Confusion related to integral of a Gaussian

Question

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

$$\int_{-\infty}^{\infty}e^{-x^{2}+bx+c}\:dx=\sqrt{\pi}e^{\frac{b^{2}}{4}+c}$$

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?

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. — Dilip Sarwate, Jan 15 '13 at 16:27

score 1 · Answer 1 · answered Jan 15 '13 at 16:55

1

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.

answered Jan 15 '13 at 16:55

Ron Gordon

138,521

You seem to be able to see a Gaussian random variable in the question. Where? – Did Jan 15 '13 at 17:39
Not explicitly, but he did ask about a mean and variance, which may define the parameters of a Gaussian function. – Ron Gordon Jan 15 '13 at 18:41
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. – Did Jan 15 '13 at 19:23
It IS necessary to get a meaningful value of a mean and variance (i.e. 1st and 2nd moments of the Gaussian). – Ron Gordon Jan 15 '13 at 19:28
...of the Gaussian which might or might not be there but nobody knows where? Not what I call crystal clear... – Did Jan 15 '13 at 19:47
1

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. – Ron Gordon Jan 15 '13 at 19:49
let us continue this discussion in chat – Ron Gordon Jan 15 '13 at 19:52

Confusion related to integral of a Gaussian

1 Answers1