# Convolution of random vectors

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.

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

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

Edit:

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.

• 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. – Danica Oct 14 '14 at 23:57
• Edited. Yep, my fault, edited the solution with two references explaining how to do it. – user45299 Oct 15 '14 at 1:14
• 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 $\sigma_a \sigma_b$. – Danica Oct 15 '14 at 1:18
• 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 ? – upol94 Oct 15 '14 at 15:44