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For two random variables $A$ and $B$. Often times, I say people write the following,

\begin{equation} E(\frac{A}{B})=\frac{E(A)}{E(B)}\{1- \frac{Cov(A,B)}{E(A)E(B)}+\frac{Var(B)}{[E(B)]^2} \}. \end{equation}

Can someone give me a clue how to derive the formula above?

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    $\begingroup$ Why not ask the people who wrote the formula? Generally speaking, $E\left[\frac AB\right]$ can be an undefined quantity as can the ratio $\frac{E[A]}{E[B]}$. For example, if $A$ and $B$ are independent standard normal random variables, then $\frac{A}{B}$ is a Cauchy random variable for which the mean is undefined (not $0$ as physicists allegedly claim) while $\frac{E[A]}{E[B]}$ is of the form $\frac 00$ which is also undefined. $\endgroup$ Commented Sep 12, 2015 at 15:08

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It seems that's really just an approximation based on a second-order taylor series expansion that isn't garaunteed to be true under all circumstances, but this paper seems to outline a pretty concise derivation of it: http://www.stat.cmu.edu/~hseltman/files/ratio.pdf

I thought that link was especially helpful because it shows how to reach a first-order taylor series approximation at $E(X/Y) = E(X)/E(Y)$ and then further improve that to reach the formula you mentioned.

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    $\begingroup$ Thanks, Eric. It is an approximation, not an identity. Now, I see it. $\endgroup$
    – Jie Wei
    Commented Sep 13, 2015 at 12:52

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