91Ó°ÊÓ

Referring to exercise 19, suppose that\({X_1},...,{X_n}\) from a random sample from the exponential distribution with an unknown parameter \(\beta \).

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

\(f\left( {x\left| \beta \right.} \right) = \beta \exp \left( { - \beta x} \right)\)

a.

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

\(f\left( {x\left| \beta \right.} \right) = \beta \exp \left( { - \beta x} \right)\)

The mean of the exponential distribution is,

\(E\left( x \right) = \frac{1}{\beta }\)

Let,

\(\alpha \left( \beta \right) = \frac{1}{\beta }\)

Then,

\(\alpha '\left( \beta \right) = - \frac{1}{{{\beta ^2}}}\)

It is known that\({\hat \beta _n}\)is approximately normal with mean\(\beta \)and variance\(\frac{{{\beta ^2}}}{n}\)

The delta method is a general technique for calculating the variance of a function of known variance asymptotically normal random variables. The delta technique is used in this instance to construct a closed-form solution for the margin's standard errors by making use of the fact that the margin is (typically) an endlessly differentiable function of the data, X, and the vector of\(\beta s\).

Therefore, \(\frac{1}{{{{\hat \beta }_n}}}\) will be approximately normal with mean \(\frac{1}{\beta }\) and the variance \({\left( {\alpha '\left( \beta \right)} \right)^2}\left( {\frac{{{\beta ^2}}}{n}} \right) = \frac{1}{{\left( {n{\beta ^2}} \right)}}\)

Equivalently, the asymptotic distribution of \({\left( {n{\beta ^2}} \right)^{\frac{1}{2}}}\left( {\frac{1}{{{{\hat \beta }_n}}} - \frac{1}{\beta }} \right)\) is standard normal distribution.

b.

Since the mean of the exponential distribution is\(\frac{1}{\beta }\)and variance is\(\frac{1}{{{\beta ^2}}}\)

It follows directly from the central limit theorem that the asymptotic distribution of

\({\bar X_n} = \frac{1}{{{{\hat \beta }_n}}}\)is exactly found in part (a).

Hence, (Proved)