If $X$ and $Y$ are independent random variables, is the following true? Is there an easy way to show this?
$$E\left[\frac{X}{X+Y}\right]=\frac{E[X]}{E[{X+Y}]}=\frac{E[X]}{E[X]+E[Y]}$$
If this is not true in general, are there special cases when it is true (e.g. what $X$ and $Y$ are strictly positive and finite)?
The second part of the equality is always true provided the expected values exist even if $X$ and $Y$ are not independent because $E[X+Y]=EX+EY$.
The first part of the equality is not true in general and not true if $X$ and $Y$ are strictly positive and finite.
Take the example:
$X$ is equally likely to be $1$ or $2$
$Y$ is $1$ with probability 0.25 and $2$ with probability 0.75
Verify that
$$E\left[\frac{X}{X+Y}\right]=\frac{31}{96}$$
but $\frac{EX}{EX+EY}=\frac{1.5}{1.5+1.75}=\frac{6}{13}$
You can also just try different positive continuous distributions such as log-Normal and you can easily find counterexamples.
However, if $X$ and $Y$ have the same distribution, then $$1=E\left[\frac{X+Y}{X+Y}\right]=E\left[\frac{X}{X+Y}\right]+E\left[\frac{Y}{X+Y}\right]=2E\left[\frac{X}{X+Y}\right]$$ So, $E\left[\frac{X}{X+Y}\right]=\frac{1}{2}=\frac{EX}{EX+EY}$. This is true even if $X$ and $Y$ are not independent.
