As degrees of freedom increase, why does the $\chi^2$ PDF approach a normal PDF?

Question

I am working with the $\chi^2$ pdf, and I have read several places that describe when the degrees of freedom increases, the $\chi^2$ pdf approaces a normal pdf.

Can anyone in detail explain to me, why this is the case.

Thanks in advance.

Answer 1

It is simply because of the Central Limit Theorem.

If $X \sim Chisq(5),$ then we can say that $X = Z_1^2 + Z_2^2 + \dots + Z_5^2,$ where the $Z_i$ are iid standard normal. As the df increase more $Z_i^2$s are being added together. Although $Z_i^2$ is a non-normal, right-skewed distribution it has a finite variance and so the CLT applies. Because of the skewness of $Z_i^2$ convergence is rather slow.

Here is a plot of the density of $Chisq(100)$ (solid curve), with the density of $Norm(100, \sqrt{200}).$ The chi-squared density is still slightly right-skewed.

Tangentially related comment: The probability that an $n$-variate uncorrelated standard normal observation falls inside the $n$-dimensional unit ball (centered at the origin) is given by $P(X \le 1)$ where $X \sim Chisq(df=n).$ This probability decreases markedly as $n$ increases. There is a 'lot of room' in higher dimensional spaces. Here is a plot of probabilities for $n = 1, 2, \dots, 10.$