Question

Suppose that\({{\bf{X}}_{\bf{1}}}{\bf{,}}...{{\bf{X}}_{\bf{n}}}\)form a random sample from a distribution for which the p.d.f. is as follows:

\({\bf{f}}\left( {{\bf{x}}\left| {\bf{\theta }} \right.} \right){\bf{ = }}\left\{ {\begin{align}{}{{\bf{\theta }}{{\bf{x}}^{{\bf{\theta - 1}}}}}&{{\bf{for}}\,\,{\bf{0 < x < 1,}}}\\{\bf{0}}&{{\bf{otherwise,}}}\end{align}} \right.\)

where the value of θ is unknown (θ > 0). Determine the asymptotic distribution of the M.L.E. of θ. (Note: The M.L.E. was found in Exercise 9 of Sec. 7.5.)

Short answer

The asymptotic distribution of \(\theta \) is follows a standard normal distribution.

Step-by-step solution

Given information

Suppose that \({X_1},...{X_n}\) from a random sample from a distribution for which p.d.f. is,

\(f\left( {x\left| \theta \right.} \right) = \left\{ {\begin{align}{}{\theta {x^{\theta - 1}}}&{for\,\,0 < x < 1,}\\0&{otherwise,}\end{align}} \right.\)

Finding the asymptotic distribution of \({\bf{\theta }}\) 

The p.d.f. is given by,

\(f\left( {x\left| \theta \right.} \right) = \left\{ {\begin{align}{{}{}}{\theta {x^{\theta - 1}}}&{for\,\,0 < x < 1,}\\0&{otherwise,}\end{align}} \right.\)

Then,

Taking log in left side.

\(\log f\left( {x\left| \theta \right.} \right) = \lambda \left( {x\left| \theta \right.} \right)\)

So,

\(\lambda \left( {x\left| \theta \right.} \right) = \log \theta + \left( {\theta - 1} \right)\log x\)

Then,

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

The Fisher information is,

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

It follows that the asymptotic distribution of \(\frac{{{n^{\frac{1}{2}}}}}{\theta }\left( {{{\hat \theta }_n} - \theta } \right)\) is standard normal distribution.

Therefore, the asymptotic distribution of \(\theta \) is follows a standard normal distribution.