Find unbiased estimator of $\theta$ when $f(x;\theta )=\frac{2x}{\theta }e^{\frac{-x^{2}}{\theta }}$

Question

Let $f(x;\theta )=\frac{2x}{\theta }e^{\frac{-x^{2}}{\theta }}$ be a probability density function of sample $(X_{1},X_{2},...,X_{n})$ where $x\geq 0$. Find unbiased estimator for $\theta$.

Here is my solution : $E\left [ \bar{X} \right ] = E\left [ X \right ]=\frac{1}{\theta }\int_{0}^{\infty }2x^{2}e^{\frac{-x^{2}}{\theta }}dx=\frac{1}{\theta }\int_{0}^{\infty }xe^{\frac{-x^{2}}{\theta }}2xdx$

Let $$ t=\frac{x^{2}}{\theta } (t=\frac{x^{2}}{\theta }\Rightarrow \theta t=x^{2}\Rightarrow x=\sqrt{\theta t},x\geq 0) , dt=\frac{2x}{\theta }d\theta \Rightarrow \theta dt=2xdx$$

$=\frac{1}{\theta }\int_{0}^{\infty }(\theta )^{\frac{1}{2}}e^{-t}\theta dt=\frac{\theta ^{\frac{3}{2}}}{\theta }\int_{0}^{\infty }t^{\frac{1}{2}}e^{-t}dt=\sqrt{\theta }.\Gamma (\frac{3}{2})=\sqrt{\theta }.\frac{\sqrt{\pi}}{2}$

$\Rightarrow E\left ( \bar{X} \right )=\sqrt{\theta }.\frac{\sqrt{\pi}}{2}\Rightarrow E(\frac{4\bar{(X)}^{2}}{\pi})=\theta $

Hence unbiased estimator for $\theta$ is $\hat{\theta }=\frac{4\bar{(X)}^{2}}{\pi}$

Is my solution right? I am not sure the calculation. Any idea will be appreciated.

Answer 1

You correctly calculated that $E(x)=\frac{1}{2}\sqrt{\theta \pi}$. But as noted in the comments this does not allow you to compute $E(x^2)$. If you actually calculate that (wolfram alpha), you will find $$ E(x^2) = \int x^2\ f(x;\theta)\ dx=\theta.$$ which means that $x^2$ is an unbiased estimator of $\theta$.

Answer 2

@softglance: The estimator you found is a nice estimator and it is exactly the Method of Moment estimator for $\theta$

Unfortunately it is not very easy to calculate its expectation.

A clever approach is the one showed by @Simon (+1) but it consider only one observation.

To calculate its expectation no integrals are needed, just observe that your density is a known one, a Rayleigh with

$$\mathbb{E}[X]=\sqrt{\theta}\frac{\sqrt{\pi}}{2}$$

$$\mathbb{V}[X]=\frac{4-\pi}{4}\theta$$

thus

$$\mathbb{E}[X_1^2]=\mathbb{V}[X]+\mathbb{E}^2[X]=\theta$$

If you want to get an unbiased estimator for $\theta$ using all you $n-$ sized random sample simply use

$$T=\frac{\sum_iX_i^2}{n}$$

that is obviously unbiased.

Another way is to remember that $Y=\sum_iX_i^2\sim Gamma\left(n;\frac{1}{\theta}\right)$

with expectation $E(Y)=n\theta$ and thus if you choose $T=\frac{Y}{n}$ you get an unbiased estimator