}}{24}$

# Compound Poisson distribution

The compound distribution is a model for describing the aggregate claims arised in a group of independent insureds. Let $N$ be the number of claims generated by a portfolio of insurance policies in a fixed time period. Suppose $X_1$ is the amount of the first claim, $X_2$ is the amount of the second claim and so on. Then $Y=X_1+X_2+ \cdots + X_N$ represents the total aggregate claims generated by this portfolio of policies in the given fixed time period. In order to make this model more tractable, we make the following assumptions:

• $X_1,X_2, \cdots$ are independent and identically distributed.
• Each $X_i$ is independent of the number of claims $N$.

The number of claims $N$ is associated with the claim frequency in the given portfolio of policies. The common distribution of $X_1,X_2, \cdots$ is denoted by $X$. Note that $X$ models the amount of a random claim generated in this portfolio of insurance policies. See these two posts for an introduction to compound distributions (An introduction to compound distributions, Some examples of compound distributions).

When the claim frequency $N$ follows a Poisson distribution with a constant parameter $\lambda$, the aggreagte claims $Y$ is said to have a compound Poisson distribution. After a general discussion of the compound Poisson distribution, we discuss the property that an independent sum of compound Poisson distributions is also a compound Poisson distribution. We also present an example to illustrate basic calculations.

Compound Poisson – General Properties

Distribution Function
$\displaystyle F_Y(y)=\sum \limits_{n=0}^{\infty} F^{*n}(y) \frac{\lambda^n e^{-\lambda}}{n!}$

where $\lambda=E[N]$, $F$ is the common distribution function of $X_i$ and $F^{*n}$ is the n-fold convolution of $F$.

Mean and Variance
$\displaystyle E[Y]=E[N] E[X]= \lambda E[X]$

$\displaystyle Var[Y]=\lambda E[X^2]$

Moment Generating Function and Cumulant Generating Function
$\displaystyle M_Y(t)=e^{\lambda (M_X(t)-1)}$

$\displaystyle \Psi_Y(t)=ln M_Y(t)=\lambda (M_X(t)-1)$

Note that the moment generating function of the Poisson $N$ is $M_N(t)=e^{\lambda (e^t - 1)}$. For a compound distribution $Y$ in general, $M_Y(t)=M_N[ln M_X(t)]$.

Skewness
$\displaystyle E[(Y-\mu_Y)^3]=\Psi_Y^{(3)}(0)=\lambda E[X^3]$

$\displaystyle \gamma_Y=\frac{E[(Y-\mu_Y)^3]}{Var[Y]^{\frac{3}{2}}}=\frac{1}{\sqrt{\lambda}} \frac{E[X^3]}{E[X^2]^{\frac{3}{2}}}$

Independent Sum of Compound Poisson Distributions
First, we state the results. Suppose that $Y_1,Y_2, \cdots, Y_k$ are independent random variables such that each $Y_i$ has a compound Poisson distribution with $\lambda_i$ being the Poisson parameter for the number of claim variable and $F_i$ being the distribution function for the individual claim amount. Then $Y=Y_1+Y_2+ \cdots +Y_k$ has a compound Poisson distribution with:

• the Poisson parameter: $\displaystyle \lambda=\sum \limits_{i=1}^{k} \lambda_i$
• the distribution function: $\displaystyle F_Y(y)=\sum \limits_{i=1}^{k} \frac{\lambda_i}{\lambda} \thinspace F_i(y)$

The above result has an insurance interpretation. Suppose we have $k$ independent blocks of insurance policies such that the aggregate claims $Y_i$ for the $i^{th}$ block has a compound Poisson distribution. Then $Y=Y_1+Y_2+ \cdots +Y_k$ is the aggregate claims for the combined block during the fixed policy period and also has a compound Poisson distribution with the parameters stated in the above two bullet points.

To get a further intuitive understanding about the parameters of the combined block, consider $N_i$ as the Poisson number of claims in the $i^{th}$ block of insurance policies. It is a well known fact in probability theory (see [1]) that the indpendent sum of Poisson variables is also a Poisson random variable. Thus the total number of claims in the combined block is $N=N_1+N_2+ \cdots +N_k$ and has a Poisson distribution with parameter $\lambda=\lambda_1 + \cdots + \lambda_k$.

How do we describe the distribution of an individual claim amount in the combined insurance block? Given a claim from the combined block, since we do not know which of the constituent blocks it is from, this suggests that an individual claim amount is a mixture of the individual claim amount distributions from the $k$ blocks with mixing weights $\displaystyle \frac{\lambda_1}{\lambda},\frac{\lambda_2}{\lambda}, \cdots, \frac{\lambda_k}{\lambda}$. These mixing weights make intuitive sense. If insurance bock $i$ has a higher claim frequency $\lambda_i$, then it is more likely that a randomly selected claim from the combined block comes from block $i$. Of course, this discussion is not a proof. But looking at the insurance model is a helpful way of understanding the independent sum of compound Poisson distributions.

