Sum of squares of iid Normal distributions

Question

What distribution for iid Standardized Normal Distributions $X_i \sim N(0,1),i=1,\ldots,n$ has this distribution $$X_1^2+\ldots+X_n^2?$$

What role does independence play there? Thank you.

Answer 1

Comment: The following information is scattered in various of the references mentioned in previous Comments. Here, one one page, is the outline of a proof that for $n$ independent $X_i \sim \mathsf{Normal}(0,1),$ one has $Q = \sum_{i=1}^n X_i^2 \sim \mathsf{Chisq}(n).$

First, show that $X_1^2 \sim \mathsf{Chisq}(1).$ One method is to find the density of $|X_i|,$ with support $(0, \infty).$ Then use elementary transformation to find that $X_1^2$ has the PDF of $\mathsf{Chisq}(1).$

Next, use PDF's to find the MGF's $M_X(t) = \frac{1}{(1-2t)^{1/2}}$ of $X_1$ and $M_Q(t) = \frac{1}{(1-2t)^{n/2}}$ of $\mathsf{Chisq}(n),$ for $t < 1/2.$

Finally, use the multiplicative property of MGFs to recognize that $[M_X(t)]^n = M_Q(t).$