Independence of real and imaginary part of the product of two independent normal variables

Question

Let $X_1,X_2,Y_1,Y_2$ be iid standard normal variables $N(0,1).$ Let $X=X_1+iX_2,$ $Y=Y_1+iY_2$ and $Z=XY.$ We have : $Z=(X_1Y_1 - X_2Y_2) + i(X_1Y_2 + X_2Y_1).$ From https://en.wikipedia.org/wiki/Laplace_distribution#Related_distributions : $\Re(Z)$ ~ $\mathrm{Laplace}(0,1)$ and $\Im(Z)$ ~ $\mathrm{Laplace}(0,1).$ Furthermore, we can check that $\Re(Z)$ and $\Im(Z)$ are not correlated. My question is: are they independent?

I found several links related to the variable $Z$, but none answered the question :

• What processes could generate Laplace-distributed (double exponential) data or parameters?
• https://mathoverflow.net/questions/346581/the-distribution-of-the-sum-of-inner-products-of-two-independent-complex-normal
• https://mathoverflow.net/questions/258999/what-is-the-probability-density-function-pdf-of-the-dot-product-of-m-complex-n
• https://mathoverflow.net/questions/296632/is-the-distribution-of-the-real-part-of-product-of-two-independent-complex-varia (A comment suggests they are not independent but does not explain why.)

Answer 1

$X_1Y_1 - X_2Y_2$ and $X_1Y_2 + X_2Y_1$ are not independent, for if they were, then \begin{align*} E[(X_1Y_1 - X_2Y_2)^2(X_1Y_2 + X_2Y_1)^2] = E[(X_1Y_1 - X_2Y_2)^2] \times E[(X_1Y_2 + X_2Y_1)^2]. \tag{1}\label{1} \end{align*}

But you can verify that the left hand side of $\eqref{1}$ is $8$ (by breaking the product then evaluating expectations term-by-term) whereas the right hand side of $\eqref{1}$ is $4$.

Details

\begin{align*} & E[(X_1Y_1 - X_2Y_2)^2] = E[X_1^2Y_1^2] + E[X_2^2Y_2^2] = 1 + 1 = 2, \\ & E[(X_1Y_2 + X_2Y_1)^2] = E[X_1^2Y_2^2] + E[X_2^2Y_1^2] = 1 + 1 = 2, \\ & E[(X_1Y_1 - X_2Y_2)^2(X_1Y_2 + X_2Y_1)^2] \\ =& E[X_1^4Y_1^2Y_2^2] + E[X_1^2Y_2^2Y_1^4] + E[X_1^2X_2^2Y_2^4] + E[X_2^4Y_2^2Y_1^2] - 4E[X_1^2X_2^2Y_1^2Y_2^2] \\ =& 4 \times 3 - 4 = 8. \end{align*}