Finding pdfs of $\frac1{X^2}$ and $\frac{1}{4}\left(\frac1{X^2}+\frac1{W^2}\right)$ where $X,W$ are independent $N(0,1)$ $X,W$ are independent random variables, both $N(0,1)$, i.e. $f_X(x)=f_W(x)= \frac{1}{{\sqrt{2\pi}}}e^{-\frac{x^2}{2}}$.


*

*Find PDF of $Y:=\frac{1}{X^2}$

*Find PDF of $\frac{1}{4}\left(\frac{1}{X^2}+\frac{1}{W^2}\right)$
For the first case $a)$ I have:
\begin{eqnarray*}
 F_Y(y)=P(Y \leq y)=P(\frac{1}{X^2}\leq y) \:=
\end{eqnarray*}
\begin{Bmatrix}P(\frac{1}{X^2}\leq y) \: y<0
 & \\ P(\frac{1}{X^2} \leq y), \: y>0
 & 
\end{Bmatrix}
\begin{Bmatrix} 0,  \: y<0
 & \\ P(\frac{1}{X^2} \geq y), \: y>0
 & 
\end{Bmatrix}
$$ P(X^2 \geq \frac{1}{y})= P(X \leq -\frac{1}{\sqrt{y}} \cup X\geq \frac{1}{\sqrt{y}})=$$
$$ = P(X\leq -\frac{1}{\sqrt{y}}) + 1 - P(X\leq \frac{1}{\sqrt{y}}))= $$
$$= F_X(-\frac{1}{\sqrt{y}}) +1 - F_X(\frac{1}{\sqrt{y}}) $$
we have therefore:
$$ F_Y(y)= \left\{\begin{matrix}F_X(-\frac{1}{\sqrt{y}}) +1 - F_X(\frac{1}{\sqrt{y}}), \: y>0
 & \\ 0, \: y<0
 & 
\end{matrix}\right. $$
for $y>0$ we have:
$$ f_Y(y) = f_X(-\frac{1}{\sqrt{y}}))*(-\frac{1}{\sqrt{y}}))´ - f_X(\frac{1}{\sqrt{y}}))*(\frac{1}{\sqrt{y}}))´ $$
$$ = f_X(-\frac{1}{\sqrt{y}}))*\frac{y^{\frac{-3}{2}}}{2} - f_X(\frac{1}{\sqrt{y}}))*(-\frac{y^{\frac{-3}{2}}}{2})$$
$$ =\frac{y^{\frac{-3}{2}}}{2}*(f_X(-\frac{1}{\sqrt{y}})) - f_X(\frac{1}{\sqrt{y}}))) $$
$$ = \frac{1}{2} * y^{\frac{-3}{2}}(\frac{1}{\sqrt{2\pi }}e^{-\frac{1}{2y}} + \frac{1}{\sqrt{2\pi }}e^{-\frac{1}{2y}}) =  y^{\frac{-3}{2}}*\frac{1}{\sqrt{2\pi }}*e^{-\frac{1}{2y}}$$
For the second part, there is a hint that we should use polar coordinates and that the substitution $t=\cot(2 \theta)$ could be helpful. 
So since we have the PDF of both X and W (if what I have in (a) is correct), I tried using the convolution formula and calculating the integral, but I haven't been very successful, so I'd appreciate some help.
 A: Your work for the first problem is correct.
Let $\Phi(\cdot)$ and $\phi(\cdot)$ be the cdf and pdf of standard normal distribution, as usual.
Since $X$ is standard normal, you have for every $y>0$,
$$P\left(\frac1{X^2}\le y\right)=\cdots=1-\Phi\left(\frac1{\sqrt y}\right)+\Phi\left(-\frac1{\sqrt y}\right)=2\left(1-\Phi\left(\frac1{\sqrt y}\right)\right)$$
So the pdf of $Y=\frac1{X^2}$ must be
$$f_Y(y)=\frac1{y^{3/2}}\phi\left(\frac1{\sqrt y}\right)=\frac1{\sqrt{2\pi}y^{3/2}}e^{-1/2y}\mathbf1_{y>0}$$
You can also use change of variables to directly say
\begin{align}
f_Y(y)&=\phi\left(\frac1{\sqrt y}\right)\left|\frac{\mathrm d}{\mathrm dy}\left(\frac1{\sqrt y}\right)\right|+\phi\left(\frac{-1}{\sqrt y}\right)\left|\frac{\mathrm d}{\mathrm dy}\left(\frac{-1}{\sqrt y}\right)\right|
\\&=2\phi\left(\frac1{\sqrt y}\right)\left|\frac{\mathrm d}{\mathrm dy}\left(\frac1{\sqrt y}\right)\right|
\\&=\frac1{y^{3/2}}\phi\left(\frac1{\sqrt y}\right)
\end{align}
For the second problem, transform to polar coordinates $(X,W)\mapsto (R,\Theta)$ such that $X=R\cos\Theta$ and $W=R\sin\Theta$. Then 
$$Z=\frac14\left(\frac1{X^2}+\frac1{W^2}\right)=\frac{X^2+W^2}{4X^2W^2}=\frac1{(R\sin(2\Theta))^2}$$
All you have to do now is show that $R\sin(2\Theta)$ also has a standard normal distribution, so that $$Z\stackrel{d}=\frac1{X^2}=Y$$
