Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We know that $\text{exp}(-\alpha |x|^2)$ is a fixed point for the unitary Fourier transform if $\text{Re } \alpha > 0$.

I know many arguments to show this (contour-integration and differentiation).

Is there a not an elegant way where we can exploit that fact that the Gaussian is rotationally symmetric? A sketch would be fine.

share|cite|improve this question
up vote 10 down vote accepted

Here's a different argument. Take a zero-mean Gaussian random variable $X$. We know that the sum of $n$ copies of $X$, scaled down by a factor of $\sqrt{n}$, is distributed the same as $X$. On the other hand, by the characteristic function argument (which can be used to prove the central limit theorem) we know that the Fourier transform of $(X_1+\cdots+X_n)/\sqrt{n}$ converges to $\exp(- \sigma^2x^2/2)$.

Edit: an even simpler way. Again take $X$ to be a zero-mean Gaussian. We know that $(X+X)/\sqrt{2}$ is equidistributed with $X$. We immediately deduce that all higher-order cummulants are nil, and since the second cummulant is the variance, we get that the Fourier transform is $\exp(-\sigma^2x^2/2)$.

share|cite|improve this answer
Nice! I like stochastic arguments in analysis. – Jonas Teuwen Jan 14 '11 at 23:56
Well, the usual way is rather to prove that the characteristic function of a Gaussian random variable is a Gaussian function and use this to prove the central limit theorem. – Rasmus Jan 15 '11 at 1:12
A better way to think about the usual proof is that we find the characteristic function of the limiting random variable, and then we verify this c.f. corresponds to a Gaussian. Alternatively, we could find the Gaussian distribution by applying an inverse Fourier transform. – Yuval Filmus Jan 15 '11 at 1:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.