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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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    no, i got this from past papers.2011-05-19

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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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    @cardinal, sometimes downvotes really, really do not make sense :)2011-05-23
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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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    @Luboš Motl, to be clear, I wrote that it can be reformulated leaving an integrand that is an easily recognizable probability density function. That obviously *does* necessarily involve being able to recognize the integrand as such. But, for a student in a class at the level that @amul28 appears to be, that should be a part of their toolbox at that stage.2011-05-19