91Ó°ÊÓ

Suppose that X_{1,…â¶Ä¦.,}X_nfrom a random sample from an exponential distribution with an unknown parameter β.

The p.d.f of an exponential distribution is,

f (x|β) = βexp(-βx)

As given in exercise 11 of section 8.8, with d(x) = x

Therefore \begin{aligned}T=\sum_{i-1}^{n}X{i}end{aligned}

Know that,

E(X_i) = 1/β and Var (X_i) = 1/β²

Hence,

\begin{aligned}E(T)=E\left(\sum_{i-1}^{n}\right)=\sum_{i-1}^{n}E(X_{i})=\frac{n}{\beta}\end{aligned}

\begin{aligned}Var(T)=Var\left(\sum_{i-1}^{n}X_{i}\right)\end{aligned}

\begin{aligned}Var\left(\sum_{i-1}^{n}X_{i}\right)\end{aligned}

=²Ô/β²

Since any linear function of T will also be an efficient estimator, it follows that x̄_n =T/n will be an efficient estimator of 1/β

So,

·¡(³æÌ„_n) = E(T/n)

= n/nβ

= 1/β

Var (x̄_n) = Var (T/n)

= 1/n² Var (T)

= n/n² β

= 1/nβ

Therefore, the mean of the estimator is 1/β and the variance of the estimator is 1/nβ.