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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?

Thanks in advance.

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    $\begingroup$ This is the same as $E(\dfrac{X_2}{X_1+X_2})$ and their sum is 1. $\endgroup$ – Omran Kouba Dec 3 '15 at 22:49
  • $\begingroup$ Yes. Use Linearity of Expectation, and Symmetry. $\endgroup$ – Graham Kemp Dec 3 '15 at 23:14
  • $\begingroup$ @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 ? $\endgroup$ – Tobias Dec 4 '15 at 9:21
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    $\begingroup$ @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. $\endgroup$ – Omran Kouba Dec 4 '15 at 10:11
  • $\begingroup$ @OmranKouba Ah, i get it now.Thanks alot for your help. I will post the answer to this problem later. $\endgroup$ – Tobias Dec 4 '15 at 19:15
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$$ 2E\left(\frac{X_1}{X_1+X_2}\right) = E\left(\frac{X_1}{X_1+X_2}\right) + E\left(\frac{X_2}{X_1+X_2}\right) = E\left(\frac{X_1+X_2}{X_1+X_2}\right) = 1 \\ \implies 2E\left(\frac{X_1}{X_1+X_2}\right) = 1 \iff E\left(\frac{X_1}{X_1+X_2}\right) = \frac{1}{2} $$

The proof uses the linearity of expectation.

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