Show that $Y\sim\Gamma(\frac{1}{2},\frac{1}{2})$

Question

let $X\sim N(0,1)$ be a standard normally distributed stochastic variable and let $Y=X^2$.

Show that $Y\sim\Gamma(\frac{1}{2},\frac{1}{2})$, ie. $Y\sim\chi^2(1)$.

Answer 1

Use mgf:

$$E[e^{tY}] = E[e^{tX^2}] = \int_{\mathbb R} e^{tx^2} f_X(x) dx$$

$$ = \int_{\mathbb R} \frac{e^{(t-1/2)(x^2)}}{\sqrt{2\pi}} dx = \frac{1}{\sqrt{1-2t}}$$

Hence $Y \sim \chi^2(1)$

Answer 2

Outline/Hint: Notice that the range of $X$ is $(-\infty,\infty)$. That makes $Y = X^2$ a many-to-one relationship. Thus, use $$f_Y(y) = \frac{f_X(-\sqrt y)}{\left|\frac{dy}{dx}\right|}+\frac{f_X(\sqrt y)}{\left|\frac{dy}{dx}\right|}$$

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Then, $$f_Y(y) = \frac{\frac{1}{\sqrt{2\pi}}\exp\left\{-\frac{1}{2}(-\sqrt y)^2\right\}}{|-2\sqrt y|}+\frac{\frac{1}{\sqrt{2\pi}}\exp\left\{-\frac{1}{2}(\sqrt y)^2\right\}}{|2\sqrt y|} = \frac{1}{\sqrt{2\pi y}}e^{-\frac{1}{2}y^2},$$ which is the density of $\chi^2(1)$.