# Is $\frac{m-1}{x}$ an unbiased estimator of $\theta$ for given pdf?

Let $X$ be a continuous random variable with pdf, $$f(x;\theta)=\frac {\theta^m.x^{m-1}e^{-\theta x}} {(m-1)!} ; x\geq0, \theta>0$$ Is $\frac{m-1}{x}$ an unbiased estimator of $\theta$ for given pdf?

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2
If this is a homework question, please indicate what have you tried/how you got this result. – Dennis Gulko May 19 '11 at 11:12
no, i got this from past papers. – amul28 May 19 '11 at 11:43

## 2 Answers

For $m \geq 2$, $$\begin{align} \newcommand{\e}{\mathbb{E}}\newcommand{\rd}{\,\mathrm{d}} \e \frac{m-1}{X} = \int_0^\infty \frac{m-1}{x} \frac{\theta^m x^{m-1}}{(m-1)!} e^{-\theta x} \rd x = \theta \int_0^\infty \frac{\theta^{m-1} x^{m-2}}{(m-2)!} e^{-\theta x} \rd x = \theta \end{align}$$ since the integrand after the second equality is simply a gamma pdf with parameters $m-1$ and $\theta$ and so must integrate to one.

This is closely related to your previous question, by the way.

This "trick" pops up quite frequently in introductory mathematical statistics, so it's worth learning and being able to recognize when it can be applied.

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yes, even i understood that after long thinking. Thank you. – amul28 May 19 '11 at 13:08
Apologies but how is it different, if it is, from a copy of my answer that was posted 1 hour earlier? ;-) – Luboš Motl May 19 '11 at 14:18
@Luboš Motl, The point of my post was to demonstrate a very common trick for solving these problems. Mainly, factoring out certain terms to leave an integral of a probability density. This is not the only place that such a trick is useful. The OP was interested in an analytical means for solving the problem "by hand", as stated in his comment to you. This provides that, whereas the use of Mathematica does not. – cardinal May 19 '11 at 14:58
The downvote seems a little harsh, too. Particularly, as I was trying to demonstrate a (simple) complementary technique. – cardinal May 19 '11 at 14:59
@Luboš, if the answer has additional details, and in this case it certainly does, it is ok to post it. Your answer is a good one too, but it has a drawback, since it relies on having Mathematica, which is not cheap. – mpiktas May 23 '11 at 10:15
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it's an unbiased estimator if the expectation value of $(m-1)/x$ is equal to $\theta$. The expectation value is calculated as the integral over $x$ $$\int_0^\infty{\rm d}x \frac{m-1}{x} f(x;\theta) = \theta$$ so that it works, the answer is Yes. After a few trivial substitutions, the integral is just the Euler integral for the Gamma functions. In Mathematica, you may type

Integrate[(m - 1)/x theta^m x^(m - 1) Exp[-theta x] / Gamma[m], {x, 0, Infinity}]

and decode the result $\theta$ from the output, while accepting the sensible inequalities. Mathematica didn't exactly cancel the Gamma functions (or factorials) in an elegant way but they do cancel.

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thanks, but i want to know any other easy way of solving this by hand, because finding the integral by hand is very tedious! – amul28 May 19 '11 at 11:47
If it's not for a homework then what is the reason why can't you do it with a Mathematica, moreover an integral is quite trivial through the definition of Gamma function. – Ilya May 19 '11 at 12:35
because this is a previous exam paper so i want to know how to solve it. – amul28 May 19 '11 at 13:06
I didn't do it with Mathematica first, it's just an ordinary Euler integral! Cardinal didn't do any extra steps to calculate the Euler integral, either. He just wrote that it's an elementary integral just like I did. – Luboš Motl May 19 '11 at 14:19
To prove that the Euler integral $\int_0^\infty dt\exp(-t)t^n$ is equal to $n!=\Gamma(n+1)$ for integer $n$, just integrate it by parts several times, so that the exponent above $t$ is always reduced by one. In this way, you will collect factors of $n, n-1$, and so on, and finally you collect the whole $n!$ and the remaining integral will be $\int_0^\infty dt \exp(-t)$ which is easily integrated to one. – Luboš Motl May 19 '11 at 14:21
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