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I'm having a bit of trouble reading this:

Let $X_i$s be id r.v.s from the Exponential distribution

and

$$\gamma=\frac{1}{\lambda}=E(X_i)$$

then

$$\gamma_{est}=\bar{X}$$ and (specifically unintuitively) $$Var(\gamma_{est})=\frac{\gamma^2}{n}$$

(Notice $Var(X_i)=\gamma^2$)

Can someone clear this up?

It looks like $$Var(\gamma_{est})=\frac{\gamma^2}{n}=\frac{Var(X_i)}{n}$$ but why is this?


Is it perhaps because of this calculation https://onlinecourses.science.psu.edu/stat414/node/167 (bottom page). $Var(X_i)=\sigma^2$

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  • $\begingroup$ Hint: First compute or look up the variance of the $X_i$. Can you compute the variance of $\frac{1}{2}X_1 + \frac{1}{2}X_2$? Now generalize. $\endgroup$ – whuber Mar 4 '16 at 0:56
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    $\begingroup$ @whuber So it's about the link I posted? $\endgroup$ – mavavilj Mar 4 '16 at 11:19
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It is because of the bottom page of the link OP posted, as OP suspected.

$$Var[\overline{X}] = \frac{1}{n^2} Var[\sum_i X_i] = \frac{1}{n^2} \sum_i Var[X_i] = \frac{1}{n^2} \sum_i \sigma^2 = \frac{n}{n^2} \sigma^2 = \frac{1}{n} \sigma^2$$

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