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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

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    $\begingroup$ 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. $\endgroup$
    – Glen_b
    Commented Nov 10, 2019 at 7:50
  • $\begingroup$ To define a moment estimator of $\gamma$, you first need to express $\gamma$ as a moment. $\endgroup$
    – Xi'an
    Commented Nov 10, 2019 at 16:35
  • $\begingroup$ 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 $\theta$. $\endgroup$
    – john_w
    Commented Nov 10, 2019 at 19:57
  • $\begingroup$ 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. $\endgroup$
    – john_w
    Commented Nov 10, 2019 at 20:02
  • $\begingroup$ 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? $\endgroup$
    – john_w
    Commented Nov 11, 2019 at 1:50

1 Answer 1

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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 $T$?
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