Distribution of difference of two random variables with chi-squared distribution

Question

Supose that we have two random variables $X \sim \chi_k^2$ and $Y \sim \chi_k^2$, with the same degrees of freedom.

A chi-squared distribution cannot have zero degrees of freedom, so what would be the distribution of $X - Y$ ?

Answer 1

This is not a chi-squared density, $X-Y$ will have support on $(-\infty, +\infty)$.

If the two variables are independent, it has mean 0 and variance $4k$.

If $k$ is large, its density is well approximated a normal variable with 0 mean and $4k$ variance. In the general case, its MGF has a closed form: $$E(\exp(t(X-Y))) = (1-4t^2)^{-k/2}$$

If the 2 variables have dependence, the nature of that dependence needs to be explicited.

Answer 2

Since $X$ and $Y$ can also be considered to be Gamma random variables with (order, rate) parameters $\left(\frac k2, \frac 12\right)$, then, as Sebapi points out, $X-Y$ has support $(-\infty,\infty)$. Furthermore, if $X$ and $Y$ are assumed to be independent, then the pdf of $X-Y$ is also symmetric about $0$ and has mean zero.

For independent $X$ and $Y$, the pdf of their difference $X-Y$ is the cross-correlation of the pdfs of $X$ and $Y$ (not the convolution of the pdfs as happens for the sum $X+Y$) and in this case, since the pdfs are identical, the cross-correlation is actually an autocorrelation and so the pdf of $X-Y$ is symmetric about $0$ as claimed. Furthermore, for the case here of $X$ and $Y$ being Gamma random variables, an explicit form can be deduced from this answer of mine here on stats.SE.

When $k$ is an odd number, then, as noted in my answer cited above, it is possible to write the pdf $f_{X-Y}(z)$ of $X-Y$ in terms of polynomial, exponential, and Bessel functions of $z$. However when $k$ is an even number, then for $z \geq 0$, $f_{X-Y}(z)$ is proportional to a mixture of Gamma pdfs with (order, rate) parameters $\Gamma\left(1, \frac 12\right), \Gamma\left(2, \frac 12\right), \ldots, \Gamma\left(\frac k2, \frac 12\right)$ pdfs (equivalently, $\chi_2^2, \chi_4^2, \ldots, \chi_{k}^2$ pdfs) and of course, since $f_{X-Y}(z)$ is an even function of $z$, the same curve "flipped over" is the pdf curve on the negative axis.

Those worried that Gamma pdfs have value $0$ at the origin whereas autocorrelation functions have a maximum at the origin and so something is awry in the above claims should relax. A $\chi_2^2$ pdf is an exponential pdf which has nonzero value (or a limiting value that is nonzero at the origin for those whose exponential random variables take on values only on the positive real line) and so that mixture pdf is indeed nonzero at the origin, and because of the weights in the mixture, indeed a maximum at the origin.

Answer 3

Using the following links and parameter correspondence between the characteristic function(chf) of difference of $\Gamma(\alpha,\nu_{\Gamma})$ and $VG(\sigma,\nu)$ symm. variance gamma rvs we can obtain density of $X-Y$ indirectly, alternative to the method of chf inversion, \begin{equation} \phi_{VG}(u,\sigma,\nu)=\left(\frac{1}{1+\frac{\nu u^{2}\sigma^{2}}{2}}\right)^{\frac{1}{\nu}} \end{equation} \begin{equation} \phi_{\Gamma}(u,\alpha,\nu_{G})=\left(\frac{1}{1+\nu_{\Gamma}^{2}}\right)^{\alpha} \end{equation} \begin{equation} \sigma=\nu_{G}\sqrt{2\alpha}\\ \nu=\frac{1}{\alpha} \end{equation} and the fact that $\chi^{2}(k)\sim\Gamma(\frac{k}{2},2)$, we can see that final density will be almost the same as @whuber's formula above with minor differences. \begin{equation} f_{X-Y}=\frac{2^{-k}|x|^{\frac{k-1}{2}}K_{k-\frac{1}{2}}\left(\frac{|x|}{2} \right)}{\sqrt{\pi}\Gamma\left(\frac{k}{2}\right)} \end{equation} Following MC experiments with 2 and 10 df and 100K simulations respectively, provides a good match.