# $X_1$, $X_2$ i.i.d RVs, $X_1$ is uniformly distributed. Show $E\left(\frac{X_1}{X_1+X_2}\right)=\frac{1}{2}$

Let $X_1$, $X_2$ be two i.i.d. random variables and $X_1$ is uniformly distributed (discrete) on the set $\{1,2,3\}.$ Show that:

$$E\left(\frac{X_1}{X_1+X_2}\right)=\frac{1}{2}$$

Can someone give me a hint how to start?

• This is the same as $E(\dfrac{X_2}{X_1+X_2})$ and their sum is 1. – Omran Kouba Dec 3 '15 at 22:49
• @OmranKouba I'm sorry but i dont really get why $E(\frac{X_2}{X_1+X_2})$ is the same as $E(\frac{X_1}{X_1+X_2})$ and how this might help me solve this problem. Can you please explain this a little bit more ? – Tobias Dec 4 '15 at 9:21
• @TobiasD, because $X_1$ and $X_2$ are independent and identically distributed. In your case $E(\frac{X_1}{X_1+X_2})=\frac{1}{9}\sum_{k=1}^3\sum_{j=1}^3\frac{j}{j+k}$ which can be calculated directly withought the trick I mentioned before. – Omran Kouba Dec 4 '15 at 10:11
$$2E\left(\frac{X_1}{X_1+X_2}\right) = E\left(\frac{X_1}{X_1+X_2}\right) + E\left(\frac{X_2}{X_1+X_2}\right) = E\left(\frac{X_1+X_2}{X_1+X_2}\right) = 1 \\ \implies 2E\left(\frac{X_1}{X_1+X_2}\right) = 1 \iff E\left(\frac{X_1}{X_1+X_2}\right) = \frac{1}{2}$$