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Math Help - [SOLVED] Finding the integration result

  1. #1
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    [SOLVED] Finding the integration result

    Actually I was trying to find the expected value, nevertheless, hope someone can give me some idea of deriving the result:

    given: f(x|\theta)=\frac{x}{\theta ^2}e^{(-x^2)/(2\theta ^2)} x\geq 0

    I want to solve for: \int_{0}^{\infty}\frac{x^2}{\theta ^2}e^{(-x^2)/(2\theta ^2)}.dx .
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by noob mathematician View Post
    Actually I was trying to find the expected value, nevertheless, hope someone can give me some idea of deriving the result:

    given: f(x|\theta)=\frac{x}{\theta ^2}e^\frac{-x^2}{(2\theta ^2)} x\geq 0

    I want to solve for: \int_{0}^{\infty}\frac{x^2}{\theta ^2}e^\frac{-x^2}{(2\theta ^2)} .
    Make the substitution u=x^2/(2\theta^2).

    Then \,du=x/ \theta^2\,dx.

    So \int_0^{\infty}\frac{x^2}{\theta^2}e^{-x^2/(2\theta^2)}\,dx=\sqrt{2}\theta\int_0^{\infty}u^{1/2}e^{-u}\,du=\sqrt{2}\theta\Gamma\!\left(\tfrac{3}{2}\ri  ght)=\tfrac{1}{2}\sqrt{2}\theta\Gamma\!\left(\tfra  c{1}{2}\right)=\tfrac{1}{2}\sqrt{2\pi}\theta.

    Does this make sense?
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  3. #3
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    Smile

    Quote Originally Posted by Chris L T521 View Post
    Make the substitution u=x^2/(2\theta^2).

    Then \,du=x/ \theta^2\,dx.

    So \int_0^{\infty}\frac{x^2}{\theta^2}e^{-x^2/(2\theta^2)}\,dx=\sqrt{2}\theta\int_0^{\infty}u^{1/2}e^{-u}\,du=\sqrt{2}\theta\Gamma\!\left(\tfrac{3}{2}\ri  ght)=\tfrac{1}{2}\sqrt{2}\theta\Gamma\!\left(\tfra  c{1}{2}\right)=\tfrac{1}{2}\sqrt{2\pi}\theta.

    Does this make sense?
    Thanks make full sense!
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