Suppose $s^2$ is the sample variance of a sample$(\text{of size }n)$ from a normal population with mean $\mu$ and variance $\sigma^2. \text{Here }s^2=\frac{\sum_{i=1}^n(x_i-\overline{x})^2}{n-1} $
As we know $\frac{(n-1)s^2}{\sigma^2}\sim \chi^2_{(n-1)}$ here . Let $x\sim \chi^2_{(n-1)}$ then $E[x]=n-1$ and $Var[x]=2(n-1)$ \begin{align*} E\left[\frac{(n-1)s^2}{\sigma^2}\right]=E[x]&=n-1\\ \implies \frac{(n-1)}{\sigma^2}E[s^2] &= n-1\\ \implies E[s^2]&=\sigma^2 \end{align*} Similarly, \begin{align*} Var\left[\frac{(n-1)s^2}{\sigma^2}\right]=Var[x]&=2(n-1)\\ \implies \frac{(n-1)^2}{\sigma^4}E[s^2] &=2(n-1)\\ \implies Var[s^2]&=\color{red}{\frac{2\sigma^4}{(n-1)} } \end{align*} Now we need a function of $s^2$ whose variance will be independent of $\sigma^2.$ Let $f(s^2)$ be the required transformation.

The required transformation is $$f(s^2)\stackrel{?}{=}\ln{s^2}$$

Question: How they get that transformation $?$
Similar things happen with Poisson variate and Binomial proportion with square root and $\sin^{-1}$ transformation. So I need a general approach to get that transformation which will make variance independent of population parameter.

Thanks for your time. Thanks in advance .

  • $\begingroup$ Search for Variance-stabilizing_transformation. $\endgroup$ Commented Nov 6, 2019 at 14:33
  • $\begingroup$ "The aim behind the choice of a variance-stabilizing transformation is to find a simple function f to apply to values x in a data set to create new values y=f(x) such that the variability of the values y is not related to their mean value." But I need the variability of the values y is not related to their variance(population) value? @StubbornAtom Sir $\endgroup$ Commented Nov 6, 2019 at 14:45
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    $\begingroup$ The wiki article might not be entirely clear. I gave you the keyword to look for the topic. (And I am no 'sir'.) $\endgroup$ Commented Nov 6, 2019 at 14:51

1 Answer 1


Suppose $X_1,X_2,\ldots,X_n$ are i.i.d $N(\mu,\sigma^2)$ and define the sample variance as $$S^2=\frac{1}{n}\sum\limits_{i=1}^n (X_i-\overline X)^2$$

Here we are concerned with the large-sample behaviour of $S^2$, so it does not matter if we take $n$ as the divisor above instead of $n-1$.

By CLT it can be argued that $$\sqrt n(S^2-\sigma^2)\stackrel{L}\longrightarrow N(0,2\sigma^4)\tag{1}$$

And by 'Delta-method', for a real-valued function $g$ such that $g'(\sigma^2)$ exists and $g'(\sigma^2)\ne 0$, it follows from $(1)$ that

$$\sqrt n(g(S^2)-g(\sigma^2))\stackrel{L}\longrightarrow N(0,2\sigma^4[g'(\sigma^2)]^2)\tag{2}$$

You have to solve for a $g$ such that asymptotic variance of $g(S^2)$ is free of $\sigma^2$.

So for some non-zero constant $c$, set $$\sqrt 2\sigma^2 g'(\sigma^2)=c$$

Therefore, $$\int g'(\sigma^2)\,d\sigma^2=\frac{c}{\sqrt 2}\int\frac{d\sigma^2}{\sigma^2}$$

Choosing $c=\sqrt 2$ and taking constant of integration to be zero, we have the required transformation $$g(\sigma^2)=\ln\sigma^2\quad,\text{ i.e. }\quad \color{blue}{g(S^2)=\ln S^2}$$

From $(2)$ you end up with $$\sqrt n(\ln(S^2)-\ln(\sigma^2))\stackrel{L}\longrightarrow N(0,2)$$

This is the application of variance stabilizing transformation on the sample variance from a normal population. The other transformations that you mention for the Poisson mean and Binomial proportion are derived similarly.

Result $(1)$ and many more can be found in A Course in Large Sample Theory by Thomas Ferguson. The relevant part is on page 46 of the first edition (1996):

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The main result used in this section is Cramer's Theorem on the asymptotic normality of functions of sample moments, studied through a Taylor-series expansion to one term.

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    $\begingroup$ what I know is CLT tells us $\color{red}{\sqrt{n}(\overline{x}-\mu)\text{ converge in distribution to a normal }N(0,\sigma^2)}$ And I know how to prove this but $\sqrt n(s^2-\sigma^2)\stackrel{L}\longrightarrow N(0,2\sigma^4)$ is looks new to me. So everything will be meaningful to me if you provide a source where I can get that proof $?$ I will accept your answer then. Thanks again @StubbornAtom $\endgroup$ Commented Nov 6, 2019 at 16:07
  • $\begingroup$ @emonhossain Updated. $\endgroup$ Commented Nov 6, 2019 at 18:07
  • $\begingroup$ Really you help me a lot. Thanks @StubbornAtom $\endgroup$ Commented Nov 6, 2019 at 18:33

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