Let $X_1,X_2,X_3$ be iid. U($0,1$) random variables. Then what will be the value of $E(\frac{X_1+X_2}{X_1+X_2+X_3}$) ?


closed as off-topic by Namaste, Saad, StubbornAtom, Lord_Farin, Did Jan 11 at 20:18

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So, convince yourself that random variables $Y_i$, defined through $Y_i\triangleq \displaystyle\frac{X_i}{X_1+X_2+X_3}$ for $i =1,2,3$ are identically distributed, due to symmetry. Hence, $\mathbb{E}[Y_1]=\mathbb{E}[Y_2]=\mathbb{E}[Y_3]$. Now, $Y_1+Y_2+Y_3=1$, implying that $\mathbb{E}[Y_1+Y_2+Y_3]=1\implies \mathbb{E}[Y_i]=1/3$ for every $i $.

From here, your answer is $2/3$.

  • $\begingroup$ Downvoter, please step ahead and let me know why you did downvote my post. $\endgroup$ – Aaron Jan 17 at 15:47

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