Results 1 to 2 of 2

Math Help - Expected value

  1. #1
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5

    Expected value

    \text{E}[L]=\text{E}[30X+2X^2]=30\text{E}[X]+2\text{E}[X^2]

    This is a Gamma Distrib with alpha = 3.5 and beta = 1.5

    \displaystyle p(x)=\frac{1}{\Gamma(3.5)1.5^{3.5}}x^{2.5}e^{-x/1.5^{-1}}

    \displaystyle\lim_{b\to\infty}\left[\frac{30}{\Gamma(3.5)1.5^{3.5}}\int_0^bx^{3.5}e^{-x/1.5^{-1}} \ dx+\frac{2}{\Gamma(3.5)1.5^{3.5}}\int_0^bx^{4.5}e^  {-x/1.5^{-1}} \ dx\right]

    \displaystyle =\frac{30\Gamma(4.5)}{\Gamma(3.5)1.5^{8}}+\frac{2\  Gamma(5.5)}{\Gamma(3.5)1.5^{9}}

    \displaystyle =\frac{30\cdot 3.5}{1.5^8}+\frac{2\cdot 4.5\cdot 3.5}{1.5^9}=4.91632

    The book says the solution is 228.4 though. I don't understand why I am wrong.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,111
    Thanks
    2
    You seem a little confused on some notation and some algebra.

    1) x/1.5^{-1}? Too much denominatoring going on in there. Try x/1.5.

    2) Just a couple of misconstrued items. You should get:

    30\cdot \frac{\Gamma(4.5)\cdot 1.5}{\Gamma(3.5)} + 2\cdot \frac{\Gamma(5.5)\cdot 1.5^{2}}{\Gamma(3.5)} = 30\cdot 3.5 \cdot 1.5 + 2\cdot 4.5 \cdot 3.5 \cdot 1.5^{2} = 228.375

    This is a tough place to be having an algebra breakdown. Be more careful.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Expected Value
    Posted in the Statistics Forum
    Replies: 1
    Last Post: April 25th 2011, 04:51 PM
  2. Expected value/ st dev/var
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: September 30th 2010, 03:55 AM
  3. Expected Value
    Posted in the Advanced Statistics Forum
    Replies: 9
    Last Post: September 20th 2010, 12:52 AM
  4. Expected Value
    Posted in the Statistics Forum
    Replies: 3
    Last Post: April 28th 2009, 05:05 PM
  5. expected value
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: January 16th 2009, 09:51 AM

/mathhelpforum @mathhelpforum