Sum of Squares of Standard Gaussian Random Variables has Chi-Squared Distribution

Theorem
Let $X_1, X_2, \ldots, X_n$ be independent random variables.

Let $X_i \sim \Gaussian 0 1$ for $1 \le i \le n$ where $\Gaussian 0 1$ is the standard Gaussian Distribution.

Then:


 * $\displaystyle \sum_{i \mathop = 1}^n X^2_i \sim \chi^2_n$

where $\chi^2_n$ is the chi-squared distribution with $n$ degrees of freedom.

Proof
By Square of Standard Gaussian Random Variable has Chi-Squared Distribution, we have:


 * $X^2_i \sim \chi^2_1$

for $1 \le i \le n$.

So, by Sum of Chi-Squared Random Variables, we have:


 * $\displaystyle \sum_{i \mathop = 1}^n X^2_i \sim \chi^2_{1 + 1 + 1 \ldots} = \chi^2_n$