# Gaussian center not average value?

1. May 19, 2012

### DocZaius

I put this under statistics because of your knowledge of the Gaussian.

I have run into an elementary problem. I was considering what the average value, <x>, is for a Gaussian with an x offset, and got results which don't make sense to me.

First, it is obvious that for P1(x)=e^(-(x^2)) the average is 0. The integral of x*P1(x) from -inf to inf is zero.

So I decided to center the Gaussian at x=2: P2(x)=e^(-(x-2)^2). I was expecting an average of 2, but instead got 2 times the square root of pi! The integral of x*P2(x) from -inf to inf is 2*sqrt(pi)...

I thought, well perhaps this has to do with the infinities lying on either side, and the plus side starting at higher values than the minus side...So I decided to see the average for the Gaussian centered at 2 from x=0 to x=4. Surely that must be 2, it is clearly symmetric in that interval. Turns out that x*P2(x) from 0 to 4 is (according to Wolfram Alpha) still 2*sqrt(pi).

Can anyone explain why this last Gaussian's average x isn't 2?

Thanks.

2. May 19, 2012

### D H

Staff Emeritus
You have the wrong PDF. The PDF for a normal distribution with mean $\mu$ and standard deviation $\sigma$ is given by
$$p(x) = \frac 1 {\sigma \sqrt{2\pi}} \exp\left(-\frac 1 2 \left(\frac {x-\mu}{\sigma}\right)^2\right)$$
A couple of examples for a mean of 2:

1. $\sigma=1$:
$$p(x) = \frac 1 {\sqrt{2\pi}} \exp\left(-\frac 1 2 (x-2)^2\right)$$

1. $\sigma=1/\surd 2$:
$$p(x) = \frac 1 {\sqrt{\pi}} \exp\left(-(x-2)^2\right)$$

Last edited: May 19, 2012
3. May 19, 2012

### DocZaius

I am attempting to look at the average x for the function I brought up. (It's from a quantum textbook). I understand it is different from the one you wrote. Can you explain my problems with the function I wrote?

My problem is that the intuitive idea that a Gaussian centered at x=a would have an average x of a is clearly wrong, but in the case where I look at a symmetric interval around it, I don't understand how that can be the case.

4. May 19, 2012

### D H

Staff Emeritus
Your function is not a PDF. Calculate $\int_{-\infty}^{\infty} p(x)\,dx$ for your function. It won't be one.

5. May 19, 2012

### DocZaius

Is the concept of "average x" only applicable to functions whose integrals at infinity are 1? I don't see why that must be the case.

I am aware that my function does not count as a probability function since it doesn't integrate to 1 at infinity, but it seems to me that one could still devise an "average x" from it by weighing all its x's with f(x) inside the integrand.

6. May 19, 2012

### D H

Staff Emeritus
If your function f(x) is not normalized (i.e., $\int_D f(x)\,dx \ne 1$ where the integral is over some applicable domain such as -∞ to +∞) you can still compute an average, but you will need to account for the fact that the function is not normalized. In your case,
$$\bar x \equiv \frac{\int_{-\infty}^{\infty} xf(x)\,dx}{\int_{-\infty}^{\infty} f(x)\,dx}$$

7. May 19, 2012

Thanks!