There is a 4D Gaussian function $G(u,s)=G(x|c,\mu,\Sigma )$ where $x=\begin{bmatrix}u\\ s\end{bmatrix}$,$u$ and $s$ is all 2D vector. Now I want to blur (convolve) it along with $u$ by another 2D Gaussian $g(u|c',\mu',\Sigma')$.
I know that I can write $G(u,s)$ as a 2D Gaussian $g[s](u|c_0,\mu_0,\Sigma_0)$ by fixing $s$ and convolve it by $g(u|c',\mu',\Sigma')$. To do like this means that we calculate a convolution between two 2D Gaussian functions:
$g[s](u|c_0,\mu_0,\Sigma_0)\bigotimes g(u|c',\mu',\Sigma')=g[s](u|c_1,\mu_1,\Sigma_1)=F(u,s)$
But is the convolved result $F(u,s)$ still a 4D Gaussian? If true, what is the new $c$, $\mu$, $\Sigma$?
I am thinking about another method to deal with this question is that we regard the convolution kernel $g(u|c',\mu',\Sigma')$ as a 4D function by extending its covariance matrix to a 4x4 matrix $\begin{bmatrix} \Sigma'& O \\ O & O \end{bmatrix}$. But the question is that the detaminate of this matrix is 0.
