Finding the distribution of a function of a normal random variable Question:  A particle's velocity $V$ is normally distributed with mean 0 and variance $\sigma^2$. The particle's energy is given by $W=m\frac{V^2}{2}$, where $m>0$ is a constant.  
(a) What is $\text{E}[W]$ in terms of $m$ and $\sigma^2$?
(b) What is the PDF of $W$?
I'll post my best attempt below but I'm curious if there is an easier approach to (b). I'm trying to get better at this techniques for this sort of thing.
 A: (a)
$\begin{align}\text{E}[W]&=\text{E}\left[\frac{mV^2}{2}\right] \\
&=\frac{m}{2}\text{E}[V^2] \\
&=\frac{m}{2}\left(\text{Var}(V)+\text{E}[V]^2 \right) \\
&=\frac{m}{2}(\sigma^2 + 0) \\
&=\frac{m \sigma^2}{2}
\end{align}$ 
(b) To get the PDF for $W$, start with CDF.  The CDF is zero for $w<0$ so compute it for $w\ge 0$.   
$\begin{align}F_W(w)&=P(W\le w) \\
&=P\left( \frac{mV^2}{2} \le w \right) \\
&=P\left( V^2 \le \frac{2w}{m} \right) \\
&=P\left( |V| \le \sqrt{\frac{2w}{m}} \right) \\
&=P\left( -\sqrt{\frac{2w}{m}} \le V \le \sqrt{\frac{2w}{m}} \right) \\
&=F_V \left( \sqrt{\frac{2w}{m}} \right) - F_V \left( -\sqrt{\frac{2w}{m}} \right)
\end{align}$ 
Now the CDF is done since $V\sim \text{Normal}(0,\sigma^2)$ so $F_V$ is known. 
Now take the derivative with respect to $w$...
\begin{align} f_W(w) &= \frac{d}{dw}F_W(w) \\
&= f_V \left( \sqrt{\frac{2w}{m}} \right)\frac{1}{2\sqrt{w}}\sqrt{\frac{2}{m}} + f_V \left( -\sqrt{\frac{2w}{m}} \right)\frac{1}{2\sqrt{w}}\sqrt{\frac{2}{m}}\\
&=\frac{1}{\sqrt{2mw}}\left[ f_V \left( \sqrt{\frac{2w}{m}} \right) +f_V \left( -\sqrt{\frac{2w}{m}} \right) \right] \\
&=\frac{1}{\sqrt{2mw}}\left[ \dfrac{1}{\sqrt{2\pi \sigma^2}\text{e}^{-\frac{2w}{2m \sigma^2}}} + \frac{1}{\sqrt{2\pi \sigma^2}\text{e}^{-\frac{2w}{2m \sigma^2}}} \right] \\
&= \frac{1}{\sqrt{mw \pi \sigma^2}}\text{e}^{-\dfrac{w}{m \sigma^2}} 
\end{align}

Update:  @DilipSarwate (user profile) has kindly pointed out that $W$ follows a scaled Chi-squared distribution, specifically a scaled $\chi^2(1)$ pdf.  
A: Partial confirmation, using simulation in R: Let $\sigma^2 = 4, \sigma=2, m = 10.$
set.seed(1017);  B = 10^6;  sg = 2;  m = 10
v = rnorm(B, 0, sg);  w = .5*m*v^2
mean(w);  m*sg^2/2
[1] 20.02141   # aprx energy = 20
[1] 20         # exact energy
hist(w, prob=T, col="skyblue2", main="Histogram of Kinetic Energy")
curve((pi*m*x)^-.5/sg*exp(-x/(m*sg^2)), 0, 400, add=T, col="red", lwd=2)
   # note: curve function requires argument 'x'


Everything seems OK, methods and results (+1). Your expectation is within the margin of simulation error for the constants I used, and your density function seems a good match for
the histogram. I think kinetic energy has a gamma distribution with shape parameter 1/2. (Maybe see Wikipedia on 'gamma distribution'.)
Possible alternative method for PDF. You might find it easier to get the PDF of $W$ from the normal PDF of $V$ using the 'PDF method' (also called 'transformation' method)
shown in many elementary or intermediate probability texts. I think it's especially easy here because the transformation and  the Jacobian are simple.
A: This is something I have been thinking about, here is what I have come up with:
Q: 
What is the $\mathrm{pdf}$ of the random variable $Y$, $f_Y(y)$, given $y=U(x)$ and given the $\mathrm{pdf}$ of $X$, $f_X(x)$?A: 
$f_Y(y)=\frac{\mathrm{d} }{\mathrm{d} y} F_X \left (U^{-1}(y)  \right )$
Where $F_X(x)$ is the $\mathrm{cdf}$ of $X$, and $U^{-1}(y)$ is the inverse function of $y=U(x)$.

Q: 
What is $f_Y(y)$, given $y=x^n$, $n\in\mathbb{N}$ and $X\sim N(\mu, \sigma)$?A: 
$\sqrt{\frac{1}{2\pi\sigma^2\,n^2}}\,y^{-1+1/n}\,\mathrm{exp}\left ( -\frac{(y^{1/n}-\mu)^2}{2\sigma^2} \right ) \times
\left\{\begin{matrix}
2, & n~\mathrm{even}, & 0 < y < \infty\\ 
1, & n~\mathrm{odd}, & -\infty < y < \infty
\end{matrix}\right.$
Which agrees with your answer when $\omega\rightarrow 1$.
