X and Y are independent and uniformly distributed on the interval (0, 1). If U = X + Y , and V =X/Y
find the joint density for U and V and the marginal densities for U and V.
Given that $$f_{UV}(uv) = f_X(x)f_Y(y)*|J|$$
Where i get the $|J| =\frac{vu+1}{(1+v)^{3}} $
I get the join distribution to be
$$f_{UV}(uv) = \frac{vu+1}{(1+v)^{3}}, 0<u<1+v, \frac{u}{1+v}<u<1+\frac{u}{1+v}$$
Im trying to get the marginal for U and V but im having trouble setting my limits.
So far i have$$f_V(v) = \int_0^{1+v} f_{UV}(uv) du$$
and im not sure how to set the limits for$$f_U(u) = \int f_{UV}(uv) dv$$
How can i find the right limits for it. Im getting a bit confused with it.
Best Answer
The transformation is $Y=U/(V+1), X=UV/(V+1)$
So the Jacobian should be:
$$\begin{align}J_{U,V}&=\begin{bmatrix}1/(V+1) & -U/(V+1)^2 \\ V/(V+1) & U/(V+1)^2\end{bmatrix} \\ \lVert J_{U,V}\rVert&=\left\lvert\dfrac{U}{(V+1)^2}\right\rvert\end{align}$$ …
The support for $U,V$ should be $0<U<2$ (since $U=X+Y$) and also $0<U/(V+1)<1$ and $0<UV/(V+1)<1$ (from the support).
So $$f_{\small U,V}(u,v) = \dfrac{u}{(v+1)^2}(\mathbf 1_{0<v<1,0<u<v+1}+\mathbf 1_{1\leqslant v, 0<u<(v+1)/v})$$
…
It would seem easier to find the marginals from first principles. For instance:
$$\begin{align}f_{\small X,X+Y}(x,u) &= f_{\small X}(x) f_{\small Y}(u-x)\\&= (\mathbf 1_{0<u<1, 0<x<u}+\mathbf 1_{1\leqslant u< 2, u-1<x<1})\\[2ex]\therefore\quad f_{\small U}(u)&= u\mathbf 1_{0<u<1}+(2-u)\mathbf 1_{1\leqslant u < 2}\end{align}$$