Is $E\left(\frac{X}{X+Y}\right)=\frac12$ when $X$ and $Y$ follow the same probability distribution? I had an exercise regarding random variables, and I tried to figure if the following is true:

If $X$ and $Y$ follow the same probability distribution, then $E(\frac{X}{X+Y})=E(\frac{1}{2})=\frac{1}{2}$.

I'm skeptical about this, since I have tried to prove it and got nowhere.

Even if $X$ and $Y$ are independent, the result does not necessarily hold. 
As a simple example, if $X$ and $Y$ are IID standard Normal, then $E[X]=E[Y]=0$, but $E[\frac{X}{X+Y}]$ is not equal to $\frac12$  (the expectation does not converge).