To see why the stated result is true, let $M_i(t)$ be the moment generating function of the individual claim amount in the $i^{th}$ block of policies. Then the mgf of the aggregate claims $Y_i$ is $\displaystyle M_{Y_i}(t)=e^{\lambda_i (M_i(t)-1)}$. Consequently, the mgf of the independent sum $Y=Y_1+ \cdots + Y_k$ is:

$\displaystyle M_Y(t)=\prod \limits_{i=0}^{k} e^{\lambda_i (M_i(t)-1)}= e^{\sum \limits_{i=0}^{k} \lambda_i(M_i(t)-1)} \displaystyle = e^{\lambda \biggl[\sum \limits_{i=0}^{k} \frac{\lambda_i}{\lambda} M_i(t) - 1 \biggr]}$

The mgf of $Y$ has the form of a compound Poisson distribution where the Poisson parameter is $\lambda=\lambda_1 + \cdots + \lambda_k$. Note that the component $\displaystyle \sum \limits_{i=0}^{k} \frac{\lambda_i}{\lambda}M_i(t)$ in the exponent is the mgf of the claim amount distribution. Since it is the weighted average of the individual claim amount mgf’s, this indicates that the distribution function of $Y$ is the mixture of the distribution functions $F_i$.

Example
Suppose that an insurance company acquired two portfolios of insurance policies and combined them into a single block. For each portfolio the aggregate claims variable has a compound Poisson distribution. For one of the portfolios, the Poisson parameter is $\lambda_1$ and the individual claim amount has an exponential distribution with parameter $\delta_1$. The corresponding Poisson and exponential parameters for the other portfolio are $\lambda_2$ and $\delta_2$, respectively. Discuss the distribution for the aggregate claims $Y=Y_1+Y_2$ of the combined portfolio.

The aggregate claims $Y$ of the combined portfolio has a compound Poisson distribution with Poisson parameter $\lambda=\lambda_1+\lambda_2$. The amount of a random claim $X$ in the combined portfolio has the following distribution function and density function:

$\displaystyle F_X(x)=\frac{\lambda_1}{\lambda} (1-e^{-\delta_1 x})+\frac{\lambda_2}{\lambda} (1-e^{-\delta_2 x})$

$\displaystyle f_X(x)=\frac{\lambda_1}{\lambda} (\delta_1 \thinspace e^{-\delta_1 x})+\frac{\lambda_2}{\lambda} (\delta_2 \thinspace e^{-\delta_2 x})$

The rest of the discussion mirrors the general discussion earlier in this post.

Distribution Function
As in the general case, $\displaystyle F_Y(y)=\sum \limits_{n=0}^{\infty} F^{*n}(y) \frac{\lambda^n e^{-\lambda}}{n!}$

where $\lambda=\lambda_1 +\lambda_2$, $F=F_X$ and $F^{*n}$ is the n-fold convolution of $F_X$.

Mean and Variance
$\displaystyle E[Y]=\frac{\lambda_1}{\delta_1}+\frac{\lambda_2}{\delta_2}$

$\displaystyle Var[Y]=\frac{2 \lambda_1}{\delta_1^2}+\frac{2 \lambda_2}{\delta_2^2}$

Moment Generating Function and Cumulant Generating Function
To obtain the mgf and cgf of the aggregate claims $Y$, consider $\lambda [M_X(t)-1]$. Note that $M_X(t)$ is the weighted average of the two exponential mgfs of the two portfolios of insurance policies. Thus we have:

$\displaystyle M_X(t)=\frac{\lambda_1}{\lambda} \frac{\delta_1}{\delta_1 - t}+\frac{\lambda_2}{\lambda} \frac{\delta_2}{\delta_2 - t}$

$\displaystyle \lambda [M_X(t)-1]=\frac{\lambda_1 t}{\delta_1 - t}+\frac{\lambda_2 t}{\delta_2 - t}$

$\displaystyle M_Y(t)=e^{\lambda (M_X(t)-1)}=e^{\frac{\lambda_1 t}{\delta_1 - t}+\frac{\lambda_2 t}{\delta_2 - t}}$

$\displaystyle \Psi_Y(t)=\frac{\lambda_1 t}{\delta_1 -t}+\frac{\lambda_2 t}{\delta_2 -t}$

Skewness
Note that $\displaystyle E[(Y-\mu_Y)^3]=\Psi_Y^{(3)}(0)=\frac{6 \lambda_1}{\delta_1^3}+\frac{6 \lambda_2}{\delta_2^3}$

$\displaystyle \gamma_Y=\displaystyle \frac{\frac{6 \lambda_1}{\delta_1^3}+\frac{6 \lambda_2}{\delta_2^3}}{(\frac{2 \lambda_1}{\delta_1^2}+\frac{2 \lambda_2}{\delta_2^2})^{\frac{3}{2}}}$

Reference

1. Hogg R. V. and Tanis E. A., Probability and Statistical Inference, Second Edition, Macmillan Publishing Co., New York, 1983.

# Some examples of compound distributions

We present two examples of compound distributions to illustrate the general formulas presented in the previous post (An introduction to compound distributions).

For the examples below, let $N$ be the number of claims generated by either an individual insured or a group of independent insureds. Let $X$ be the individual claim amount. We consider the random sum $Y=X_1+ \cdots + X_N$. We discuss the following properties of the aggregate claims random variable $Y$:

1. The distribution function $F_Y$
2. The mean and higher moments: $E[Y]$ and $E[Y^n]$
3. The variance: $Var[Y]$
4. The moment generating function and cumulant generating function:$M_Y(t)$ and $\Psi_Y(t)$.
5. Skewness: $\gamma_Y$.

Example 1
The number of claims for an individual insurance policy in a policy period is modeled by the binomial distribution with parameter $n=2$ and $p$. The individual claim, when it occurs, is modeled by the exponential distribution with parameter $\lambda$ (i.e. the mean individual claim amount is $\frac{1}{\lambda}$).

