I am told that $A$ and $B$ are random variables, and that $\mathbf{E}(A|B) = \gamma B$. Define $D = A/B$. Using the law of iterated expectations it can be shown that $\mathbf{E}(D) = \gamma$. Now define $Z = \overline{A}/\overline{B}$ where the overline indicates a sample average, where $\{(A_i,B_i)|i=1,...,n\}$ is a random sample. This is my proof that $\mathbf{E}(Z)$ is also = $\gamma$, and I'm wondering whether I have taken a fatal shortcut or misused a property to get the result:
$Z = \frac{n^{-1}\sum^nA_i}{n^{-1}\sum^{n}B_i}$
$\mathbf{E}(Z |B) = \mathbf{E}(\frac{\sum^nA_i}{\sum^{n}B_i}|B) = \frac{1}{\sum^{n}B_i} \cdot \mathbf{E}(\sum^{n}A_i|B) = \frac{1}{\sum^{n}B_i} \cdot (\mathbf{E}(A_1|B)+...+\mathbf{E}(A_n|B)) $
This is where I'm not sure what is correct. Can I conclude that:
$=\frac{1}{nB} \cdot (n \gamma B) = \gamma$?
How do the summation and expectation operators work here? If I know $\mathbf{E}(A|B) = \gamma B$, do I know $\mathbf{E}(A_i|B)$? Thank you in advance.