Convolution of two chi square independent random variables How do you use the convolution formula in order to prove that the sum of two chi square random variables with degrees of freedom m and n respectively results in a chi square distribution? Thank you
 A: If just getting the bottom-line conclusion is the purpose, then there are quicker ways than computing the convolution of density functions, but I take it the question is precisely about finding the convolution.
The chi-square distribution with $j$ degrees of freedom is
$$
f_j(x)\,dx = \frac 1 {\Gamma(j/2)} \left( \frac x 2\right)^{j/2-1} e^{-x/2} \left( \frac{dx} 2 \right) \text{ for } x\ge0.
$$
And then:
\begin{align}
& (f_j*f_k)(x) \\[8pt] = {} & \int_0^x f_j(u)f_k(x-u) \, du \\[8pt]
= {} & \frac 1 {\Gamma(j/2)\Gamma(k/2)} \int_0^x \left( \frac u 2 \right)^{j/2-1} \left( \frac{x-u} 2 \right)^{k/2-1} e^{-x/2} \, \frac{du} 4
\end{align}
Here you see that $e^{-u/2} \cdot e^{-(x-u)/2}$ became $e^{-x/2},$ and that quantity does not change as $u$ goes from $0$ to $x,$ so we can pull it out, thus:
$$
\frac {e^{-x/2}} {\Gamma(j/2)\Gamma(k/2)} \cdot\frac 1 {2^{(j+k)/2}} \int_0^x u^{j/2-1} (x-u)^{k/2-1} \, du \tag 1
$$
Now a simple substitution:
\begin{align}
v & = u/x \\
dv & = du/x
\end{align}
As $u$ goes from $0$ to $x,$ so $v$ goes from $0$ to $1.$
The expression on line $(1)$ above becomes
\begin{align}
& \frac {e^{-x/2}} {\Gamma(j/2)\Gamma(k/2)} \cdot \frac 1 {2^{(j+k)/2}} \int_0^1 \big( xv \big)^{j/2-1} \big( x(1-v) \big)^{k/2-1} \, \big(x\,dv\big) \\[8pt]
= {} & \frac {e^{-x/2} } {\Gamma(j/2)\Gamma(k/2)} \cdot \left( \frac x 2 \right)^{(j+k)/2-1} \cdot \frac 1 2 \underbrace{\int_0^1 v^{j/2-1} (1-v)^{k/2-1} \, dv}_\text{No “$x$'' appears here!}
\end{align}
I'll leave the remaining details to you.
(This also illustrates why the gamma function should be defined by $$ \Gamma(\alpha) = \text{an expression in which $(\alpha-1)$ appears.} $$ The reason is that, as you see above, where $j$ and $k$ occurred in the two factors, we now have $j+k$ in the result.)