The distribution function $F_Y$ is the weighted average of a point mass at $y=0$, the exponential distribution and the Erlang-2 distribution function. For $x \ge 0$, we have:

$\displaystyle F_Y(x)=(1-p)^2+2p(1-p)(1-e^{-\lambda x})+p^2(1-\lambda x e^{-\lambda x}-e^{-\lambda x})$

The mean and variance are are follows:

$\displaystyle E[Y]=E[N] \thinspace E[X]=\frac{2p}{\lambda}$

$\displaystyle Var[Y]=E[N] \thinspace Var[X]+Var[N] \thinspace E[X]^2$

$\displaystyle =\frac{2p}{\lambda^2}+\frac{2p(1-p)}{\lambda^2}=\frac{4p-2p^2}{\lambda^2}$

The following calculates the higher moments:

$\displaystyle E[Y^n]=(1-p)^2 0 + 2p(1-p) \frac{n!}{\lambda^n}+p^2 \frac{(n+1)!}{\lambda^n}$

$\displaystyle = \frac{2p(1-p)n!+p^2(n+1)!}{\lambda^n}$

The moment generating function $M_Y(t)=M_N[ln \thinspace M_X(t)]$. So we have:

$\displaystyle M_Y(t)=\biggl(1-p+p \frac{\lambda}{\lambda -t}\biggr)^2$

$\displaystyle =(1-p)^2+2p(1-p) \frac{\lambda}{\lambda -t}+p^2 \biggl(\frac{\lambda}{\lambda -t}\biggr)^2$

Note that $\displaystyle M_N(t)=(1-p+p e^{t})^2$ and $\displaystyle M_X(t)=\frac{\lambda}{\lambda -t}$.

For the cumulant generating function, we have:

$\displaystyle \Psi_Y(t)=ln M_Y(t)=2 ln\biggl(1-p+p \frac{\lambda}{\lambda -t}\biggr)$

For the measure of skewness, we rely on the cumulant generating function. Finding the third derivative of $\Psi_Y(t)$ and then evaluate at $t=0$, we have:

$\displaystyle \Psi_Y^{(3)}(0)=\frac{12p-12p^2+4p^3}{\lambda^3}$

$\displaystyle \gamma_Y=\frac{\Psi_Y^{(3)}(0)}{Var(Y)^{\frac{3}{2}}}=\frac{12p-12p^2+4p^3}{(4p-2p^2)^{\frac{3}{2}}}$

Example 2
In this example, the number of claims $N$ follows a geometric distribution. The individual claim amount $X$ follows an exponential distribution with parameter $\lambda$.

One of the most interesting facts about this example is the moment generating function. Note that $\displaystyle M_N(t)=\frac{p}{1-(1-p)e^t}$. The following shows the derivation of $M_Y(t)$:

$\displaystyle M_Y(t)=M_N[ln \thinspace M_X(t)]=\frac{p}{1-(1-p) e^{ln M_X(t)}}$

$\displaystyle =\frac{p}{1-(1-p) \frac{\lambda}{\lambda -t}}=\cdots=p+(1-p) \frac{\lambda p}{\lambda p-t}$

The moment generating function is the weighted average of a point mass at $y=0$ and the mgf of an exponential distribution with parameter $\lambda p$. Thus this example of compound geometric distribution is equivalent to a mixture of a point mass and an exponential distribution. We make use of this fact and derive the following basic properties.

Distribution Function
$\displaystyle F_Y(y)=p+(1-p) (1-e^{\lambda p y})=1-(1-p) e^{-\lambda p y}$ for $y \ge 0$

Density Function
$\displaystyle f_Y(y)=\left\{\begin{matrix}p&\thinspace y=0\\{(1-p) \lambda p e^{-\lambda p y}}&\thinspace 0 < y\end{matrix}\right.$

Mean and Higher Moments
$\displaystyle E[Y]=(1-p) \frac{1}{\lambda p}=\frac{1-p}{p} \frac{1}{\lambda}=E[N] E[X]$

$\displaystyle E[Y^n]=p 0 + (1-p) \frac{n!}{(\lambda p)^n}=(1-p) \frac{n!}{(\lambda p)^n}$

Variance
$\displaystyle Var[Y]=\frac{2(1-p)}{\lambda^2 p^2}-\frac{(1-p)^2}{\lambda^2 p^2}=\frac{1-p^2}{\lambda^2 p^2}$

Cumulant Generating Function
$\displaystyle \Psi_Y(t)=ln \thinspace M_Y(t)=ln\biggl(p+(1-p) \frac{\lambda p}{\lambda p-t}\biggr)$

Skewness
$\displaystyle E\biggl[\biggl(Y-\mu_Y\biggr)^3\biggr]=\Psi_Y^{(3)}(0)=\frac{2-2p^3}{\lambda^3 p^3}$

$\displaystyle \gamma_Y=\frac{\Psi_Y^{(3)}(0)}{(Var[Y])^{\frac{3}{2}}}=\frac{2-2p^3}{(1-p^2)^{\frac{3}{2}}}$

# More insurance examples of mixed distributions

Four posts have already been devoted to describing three models for “per loss” insurance payout. These are mixed distributions modeling the amount the insurer pays out for each random loss. They can also be viewed as mixtures. We now turn our attention to the mixed distributions modeling the “per period” payout for an insurance policy. That is, the mixed distributions we describe here are to model the total amount of losses paid out for each insurance policy in a given policy period. This involves the uncertain random losses as well as uncertain claim frequency. In other words, there is a possiblity of having no losses. When there are losses in a policy period, the number of losses can be uncertain (there can be only one loss or multiple losses). The links to the previous posts on mixed distributions are found at the end of this post.

