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It is given that $X_1,X_2,X_3 \sim^{\text{i.i.d}} R(0,1)$. Find $E(\frac{X_1+X_2}{X_1+X_2+X_3})$. Now, I have tried this for a while and I somehow feel that if I can show that $\frac{X_1+X_2}{X_1+X_2+X_3} $ and $X_1+X_2+X_3$ are independent, then I can write $E(\frac{X_1+X_2}{X_1+X_2+X_3})=\frac{E(X_1+X_2)}{E(X_1+X_2+X_3)}$, after which it is easy. Please help!

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    $\begingroup$ No, you could not write that, even if they were independent; which they are not in any case. $\endgroup$ – Graham Kemp Mar 4 '18 at 2:57
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Hint: $$\mathsf E(\tfrac{X_1+X_2}{X_1+X_2+X_3}) ~{=\mathsf E(\tfrac{X_1}{X_1+X_2+X_3})+\mathsf E(\tfrac{X_2}{X_1+X_2+X_3})\\= 1-\mathsf E(\tfrac{X_3}{X_1+X_2+X_3})}$$

Now, $X_1,X_2,X_3$ are identically and independently distributed so...

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  • $\begingroup$ Yes, I have done this step..but what after this? $\endgroup$ – Legend Killer Mar 4 '18 at 3:01
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    $\begingroup$ $e+e=1-e\implies e=?\quad$ +1 $\endgroup$ – Macavity Mar 4 '18 at 3:05
  • $\begingroup$ The fact that distributions of $X_1,X_2,X_3$ are identical does not guarantee that these expectations are the same. Without independence it is wrong in general. $\endgroup$ – NCh Mar 4 '18 at 6:27

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