Question

(ii) Suppose X N(0,1) has the standard normal distribution. Find the probability distribution function ofY - 2

Answer 1

here as we know that pdf of X :f(x)=$\frac{1}{\sqrt{2\pi }}e^{-x^{2}/2}$  

also as Y=1/X²

X =-/+ 1/$\sqrt{y}$

|dx/dy| =(1/2)y^-3/2

hence from transformation method:

pdf of Y f(y)=|dx/dy|*(f_x(g(y))+f_x(-g(y)))

=$(1/2)y^{-3/2} *(\frac{1}{\sqrt{2\pi }}e^{-(1/\sqrt{y})^{2}/2}+\frac{1}{\sqrt{2\pi }}e^{-(-1/\sqrt{y})^{2}/2})$

f(y)= $y^{-3/2} *(\frac{1}{\sqrt{2\pi }}e^{-\frac{1}{2y}})$ for 0 <y< $\infty$