# Finding expectation of random variable $Y = \frac{A}{B}$

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 $$E( \frac{1}{\sum_{i=1}^{10} X_i})$$

• 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...
– Did
Sep 15, 2017 at 14:46
• @Did each of $u_{i} = 1/10$ by symmetry ? Sep 15, 2017 at 16:24
• @BAYMAX Yes. 
– Did
Sep 15, 2017 at 17:35
• Also related: 1, 2. Sep 15, 2017 at 19:20