I want to calculate the convolution $F * G$ of two Gaussian functions without resorting to Fouritertransforms:

$F(t) := \exp(-at^2), G(t) := \exp(-bt^2) \qquad a,b>0$

But intuitively I expected the convolution to result again in a non constant function. Can anyone find my mistake / confirm that this calculation is correct?

Let $\Omega = \mathbb R$, then

$\begin{align*} (F*G)(x) &= \int_\Omega F(t)G(x-t)dt &\\ & = \int_\Omega e^{-at^2-b(x-t)^2} dt \qquad\qquad \quad \text{substitute }u = t+\frac{1}{2} x \implies "du=dt" \\ &=\int_\Omega e^{-a(u-\frac{1}{2}x)^2-b(\frac{1}{2}x-u)^2}dt \qquad \text{substitute }v = u-\frac{1}{2} x \implies "du=dv"\\ &=\int_\Omega e^{-(a+b)v^2} dv \qquad\qquad \qquad\,\, \text{substitute } w = \sqrt{a+b}v \implies"dw = \sqrt{a+b}dv" \\ &=\frac{1}{\sqrt{a+b}}\int_\Omega e^{-w^2}dw \\ &=\frac{\sqrt{\pi}}{\sqrt{a+b}} \end{align*}$

  • 1
    $\begingroup$ Hi @flawr. There is a mistake after the first substitution : indeed, if $t=u-x/2$, then $x-t=x-(u-x/2)=3x/2-u$. You can also see that there is something that goes wrong by seeing that $v=u-x/2=(t+x/2)-x/2=t$. You should end up with a new gaussian : take the Fourier tranform of the convolution to get the product of two new gaussians (as the Fourier transform of a gaussian is still a gaussian), then take the inverse Fourier transform to get another gaussian. $\endgroup$ – Nicolas Apr 16 '16 at 15:54
  • $\begingroup$ Oh I see now, thank you. $\endgroup$ – flawr Apr 16 '16 at 16:37

You seem to have lost the constant term modified by the completion of the square: $$ \begin{align} \int_{-\infty}^\infty e^{-a(x-t)^2}e^{-bt^2}\,\mathrm{d}t &=\int_{-\infty}^\infty e^{-ax^2+2axt-at^2-bt^2}\,\mathrm{d}t\\ &=e^{-ax^2+\frac{a^2x^2}{a+b}}\int_{-\infty}^\infty e^{-\frac{a^2x^2}{a+b}+2axt-(a+b)t^2}\,\mathrm{d}t\\ &=e^{\frac{-abx^2}{a+b}}\int_{-\infty}^\infty e^{-(a+b)\left(t-\frac{ax}{a+b}\right)^2}\,\mathrm{d}t\\ &=\sqrt{\tfrac\pi{a+b}}\,e^{\frac{-abx^2}{a+b}} \end{align} $$

  • $\begingroup$ Which, BTW, should match the result from the Central Limit Theorem that $N(\mu_1,\sigma_1) * N(\mu_2,\sigma_2) = N(\mu_1+\mu_2,\sqrt{\sigma_1^2+\sigma_2^2})$ $\endgroup$ – Acccumulation Oct 8 '19 at 14:50
  • $\begingroup$ Which, I believe, it does. That might indicate that this question might merit the probability-distributions tag. $\endgroup$ – robjohn Oct 8 '19 at 15:04

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