The following is the general setting of the insurance problem we discuss in this post.

1. The random variable $X$ is the size of the random loss that is covered in an insurance contract. We assumme that $X$ is a continuous random variable. Naturally, the support of $X$ is the set of nonnegative numbers (or some appropriate subset).
2. Let $Z$ be the “per loss” payout paid to the insured by the insurer. The variable $Z$ could refect the coverage modification such as deductible and/or policy cap or other policy provisions that are applicable in the insurance contract.
3. Let $N$ be the number of claims in a given policy period. In this post, we assume that $N$ has only two possibilities: $N=0$ or $N=1$. In other words, each policy has at most one claim in a period. Let $p=P[N=1]$.
4. Let $Y$ be the total amount paid to the insured by the insurer during a fixed policy period.

The total claims variable $Y$ is the mixture of $Y \lvert N=0$ and $Y \lvert N=1$. The conditional variable $Y \lvert N=0$ is a point mass representing “no loss”. On the other hand, we assume that $[Y \lvert N=1]=Z$. Thus $Y$ is a mixture of a point mass at the origin and the “per loss” payout variable $Z$

We first have a general discussion of the stated insurance setting. Then we discuss several different cases based on four coverage modifications that can be applied in the insurance contract. In each case, we illustrate with the exponential distribution. The four cases are:

• Case 1. $Z=X$. There is no coverage modification. The insurer pays the entire loss amount.
• Case 2. The insurance contract has a cap and the cap amount is $m$.
• Case 3. The insurance contract is an excess-of-loss policy. The deductible amount is $d$.
• Case 4. The insurance contract has a deductible $d$ and a policy cap $m$ where $d.

General Discussion
The total payout $Y$ is the mixture of a point mass at $y=0$ and the “per loss” payout $Z$. The following is the distribution $F_Y(y)$:

$\displaystyle F_Y(y)=(1-p) \thinspace F_U(y)+p \thinspace F_Z(y)$ where $\displaystyle F_U(x)=\left\{\begin{matrix}0&\thinspace x<0\\{1}&\thinspace 0 \le x\end{matrix}\right.$

Since the distribution of $Y$ is a mixture, we have a wealth of information available for us. For example, the following lists the mean, higher moments, variance, the moment generating function and the skewness.

• $\displaystyle E[Y]=p \thinspace E[Z]$
• $\displaystyle E[Y^n]=p \thinspace E[Z^n]$ for al integers $n>1$
• $\displaystyle Var[Y]=pE[Z^2]-p^2 E[Z]^2$
• $\displaystyle M_Y(t)=(1-p)+p \thinspace M_Z(t)$
• $\displaystyle \gamma_Y=p \thinspace \gamma_Z$

The Derivations:
$\displaystyle E[Y]=(1-p) \thinspace 0+p \thinspace E[Z]=p \thinspace E[Z]$

$\displaystyle E[Y^n]=(1-p) \thinspace 0^n+p \thinspace E[Z^n]=p \thinspace E[Z^n]$ for al integers $n>1$

$\displaystyle Var[Y]=E[Y^2]-E[Y]^2=pE[Z^2]-p^2 E[Z]^2$

$\displaystyle M_Y(t)=(1-p) \thinspace e^0+p \thinspace M_Z(t)=(1-p)+p \thinspace M_Z(t)$

$\displaystyle \gamma_Y=(1-p) \thinspace 0+p \thinspace \gamma_Z=p \thinspace \gamma_Z$

The following is another way to derive $Var[Y]$ using the total variance formula:

$\displaystyle Var[Y]=E_N[Var(Y \lvert N)]+Var_N[E(Y \lvert N)]$

$\displaystyle =(1-p)0+pVar[Z] + E_N[E(Y \lvert N)^2]-E_N[E(Y \lvert N)]^2$

$\displaystyle =pVar[Z] + (1-p)0^2+pE[Z]^2-p^2 E[Z]^2$

$\displaystyle =pE[Z^2]-p E[Z]^2 +pE[Z]^2-p^2 E[Z]^2$

$\displaystyle =pE[Z^2]-p^2 E[Z]^2$

The above derivations are based on the idea of mixtures. The two conditional variables are $Y \lvert N=0$ and $\lbrace{Y \lvert N=1}\rbrace=Z$. The mixing weights are $P[N=0]$ and $P[N=1]$. For more basic information on distributions that are mixtures, see this post (Basic properties of mixtures).

We now discuss the four specific cases based on the variations on the coverage modifications that can be placed on the “per loss” variable $Z$.

Case 1
This is the case that the insurance policy has no coverage modification. The insurer pays the entire random loss. Thus $Z=X$. The following is the payout rule of $Y$:

$\displaystyle Y=\left\{\begin{matrix}0&\thinspace \text{no loss occurs}\\{Z=X}&\thinspace \text{a loss occurs}\end{matrix}\right.$

This is a mixed distribution consisting of a point mass at the origin (no loss) and the random loss $X$. In this case, the “per loss” variable $Z=X$. Thus $Y$ is a mixture of of the following two distributions.

