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I have to calculate the pdf of $Z = X*Y$, where $X \in \mathcal{C}(\mu_x,\Sigma_x)$ and $Y \in \mathcal{C}(\mu_y,\Sigma_y)$, where $\mathcal{C}$ is a complex distribution. It can be assumed that $\mu_x$ and $\mu_y$ are real and $\Sigma_x$ and $\Sigma_y$ are equal to $\sigma^2 \left[\begin{matrix}1 & 0\\0 & 1\end{matrix}\right]$, i.e. the real and imaginary part have equal variance and are independently distributed. Also, X and Y are independent.

From the literature (Papers referencing Rohatgi 76) and from the book Rohatgi-2001 I have found that the real-case can be calculated using $$ p_{XY}(v) = \int_{-\infty}^\infty \frac{1}{|x|}p_X(x)p_Y\left(\frac{v}{x}\right) dx ~~~~~(1) $$

My guess would be a solution similar to this along the form $$ p_{XY}(z) = \int_{-\infty}^\infty \int_{-\infty}^\infty\frac{1}{(a+ib)(a-ib)}p_X(a+ib)p_Y\left(\frac{z}{a+ib}\right) da db $$ but I have not been able to find any sources for this. Also, when I looked in Rohatgi 2001 on p. 141 for the 1D case derivation for hints, it only said that the proof was left as an exercise. This is all well and fine if you know someone to help you.

So... Can anyone give me a link/solution to my problem or perhaps a link to a rigorous derivation of (1) I would be very thankful. I am "simply" looking for a solution in integral form that I can solve so I don't have to use excessive Monte Carlo simulation.

Thank you in advance to all who post.

/Henrik Andresen

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I dont know if this is still relevant, but there is a paper about this problem:

O'Donoughue, N.; Moura, J.M.F., "On the Product of Independent Complex Gaussians," in Signal Processing, IEEE Transactions on , vol.60, no.3, pp.1050-1063, March 2012

Best, Rick

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This does not answer the question. You could have provided this useful information as a link once you have enough reputation – Shailesh Feb 2 at 12:39

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