Show that the random variables $Z = XY \sim N(0,\alpha^2)$. The random variables $X$ and $Y$ are independent with 
$$ f_X (x) = x/α^2 e^{-x^2/2α^2} \text{ for } x > 0 $$ and 
$$ f_Y(y) = \frac {1} {\pi \sqrt{1-y^2}}    \text{ for } |y|<1 $$ and is zero otherwise.
Show that the random variable $Z = XY$ is $N(0,\alpha^2)$.
My thinking on how to solve this is by finding the joint density $f_{X,Y}(x,y)$ of $X$ and $Y$ given that the individual densities $f_X(x)$ and $f_Y(y)$ are given. 
If the joint density was given, I would differentiate the cdf of $X$ with respect to $x$ to get the marginal density (similarly with $Y$).  However, I'm not really sure how to do it "in reverse" for lack of a better term. Not certain this is the correct approach. 
 A: Take $U=XY$ and $V=X$ then $Y=U/V$. (Note: $V^2 >U^2$) The  Jacobian will be
$
   J=
  \left[ {\begin{array}{cc}
   0 & 1/v \\       1 & -u/v^2 \      \end{array} } \right]
$
i.e. the |$Det(J)|=1/v$ by which you'll multiply the joint density function. 
Since $X$ and $Y$ are independent,
$$f_{X,Y}(x,y)=\frac{1}{\alpha^2 \pi} \frac{1}{\sqrt{1-y^2}}\times x\times \exp{(-x^2/2\alpha^2)} I(x>0, |y|<1)$$
$$f_{U,V}(u,v)=\frac{1}{\alpha^2 \pi} \frac{1}{\sqrt{1-\frac{u^2}{v^2}}}\times v \times \exp{(-v^2/2\alpha^2)} \times \frac{1}{v}=\frac{1}{\alpha^2 \pi}\frac{v \times exp(-v^2/2\alpha^2)}{\sqrt{v^2-u^2}}I(v>0, v^2>u^2)$$
Now simply integrate $v$ out of the above joint density. Note Range of $V^2$ is $(U^2,\infty)$. Substitute $y^2=v^2-u^2$ which implies $v.dv=y.dy$ and range of $y$ will be $(0,\infty)$. 
$$\int_{|u|}^{\infty} f(u,v).dv=\frac{1}{\alpha^2 \pi}\int_{|u|}^{\infty} \frac{ exp(-v^2/2\alpha^2)}{\sqrt{v^2-u^2}}v \times dv=\frac{1}{\alpha^2 \pi}\int_{0} ^{\infty}\frac{\exp\{-(u^2+y^2)/2\alpha^2\}}{y}y.dy=\frac{1}{\alpha^2 \pi}e^{-u^2/2\alpha^2}\int_{0} ^{\infty}e^{-y^2/2\alpha^2}dy$$
I think now you can figure it out. If you've any further query let's know.
A: Along the same lines as Batminovski's answer but with a different flavoring: $X$ has a Rayleigh distribution (familiar to wireless
communications engineers as the distribution of the amplitude of
a sinusoidal signal that has undergone Rayleigh fading) while $Y$ has the
distribution of $\cos(\Theta)$ where $\Theta \sim U[0,2\pi)$ is a
uniformly distributed random variable. Thus, $XY = R\cos(\Theta)$
has normal distribution $N(0,\sigma^2)$ and $XW = R\sin(\Theta)$
also has normal distribution $N(0,\sigma^2)$. Furthermore, $XY$ and $XW$
are independent random variables. This result is the basis of
what is known as the Box-Muller method for generating samples from
a normal distribution.
A: Physics Argument:
Consider a two-dimensional ideal gas with molecular mass $m$ at temperature $T$.  The velocity $X$- and $Y$-components of a gas molecule, called $V_x$ and $V_y$, respectively, are independent with the same $\mathcal{N}\left(0,\alpha^2\right)$-distribution, where $\alpha:=\sqrt{\frac{k_\text{B}T}{m}}$ and $k_\text{B}$ is the Boltzmann constant.  
Let $V:=\sqrt{V_x^2+V_y^2}$ be the speed of a gas molecule.  It can be easily seen that $V$ has the probabilistic density function $f_V$ given by $f_V(v)=\frac{v}{\alpha^2}\exp\left(-\frac{v^2}{2\alpha^2}\right)$ for $v\geq 0$, and $f_V(v)=0$ otherwise.  This distribution is known as the two-dimensional Maxwell-Boltzmann distribution with parameter $\alpha$.  
The angle the velocity of a gas molecule made to the $X$-axis is denoted by $\Theta$.  Then, $\Theta$ is uniformly distributed on $[0,2\pi)$.  Note that $C:=\cos(\Theta)$ has the probabilistic density $f_C$, where $f_C(c)=\frac{1}{\pi\sqrt{1-c^2}}$ for $c\in(-1,+1)$, and $f_C(c)=0$ if $|c|\geq 1$.  
The speed $V$ of a gas molecule is independent from its direction $\Theta$, whence $V$ and $C$ are independent random variables.  However, $V_x=VC$ has the $\mathcal{N}\left(0,\alpha^2\right)$-distribution.
