# Which transformation needed to make variance independent of population parameter?

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.

• Search for Variance-stabilizing_transformation. – StubbornAtom Nov 6 '19 at 14:33
• "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 – emonhossain Nov 6 '19 at 14:45
• The wiki article might not be entirely clear. I gave you the keyword to look for the topic. (And I am no 'sir'.) – StubbornAtom Nov 6 '19 at 14:51

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):

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.

• 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 – emonhossain Nov 6 '19 at 16:07
• @emonhossain Updated. – StubbornAtom Nov 6 '19 at 18:07
• Really you help me a lot. Thanks @StubbornAtom – emonhossain Nov 6 '19 at 18:33