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Math Help - expected value

  1. #16
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    Re: expected value

    so i need to integrate y^3*(e^-(y^2/2v))/(2piv)^0.5 between + and - infinity and show its 0. How to compute the integral?
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  2. #17
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    Re: expected value

    Quote Originally Posted by Duke View Post
    so i need to integrate y^3*(e^-(y^2/2v))/(2piv)^0.5 between + and - infinity and show its 0. How to compute the integral?
    You don't, you look at the symmetry/antisymmetry about y=0.

    Call the integrand I(y) so:

    I(y)=y^3*(e^-(y^2/2v))/(2piv)^0.5

    Now what is I(-y) in relation to I(y)?

    CB
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  3. #18
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    Re: expected value

    I(-y)= -I(y) due to the y^3.
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  4. #19
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    Re: expected value

    Quote Originally Posted by Duke View Post
    I(-y)= -I(y) due to the y^3.
    Then the integral we may write

    \int_{-\infty}^{\infty} I(y)dy =\int_{-\infty}^0 I(y) dy + \int_{0}^{\infty} I(y) dy= ..

    or you may just know what that means for the integral.

    CB
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  5. #20
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    Re: expected value

    =0
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  6. #21
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    Re: expected value

    Quote Originally Posted by Duke View Post
    =0
    Yes.

    CB
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