Connection between Normal,Gamma and beta distribution if $X,Y,Z\sim {N}(0,1)$ then $(X^2+Y^2+Z^2)\sim \text{Gamma} (0.5\cdot 3 , 0.5)=\text{Gamma} (1.5,0.5)$
Also, we define $T$, such that $T= \frac{X^2 }{ X^2+Y^2+Z^2} $
I don't understand why then $T\sim\text{Beta} (0.5,1)$ ?
I'll be very happy if someone can explain this for me :-)
Thanks
 A: Just to mention once again a fully automated approach to compute the density $g$ of $T$, since $g$ is a function such that, for every bounded measurable function $u$,
$$
E[u(T)]=\int u(t)g(t)\mathrm dt.
$$
Now, since $T=X^2/(X^2+Y^2+Z^2)$ where $(X,Y,Z)$ are i.i.d. with common density $f$, one knows that
$$
E[u(T)]=\int\!\!\!\int\!\!\!\int  u(x^2/(x^2+y^2+z^2))f(x)f(y)f(z)\mathrm dx\mathrm dy\mathrm dz,
$$
hence one wants to transform the second RHS into the first RHS and one knows that a change of variables will do this. 
Spherical coordinates can be chosen as $x=r\cos a$, $y=r\sin a\cos b$, $z=\sin a\sin b$, with $r\gt0$, $a$ in $[0,\pi]$ and $b$ in $[-\pi,\pi]$. Then, $x^2+y^2+z^2=r^2$ and $\mathrm dx\mathrm dy\mathrm dz=r^2\sin a\mathrm dr\mathrm da\mathrm db$, hence
$$
E[u(T)]=\int\!\!\!\int\!\!\!\int  u(\cos^2a)\mathrm e^{-r^2/2}r^2\sin a\mathrm dr\mathrm da\mathrm db,
$$
and, up to a multiplicative constant,
$$
E[u(T)]\propto\int_0^\pi u(\cos^2a)\sin a\mathrm da\propto\int_0^{\pi/2} u(\cos^2a)\sin a\mathrm da.
$$
For the second change of variable, the only interesting choice is $t=\cos^2a$, then $\sin a\mathrm da\propto\mathrm d(\cos a)\propto\mathrm d(\sqrt{t})\propto\mathrm dt/\sqrt{t}$, hence
$$
E[u(T)]\propto\int_0^1 u(t)\frac{\mathrm dt}{\sqrt{t}},
$$
which shows that
$$
g(t)\propto\frac{\mathbf 1_{0\lt t\lt1}}{\sqrt{t}}.
$$
Surely you can finish...
