Is $E\left(\frac{X}{X+Y}\right)=\frac12$ when $X$ and $Y$ follow the same probability distribution?

Question

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.

Answer 1

This is not true in general, even when every expectation involved exists: For a counterexample, assume that $(X,Y)$ is uniformly distributed on the set $\{(1,2),(2,3),(3,1)\}$, then $X$ and $Y$ are both uniformly distributed on the set $\{1,2,3\}$ while $Z=\frac{X}{X+Y}$ is uniformly distributed on the set $\left\{\frac13,\frac25,\frac34\right\}$ hence $E(Z)=\frac13\left(\frac13+\frac25+\frac34\right)=\frac{89}{180}\ne\frac12$.

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If $(X,Y)$ is assumed independent and identically distributed, and almost surely positive, then the result holds because, then, the two random couples $(X,Y)$ and $(Y,X)$ have the same distribution hence $Z=\frac{X}{X+Y}$ and $1-Z=\frac{Y}{Y+X}$ are identically distributed hence $E(Z)=E(1-Z)$, qed.

Answer 2

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).