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Math Help - Expected value of X

  1. #1
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    Expected value of X

    I am trying to find the expected value of X.

    <br />
f(x) = 2x e^{ - x^2 } dx,<br />
for all x \in R = (0,\infty )<br />





    <br />
u = E[X] = \int_R^{} {x*f(x)} dx = 2\int\limits_0^\infty  {x^2 e^{ - x^2 } dx} <br />

    I could not find any integral table to make this any easier and my trusty ti-92+ would not integrate this either.


    So, I have tried to solve by letting
    <br />
u = x^2 and\_du = 2xdx<br />
    Again, I have gotten nowhere.

    Any suggestions would be appreciated.

    Thanks.

    Note:
    I just found a table that lists the integral = .5*(pi)^1/2.
    Last edited by kid funky fried; October 1st 2008 at 06:31 PM.
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  2. #2
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    Quote Originally Posted by kid funky fried View Post
    I am trying to find the expected value of X.

    <br />
f(x) = 2x e^{ - x^2 } dx,<br />
for all x \in R = (0,\infty )<br />





    <br />
u = E[X] = \int_R^{} {x*f(x)} dx = 2\int\limits_0^\infty {x^2 e^{ - x^2 } dx} <br />

    I could not find any integral table to make this any easier and my trusty ti-92+ would not integrate this either.


    So, I have tried to solve by letting
    <br />
u = x^2 and\_du = 2xdx<br />
    Again, I have gotten nowhere.

    Any suggestions would be appreciated.

    Thanks.

    Note:
    I just found a table that lists the integral = .5*(pi)^1/2.
    It can be reduced to a 'standard' definite integral using integration by parts. I have no time now but will post later.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    It can be reduced to a 'standard' definite integral using integration by parts. I have no time now but will post later.
    You might find this thread has what you want: http://www.mathhelpforum.com/math-he...-function.html

    After reading this, let me know if you still need help with your specific calculation.
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  4. #4
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    Mr. Fantastic,

    I briefly read the thread....it seems advanced for the class I am in (intro to prob).

    So, to answer your question, i guess i am lost...i have given this problem hours of thought and feel like i have gotten nowhere.
    this is a homework problem and many classmates and i are baffled.
    i will review the thread again to see if i can understand how it applies.
    thx for your help.
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  5. #5
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    Quote Originally Posted by kid funky fried View Post
    Mr. Fantastic,

    I briefly read the thread....it seems advanced for the class I am in (intro to prob).

    So, to answer your question, i guess i am lost...i have given this problem hours of thought and feel like i have gotten nowhere.
    this is a homework problem and many classmates and i are baffled.
    i will review the thread again to see if i can understand how it applies.
    thx for your help.
    E(X) = 2 \int_{0}^{+\infty} x^2 e^{-x^2} \, dx.

    Substitute u = x^2 \Rightarrow dx = \frac{du}{dx}:

    E(X) = \int_{0}^{+\infty} u^{1/2} e^{-u} \, du.


    There is a well known special function called the Gamma function:

    Definition: \Gamma(n) = \int_{0}^{+\infty} u^{n-1} e^{-u} \, du.

    Special property: \Gamma (n+1) = n \Gamma (n).


    Therefore E(X) = \Gamma \left(\frac{3}{2}\right) = \frac{1}{2} \Gamma \left(\frac{1}{2}\right).

    It's well known that \Gamma \left(\frac{1}{2}\right) = \sqrt{\pi}.

    Therefore E(X) = \frac{1}{2} \, \sqrt{\pi}.
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