$\displaystyle F_U(x)=\left\{\begin{matrix}0&\thinspace x<0\\{1}&\thinspace 0 \le x\end{matrix}\right.$

$\displaystyle F_Z(x)=\left\{\begin{matrix}0&\thinspace x<0\\{F_X(x)}&\thinspace 0 \le x\end{matrix}\right.$

Case 1 – Distribution Function
The following shows $F_Y$ as a mixture, the explicit rule of $F_Y$ and the density of $Y$.

$\displaystyle F_Y(x)=(1-p) \thinspace F_U(x) + p \thinspace F_Z(x)$.

$\displaystyle F_Y(x)=\left\{\begin{matrix}0&\thinspace x<0\\{1-p+p \thinspace F_X(x)}&\thinspace 0 \le x\end{matrix}\right.$

$\displaystyle f_Y(x)=\left\{\begin{matrix}1-p&\thinspace x=0\\{p \thinspace f_X(x)}&\thinspace 0 < x\end{matrix}\right.$

Case 1 – Basic Properties
Using basic properties of mixtures stated in the general case, we obtain the following:

$\displaystyle E[Y]=p \thinspace E[X]$

$\displaystyle E[Y^n]=p \thinspace E[X^n]$ for all integers $n>1$

$\displaystyle Var[Y]=p \thinspace E[X^2] - p^2 E[X]^2$

$\displaystyle M_Y(t)=1-p + p \thinspace M_X(t)$

$\displaystyle \gamma_Y=p \thinspace \gamma_X$

Case 1 – Exponential Example
If the unmodified random loss has an exponential distribution, we have the following results:

$\displaystyle E[Y]=\frac{p}{\lambda}$

$\displaystyle E[Y^n]=\frac{p \thinspace n!}{\lambda^n}$ for all integers $n>1$

$\displaystyle Var[Y]=\frac{2p-p^2}{\lambda^2}$

$\displaystyle M_Y(t)=1-p+\frac{p \thinspace \lambda}{\lambda - t}$

$\displaystyle \gamma_Y=2 \thinspace p$

Case 2
This is the case that the insurance policy has a policy cap. The “per loss” payout amount is capped at the amount $m$. The following is the payout rule of $Y$:

$\displaystyle Y=\left\{\begin{matrix}0&\thinspace \text{no loss occurs}\\{Z}&\thinspace \text{a loss occurs}\end{matrix}\right.$

$\displaystyle Z=\left\{\begin{matrix}X&\thinspace X

$\displaystyle Y=\left\{\begin{matrix}0&\thinspace \text{no loss occurs}\\{X}&\thinspace \text{a loss occurs and } X

Case 2 – Per Loss Variable $Z$
The following lists out the information we need for $Z$. For more information about the “per loss” payout for an insurance contract with a policy cap, see the post An insurance example of a mixed distribution – I.

$\displaystyle F_Z(x)=\left\{\begin{matrix}0&\thinspace x<0\\{F_X(x)}&\thinspace 0 \le x

$\displaystyle f_Z(x)=\left\{\begin{matrix}f_X(x)&\thinspace X

$\displaystyle E[Z]=\int_0^{m} x \thinspace f_X(x) \thinspace dx + m \thinspace [1-F_X(m)]$

$\displaystyle E[Z^n]=\int_0^{m} x^n \thinspace f_X(x) \thinspace dx + m^n \thinspace [1-F_X(m)]$ for all integers $n > 1$

$\displaystyle M_Z(t)=\int_0^m e^{tx}f_X(x)dx+e^{tm}[1-F_X(m)]$

$\displaystyle \gamma_Z=\int_0^m \biggl(\frac{z-\mu_Z}{\sigma_Z}\biggr)^3f_X(z)dz+\biggl(\frac{m-\mu_Z}{\sigma_Z}\biggr)^3 [1-F_X(m)]$

Case 2 – Distribution Function
Since $Z$ is a mixture, the distribution of $Y$ is a mixture of a point mass at the origin (no loss) and the mixture $Z$. As in the general case discussed above, the distribution function $F_Y$ is a weighted average of $F_U$ and $F_Z$ where $F_U$ is the distribution function of the point mass at $y=0$. The following shows the distribution function and the density function of $Y$.

$\displaystyle F_U(x)=\left\{\begin{matrix}0&\thinspace x<0\\{1}&\thinspace 0 \le x\end{matrix}\right.$

$\displaystyle F_Y(y)=(1-p) \thinspace F_U(y) + p \thinspace F_Z(y)$.

$\displaystyle F_Y(y)=\left\{\begin{matrix}0&\thinspace y<0\\{1-p+pF_X(y)}&\thinspace 0 \le y

$\displaystyle f_Y(y)=\left\{\begin{matrix}1-p&y=0\\{pf_X(y)}&\thinspace 0

Case 2 – Basic Properties
To obtain the basic properties such as $E[Y]$, $E[Y^2]$, $M_Y(t)$ and $\gamma_Y$, just take the weighted average of the point mass and the “per loss” $Z$ of this case. In other words, they are obtained by weighting the point mass (of no loss) with the “per loss variable $Z$.

