Given that random variable Y Y Y is defined as Y = X 2 , Y = X^2, Y = X 2 , where X ∼ N ( 0 , 1 ) . X\sim N(0, 1). X ∼ N ( 0 , 1 ) .
Then for y < 0 , y<0, y < 0 , P ( Y < y ) = 0 P(Y<y)=0 P ( Y < y ) = 0 and
for y ≥ 0 , y\geq 0, y ≥ 0 ,
P ( Y < y ) = P ( X 2 < y ) = P ( − y < X < y ) P(Y<y)=P(X^2<y)=P(-\sqrt{y}<X<\sqrt{y}) P ( Y < y ) = P ( X 2 < y ) = P ( − y < X < y )
= F X ( y ) − F X ( − x ) = F X ( y ) − ( 1 − F X ( y ) ) =F_X(\sqrt{y})-F_X(-\sqrt{x})=F_X(\sqrt{y})-(1-F_X(\sqrt{y})) = F X ( y ) − F X ( − x ) = F X ( y ) − ( 1 − F X ( y ))
= 2 F X ( y ) − 1 =2F_X(\sqrt{y})-1 = 2 F X ( y ) − 1
f Y ( y ) = d d y ( 2 F X ( y ) − 1 ) = 2 d F X ( y ) d y f_Y(y)=\dfrac{d}{dy}(2F_X(\sqrt{y})-1)=2\dfrac{dF_X(\sqrt{y})}{dy} f Y ( y ) = d y d ( 2 F X ( y ) − 1 ) = 2 d y d F X ( y )
= 2 d d y ( ∫ − ∞ y 1 2 π e − t 2 / 2 d t ) =2\dfrac{d}{dy}\bigg(\displaystyle\int_{-\infin}^{\sqrt{y}}\dfrac{1}{\sqrt{2\pi}}e^{-t^2/2}dt\bigg) = 2 d y d ( ∫ − ∞ y 2 π 1 e − t 2 /2 d t )
= 2 ( 1 2 π ) e − ( y ) 2 / 2 ( 1 2 y ) =2(\dfrac{1}{\sqrt{2\pi}})e^{-(\sqrt{y})^2/2}(\dfrac{1}{2\sqrt{y}}) = 2 ( 2 π 1 ) e − ( y ) 2 /2 ( 2 y 1 )
= 1 2 ( 1 π ) e − y / 2 ( 1 y ) =\dfrac{1}{\sqrt{2}}(\dfrac{1}{\sqrt{\pi}})e^{-y/2}(\dfrac{1}{\sqrt{y}}) = 2 1 ( π 1 ) e − y /2 ( y 1 ) Use π = Γ ( 1 2 ) . \sqrt{\pi}=\Gamma(\dfrac{1}{2}). π = Γ ( 2 1 ) . Then
f Y ( y ) = 1 2 ⋅ Γ ( 1 2 ) y − 1 / 2 e − y / 2 f_Y(y)=\dfrac{1}{\sqrt{2}\cdot\Gamma(\dfrac{1}{2})}y^{-1/2}e^{-y/2} f Y ( y ) = 2 ⋅ Γ ( 2 1 ) 1 y − 1/2 e − y /2 where F F F and f f f are the cdf and pdf of the corresponding random variables.
Then Y = X 2 ∼ χ 1 2 Y=X^2\sim\chi_1^2 Y = X 2 ∼ χ 1 2 .
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