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:
$$E\left[\frac{a X_{1} + b X_{2} + c X_{3} + d X_{4}}{e X_{1} + f X_{2}}\right] = E \left[\frac{a X_{1} + b X_{2}}{e X_{1} + f X_{2}}\right]?$$
By linearity of expectation operator, we have: \begin{equation} 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] \quad (1) \end{equation} Note that the random variables $cX_3 + dX_4$ and $\frac{1}{eX_1 + fX_2}$ are independent, thus, $$E \left[\frac{c X_{3} + d X_{4}}{e X_{1} + f X_{2}}\right] = E \left[\frac{1}{e X_{1} + f X_{2}}\right] \times E(cX_3 + dX_4)$$ in which $E(cX_3 + dX_4) = cE(X_3) + dE(X_4) = 0$, by assumption. Hence the second term in $(1)$ vanishes and the desired result follows.