Unbiasedness and Minimum Variance of Estimators of Exponential Distribution

Question

We have an exponential random variable $X$ and we take two samples $X_1$ and $X_2$.

$f(x,\theta) = (\frac{1}{\theta})\times e^{(\frac {-x}{\theta})}, x\gt0$,

$\theta$ being an unknown parameter. We have two estimators

$$\hat\theta_1 = \frac{X_1}{2}+\frac{X_2}{2} = \bar X$$ and

$$\hat\theta_2=\frac {4\times(X_1\times X_2)^{\frac12}}{\pi}.$$

In terms of unbiasedness and minimum variance, which one of the estimators is better?

What I did is this:

I substitute the first value and then simply integrated

$E(\hat\theta_1) = \frac{X_1}{2}+\frac{X_2}{2}$, I got the same value when I integrated it

I can't do anything for the second estimators

Any help is highly appreciated.

Answer 1

The mean and variance for the first estimator can be found by handling the two random variables $X_1$ and $X_2$ separately.

But to handle the second estimator, you need to start by writing the joint density function of $X_1$ and $X_2$. That should be easy because these are independent observation The integral for the expectation of your second estimator $\hat \theta_2$ will be the product of the joint density function. put the factors with $x_1$ together and the factors with $x_2$ together. You should be able to find your way from there to showing that $\hat \theta_2$ is unbiased. Then you need to find its variance.

If you know about sufficient statistics, you know that $X_1 + X_2$ is a sufficient statistic for $\theta$ and so $\hat \theta_1$ is going to win this contest!

Below is a brief simulation for the case $\theta = 3$. (I'm using R, which uses rates instead of means to parameterize exponentials, hence the argument 1/3.) When you get your results you can see if they agree (roughly) with simulated values, when $\theta = 3.$

 theta = 3
 x1 = rexp(10^6, 1/theta); x2 = rexp(10^6, 1/theta)
 th1.hat = (x1 + x2)/2;  th2.hat = 4*sqrt(x1*x2)/pi
 mean(th1.hat);  mean(th2.hat)
 ## 3.000196   # both estimates seem
 ## 3.000815   #   to be unbiased
 var(th1.hat); var(th2.hat)
 ## 4.482319   # but the second clearly
 ## 5.569243   #   has the larger variance