Let A and B be statistically independent, identically distributed (iid) random variables having chi-square distribution with four degrees of freedom.

  • calculate the joint pdf for $X=\frac{A-B}{A+B}$ and $Y=A+B$
  • calculate the marginal pdfs of X and Y

So I did the following:

The chi-square distribution with four degrees of freedom is given by $f_X(x) = \frac{1}{4}x e^{-x/2}$

Since A and B are statistically independent $f_{A,B} = f_A(a) \cdot f_B(b)=\frac{1}{16}a b e^{-\frac{a+b}{2}}$

So since $X=\frac{A-B}{A+B}$ and $Y=A+B$ I expressed that as $A=\frac{1}{2}(Y+XY)$ and $B=\frac{1}{2}(Y-XY)$

Now for the transformation I have to calculate the det. of the Jacobian matrix

$\det\begin{pmatrix} \frac{\partial \frac{1}{2}(Y+XY)}{\partial X} & \frac{\partial \frac{1}{2}(Y+XY)}{\partial Y} \\ \frac{\partial \frac{1}{2}(Y-XY)}{\partial X} & \frac{\partial \frac{1}{2}(Y-XY)}{\partial Y}\end{pmatrix} = \det\begin{pmatrix} 1/2 Y & 1/2[1+X] \\ -1/2 Y & 1/2[1-X]\end{pmatrix} = \frac{1}{2} Y$

So the joint pdf will be $f_{X,Y}(x,y) = f_{A,B}\left(\frac{1}{2}[y+xy], \frac{1}{2} [y-xy]\right) \cdot \frac{1}{2} Y = \frac{1}{128}y^3(1-x^2)e^{-y/2}$

When I now try to calculate the marginal density $f_Y(y) = \int_0^\infty \frac{1}{128}y^3(1-x^2)e^{-y/2} \, dx$ i have the problem that it doesn't converge

I would be thankful if someone could show me where my mistake is .

  • $\begingroup$ If a chi-square distribution with four degrees of freedom is the sum of the squares of four standard Gaussians, then the sum of two independent versions will be the sum of the squares of eight standard Gaussians, i.e. a chi-square distribution with eight degrees of freedom $\endgroup$ – Henry Dec 12 '17 at 0:01
  • $\begingroup$ @Henry I'm not sure if I understand you right, how does this help me? it is not just the sum, what i am looking for? $\endgroup$ – Felix Knorr Dec 12 '17 at 11:35
  • $\begingroup$ $Y=A+B$ looks like a sum to me, so I would have thought that the density of $Y$ should be something like $\frac{1}{96} y^3 \exp(-y/2)$ $\endgroup$ – Henry Dec 12 '17 at 12:49

Thanks to @Henry I found out my mistakes, I used the wrong integration borders,

just for completeness i will answer my own question:

$f_X(x) = \int\limits_0^{\infty} \frac{1}{128}y^3(1-x^2)e^{-y/2}\, dy = \frac{3}{4} (1-x^2)$

$f_Y(y) = \int\limits_{-1}^{1} \frac{1}{128}y^3(1-x^2)e^{-y/2}\, dx = \frac{1}{96}y^3e^{-y/2}$

So we see $X$ and $Y$ are again statistically indepent since $f_{X,Y}(x,y) = f_X(x) \cdot F_Y(y)$

Thanks again Henry

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