# inverse of an exponential distribution

I have a question regarding this.

Say I have $$X_1, ..., X_n$$ be random sample from an exponential distribution i.e. $$Exp(\theta)$$, and let $$\gamma = \theta^2$$. Let denote $$\gamma^{mme}$$ as the method of moment estimator of $$\gamma$$. How do we show that the MME is a biased estimator of $$\gamma$$?

I found that the method of moment estimator of $$\gamma = \theta^2$$ is $$(\frac{1}{\bar{x}})^2$$ ( hopefull this is correct).

Then I tried to find the biased of $$(\frac{1}{\bar{x}})^2$$. So I tried to find the expected value of $$(\frac{1}{\bar{x}})^2$$

But I got stuck with that. Could someone give me some hints.

thank you

• Yoou should be explicit about which parameterization you're using. It looks like $f(x;\theta) = \theta e^{-\theta x}\, \mathbb{I}_{x>0}\,,\: \text{ for } \theta>0$ but it's best to be clear. Also, in reference to your title, you're not actually inverting the distribution. Commented Nov 10, 2019 at 7:50
• To define a moment estimator of $\gamma$, you first need to express $\gamma$ as a moment. Commented Nov 10, 2019 at 16:35
• Hello I tried finding expected value of X and expected value of $X^2$, but both expected value give $1 / \theta$ as the expected value. So I am not sure how to go about getting gamma which is the power of two of $\theta$. Commented Nov 10, 2019 at 19:57
• Also the other part of the questions says to check the method of moment estimator is the same as the maximum likelihood estimator. So I got the mle to be the one I have above when I said it is the method of moment estimator. So may I know if the method of moment estimator is correct above? If not , is it possible to get some more hints. Commented Nov 10, 2019 at 20:02
• yes Glen_b, that is the form of the parameterization in my question. So could you provide more hints or comments as to how to solve the problem? Commented Nov 11, 2019 at 1:50

First, let $$T = \frac{1}{\bar X}$$ so that $$E\left(\frac{1}{\bar X^2}\right) = E(T^2) = E(T)^2 + Var(T)$$
This problem can be solved easily once we have identified the distribution of $$T$$. Here are some hints to help you find the distribution of $$T$$.
1. Show that $$Y_i = X_i/n$$ has an $$Exp(n\theta)$$ distribution.
2. Given that $$T^{-1} = \sum_{i=1}^n Y_i$$, what is the distribution of $$T^{-1}$$?
3. Given the distribution of $$T^{-1}$$, what is the distribution of $$T$$?