trying to verify pdf for distance between normally distributed points Math people:
I am trying to find the probability density function for the distance between two points in $\mathbb{R}^3$ selected independently according to the Gaussian pdf $F(\mathbf{z}) = \left(\frac{1}{\sqrt{2\pi}}\right)^3 \exp(-\frac{1}{2}|\mathbf{z}|^2)$.  I keep getting $f(t) = \frac{t^2\exp(-\frac{t^2}{4})}{2\sqrt{\pi}}$.  I found a paper (http://www.pupr.edu/hkettani/papers/HICS2008.pdf) which, if I understand correctly, states that the actual pdf is $\sqrt{\frac{2}{\pi}} t^2\exp(-\frac{t^2}{2})$ (see p. 10, second "$f(x)$").  That paper actually proves a much more general result and gives this as a special case.  I am reluctant to just take their word for it, since I used only Calc 3 techniques and their arguments are much more sophisticated and difficult.  Has anyone seen this problem before, perhaps as a calculus or probability exercise?  Assuming that either their formula or mine is right, is there an easy way to test which is right experimentally (generating random vectors, etc.)?
Stefan (STack Exchange FAN)

The square of the distance is $r^2=(X_1-X_2)^2+(Y_1-Y_2)^2+(Z_1-Z_2)^2$, where the $X_i$s, $Y_i$s, and $Z_i$s are independent standard normal r.v.s.  So, $r^2/2$ is $\chi^2$ distributed with $3$ degrees of freedom, and as you say, the distance $r$  has density $r^2\exp(-\frac{r^2}{4})/(2\sqrt{\pi}).$  p. 10 of the paper you mention is treating a different case, where each component of the difference between the two points is standard normal.  
