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I am trying to figure out the distribution of $Z=Y-X$ where $Y$ and $X$ and both just $\chi_1^2$ random variables. I am confused by the negative, does anyone know what the distribution of the negative of a $\chi_1^2$ random variable is?

MathStudent
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  • $\chi_1^2\sim N^2$ where $N$ is standard normal. So $Z\sim N_1^2-N_2^2=(N_1-N_2)(N_1+N_2)\sim 2N_3N_4$. Choose representation that makes it easier for you to work with. – A.S. Nov 05 '15 at 16:34
  • Funny story, the reason why I am trying to find the distribution of $Z=Y-X$ is because I really want to find the distribution of the product of two standard normal random variables. I am really interested in seeing if there is a density I can work with easily, so I am going to have to convolve something at some point. – MathStudent Nov 05 '15 at 16:37
  • So that is the distribution of the product of those random variables? – MathStudent Nov 05 '15 at 16:52
  • Or the density rather – MathStudent Nov 05 '15 at 16:54
  • Then just do a standard transformation of variables $(N_1,N_2)\to (U,V)=(N_1^2-N_2^2,N_1^2+N_2^2)$. The inverse is easily obtained by summing and differencing the two equations and the Jacobian works out to a nice $\frac 1 {4\sqrt{V^2-U^2}}$. $f_{U,V}(u,v)=\frac 1 {2\pi}e^{-\frac 1 2 V}\frac 1 {\sqrt{V^2-U^2}}$ (multiplied by $4$ for four different combinations of signs on $N_1,N_2$). Not sure how to integrate $V$ out. though. – A.S. Nov 05 '15 at 17:03
  • See http://math.stackexchange.com/a/85525/274197 – A.S. Nov 05 '15 at 17:11
  • And http://mathworld.wolfram.com/NormalProductDistribution.html – A.S. Nov 05 '15 at 17:30

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