Suppose that (X,Y) has density function
$$f(x,y)=\frac{2}{(1+x+y)^3}\quad x,y\gt0 \quad$$
Determine the distribution of
(a) X + Y
(b) X − Y
My attempt at solving (a): $$U=X+Y,V=Y$$ $$f_{U,V}(u,v)=f_{X,Y}(u-v,v)|J|=\frac{2}{(1+u-v+v)^3}*1=\frac{2}{(1+u)^3}$$ $$f_{U}(u)=\int f_{U,V}(u,v)dv=\frac{2v}{(1+u)^3}$$
Answer in the book is $\frac{2u}{(1+u)^3}$
My attempt at solving (b): $$U=X-Y,V=Y$$ $$f_{U,V}(u,v)=f_{X,Y}(u+v,v)|J|=\frac{2}{(1+u+v+v)^3}*1=\frac{2}{(1+u+2v)^3}$$ $$f_{U}(u)=\int f_{U,V}(u,v)dv=\frac{-1}{2(2v+u+1)^2}$$
Answer in the book is $\frac{1}{2(1+|u|)^2}$
What did i do wrong? any feedback is appreciated!
Look to the bounds of the integrals. They are not indefinite. Where do your $U,V$ "live"?
The support for $(X,Y)$ is $\{(x,y): 0<x, 0<y\}$
So setting $U=X+Y$, then the support for $(U,Y)$ is $\{(u,y): 0<y<u\}$ . Because is $X$ is strictly positive then surely: $Y<X+Y$ . Thus you must have:
$$f_U(u)=\int_{0}^u \frac 2{(1+(u-y)+y)^3}\mathsf d y=\dfrac{2u}{(1+u)^3}$$
Likewise setting $V=X-Y$, then the support for $(V,Y)$ is $\{(v,y):0< y, -y<v\}$.
$$f_V(v) = \int_{\max\{0,-v\}}^\infty \dfrac{2}{(1+(v+y)+y))^3}\mathsf d y$$
So...