Trying to solve this question:

Let $f(x)=e^{-x^2}$ be a Gaussian. Compute explicitly $(f*f)(x)$.

Using the definition of the convolution, and given the fact that the convolution of 2 Gaussians is another Gaussian, I got \begin{align*} (f*f)(x) &= \int_{-\infty}^{\infty}f(x-y)f(y)\,dy\\ &=\int_{-\infty}^{\infty}e^{-(x-y)^2}e^{-y^2}\,dy \end{align*} but I'm not sure how to proceed from here.

Any tips would be appreciated!

  • $\begingroup$ Where did you use the fact that convolution of two Gaussians is a Gaussian? $\endgroup$ – Gae. S. Oct 7 '19 at 21:50
  • $\begingroup$ don't think i used it in the questions, but it is a well known property $\endgroup$ – nickoba Oct 7 '19 at 21:55
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    $\begingroup$ You could rewrite the integrand so you can complete the square involving $y$ $\endgroup$ – Henry Oct 7 '19 at 21:55
  • $\begingroup$ math.stackexchange.com/questions/1745174/… $\endgroup$ – cmk Oct 7 '19 at 22:07
  • $\begingroup$ Next step: $\displaystyle\int_{-\infty}^\infty e^{-x^2+2xy-2y^2}\,\mathrm{d}y$ $\endgroup$ – robjohn Oct 8 '19 at 7:13

First, complete the square to get $-a((y+b)^2+cx^2)$, then you could take $e^{-acx^2}$ beyond the sign of the integral since integration goes over $y$ and change the integration variable to $(y+b)$. Finally, use the well-known formula for the Gaussian integral. As an answer, I've got $\sqrt{\frac{\pi}{2}}\cdot e^{-\frac{x^2}{2}}$


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