0

Given a vector with elements sampled from a Gaussian distribution, and also given a matrix with elements sampled from the other independent Gaussian distribution.

What is the probability distribution of elements in the vector obtained by the product of the given matrix and vector?

For example, $$ X=\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{bmatrix} $$, where $x_{i}$ is a Gaussian random variable $\sim \mathcal{N}_{1}(\mu=0,\sigma_{1}^2)$.

$$ A=\begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{bmatrix} $$, where $a_{ij}$ is also a Gaussian random variable $\sim \mathcal{N}_{2}(\mu=0,\sigma_{2}^2)$.

The question is, assuming $\mathcal{N}_{1}$ and $\mathcal{N}_{2}$ are independent, what is the probability density distribution of the elements in $AX$?

Rocky Tseng
  • 11
  • If you assume $A$ is independent of $X$, then the distribution of $AX$ is easily described by conditioning over $A$, and then integrating. Whether this yields a form that is useful to you is a different story. – Andrew Oct 21 '23 at 14:48
  • @Andrew Could you elaborate on what "conditioning" and "integrating" mean? (I know integration, but I have no idea how integration involves this problem) – Rocky Tseng Oct 21 '23 at 14:52
  • Conditional on $A$, $AX\sim \mathcal N(0, \sigma_1^2 AA^T)$. You can recover the distribution of $AX$ by integrating – Andrew Oct 21 '23 at 14:54

1 Answers1

0

Fix $i$ and let $\sigma_{1}z_{j}=x_{j}$ and $\sigma_{2}z_{ij}=a_{ij}$. Then, $$ \frac{2}{\sigma_{1}\sigma_{2}}a_{ij}x_{j}=\left(\frac{1}{\sqrt{2}}\left(z_{j}+z_{ij}\right)\right)^{2}-\left(\frac{1}{\sqrt{2}}\left(z_{j}-z_{ij}\right)\right)^{2} $$ is a sum of chi-squared random variables. See Is the product of two Gaussian random variables also a Gaussian?

parsiad
  • 25,154
  • Thanks for your prompt reply! Allow me to summarize what I currently know: Since $z_{j}$ and $z_{ij}\sim\mathcal{N}(0,1)$ thus $(z_{j}\pm z_{ij})\sim\mathcal{N}$. The square of a Gaussian random variable $\sim\chi_{1}^2$, so $(z_{j}+z_{ij})^2-(z_{j}-z_{ij})^2$ follows the subtraction of two chi-squared distribution. Correct? – Rocky Tseng Oct 22 '23 at 00:33
  • @RockyTseng: Yes, but keep in mind that the two are not (generally) independent – parsiad Oct 22 '23 at 20:25