If I have the following random variables : $X_1,...,X_n$ with IID distribution of Exp($\lambda$) , $\lambda > 0 $. I want to show two things;
- The mean $\overline{X}_n$ is an unbiased estimator of $\frac{1}{\lambda}$
- That $\frac{1}{\bar{X}_n}$ is an unbiased estimator of $\lambda$
Now,
I know in general I should check if $E(\mu)* T = g(\mu) $ for every $\mu \in \Theta $
But not sure if in the first part I should basically integrate the pdf of exponential distribution with the mean? Because I always get a diverging $lim$ at the end for -infinity.
Also, what confuses me that the IID property here, does it mean this for the mean:
$E(\bar{X_n}) = \frac{n*\frac{1}{\lambda}}{n} $ as $\frac{1}{\lambda}$ is the expected value for exponential dist?
Finally, is it a viable option to use n =1 and then apply induction?
Any advice is greatly appreciated!
Your last line is the correct reasoning for the first question.
For the second question unfortunately $E[1/\bar{X}] \ne 1 / E[\bar{X}]$ in general so you should compute it directly.
You can show that $\bar{X}_n \sim \text{Gamma}(n, n\lambda)$. Since the density of this gamma distribution is $f(x) = (\lambda n)^n x^{n-1} e^{-\lambda n x} / (n-1)!$, we obtain $$E[1/\bar{X}] = \int_0^\infty \frac{1}{x} f(x) \, dx = \int_0^\infty (\lambda n)^n x^{n-2} e^{-\lambda n x}/(n-1)! \, dx = \frac{\lambda n}{n-1} \underbrace{\int_0^\infty (\lambda n)^{n-1} x^{n-2} e^{-\lambda n x} / (n-2)! \, dx}_{=1}$$ so this is not an unbiased estimator of $\lambda$.