# Convolution of Gaussians is Gaussian

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.