Results 1 to 3 of 3

Thread: Probabilty Density Function

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    2

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    5
    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 $\displaystyle E(X)=\int_0^{\infty}x^2e^{-\frac{x^2}{2}}\,dx$

    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 $\displaystyle u=\frac{x^2}{2}\implies\,du=x\,dx$

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

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

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

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

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

    --OR--

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

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

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

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

    Does this make sense?

    --Chris
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2008
    Posts
    2
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding a Cumulative Distribution Function and Density Function
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: Nov 17th 2011, 09:41 AM
  2. Replies: 4
    Last Post: Oct 27th 2010, 05:41 AM
  3. Density Function
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Apr 27th 2010, 10:59 PM
  4. Replies: 1
    Last Post: Apr 10th 2009, 10:59 AM
  5. p.g.f / density function
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: Jan 17th 2009, 02:20 AM

Search Tags


/mathhelpforum @mathhelpforum