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 .
Best Answer
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