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How to derive to this point

  1. Apr 28, 2008 #1
    hello, can someone look at my problem and tell me how to get to the arrow?
    thank you so much..
    i am studying for my midterm and i have no clue where it come from.
    thanks
     

    Attached Files:

  2. jcsd
  3. Apr 28, 2008 #2

    uart

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    That first line you point to is basically just the definition of how you find the "expected value" of a function.

    That is, if a random variable X has a pdf (probablity density function) of f(x) then the expected value of x can be written as,

    [tex] E(x) = \int_{-\infty}^{+\infty} x f(x) dx [/tex]

    And more generally the expected value of a function of x can be written as,

    [tex] E(\, \phi(x)\, ) = \int_{-\infty}^{+\infty} \phi(x) f(x) dx [/tex]
     
    Last edited: Apr 28, 2008
  4. Apr 28, 2008 #3

    uart

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    BTW, there's an error in that derivation. The x^2 term that appears in the first square bracketed term in line 4 (2nd line under the completing the square heading) should be just "x" (not squared). This error is carried all the way through the derivation BTW.
     
  5. Apr 28, 2008 #4
    yeah.i understand that part..
    but how does exp(-o^2w^2/2) = exp(-x^2/2o^2)?
    the arrow....
     
  6. Apr 30, 2008 #5

    uart

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    What? It never says that anywhere. The question says to prove that,

    [tex]\Phi(\omega) = exp(-\sigma^2 \omega^2 /2)[/tex]

    You must be misreading it because nowhere does it claim the thing you state.
     
  7. Apr 30, 2008 #6

    HallsofIvy

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    It doesn't say that. It says that you are to prove that the [/b]characteristic function[/b] for the the Gaussian distribution is
    [tex]e^{\frac{-\sigma^2\omega^2}{2}}[/itex]
    and them immediately uses the fact that the Gaussian distribution (with mean 0) itself can be written as
    [tex]e^{\frac{-x^2}{2\sigma^2}[/tex]

    Those are two completely different functions.
     
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