Question

LetX1, . . . , Xnbe a random sample from the exponential distribution with parameterθ. Find the c.d.f. for the sampling distribution of the M.L.E. ofθ. (The M.L.E. itself was found in Exercise 7 in Sec. 7.5.)

Short answer

The c.d.f of the sampling distribution of the M.L.E of\(\theta \)is,

\(H\left( t \right) = 1 - G\left( {\frac{n}{t}} \right)\)

Step-by-step solution

Given information

Referring to Exercise 7 in Sec. 7.5

Finding sample size

The MLE is \(\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over \theta } = n/T\) , where T was shown to have gamma distribution with parameters n and \(\theta \)

Let \(G\left( \cdot \right)\) denote the cdf of the sampling distribution of T. Let \(H\left( \cdot \right)\) be the cdf of the sampling distribution of \(\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}} \over \theta } \).

Then\(H\left( t \right) = 0\,for\,t \le 0,\,and\,for\,t > 0\)

\(\begin{align}H\left( t \right) &= \Pr \left( {\hat \theta \le t} \right)\\ &= \Pr \left( {\frac{n}{T} \le t} \right)\\ &= \Pr \left( {T \ge \frac{n}{t}} \right)\\ &= 1 - G\left( {\frac{n}{t}} \right)\end{align}\)

The c.d.f of the sampling distribution of the M.L.E of\(\theta \)is,

\(H\left( t \right) = 1 - G\left( {\frac{n}{t}} \right)\)