Suppose, I have two random vectors $A=[A_1, A_2, \dots A_k]$ and $B=[B_1, B_2, \dots B_m]$. What could be the joint PDF $f_{\mathbf{y}}(y_1,y_2,\dots y_N)$ where

$\mathbf{y}=A \ast B$, here $\ast$ represents convolution. In this example, all the components of $A$ and $B$ are independent, zero mean complex Gaussian random variables with standard deviation $\sigma_a$ and $\sigma_b$, respectively. You can mention the steps and/or refer to any paper that has some hints. Thank you in advance.

  • $\begingroup$ Is this for a course? $\endgroup$ – gung - Reinstate Monica Oct 21 '14 at 20:51
  • $\begingroup$ No, for a research. $\endgroup$ – upol94 Oct 21 '14 at 20:52

To calculate the convolution of two vectors, I believe you should apply the formula for discrete convolution. Then if,

$$ A=(A_1,A_2,…,A_k)^T \text{ and } B=(B_1,B_2,…,B_m)^T$$

Then the discrete convolution is a vector such that every component $i$ is

$$(A\ast B)_i = \sum_{i = 1}^m A_{i-j}B_j \text{ }i=1,2,3...$$

$A_{i-j}$ and $B_{j}$ are Gaussian random variables with $\mu = 0$. That is, its PDF is

$$A_{i-j} \sim N(0, \sigma_A^2)$$ $$B_{j} \sim N(0, \sigma_B^2) $$


The PDF of the product is discussed in the following paper. Further information can be found in section 3 and 4 of this paper. The pdf of the two normals is neither normal or symmetric. You should use numerical integration to find its value.

  • $\begingroup$ 1)$\newcommand{\E}{\mathbb{E}} \DeclareMathOperator{\Var}{Var}$ Your formula for the product of Gaussians is, I believe, for the product of the pdfs of Gaussians, not for the pdf of the product of two Gaussian random variates. Note that $\Var(A_i B_j) = \E((A_i B_j)^2) - \E(A_i B_j)^2 = \E(A_i^2) \E(B_j^2) - \E(A_i)^2 \E(B_j)^2 = \sigma_a^2 \sigma_b^2$, which disagrees with your claim. 2) This only addresses the marginal distribution of each component of $y$, and not the joint; different components of $y$ will not be independent. $\endgroup$ – Danica Oct 14 '14 at 23:57
  • $\begingroup$ Edited. Yep, my fault, edited the solution with two references explaining how to do it. $\endgroup$ – user45299 Oct 15 '14 at 1:14
  • $\begingroup$ If the normals are mean 0 (as they are here) then the pdf will have the (symmetric) normal product distribution, which has a pdf in terms of the modified Bessel function of the second kind, and whose characteristic function is (from a simple extension of this argument) $\left(1 + \sigma_a^2 \sigma_b^2 t^2\right)^{-1/2}$. The sum of two of these happens to be a Laplace distribution with mean 0 and scale $\sigma_a \sigma_b$. $\endgroup$ – Danica Oct 15 '14 at 1:18
  • $\begingroup$ Actually, in my case $A$ and $B$ are complex Gaussian, So when I separate the real and imaginary part, they are correlated as well. Is there any easier way ? $\endgroup$ – upol94 Oct 15 '14 at 15:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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