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$X1_, X_2$ are two independent random variables whose expected values are known. I am trying to find the expected value of $(X_1/(X_1+X_2))$. Since $X_1$ and $X_1+X_2$ are dependent, I tried to use the formula for covariance to calculate the expected value of the product but this results in variance ($X_1$) being part of the resultant expression which is unknown. Is there any other way of finding the expected value of the expression $(X_1/(X_1+X_2))$?

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    $\begingroup$ there's no general closed form formula $\endgroup$ – Aksakal May 30 '18 at 23:22
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    $\begingroup$ Can it be approximated? $\endgroup$ – gaganso May 31 '18 at 2:43
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    $\begingroup$ We need some more information. Are $X_1, X_2$ positive random variables? In that case the ratio are $\le 1$. If not, the ratio could even have infinite expectation (or undefined). See stats.stackexchange.com/questions/299722/… $\endgroup$ – kjetil b halvorsen Jul 6 '18 at 23:01
  • $\begingroup$ Yes, they are positive random variables. $\endgroup$ – gaganso Jul 7 '18 at 4:50
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An extended comment:

Let $\mathcal G(a,b)$ be the gamma density with pdf $f(x)\propto e^{-ax}x^{b-1}\mathbf1_{x>0}$.

Consider independent random variables $X_1\sim\mathcal G(a,b)$ and $X_2\sim\mathcal G(a,c)$.

Then it can be shown that $X_1+X_2$ is independent of $\frac{X_1}{X_1+X_2}$.

In fact, $X_1+X_2\sim\mathcal G(a,b+c)$ and $\frac{X_1}{X_1+X_2}\sim\mathcal{Be}(b,c)$, the beta distribution of the first kind.

This is a standard relation between beta and gamma variables.

Now,

\begin{align}E(X_1)&=E\left(\frac{X_1}{X_1+X_2}\cdot X_1+X_2\right) \\&=E\left(\frac{X_1}{X_1+X_2}\right)E(X_1+X_2) \end{align}

Hence, $$E\left(\frac{X_1}{X_1+X_2}\right)=\frac{E(X_1)}{E(X_1)+E(X_2)}$$

While I do not know a general formula for the expectation you ask for, in some special cases like the above, we can use well-known relations between functions of the random variables under consideration and use their independence and proceed. Here, of course all the expectations exist which may not be the case in general.

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