$V$ is a normal rv with mean $0$ and variance $2$. Find the PDF of the function $Z = \frac{mV^2}{2}$

Question

How can I find the PDF of the function $Z=\frac{mV^2}{2}$, in terms of $\sigma$ and $\mu$, where $\mu$ is a constant greater than $0$?

Answer 1

This is a hint. You can use a change of variable or the cdf method $$P(Z<z) = P\left(\frac{m}{2}V^2<z\right).$$ This method seems more popular.

Since the title says $\mu = 0, \sigma^2 = 2$, then I know that $$Y = \left(\frac{V}{\sqrt 2}\right)^2\sim \text{Gamma}(1/2,1/2).$$ From there, I can tell you that $$Z = \frac{\sqrt 2 m}{2}Y \sim \text{Gamma}\left(\frac{1}{2},\frac{1}{\sqrt 2 m}\right). $$ You can compare your work against this.

Answer 2

If $X\sim N(0,1)$, then $X^2\sim\chi^2(1)$, where $\chi^2(1)$ is the chi-squared distribution with one degree of freedom.

If the random variable $Y$ has a density function $f_Y(y)$, then the random variable $cY$ has the density function $$ f_{cY}(x)=\frac 1cf_Y\Big(\frac xc\Bigr) $$ (see here for more details).