# ($X_i$ ~Exp distr.): $\gamma=\frac{1}{\lambda}=E(X_i)$ $\implies$ $\gamma_{est}=\bar{X}$ and $Var(\gamma_{est})=\frac{\gamma^2}{n}$

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). $Var(X_i)=\sigma^2$

• 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. – whuber Mar 4 '16 at 0:56
• @whuber So it's about the link I posted? – mavavilj Mar 4 '16 at 11:19

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