Case 2 – Exponential Example
If the unmodified loss $X$ has an exponential distribution, we have the following results:

$\displaystyle E[Y]=\frac{p}{\lambda}(1-e^{-\lambda m})$

$\displaystyle E[Y^2]=p\biggl(\frac{2}{\lambda^2}-\frac{2m}{\lambda}e^{-\lambda m}-\frac{2}{\lambda^2}e^{-\lambda m}\biggr)$

$\displaystyle Var[Y]=p \thinspace E[Z^2] - p^2 E[Z]^2$

$\displaystyle M_Y(t)=1-p+pM_Z(t)$ where

$\displaystyle M_Z(t)=\int_0^m e^{tx} \lambda e^{-\lambda x}dx+e^{tm} e^{-\lambda m}$

$\displaystyle =\frac{\lambda}{\lambda -t}-\frac{\lambda}{\lambda -t} e^{-(\lambda-t)m}+e^{-(\lambda-t)m}$

Case 3
This is the case that the insurance policy is an excess-of-loss policy. The insurer agrees to pay the insured the amount of the random loss $X$ in excess of a fixed amount $d$. The following is the payout rule of $Y$:

$\displaystyle Y=\left\{\begin{matrix}0&\thinspace \text{no loss occurs}\\{Z}&\thinspace \text{a loss occurs}\end{matrix}\right.$

$\displaystyle Z=\left\{\begin{matrix}0&\thinspace X

$\displaystyle Y=\left\{\begin{matrix}0&\thinspace \text{no loss occurs}\\{0}&\thinspace \text{a loss occurs and } X

Case 3 – Per Loss Variable $Z$
The following lists out the information we need for $Z$. For more information about the “per loss” payout for an insurance contract with a deductible, see the post An insurance example of a mixed distribution – II.

$\displaystyle F_Z(y)=\left\{\begin{matrix}0&\thinspace y<0\\{F_X(y+d)}&\thinspace y \ge 0\end{matrix}\right.$

$\displaystyle f_Z(y)=\left\{\begin{matrix}F_X(d)&\thinspace y=0\\{f_X(y+d)}&\thinspace y > 0\end{matrix}\right.$

$\displaystyle E[Z]=\int_o^{\infty} y \thinspace f_X(y+d) \thinspace dy$

$\displaystyle E[Z^n]=\int_o^{\infty} y^n \thinspace f_X(y+d) \thinspace dy$ for all integer $n>1$

$\displaystyle M_Z(t)=F_X(d) e^{0} + \int_0^{\infty} e^{tz}f_X(z+d) dz$

$\displaystyle =F_X(d) + e^{-td} \int_d^{\infty} e^{tw} f_X(w) dw$

$\displaystyle \gamma_Z=F_X(d) \biggl(\frac{0-\mu_Z}{\sigma_Z}\biggr)^3+\int_0^{\infty} \biggl(\frac{z-\mu_Z}{\sigma_Z}\biggr)^3 f_X(z+d) dz$

Case 3 – Distribution Function
Since $Z$ is a mixture, the distribution of $Y$ is a mixture of a point mass at the origin (no loss) and the mixture $Z$. As in the general case discussed above, the distribution function $F_Y$ is a weighted average of $F_U$ and $F_Z$ where $F_U$ is the distribution function of the point mass at $y=0$. The following shows the distribution function and the density function of $Y$.

$\displaystyle F_U(x)=\left\{\begin{matrix}0&\thinspace x<0\\{1}&\thinspace 0 \le x\end{matrix}\right.$

$\displaystyle F_Y(y)=(1-p) \thinspace F_U(y) + p \thinspace F_Z(y)$.

$\displaystyle F_Y(y)=\left\{\begin{matrix}0&\thinspace y<0\\{1-p+pF_X(y+d)}&\thinspace 0 \le y\end{matrix}\right.$

$\displaystyle f_Y(y)=\left\{\begin{matrix}1-p+pF_X(d)&y=0\\{pf_X(y+d)}&\thinspace 0

Note that the point mass of $Y$ is made up of two point masses, one from having no loss and one from having losses less than the deductible.

Case 3 – Basic Properties
The basic properties of $Y$ as a mixture are obtained by applying the general formulas with the specific information about the “per loss” $Z$ in this case. In other words, they are obtained by weighting the point mass (of no loss) with the “per loss variable $Z$.

Case 3 – Exponential Example
If the unmodified loss $X$ has an exponential distribution, then we have the following results:

$\displaystyle E[Y]=pE[Z]=p \thinspace \frac{e^{-\lambda d}}{\lambda}=p \thinspace e^{-\lambda d} E[X]$

$\displaystyle E[Y^2]=pE[Z^2]=p \thinspace \frac{2e^{-\lambda d}}{\lambda^2}=p \thinspace e^{-\lambda d}E[X^2]$

$\displaystyle Var[Y]=p \thinspace \frac{2e^{-\lambda d}}{\lambda^2}-p^2 \thinspace \frac{e^{-2\lambda d}}{\lambda^2}=pe^{-\lambda d}(2-pe^{-\lambda d})Var[X]$

$\displaystyle M_Y(t)=1-e^{-\lambda d}+e^{-\lambda d} \frac{\lambda}{\lambda -t}$

$\displaystyle = 1-e^{-\lambda d}+e^{-\lambda d} M_X(t)$

Case 4
This is the case that the insurance policy has both a policy cap and a deductible. The “per loss” payout amount is capped at the amount $m$ and is positive only when the loss is in excess of the deductible $d$. The following is the payout rule of $Y$:

$\displaystyle Y=\left\{\begin{matrix}0&\thinspace \text{no loss occurs}\\{Z}&\thinspace \text{a loss occurs}\end{matrix}\right.$

$\displaystyle Z=\left\{\begin{matrix}0&\thinspace X

$\displaystyle Y=\left\{\begin{matrix}0&\thinspace \text{no loss occurs}\\{0}&\text{a loss and }X

Case 4 – Per Loss Variable $Z$
The following lists out the information we need for $Z$. For more information about the “per loss” payout for an insurance contract with a deductible and a policy cap, see the post An insurance example of a mixed distribution – III.

