I have two continuous RVs $u$ and $v$ where $$U=\frac{X}{Y+1};~V=\frac{Y}{X+1} $$ Here $X$ and $Y$ are exponential RVs: $e^{-x}$ and $e^{-y}$ which are independent. I want to find correlation coefficient between $u$ and $v$. By definition, $$cor(U,V)=\frac{cov(U,V)}{\sigma_u\sigma_v}.$$ However, again $$cov(u,v)=\int\int F_{U,V}(u,v)-F_{U}(u)F_{V}(v)du dv$$ I am unable to compute this, specially because of the joint density. Can someone please help me?
With @Graham's guide, I calculate $$cov(u,v)=\int_{0}^{\infty}\int_{0}^{\infty}\frac{xy e^{-(x+y)}}{(1+x)(1+y)}dxdy -\int_{0}^{\infty}\int_{0}^{\infty}\frac{x e^{-(x+y)}}{(1+y)}dxdy \int_{0}^{\infty}\int_{0}^{\infty}\frac{y e^{-(x+y)}}{(1+x)}dxdy$$
$\def\Cov{\mathop{\sf Cov}}\def\E{\mathop{\sf E}}\def\d{\mathop{\rm d}}$
No. You have no need to find the joint density (or even the marginal) of $U,V$.
By definition (and totally not what you had): $$\Cov(U,V) = \E(UV)-\E(U)\E(V)$$
and by the Law of the Unconscious Statatician: $$ \E(g(X,Y)) = \int_0^\infty\int_0^\infty g(x,y) e^{-x}e^{-y}\d x\d y$$
You have $U(x,y)=x/(1+y)$ and $V(x,y)= y/(1+x)$
PS: It may help to know the Exponential Integral Function, $\def\Ei{\mathop{\rm Ei}}\Ei(t)$ is defined as: $$\Ei(t) := \int_{-t}^{\infty} \dfrac{e^{-s}}{s}~\d s$$