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Consider a simple sample $X_{1},X_{2},\ldots,X_{n}$ whose distribution is given by $X\sim Exp(\lambda)$.

(a) Determine an estimator for $\lambda$ according to the method of moments.

(b) Determine another estimator for $\lambda$ different from the previous one.

(c) Determine an estimate of $\textbf{P}(X\geq 1)$ in accordance to the method of moments.

MY ATTEMPT

As to the first case, we have \begin{align*} \frac{1}{\lambda} = \textbf{E}(X) \Rightarrow \hat{\lambda} = \frac{n}{\displaystyle\sum_{k=1}^{n}X_{k}} \end{align*}

As to the second case, we have \begin{align*} \frac{1}{\lambda^{2}} = \textbf{Var}(X) = \textbf{E}(X^{2}) - \textbf{E}(X)^{2} \Rightarrow \hat{\lambda} = \left[\frac{1}{n}\sum_{k=1}^{n}X^{2}_{k} - \left(\frac{1}{n}\sum_{k=1}^{n}X_{k}\right)^{2}\right]^{-1/2} \end{align*}

EDIT

In the third case, we have \begin{align*} \textbf{P}(X\geq 1) = 1 - \textbf{P}(X < 1) = 1 - \exp(-1\times\lambda) = 1 - \exp(-\lambda) \end{align*}

Now it suffices to substitute $\lambda$ for any of the above expressions.

Could someone double-check my reasoning? Thanks in advance!

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    $\begingroup$ Exponential distribution is continuous, but you're treating it as discrete.... is there a typo? $\endgroup$ – Lee David Chung Lin Apr 8 at 2:00
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    $\begingroup$ Indeed, you are right. I was thinking about the poisson distribution when I wrote it. Thanks for the observation. I will edit it. $\endgroup$ – user1337 Apr 8 at 2:08
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    $\begingroup$ What have you tried? You know, don't you, that this is not a site for us to do your homework for you? $\endgroup$ – David G. Stork Apr 8 at 2:22
  • $\begingroup$ Plug the estimator in $(a)$ into $e^{-\lambda}$. $\endgroup$ – d.k.o. Apr 8 at 2:27
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    $\begingroup$ Sorry, David. I know it. I just forgot to mention my attempts. $\endgroup$ – user1337 Apr 8 at 2:38

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