Question

Suppose that\({{\bf{X}}_{\bf{1}}}{\bf{,}}...{\bf{,}}{{\bf{X}}_{\bf{n}}}\)form a random sample from the exponential distribution with unknown parameter β. Show that if n is large, the distribution of the M.L.E. of β will be approximately a normal distribution with mean β and variance\(\frac{{{{\bf{\beta }}^{\bf{2}}}}}{{\bf{n}}}\).

Short answer

Proved.

Step-by-step solution

Given information

Suppose that\({X_1},...,{X_n}\) from a random sample from the exponential distribution with an unknown parameter \(\beta \).

Showing part

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

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

Then,

Taking log,

\(\begin{align}\lambda \left( {x\left| \beta \right.} \right) &= \log f\left( {x\left| \beta \right.} \right)\\ &= \log \left( \beta \right) - \beta x\end{align}\)

\(\begin{align}\lambda '\left( {x\left| \beta \right.} \right) &= \frac{1}{\beta } - x\\\lambda ''\left( {x\left| \beta \right.} \right) &= - \frac{1}{{{\beta ^2}}}\end{align}\)

The Fisher information is,

\(\begin{align}I\left( \beta \right) &= - {E_\theta }\left( { - \frac{1}{{{\beta ^2}}}} \right)\\ &= \frac{1}{{{\beta ^2}}}\end{align}\)

The distribution of the M.L.E of \(\beta \) will be approximately the normal distribution with the mean is \(\beta \) and the variance is \(\frac{1}{{{\beta ^2}}}\)

Hence, (Proved)