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Quantum statistics expectation value

  1. Sep 4, 2011 #1
    1. What is the expectation value, <x>, for the
    given distribution over the interval from – to + infinity of the function: f(x)=e^(-.5(x-mu)^2(sigma^-2))




    2. This is a statistics problem i think. I just need to know how this type of problem is worked out because it is relevant to my quantum class.



    3. I have absolutely no idea where to begin.
     
  2. jcsd
  3. Sep 4, 2011 #2

    vela

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    The expectation value of a function of a random variable X is given by
    [tex]E[g(X)] = \int_{-\infty}^\infty g(x) f(x)\,dx[/tex]
    Do you know how to integrate a Gaussian?
     
  4. Sep 5, 2011 #3
    Not exactly. My professor ran very very quickly through the gaussian integral once before but I don't quite have the math background to have taken it in (in multivariable calculus currently).
     
  5. Sep 5, 2011 #4

    vela

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  6. Sep 5, 2011 #5
    Okay so the Gaussian integral is just the integral from -to+infinity of e^(-x^2). I see what they did through to get the answer. Where does the Gaussian fit into that other integral you listed?
    Thanks for the help. I was pretty clueless when this was presented in class because I haven't seen this math before so I apologize for how slow I am.
     
  7. Sep 5, 2011 #6

    vela

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    Tell us what integral you need to calculate to find <x>.
     
  8. Sep 5, 2011 #7
    So do I just take this
    [tex]E[g(X)] = \int_{-\infty}^\infty g(x) f(x)\,dx[/tex]
    and use the function, [tex] f(x) = e^{(-(x-μ)^2/2σ^2)} [/tex] as f(x) and would g(x) be the Gaussian function? Not really sure what g(x) is unless that guess is correct.
     
  9. Sep 5, 2011 #8

    vela

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    g(X) is what you're finding the expectation value of. In this case, g(X)=X.
     
  10. Sep 5, 2011 #9
    Oh! That makes a lot more sense.
    [tex]E[x] = \int_{-\infty}^\infty xe^{(-(x-μ)^2/2σ^2)}\,dx[/tex]
    is the integral I'm left with. I see it's similar to the an example on the gaussian integral wiki page where [tex]\int_{-\infty}^\infty x^{2n}e^{-ax^2}\,dx[/tex]
    So I figured I'd go off and see if I could solve it with just some sort of u substitution but the closest I can bring the integral i have to that is [tex]e^{-μ^2}\int_{-\infty}^\infty xe^{-x^2/2σ^2}e^{xμ/σ^2}\,dx[/tex] So I doubt it's that easy and probably requires some sort of harder method. How should I go about this?
     
  11. Sep 5, 2011 #10

    vela

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    Try the substitution u = x-mu.
     
  12. Sep 5, 2011 #11
    okay so to double check so far:
    [tex]E[x] = \int_{-\infty}^\infty xe^{(-(x-μ)^2/2σ^2)}\,dx[/tex]
    [tex]u=x-μ[/tex][tex]du=dx [/tex]
    [tex]E[x] = \int_{-\infty}^\infty ue^{(-u^2/2σ^2)}\,du + μ\int_{-\infty}^\infty e^{(-u^2/2σ^2)}\,du [/tex]
    These look like 2 of the examples on wikipedia but if I just solve them I get something like
    [tex]σ^5(2π)^{1/2} + μσ(2)^{1/2}(-1/2)![/tex]
    is that right? a concern i have is that I never had to substitute x-μ=u back in
     
  13. Sep 5, 2011 #12

    vela

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    You should find the first integral is 0. You can integrate it using a substitution or simply note that the integrand is an odd function.

    What's (-1/2)! equal to?
     
  14. Sep 5, 2011 #13
    Okay I see that. I don't understand the gamma function but I take it from wikipedia that (-1/2)! is equal to Γ(1/2)
    So I guess it comes down to Γ(1/2)μσ(2)^(1/2) which is
    [tex]μσ(2)^{1/2}\int_{0}^\infty t^{-1/2}e^{-t}\,dt[/tex]
    I also don't know where to go about this integral, by parts I get left with something minus the gamma function of -.5 which doesn't seem to go anywhere.
     
  15. Sep 5, 2011 #14

    vela

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    Look at the very first integral on that Wiki page I linked to earlier.
     
  16. Sep 6, 2011 #15
    Thanks for the help! I finished and got through it in time for class. Doubled checked with a peer.
     
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