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Math Help - Expecatation of Normal Distribution

  1. #1
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    Expecatation of Normal Distribution

    Hi everybody,

    I attached the first part to derive the expectation E[X] of the Normal distribution.

    I understand that (x-)/σ is replaced by y. But why is there an additional σ at the very end of the second part before dy. Why is this sigma added? A bit confusing for me..


    Thanks for your help.
    Attached Thumbnails Attached Thumbnails Expecatation of Normal Distribution-expectation-normal-distribution.jpg  
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  2. #2
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    Re: Expecatation of Normal Distribution

    Also, I have a problem to understand the integration by parts when solving for E(X^2).
    Again I attached the solution. In don't understand how the first term in the second row is integrated:
    I would use f(x)=y^2 ; f'(x)=2y; g'(x)=exp(-0.5(y^2) ; g(x) = (2*pi)^0.5
    However I would get 0 with this approach for that term

    ...In the solution, I dont get what was used for f(x), g'(x) etc...

    Again, thanks for any help. This forum is great!
    Attached Thumbnails Attached Thumbnails Expecatation of Normal Distribution-e-x-2-normal-distribution.jpg  
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  3. #3
    Grand Panjandrum
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    Re: Expecatation of Normal Distribution

    Quote Originally Posted by Marmy View Post
    Hi everybody,

    I attached the first part to derive the expectation E[X] of the Normal distribution.

    I understand that (x-)/σ is replaced by y. But why is there an additional σ at the very end of the second part before dy. Why is this sigma added? A bit confusing for me..


    Thanks for your help.
    If:

    y=\frac{x-\mu}{\sigma}

    dy=\frac{1}{\sigma}dx

    dx=\sigma dy
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  4. #4
    Grand Panjandrum
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    Re: Expecatation of Normal Distribution

    Quote Originally Posted by Marmy View Post
    Also, I have a problem to understand the integration by parts when solving for E(X^2).
    Again I attached the solution. In don't understand how the first term in the second row is integrated:
    I would use f(x)=y^2 ; f'(x)=2y; g'(x)=exp(-0.5(y^2) ; g(x) = (2*pi)^0.5
    However I would get 0 with this approach for that term

    ...In the solution, I dont get what was used for f(x), g'(x) etc...

    Again, thanks for any help. This forum is great!
    Split:

    -y^2e^{-\frac{y^2}{2}}= y \left[-y e^{-\frac{y^2}{2}} \right]

    then the term in the square brackets is the derivative of e^{-\frac{y^2}{2}}

    CB
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  5. #5
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    Re: Expecatation of Normal Distribution

    Thanks, again very helpful!
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