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?

  • 3
    $\begingroup$ 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

1 Answer 1


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.

  • 2
    $\begingroup$ Thanks, Eric. It is an approximation, not an identity. Now, I see it. $\endgroup$
    – Jie Wei
    Commented Sep 13, 2015 at 12:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.