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Math Help - Probabilty Density Function

  1. #1
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    Probabilty Density Function

    Hi guy's
    I have been given this question.

    The Random variable X follows the distribution given by the density
    f(x) = cxe^(-x^2/2)

    determine c and E(x)

    I've taken the integral of cxe^(-x^2/2) between 0 and infinity and tried to solve for 1.

    But I'm getting the result of the intergral as C = 1

    Is this right?

    regards
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by boneill3 View Post
    Hi guy's
    I have been given this question.

    The Random variable X follows the distribution given by the density
    f(x) = cxe^(-x^2/2)

    determine c and E(x)

    I've taken the integral of cxe^(-x^2/2) between 0 and infinity and tried to solve for 1.

    But I'm getting the result of the intergral as C = 1

    Is this right?

    regards
    Yes.

    Now, the expected value would be E(X)=\int_0^{\infty}x^2e^{-\frac{x^2}{2}}\,dx

    Here are two different approaches (one requires the definition of the Gamma Function, the other requires integration by parts):

    To integrate this guy, make the substitution u=\frac{x^2}{2}\implies\,du=x\,dx

    Now, we have \int_0^{\infty}x\cdot e^{-u}\,du

    But, u=\frac{x^2}{2}\implies x=\sqrt{2u}

    Thus, we have \sqrt{2}\int_0^{\infty}u^{\frac{1}{2}}e^{-u}\,du

    Now, we can recognize \int_0^{\infty}u^{\frac{1}{2}}e^{-u}\,du as \Gamma\left(\tfrac{3}{2}\right)=\frac{\sqrt{\pi}}{  2}

    Therefore, \sqrt{2}\int_0^{\infty}u^{\frac{1}{2}}e^{-u}\,du=\sqrt{2}\cdot\frac{\sqrt{\pi}}{2}=\color{re  d}\boxed{\frac{\sqrt{2}\sqrt{\pi}}{2}}

    --OR--

    \int_0^{\infty}x^2e^{-\frac{x^2}{2}}\,dx=\int_0^{\infty}x\cdot xe^{-\frac{x^2}{2}}\,dx

    Let u=x\implies \,du=\,dx and \,dv=xe^{-\frac{x^2}{2}}\,dx\implies v=-e^{-\frac{x^2}{2}}

    Thus, \int_0^{\infty}x^2e^{-\frac{x^2}{2}}\,dx=\left.\left[-xe^{-\frac{x^2}{2}}\right]\right|_0^{\infty}+\int_0^{\infty}e^{-\frac{x^2}{2}}\,dx=\left[0+0\right]+\frac{1}{2}\sqrt{\frac{\pi}{1/2}}=\frac{1}{2}\sqrt{2\pi}=\color{red}\boxed{\frac  {\sqrt{2}\sqrt{\pi}}{2}}

    Thus, \color{red}\boxed{E(X)=\frac{\sqrt{2}\sqrt{\pi}}{2  }}

    Does this make sense?

    --Chris
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  3. #3
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    Thanks Chris,
    That makes a lot of sense.

    I found out what I was doing wrong in my own calculations for E(x) as well.

    I was taking the integral of 1*x*e^-2^2/2 instead of x*x*e^-2^2/2 =... x^2*e^-2^2/2

    Thanks alot I'll have to be more carefull in the future!
    regards
    Brendan
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