Approximation of $\left(1-x^2\right)^{\tfrac{n-3}{2}}$ to Normal distribution

Question

Giving the PDF of $[-1<x<1]$ $$ f(x) = \frac{1}{\operatorname{B}\left(\frac{n-1}{2}, \frac{1}{2}\right)} \left(1-x^2\right)^{\tfrac{n-3}{2}} $$

where $\operatorname{B}\left(\frac{n-1}{2}, \frac{1}{2}\right)$ is Beta function.

Simulations show that when $n \to \infty$, $f(x)$ approximates Normal distribution (in this case, $n>10$).

The term beta can be approximated by Gamma function for large $n$. However I don't know how to proceed the term $\left(1-x^2\right)^{\tfrac{n-3}{2}}$ that defines the Normal form.

I appreciate any helps.

Answer 1

Let $a \in \mathbb{R},x=a n^{-1/2}$ for $n \geq |a|$. Then

$$(1-x^2)^{n/2}=(1-a^2/n)^{n/2}=\left ( (1-a^2/n)^n \right )^{1/2}.$$

The inner limit is going to $e^{-a^2}$, so the outer limit is going to $e^{-a^2/2}$. Thus the PDF of the rescaled random variable $\sqrt{n} X_n$ behaves asymptotically like the standard normal PDF in a $O(n^{-1/2})$ vicinity of zero. $X_n$ itself behaves asymptotically like zero, which is why we needed this rescaling to begin with.

Also note that outside a $O(n^{-1/2})$ vicinity of zero, both distributions are effectively giving zero probability, so this does not matter to this order of approximation.