Question

Suppose that\({X_1},{X_2}...{X_n}\).form a random sample from the exponential distribution with mean\(\theta \). Let\(\overline {{X_n}} \)be the sample average. Find a variance stabilizing transformation for\(\overline {{X_n}} \).

Short answer

Variance stabilizing transformation for \(\overline {{X_n}} \) is \(\log \left( x \right)\)

Step-by-step solution

Given information

Let \({X_1},...,{X_n}\) form a random sample for the exponential distribution with mean \(\theta \)

Finding variance stabilizing transformation for \(\overline {{X_n}} \)

According to the central limit theorem,\(\overline {{X_n}} \)has approximately the normal distribution with mean\(\theta \)and variance\({\theta ^2}\).

\(\alpha \left( x \right) = \int\limits_a^x {{\theta ^{ - 1}}d\theta } \)

Choosing a=1

\(\alpha \left( x \right) = \int\limits_1^x {{\theta ^{ - 1}}d\theta } \)

Therefore

\(\alpha \left( x \right) = \log \left( x \right)\)

Hence the stabilizing transformation for \(\overline {{X_n}} \) is \(\log \left( x \right)\).