Square of Standard Gaussian Random Variable has Chi-Squared Distribution

Theorem

Let $X \sim \Gaussian 0 1$ where $\Gaussian 0 1$ is the standard Gaussian distribution.

Then $X^2 \sim \chi^2_1$ where $\chi^2_1$ is the chi-square distribution with $1$ degree of freedom.

Let $Y \sim \chi^2_1$.

We aim to show that:

$\map \Pr {Y < x^2} = \map \Pr {-x < X < x}$

for all real $x \ge 0$.

We have:

$\ds \map \Pr {Y < x^2}$

$=$

$\ds \frac 1 {\sqrt 2 \map \Gamma {1 / 2} } \int_0^{x^2} t^{\paren {1 / 2} - 1} e^{-t / 2} \rd t$

$\ds $

$=$

$\ds \frac 1 {\sqrt {2 \pi} } \int_0^{x^2} \frac {e^{-t / 2} } {\sqrt t} \rd t$

$\ds $

$=$

$\ds \frac 2 {\sqrt {2 \pi} } \int_0^x u \frac {e^{-u^2 / 2} } {\sqrt {u^2} } \rd u$

$\ds $

$=$

$\ds \frac 2 {\sqrt {2 \pi} } \int_0^x e^{-u^2 / 2} \rd u$

$\ds $

$=$

$\ds \frac 1 {\sqrt {2 \pi} } \int_{-x}^x e^{-u^2 / 2} \rd u$

$\ds $

$=$

$\ds \map \Pr {-x < X < x}$

$\blacksquare$