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

Question

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?

Answer 1

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.

Answer 2

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.