Unbiased estimator based on geometric mean for random sample on an exponential distribution

Question

Let $X_1, \ldots, X_n$ be a random sample on an exponential distribution with mean $\theta$.

Obtain an unbiased estimator for $\theta$ based on $G$, where $G$ is the geometric mean of the observations.

Hint: answer may be expressed in terms of the gamma function.

Approach: I understand that I can obtain the MLE estimate for theta by differentiating the log-likelihood functions, and letting it = 0. This gives me $\hat{\theta} = \frac{1}{n} \sum_{i=1}^n X_i$.

I don't see how to bring $G$ or the gamma function into the answer

Any advice would be appreciated

Answer 1

First find the expected value of $G$. This may involve a couple of tricks, such as knowing the expected value of a product of independent random variables is the product of their expected values, and maybe a substitution to make your integral look like the Gamma function.

Your result should be a constant multiple of $\theta$, where the constant multiple involves just the Gamma function and $n$. Then just move that constant multiple to the other side and bring it inside the expectation using linearity. Whatever is inside that expectation is just a function of the data, so it will be an unbiased estimator of $\theta$.