$\displaystyle F_Z(y)=\left\{\begin{matrix}0&\thinspace y<0\\{F_X(y+d)}&\thinspace 0 \le y < m\\{1}&m \le y\end{matrix}\right.$

$\displaystyle f_Z(y)=\left\{\begin{matrix}F_X(d)&\thinspace y=0\\{f_X(y+d)}&\thinspace 0 < y < m\\{1-F_X(d+m)}&y=m\end{matrix}\right.$

$\displaystyle E[Z]=\int_0^m y \thinspace f_X(y+d) \thinspace dy + m \thinspace [1-F_X(d+m)]$

$\displaystyle E[Z^n]=\int_0^m y^n \thinspace f_X(y+d) \thinspace dy + m^n \thinspace [1-F_X(d+m)]$ for all integer $n>1$

$\displaystyle M_Z(t)=F_X(d) e^0 + \int_0^m e^{tx} f_X(x+d) dx + e^{tm} [1-F_X(d+m)]$

$\displaystyle =F_X(d) + \int_0^m e^{tx} f_X(x+d) dx + e^{tm} [1-F_X(d+m)]$

$\displaystyle \gamma_Z=F_X(d) \biggl(\frac{0-\mu_Z}{\sigma_Z}\biggr)^3+\int_0^{\infty} \biggl(\frac{z-\mu_Z}{\sigma_Z}\biggr)^3 f_X(z+d) dz$

$\displaystyle + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [1-F_X(d+m)] \biggl(\frac{m-\mu_Z}{\sigma_Z}\biggr)^3$

Case 4 – Distribution Function
Since $Z$ is a mixture, the distribution of $Y$ is a mixture of a point mass at the origin (no loss) and the mixture $Z$. As in the general case discussed above, the distribution function $F_Y$ is a weighted average of $F_U$ and $F_Z$ where $F_U$ is the distribution function of the point mass at $y=0$. The following shows the distribution function and the density function of $Y$.

$\displaystyle F_U(x)=\left\{\begin{matrix}0&\thinspace x<0\\{1}&\thinspace 0 \le x\end{matrix}\right.$

$\displaystyle F_Y(y)=(1-p) \thinspace F_U(y) + p \thinspace F_Z(y)$.

$\displaystyle F_Y(y)=\left\{\begin{matrix}0&\thinspace y<0\\{1-p+pF_X(y+d)}&\thinspace 0 \le y

$\displaystyle f_Y(y)=\left\{\begin{matrix}1-p+pF_X(d)&y=0\\{pf_X(y+d)}&\thinspace 0

Note that the point mass of $Y$ is made up of two point masses, one from having no loss and one from having losses less than the deductible.

Case 4 – Basic Properties
The basic properties of $Y$ as a mixture are obtained by applying the general formulas with the specific information about the “per loss” $Z$ in this case. In other words, they are obtained by weighting the point mass (of no loss) with the “per loss variable $Z$.

Case 4 – Exponential Example
If the unmodified loss $X$ has an exponential distribution, then we have the following results:

$\displaystyle E[Y]=pE[Z]=p e^{-\lambda d} \frac{1}{\lambda} (1-e^{-\lambda m})=p e^{-\lambda d} (1-e^{-\lambda m}) E[X]$

Another view of $E[Y]$:
$\displaystyle E[Y]=e^{-\lambda d} E[Y_2]$ where $Y_2$ is the $Y$ in Case 2.

Also, it can be shown that:
$\displaystyle E[Y^2]=e^{-\lambda d} E[Y_2^2]$ where $Y_2$ is the $Y$ in Case 2.

Here's the links to the previous discussions of mixed distributions:
An insurance example of a mixed distribution – I
An insurance example of a mixed distribution – II
An insurance example of a mixed distribution – III
Mixed distributions as mixtures

# Mixed distributions as mixtures

A random variable $X$ is a mixture if its distribution function $F_X$ is a weighted average of a family of conditional distribution functions. The random variable $X$ is a mixed distribution if it is a distribution that has at least one probability mass (i.e. there is at least one point $a$ in the support of $X$ such that $P[X=a]>0$) and there is some interval $(a,b)$ contained in the support such that $P[X=c]=0$ for every $c \in (a,b)$. It turns out that a mixed distribution can be expressed as a mixture. Three examples of mixed distributions from insurance applications have been presented in this blog. We demonstrate that these three mixed distributions are mixtures. The links to some previous posts on mixtures can be found at the end of this post.

Example 1
Link: An insurance example of a mixed distribution – I The mixed distribution in this example is the “per loss” payout for an insurance contract that has a policy maximum.

Example 2
Link: An insurance example of a mixed distribution – II The mixed distribution in this example is the “per loss” payout for an insurance policy that has a deductible.

Example 3
Link: An insurance example of a mixed distribution – III The mixed distribution in this example is the “per loss” payout of an insurance contract where there are both a deductible and a policy maximum.

Throughout this post, let $X$ be the unmodified random loss. We assume that $X$ is a continuous random variable with support the nonnegative real numbers.

