I am a bit confused about calculating the integral of a Gaussian


Given above is the integral of a Gaussian. The integral of a Gaussian is Gaussian itself. But what is the mean and variance of this Gaussian obtained after integration?

  • $\begingroup$ You are not computing "the integral of a Gaussian", whatever that means. You are computing the integral of a function of the real variable $x$, not of a random variable. The result is the quantity on the right which can be regarded as a constant if $b$ and $c$ are known constants or as a function of two variables $b$ and $c$ if $b$ and $c$ are regarded as parameters of the integrand. So, the question you ask is meaningless: there is no mean and no variance because the result is not a random variable. $\endgroup$ – Dilip Sarwate Jan 15 '13 at 16:27
  • $\begingroup$ Sorry but where is the random variable? $\endgroup$ – Did Jan 15 '13 at 17:38

The question is only meaningful if $\Im{b} \ne 0$. Let's say that, rather, $\Re{b} = 0$ and $b = i B$. Now you can assign a mean/variance to the resulting Gaussian. This, BTW, is related to the well-known fact that a Fourier transform of a Gaussian is a Gaussian.

  • $\begingroup$ You seem to be able to see a Gaussian random variable in the question. Where? $\endgroup$ – Did Jan 15 '13 at 17:39
  • $\begingroup$ Not explicitly, but he did ask about a mean and variance, which may define the parameters of a Gaussian function. $\endgroup$ – Ron Gordon Jan 15 '13 at 18:41
  • $\begingroup$ Sorry but I do not get it. And why do you assume that $\Im b\ne0$? This is not necessary to get a convergent integral. $\endgroup$ – Did Jan 15 '13 at 19:23
  • $\begingroup$ It IS necessary to get a meaningful value of a mean and variance (i.e. 1st and 2nd moments of the Gaussian). $\endgroup$ – Ron Gordon Jan 15 '13 at 19:28
  • 1
    $\begingroup$ I get it, although I think you are asking for more rigor than the OP is asking. I understand that this is the Mathematics exchange and rigor is warranted, but in this case, I think we can let go a bit. $\endgroup$ – Ron Gordon Jan 15 '13 at 19:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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