# Expected Value of the Ratio of Independent Variables, E(X/(X+Y)) [duplicate]

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.