Answer to Question #242556 in Statistics and Probability for PRS

Question

Given that X N(0,1) then show that the pdf of Y=X² is $\chi$ ₁²

Answer 1

Given that random variable $Y$is defined as $Y = X^2,$  where $X\sim N(0, 1).$

Then for $y<0,$ $P(Y<y)=0$ and

for $y\geq 0,$

$$P(Y<y)=P(X^2<y)=P(-\sqrt{y}<X<\sqrt{y})$$

$$=F_X(\sqrt{y})-F_X(-\sqrt{x})=F_X(\sqrt{y})-(1-F_X(\sqrt{y}))$$

$$=2F_X(\sqrt{y})-1$$

$$f_Y(y)=\dfrac{d}{dy}(2F_X(\sqrt{y})-1)=2\dfrac{dF_X(\sqrt{y})}{dy}$$

$$=2\dfrac{d}{dy}\bigg(\displaystyle\int_{-\infin}^{\sqrt{y}}\dfrac{1}{\sqrt{2\pi}}e^{-t^2/2}dt\bigg)$$

$$=2(\dfrac{1}{\sqrt{2\pi}})e^{-(\sqrt{y})^2/2}(\dfrac{1}{2\sqrt{y}})$$

$$=\dfrac{1}{\sqrt{2}}(\dfrac{1}{\sqrt{\pi}})e^{-y/2}(\dfrac{1}{\sqrt{y}})$$

Use $\sqrt{\pi}=\Gamma(\dfrac{1}{2}).$ Then

$$f_Y(y)=\dfrac{1}{\sqrt{2}\cdot\Gamma(\dfrac{1}{2})}y^{-1/2}e^{-y/2}$$

where $F$ and $f$ are the cdf and pdf of the corresponding random variables.

Then $Y=X^2\sim\chi_1^2$ .