Question

Let $X$ be a random variable having probability density function $f_{X}(x)=\begin{cases}2x & 0<x<1 \\ 0 & \text{otherwise}\end{cases}$ then the density of $Y=\frac{1}{X^{\frac{1}{\alpha}}}$ is

• A. $f_{Y}(y)=\frac{\alpha}{y^{\alpha+1}}; y>1$
• B. $f_{Y}(y)=\frac{\alpha}{2y^{\frac{\alpha}{2}+1}}; y>1$
• C. $f_{Y}(y)=\frac{2\alpha}{y^{2\alpha+1}}; y>1$
• D. $f_{Y}(y)=\frac{1}{2\alpha y^{\frac{1}{2\alpha}+1}}; y>1$

Hint:

When performing a transformation $Y=g(X)$, the PDF of $Y$ is $f_Y(y) = f_X(x) |\frac{dx}{dy}|$. Always ensure you substitute $x$ entirely with its equivalent expression in $y$ and re-calculate the support (range) for the new variable.

Answer 1

The Correct Option is C