Question

A random variable $X$ has probability density function as below:
$f_X\left(x\right)=\{\begin{array}{cc}8x;& 0\le x\le \frac{1}{2}\\ 0;& otherwise\end{array}$
If random variable $Y=2+4X^2$ then the probability density function $f_Y\left(y\right)$ at $y=3$ is

• A. $\sqrt{2}$
• B. 1
• C. $\frac{1}{2\sqrt{2}}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is B 1

$$\begin{array}{cc}{f}_Y\left(y\right)& =f_x\left(x\right)\left|\frac{dx}{dy}\right|\\ Y& =2+4X^2\\ \Rightarrow X& =\sqrt{\frac{Y-2}{4}}=\frac{\sqrt{Y-2}}{2}\\ \Rightarrow \frac{dx}{dy}& =\frac{1}{4}.\frac{1}{\sqrt{y-2}}\\ \Rightarrow f_Y\left(y\right)& =\frac{f_x\left(x\right)}{4\sqrt{y-2}}=\frac{8\sqrt{y-2}}{8\sqrt{y-2}}=1;2\le y\le 3\\ \therefore f_Y\left(3\right)& =1\end{array}$$