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

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

• As a routine bookwork question, please add the self-study tag and read its tag wiki. – Glen_b Sep 21 '14 at 6:20
• Note that the question doesn't say anything about maximum likelihood estimation. It talks only about forming an unbiased estimator from the geometric mean. So your first task would be to figure out what $E(G)$ is and then figure out how you might modify it (if necessary) to say, $T=f(G)$ such that $E(T)=\theta$. – Glen_b Sep 21 '14 at 6:26

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 $\theta$.