# Copula models and the distribution of the sum of random variables without Monte Carlo

There is a vast literature on copula modelling. Using copulas I can describe the joint law of two (and more) random variables $X$ and $Y$, i.e. $F_{X,Y}(x,y)$. Very often in risk management (credit risk, operational risk, insurance) the task is to model a sum $Z=X+Y$ and find its distribution $F_Z(z) = F_{X+Y}(z)$.

I know several approaches that do not directly use copulas (e.g. commons shock models and mixed compound Poisson models) but how can I elegantly combine a copula model and a model for the sum (without Monte Carlo of course - otherwise it would be easy).

Is there some useful Fourier-transform approach? I had the feeling that in the case of Archemedian copulas there could be a chance (looking at this mixture representation as in e.g. in Embrechts, Frey, McNeil). Who has an idea? Are there any papers on this?

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Check the article "Fast computation of the distribution of the sum of two dependent random variables" by Embrechts and Puccetti (searcheable via google) – Alexey Kalmykov Jan 24 at 18:39
@AlexeyKalmykov Thanks for the link. This looks very interesting. I will read it soon. Do you know anything more applied too? If you make your comment an answer then I will accept if nothing else comes in. Thanks! – Richard Jan 25 at 8:32
Does anybody have an idea which is more copula based? Do copulas and sums don't go together well? I mean - the answer by Julian Wergieluk shows that this is not true in general - but using copulas we do not gain any insight to the sum? No (Fourier/moment/ ...) transform trick? – Richard Jan 29 at 13:16
– Richard Jan 29 at 13:25
Let's see whether a bounty attracts new ideas. – Richard Feb 7 at 13:02
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## 2 Answers

In general setting this is quite a tough problem and it looks like just switching from regular multivariate probability to copulas doesn't make it easier. In general case you need to rely on numerical methods for integration.

There is a nice overview of the problem in Copula Theory and Its Applications: Proceedings of the Workshop Held in Warsaw, 25-26 September 2009, Part I, Section 5.3, "The Calculation of the Distribution of the Sum of Risks".

If you want to avoid Monte Carlo methods, you can look at a new deterministic method AEP, which was specially designed to tackle this problem. Recursive and FFT methods are compared here.

I haven't seen any work that deals with this problem specifically for Archimedean copulas.

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– Richard Feb 13 at 21:47
@Richard Yes, there is no way to avoid numerical integration (MC or any other method). – Alexey Kalmykov Feb 13 at 22:00
So copulas do not facilitate the calculation of the law of sums. This is somewhat disappointing. In many contexts I am interested in the sum ... thanks for your links. – Richard Feb 13 at 22:05

If the density of $(X,Y)$ is known, then you may obtain the density of the sum $X+Y$ simply by applying the Jacobi's transformation formula, which describes the density of the transformed random variable $g(X,Y)$ for $g(x,y) = (x+y, x)$. Integrating out the $x$-component yields the density of $X+Y$. See Jacod/Protter Probability Essentials ch. 12 for details.

I am not sure, if this is an answer you are expecting. Perhaps you could provide us with some more specific description of your modelling task.

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thanks for your comment. This looks nice. With my question I don't have a specific application in mind. I was just wondering whether there is no useful alternative to MC in this case. – Richard Jan 25 at 11:27
By the way: do you have a link to the internet to the procedure that you propose? I don't have the book that you mention - thanks! – Richard Jan 25 at 11:28
– Julian Wergieluk Jan 25 at 20:05

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Assume the densoty of $(X,Y)$ is given by $f_{X,Y}(x,y)$. Then you propose $u=x+y$ and $v=x$ which gives $x=v$ and $y = u-v$. The applying Jacobi we get $$f_{U,V}(u,v) = f_{X,Y}(v,u-v) \cdot |-1|$$ ... now how can I proceed? I can't integrate over $u-v$ straight forward. Can you help me? Futhermore: Julian aren't you working in credit risk. Isn't there anything nice in the case of Archemidean copulas? – Richard Jan 27 at 21:17
Well thinking about it again the solution depends on the specific form. Then we must integrate over $v$ and note $u-v$ then we are done. I think I got it :) Further more the following helped me: math.uiuc.edu/~r-ash/Stat/StatLec1-5.pdf and www2.econ.iastate.edu/classes/econ671/hallam/documents/… – Richard Jan 27 at 21:25
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