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Gaussian center not average value?

  1. May 19, 2012 #1
    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?

  2. jcsd
  3. May 19, 2012 #2

    D H

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

    1. [itex]\sigma=1[/itex]:
    [tex]p(x) = \frac 1 {\sqrt{2\pi}} \exp\left(-\frac 1 2 (x-2)^2\right)[/tex]

    1. [itex]\sigma=1/\surd 2[/itex]:
    [tex]p(x) = \frac 1 {\sqrt{\pi}} \exp\left(-(x-2)^2\right)[/tex]
    Last edited: May 19, 2012
  4. May 19, 2012 #3
    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.
  5. May 19, 2012 #4

    D H

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    Your function is not a PDF. Calculate [itex]\int_{-\infty}^{\infty} p(x)\,dx[/itex] for your function. It won't be one.
  6. May 19, 2012 #5
    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.
  7. May 19, 2012 #6

    D H

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    If your function f(x) is not normalized (i.e., [itex]\int_D f(x)\,dx \ne 1[/itex] 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,
    [tex]\bar x \equiv \frac{\int_{-\infty}^{\infty} xf(x)\,dx}{\int_{-\infty}^{\infty} f(x)\,dx}[/tex]
  8. May 19, 2012 #7
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