Discussion of Example 1
Let $Y_1$ be the “per loss” insurance payout for a policy where the payout is capped at $m$. The following are the payout rule, the distribution function and the density function:

$\displaystyle Y_1=\left\{\begin{matrix}X&\thinspace X

$\displaystyle F_{Y_1}(x)=\left\{\begin{matrix}0&\thinspace x<0\\{F_X(x)}&\thinspace 0 \le x

$\displaystyle f_{Y_1}(x)=\left\{\begin{matrix}f_X(x)&\thinspace 0 < x

We show that $F_{Y_1}$ can be expressed as a weighted average of two distribution functions. One of the distributions is the random loss between $0$ and $m$. This is a limited loss and call this loss $U$. The second distribution is the point mass at $m$. Call this point mass $V$. The following are the distribution functions:

$\displaystyle F_U(x)=\left\{\begin{matrix}0&\thinspace x<0\\{\displaystyle \frac{F_X(x)}{F_X(m)}}&\thinspace 0 \le x

$\displaystyle F_V(x)=\left\{\begin{matrix}0&\thinspace x

It follows that $F_{Y_1}(x)= p \thinspace F_U(x) + (1-p) \thinspace F_V(x)$ where $p=F_X(m)$. Note that the distribution of $U$ only describes the loss within $(0,m)$. Thus the distribution function $F_U$ is obtained from $F_X$ by a scaler adjustment.

Discussion of Example 2
Let $Y_2$ be the “per loss” insurance payout for a policy where there is a deductible $d$. For each loss, the insurer pays the insured in excess of the deductible $d$. The following are the payout rule, the distribution function and the density function:

$\displaystyle Y_2=\left\{\begin{matrix}0&\thinspace X

$\displaystyle F_{Y_2}(y)=\left\{\begin{matrix}0&\thinspace y<0\\{F_X(y+d)}&\thinspace y \ge 0\end{matrix}\right.$

$\displaystyle f_{Y_2}(y)=\left\{\begin{matrix}F_X(d)&\thinspace y=0\\{f_X(y+d)}&\thinspace y > 0\end{matrix}\right.$

We show that $F_{Y_2}$ can be expressed as a weighted average of two distribution functions. One of the distributions is the random loss greater than $d$. Call this loss $U$. The second distribution is the point mass at $0$. Call this point mass $V$. The following are the distribution functions:

$\displaystyle F_U(x)=\left\{\begin{matrix}0&\thinspace x<0\\{\displaystyle \frac{F_X(x+d)-F_X(d)}{1-F_X(d)}}&\thinspace 0 \le x\end{matrix}\right.$

$\displaystyle F_V(x)=\left\{\begin{matrix}0&\thinspace x<0\\{1}&\thinspace 0 \le x\end{matrix}\right.$

It follows that $F_{Y_2}(x)= p \thinspace F_U(x) + (1-p) \thinspace F_V(x)$ where $p=1-F_X(d)$. The random variable $V$ is a point mass at the origin reflecting the case where no claim is made by the insurer. This point mass has weight $F_X(d)$. The random variable $U$ is the distribution describing the random losses that are greater than $d$.

Discussion of Example 3
Let $Y_3$ be the “per loss” insurance payout for a policy where there are both a deductible $d$ and a policy cap $m$ with $d. For each loss, the insurer pays the insured in excess of the deductible $d$ up to the policy cap $m$. The following are the payout rule, the distribution function and the density function:

$\displaystyle Y_3=\left\{\begin{matrix}0&\thinspace X

$\displaystyle F_{Y_3}(y)=\left\{\begin{matrix}0&\thinspace y<0\\{F_X(y+d)}&\thinspace 0 \le y < m\\{1}&m \le y\end{matrix}\right.$

$\displaystyle f_{Y_3}(y)=\left\{\begin{matrix}F_X(d)&\thinspace y=0\\{f_X(y+d)}&\thinspace 0 < y < m\\{1-F_X(d+m)}&y=m\end{matrix}\right.$

The distribution of $Y_3$ can be expressed as a mixture of three distributions – two point masses (one at the origin and one at $m$) and one continuous variable describing the random losses in between $0$ and $m$. Consider the following distribution functions:

$\displaystyle F_U(x)=\left\{\begin{matrix}0&\thinspace x<0\\{1}&\thinspace 0 \le x\end{matrix}\right.$

$\displaystyle F_V(x)=\left\{\begin{matrix}0&\thinspace x<0\\{\displaystyle \frac{F_X(x+d)-F_X(d)}{F_X(d+m)-F_X(d)}}&\thinspace 0 \le x

$\displaystyle F_W(x)=\left\{\begin{matrix}0&\thinspace x

The random variables $U$ and $W$ represent the point masses at $0$ and $m$, respectively. The variable $V$ describes the random losses in between $0$ and $m$. It follows that $F_{Y_3}$ is the weighted average of these three distribution functions.

$\displaystyle F_{Y_3}(x)=p_1 \thinspace F_U(x)+p_2 \thinspace F_V(x)+p_3 \thinspace F_W(x)$

The weights are: $p_1=F_X(d)$, $p_2=F_X(d+m)-F_X(d)$, and $p_3=1-F_X(d+m)$

Here’s the links to examples of mixed distributions:
Example 1 An insurance example of a mixed distribution – I
Example 2 An insurance example of a mixed distribution – II
Example 3 An insurance example of a mixed distribution – III

Here’s the links to some previous posts on mixtures:
Examples of mixtures
Basic properties of mixtures