Expectation of quotient of linear combinations of independent standard normal random variables Let $a, b, c, d, e, f$ be complex numbers with nonnegative real parts and nonnegative imaginary parts, and let $X_{1}, X_{2}, X_{3}, X_{4}$ be independent standard normal random variables. How can I verify the following:
$$E\left[\frac{a X_{1} + b X_{2} + c X_{3} + d X_{4}}{e X_{1} + f X_{2}}\right] = E \left[\frac{a X_{1} + b X_{2}}{e X_{1} + f X_{2}}\right]?$$
 A: Here is a hand-waving "proof" of the alleged result:
\begin{align}E\left[\frac{a X_{1} + b X_{2} + c X_{3} + d X_{4}}{e X_{1} + f X_{2}}\right] 
&= E \left[\frac{a X_{1} + b X_{2}}{e X_{1} + f X_{2}}\right] + E \left[\frac{c X_{3} + d X_{4}}{e X_{1} + f X_{2}}\right]\\
&= E \left[\frac{a X_{1} + b X_{2}}{e X_{1} + f X_{2}}\right]+0\\
&= E \left[\frac{a X_{1} + b X_{2}}{e X_{1} + f X_{2}}\right]
\end{align}
where the first equality follows from the linearity of expectation
while the assertion that
$$E \left[\frac{c X_{3} + d X_{4}}{e X_{1} + f X_{2}}\right]=0$$
uses the fact that $cX_3+dX_4$ and $eX_1+fX_2$ are
independent zero-mean normal random variables, and so their ratio
has a symmetric distribution (no dissension from me thus far)
and thus has expected value $0$ (which I do not agree with but 
which will likely pass muster with most people and even some readers
of stats.SE).
A: Assume that the variables are zero-mean, and jointly independent, without specifying a distribution. By linearity of the expected value,
$$E\left[\frac{a X_{1} + b X_{2} + c X_{3} + d X_{4}}{e X_{1} + f X_{2}}\right] = E\left[\frac{a X_1 + b X_2}{e X_1 + f X_2}\right] + E\left[\frac{c X_3 + d X_4}{e X_1 + f X_2}\right]$$
Let's concentrate on the second quotient. Due to independence we can distribute the expected value in the second quotient, while using linearity again:
$$E\left[\frac{c X_3 + d X_4}{e X_1 + f X_2}\right]=  \Big(c E(X_3) + d E(X_4)\Big)\cdot E\left( \frac {1}{eX_1 + fX_2}\right)$$
$$= \Big(c \cdot 0 + d \cdot 0 \Big)\cdot E\left( \frac {1}{eX_1 + fX_2}\right) = 0 \cdot E\left( \frac {1}{eX_1 + fX_2}\right)$$
since the variables are zero-mean...The fact that the coefficients are imaginary, changes nothing as regards multiplication by zero.
But what happens with the second expected value?
If indeed $X_1$ and $X_2$ are standard normal, then their sum is also normal, since they are independent. The reciprocal of a normal random variable does not have a first or a higher moment, they do not exist. So the product is not zero, and the assertion in the paper is not correct.
