A gaussian is a function of the form \(f(x)=Ae^{-(x-\mu)^2/2\sigma^2}\) for some constant \(A\) – when \(A\) is chosen to make the total integral of \(f\) equal to \(1\), you obtain the probability distribution function for a normally distributed random variable of mean \(\mu\) and variance \(\sigma^2\).

In class I mentioned the result that the convolution of two gaussian functions is again a gaussian. This is usually proved by:

- showing that the Fourier transform \(\hat{f}\) of a gaussian \(f\) is a gaussian;
- observing that the product \(fg\) of gaussians \(f\) and \(g\) is a gaussian
- using the convolution theorem to conclude that if \(f\) and \(g\) are gaussians, then so is

\[ f*g = \widehat{\hat{f}\hat{g}}. \]

However, one can also simply work out the integral. If \(f=e^{-(x-\mu)^2/\sigma^2}\) and \(g=e^{-(x-\nu)^2/\tau^2}\) then

\[ (f*g)(x) = \int e^{-(t-\mu)^2/\sigma^2-(x-t-\nu)^2/\tau^2}\,dt. \]

**Lemma:** Suppose \(Q(x,t)\) is a quadratic function in two variables of the form

\[ Q(x,t) = A(t-r)^2+B(x-t-s)^2 \]

where \(A\),\(B\),\(r\), and \(s\) are constants. Then there are constants \(A'\),\(B'\), \(r'\) and \(s'\) so that

\[ Q(x,t) = A'(x-r')^2+B'(t-x-s'). \]

**Proof:** This is an exercise in completing the square. Write \[
Q(x,t)=ax^2+bxt+ct^2+dx+et+f.
\] First complete the square to eliminate the \(et\) term. Let \(t_1=t-\frac{e}{2c}\). Then:

\[ Q(x,t)=ax^2+bx\left(\frac{e}{2c}\right)+dx+ct_1^2-\left(\frac{e}{2c}\right)^2+f+bxt_1 \]

Next complete the square to eliminate the \(xt_1\) term by setting \(t_2=t_1-\frac{b}{2c}x\) so that

\[ Q(x,t)=ax^2+(b\left(\frac{e}{2c}\right)+d)x+c(t_2-\frac{b}{2c}x)^2-\left(\frac{b}{2c}\right)^2-\left(\frac{e}{2c}\right)^2+f \]

Finally complete the square on the pure \(x\) terms to get the desired form.

**Proposition:** \((f*g)\) is a gaussian.

**Proof:** By the lemma, \((f*g)\) can be written

\[ (f*g)(x)=\int e^{-A(x-r)^2}e^{-B(t-x-s)^2}dt=e^{-A(x-r)^2}\int_{-\infty}^{\infty} e^{-B(t-x-s)^2}dt \]

Since the integral runs from \(-\infty\) to \(\infty\) its value is independent of the value of \(x\) so the integral yields a constant \(C\) leaving

\[ (f*g)(x) = Ce^{-A(x-r)^2} \]

which is a gaussian.

**Remark:** If you have more patience than I you can get the constants exactly. The full result is that if \(F\) is the gaussian distribution with mean \(\mu\) and variance \(\sigma\), and \(G\) is the gaussian distribution with mean \(\nu\) and variance \(\tau\), then \(F*G\) is the gaussian distribution with mean \(\mu+\nu\) and variance \(\sigma^2+\tau^2\).

In particular, following up on a question from class, the \(n\)-fold repeated convolution with the standard gaussian of mean zero and variance \(\sigma^2\) is equivalent to convolution with a gaussian of mean zero and variance \(n\sigma^2\).

**Remark 2:** For the probabilists, if \(X\) and \(Y\) are independent normally distributed random variables with mean \(\mu\) and \(\nu\) and variances \(\sigma\) and \(\tau\) respectively, then the distribution of \(X+Y\) is the convolution of the associated gaussian distribution functions. From this point of view the fact that the mean and variance of \(X+Y\) are \(\mu+\nu\) and \(\sigma^2+\tau^2\) respectively are natural.