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.
Thanks for your time. Thanks in advance .
 A: 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.

