# Distribution of the convolution of squared normal and chi-squared variables?

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?

• If the $Y$ wasn't squared, I'd have some specific advice. I don't think this one will be tractable (nor necessarily particularly enlightening even if it were to prove tractable). I'd be tempted to look at computational approaches, like numerical convolution or simulation, depending on exactly what you want to do with the result. – Glen_b Mar 23 '14 at 23:28
• It's very unlikely in my opinion that the integral can be done. – Dave31415 Mar 24 '14 at 1:12
• @Dave31415 For $n$ and $m$ even, the integral can be explicitly computed for positive integral values of $n$ and $m$. It will equal a linear combination of exponentials and error functions with coefficients that are polynomials in $\sqrt{t}$. The evaluation can be performed via the substitution $x=t-u^2$. For instance, with $n=2,m=4$ we obtain $\frac{1}{4}e^{-\frac{1}{8}(2\sqrt{t}+1)^2}(e^{\frac{\sqrt{t}}{2}}(\sqrt{2\pi} (4t+3)(\text{erfi}(\frac{2\sqrt{t}-1}{2\sqrt{2}})+\text{erfi}(\frac{1}{2\sqrt{2}}))+4e^\frac{1}{8})-4 e^{\frac{t}{2}+\frac{1}{8}}(2\sqrt{t}+1))$. – whuber Mar 24 '14 at 15:45
• Nice. For odd numbers, you could probably approximate it with the average of the result for bounding even numbers? Or maybe not. – Dave31415 Mar 24 '14 at 20:34
• Thanks for your replies! For some even-even cases I got a similar result involving Dawson's function, but it looks like I'll have to do some more work for a general solution... – Leo Szilard Mar 25 '14 at 21:48

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.
This is a hint only. Pearson type III can be Chi-squared. Sometimes a convolution can be found by convolving something with itself. I managed to do this for convolving ND and GD, for which I convolved a Pearson III with itself. How this works with ND$^2$ and Chi-Squared, I am not sure. But, you asked for hints, and this is a general hint. That should be enough to get you started, I hope.