the following problem came up recently while analyzing data. If the random variable X follows a normal distribution and Y follows a $\chi^2_n$ distribution (with n dof), how is $Z = X^2 + Y^2$ distributed?
Up to now I came up with the pdf of $Y^2$:
\begin{eqnarray}
\psi^2_n(x) &=& \frac{\partial F(\sqrt{x})}{\partial x} \\
&=& \left( \int_0^{\sqrt{x}} \frac{t^{n/2-1}\cdot e^{-t/2}}{2^{n/2}\Gamma(n/2)} \mathrm{d}t \right)^\prime_x \\
&=& \frac{1}{2^{n/2}\Gamma(n/2)} \cdot \left( \sqrt{x} \right)^{n/2-1} \cdot e^{-\sqrt{x}/2} \cdot \left( \sqrt{x} \right)^\prime_x \\
&=& \frac{1}{2^{n/2-1}\Gamma(n/2)} \cdot x^{n/4-1} \cdot e^{-\sqrt{x}/2}
\end{eqnarray}
as well as some simplifications for the convolution integral ($X^2$ has the pdf $\chi^2_m$ with m dof):
\begin{eqnarray}
K_{mn}(t) &:=& ( \chi^2_m \ast \psi^2_n )(t) \\
&=& \int_0^t \chi^2_m(x) \cdot \psi^2_n(t-x) \mathrm{d}x \\
&=& \left( 2^{\frac{(n+m)}{2}+1} \Gamma(\frac{m}{2}) \Gamma(\frac{n}{2}) \right)^{-1} \cdot \int_0^t (t-x)^{\frac{n}{4}-1} \cdot x^{\frac{m}{2}-1} \cdot \exp(-(\sqrt{t-x}+x)/2) \mathrm{d}x
\end{eqnarray}
Does someone see a good way of calculating this integral for any real t or does it have to be computed numerically? Or am I missing a much simpler solution?
Best Answer
In case it helps, the variable $Y^2$ is a generalised gamma random variable (see e.g., Stacy 1962). Your question is asking for the distribution of the sum of a chi-squared random variable and a generalised gamma random variable. To my knowledge, the density of the resultant variable has no closed form expression. Hence, the convolution you have obtained is an integral with no closed form solution. I think you're going to be stuck with a numerical solution for this one.
Stacy, E.W. (1962). A Generalization of the Gamma Distribution. Annals of Mathematical Statistics 33(3), pp. 1187-1192.