I am reading this example where the distribution is given by $Y=\frac{\sigma^2\chi^2_{n-1}}{n-1}.$ By CLT, $Y\sim\mathcal{N}(\sigma^2,\frac{2\sigma^4}{n-1}).$ Up to here it was all clear to me.

Then my textbook said the sitribution of $2\sqrt{Y}$, using Delta Method, is approximately $\mathcal{N}(2\sigma,\frac{2\sigma^2}{n-1}).$

However I cannot seem to use Delta Method to get the aforementioned variance. I thought the statement maybe missing out a $n$ term? I thought the variance should be $\frac{2\sigma^2}{(n-1)n}$ because we have to divide by $n$ at the end when applying Delta Method?


The statement from the textbook is correct. If we define $f(y)=2\sqrt y$, we have $f'(y)=1/\sqrt y$. By the delta method, we approximate:

$$\begin{align} \text{Var}(2\sqrt Y)&=f'(\mathbb E[Y])^2\cdot\text{Var(Y)}\\ &=\frac{1}{\sigma^2}\cdot\frac{2\sigma^4}{n-1}\\ &=\frac{2\sigma^2}{n-1} \end{align}$$

Which agrees with the textbook result.


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