Question #242556

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

1
Expert's answer
2021-10-14T11:58:05-0400

Given that random variable YYis defined as Y=X2,Y = X^2,  where XN(0,1).X\sim N(0, 1).

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

for y0,y\geq 0,

P(Y<y)=P(X2<y)=P(y<X<y)P(Y<y)=P(X^2<y)=P(-\sqrt{y}<X<\sqrt{y})

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

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

fY(y)=ddy(2FX(y)1)=2dFX(y)dyf_Y(y)=\dfrac{d}{dy}(2F_X(\sqrt{y})-1)=2\dfrac{dF_X(\sqrt{y})}{dy}

=2ddy(y12πet2/2dt)=2\dfrac{d}{dy}\bigg(\displaystyle\int_{-\infin}^{\sqrt{y}}\dfrac{1}{\sqrt{2\pi}}e^{-t^2/2}dt\bigg)


=2(12π)e(y)2/2(12y)=2(\dfrac{1}{\sqrt{2\pi}})e^{-(\sqrt{y})^2/2}(\dfrac{1}{2\sqrt{y}})

=12(1π)ey/2(1y)=\dfrac{1}{\sqrt{2}}(\dfrac{1}{\sqrt{\pi}})e^{-y/2}(\dfrac{1}{\sqrt{y}})

Use π=Γ(12).\sqrt{\pi}=\Gamma(\dfrac{1}{2}). Then


fY(y)=12Γ(12)y1/2ey/2f_Y(y)=\dfrac{1}{\sqrt{2}\cdot\Gamma(\dfrac{1}{2})}y^{-1/2}e^{-y/2}

where FF and ff are the cdf and pdf of the corresponding random variables.


Then Y=X2χ12Y=X^2\sim\chi_1^2 .




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