Gaussian center not average value?

[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.

 

[D H]

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)...

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

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)

 

[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.

 

[D H]

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

 

[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.

 

[D H]

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.

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}

 

[DocZaius]

Thanks!