Convolution of two Gaussians is a Gaussian (semi-group property)

Question

I would like to show that $$ g_{\sigma}(x, y) * g_{\tau}(x, y)=g_{\sqrt{\sigma^{2}+\tau^{2}}}\left(x, y\right) $$ where $g_{\sigma}(x, y)=\frac{1}{2 \pi \sigma^{2}} e^{-\frac{x^{2}+y^{2}}{2 \sigma^{2}}}$. My attempt so far:

Since the Fourier transform of $g_{\sigma}$ is given by $$ \hat{g}_{\sigma}(x, y) = e^{-(\alpha^2 + \beta^2)2\pi\sigma^2} $$ However I simply cannot simplify/get the product $$ \hat{g}_{\sigma}(x, y) \hat{g}_{\tau}(x, y)= e^{-(\alpha^2 + \beta^2)2\pi\sigma^2} e^{-(\alpha^2 + \beta^2)2\pi\tau^2} $$ to $g_{\sqrt{\sigma^{2}+\tau^{2}}}\left(x, y\right)$ - am I on the wrong track here, or is it just some manipulation that needs to be done?

Answer 1

You are almost there. Use $$e^Ae^B=e^{A+B}$$ In your case $$e^{-(u^2+v^2)2\pi\sigma^2}e^{-(u^2+v^2)2\pi\tau^2}=e^{-(u^2+v^2)2\pi(\sigma^2+\tau^2)}=e^{-(u^2+v^2)2\pi(\sqrt{\sigma^2+\tau^2})^2}=\mathcal{F}\{G(x, y, \sqrt{\sigma^2+\tau^2})\}$$

Answer 2

A probabilistic answer: $g_{\sigma}$ is the probability density of a two-dimensional Gaussian random vector $X_{\sigma}$ with mean zero and covariance matrix $\Sigma_{\sigma} = \sigma^{2} \text{Id}$. In general, if $f_{1}$ and $f_{2}$ are two probability densities, then $f_{1} * f_{2}$ is the probability density of the sum of two independent random variables, one with density $f_{1}$ and the other, density $f_{2}$. Hence $g_{\sigma} * g_{\tau}$ is the probability density of $X_{\sigma} + X_{\tau}$ if $X_{\sigma}$ and $X_{\tau}$ are taken to be independent.

Now the sum of two independent Gaussian vectors is itself a Gaussian vector and hence is characterized by its mean and covariance. The mean of $X_{\sigma} + X_{\tau}$ is zero and the covariance is $\sigma^{2} + \tau^{2}$ by independence. Thus, \begin{equation*} g_{\sigma} * g_{\tau} = g_{\sqrt{\sigma^{2} + \tau^{2}}}. \end{equation*}