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What substitution/manipulation was done here?

  1. Jun 9, 2015 #1
    I've attached the image with this post. It looks fairly simply to go from the first line to the second line and almost like a uv substitution, but I tried plugging it in and playing around with the integral and cannot get the exact answer. I really feel like something's just going over my head or that I'm missing a step. Regardless, any help would be great!

    Note: mu is the mean.
     

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  3. Jun 9, 2015 #2

    PeroK

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    If ##\mu## is the mean, and ##f## is the pdf, then isn't ##\mu = \int xf(x)dx##?
     
  4. Jun 9, 2015 #3

    mathman

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    Expand [itex] (x-\mu)^2 = x^2-2x\mu+\mu^2[/itex] to get three integrals and use the definition of [itex]\mu[/itex] to get the second term.
     
  5. Jun 10, 2015 #4
    By infinity limits the integral is the same for each value of ##\mu##. Are you post all informations?
     
    Last edited: Jun 10, 2015
  6. Jun 10, 2015 #5

    mathman

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    [itex]\mu[/itex] is a constant. Therefore [itex]\int_{-\infty}^{\infty}\mu^2f(x)dx=\mu^2\int_{-\infty}^{\infty}f(x)dx=\mu^2[/itex].

    Also [itex]\int_{-\infty}^{\infty}\mu xf(x)dx=\mu\int_{-\infty}^{\infty}xf(x)dx=\mu^2[/itex].
     
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