91Ó°ÊÓ

First, note that

\({Y_n} = \sum\nolimits_{i = 1}^n {{\raise0.7ex\hbox{${X_i^2}$} \!\mathord{\left/ {\vphantom {{X_i^2} n}}\right.}\!\lower0.7ex\hbox{$n$}}} \)

Asymptotically the normal distribution with mean \({\sigma ^2}\) and variance \(\frac{{2{\sigma ^2}}}{n}\).

Here we have used the fact that \(E\left( {X_i^2} \right) = {\sigma ^2}\,\,and\,\,E\left( {X_i^4} \right) = 2{\sigma ^4}\)

Let,

\(\begin{array}{c}g\left( x \right) = \frac{1}{x}\\g'\left( x \right) = - \frac{1}{{{x^2}}}\end{array}\)

So, the asymptotic distribution of\(g\left( {{Y_n}} \right)\)is the normal distribution with

\(mean = \frac{1}{{{\sigma ^2}}}\)

And

\(\begin{array}{c}{\mathop{\rm var}} iance = \left( {\frac{{2{\sigma ^4}}}{n}} \right) \times \left( {\frac{1}{{{\sigma ^8}}}} \right)\\ = \frac{2}{{n{\sigma ^4}}}\end{array}\)

Therefore, the asymptotic distribution of the statistic follows a normal distribution with \(mean = \frac{1}{{{\sigma ^2}}}\) and \({\mathop{\rm var}} iance = \frac{2}{{n{\sigma ^4}}}\)

Let,

\(h\left( \mu \right) = 2m{u^2}\)

If the asymptotic mean is,

\(Mean\left( {{Y_n}} \right) = \mu \)

The asymptotic variance is,

\(Var\left( {{Y_n}} \right) = \frac{{h\left( \mu \right)}}{n}\)

So, the variance stabilizing transformation is,

\(\begin{array}{c}\alpha \left( \mu \right) = \int_a^\mu {\frac{{dx}}{{{2^{\frac{1}{2}}}x}}} \\ = \frac{1}{{{2^{\frac{1}{2}}}}}\left[ {\log \left( x \right)} \right]_a^\mu \\ = \frac{1}{{{2^{\frac{1}{2}}}}}\log \left( \mu \right)\end{array}\)

Where\(a = 1\)to make the integral finite.

So, the asymptotic distribution of\(\frac{{\log \left( {{Y_n}} \right)}}{{{2^{\frac{1}{2}}}}}\)is the normal distribution with mean\(\frac{{2\log \left( \sigma \right)}}{{{2^{\frac{1}{2}}}}}\)and variance\(\frac{1}{n}\)

Therefore, the variance stabilizing transformation for statistics follows a normal distribution with mean\(\frac{{2\log \left( \sigma \right)}}{{{2^{\frac{1}{2}}}}}\)and variance\(\frac{1}{n}\)