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Considering $X_1$, $X_2$ ,...., $X_{10}$ be independent and identically distributed positive random variables. How can I find expectation of $$ Y = \frac{4 \cdot X_6 + 6 \cdot X_4}{\sum_{i=1}^{10} X_i} $$

I tried the following but stuck: $$ E(Y) = 4 \cdot E( \frac{X_6}{\sum_{i=1}^{10} X_i}) + 6 \cdot E(\frac{X_4}{\sum_{i=1}^{10} X_i}) $$

$$ E(Y) = 4 \cdot E( X_6) \ \cdot E( \frac{1}{\sum_{i=1}^{10} X_i}) + 6 \cdot E( X_4) \ \cdot E( \frac{1}{\sum_{i=1}^{10} X_i}) $$

How will I evaluate $$ E( \frac{1}{\sum_{i=1}^{10} X_i}) $$

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    $\begingroup$ You certainly cannot replace $$u_k=E\left(\frac{X_k}{X_1+\cdots+X_{10}}\right)$$ by $$E(X_k)E\left(\frac1{X_1+\cdots+X_{10}}\right)$$ since these are not independent. But, you might note that $$u_1+u_2+\cdots+u_{10}$$ is equal to $___$ and that by symmetry, $u_k$ and $u_\ell$ are $_______$, hence... $\endgroup$
    – Did
    Sep 15, 2017 at 14:46
  • $\begingroup$ @Did each of $u_{i} = 1/10$ by symmetry ? $\endgroup$
    – BAYMAX
    Sep 15, 2017 at 16:24
  • $\begingroup$ @BAYMAX Yes. $ $ $\endgroup$
    – Did
    Sep 15, 2017 at 17:35
  • $\begingroup$ Also related: 1, 2. $\endgroup$
    – EditPiAf
    Sep 15, 2017 at 19:20

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