Results 1 to 2 of 2

Math Help - Expected value of the logistic distribution?

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    5

    Expected value of the logistic distribution?

    How would you show that the expected value of the logistic distribution is equal to it's mean? ie E(X)=mu

    E(X) = \int_{-\infty}^{\infty} {\frac{xe^{-(x-\mu)/s}} {s\left(1+e^{-(x-\mu)/s}\right)^2}} \!

    slightly simplified:

    E(X) = e^{\mu/s}/s \int_{-\infty}^{\infty} {\frac{xe^{-x/s}} {\left(1+e^{-(x-\mu)/s}\right)^2}} \!

    Various substitutions I've tried don't seem to work, u={1+e^{-(x-\mu)/s}}, etc haven't worked to well for me.
    Last edited by evidor; October 5th 2008 at 10:03 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by evidor View Post
    How would you show that the expected value of the logistic distribution is equal to it's mean? ie E(X)=mu

    E(X) = \int_{-\infty}^{\infty} {\frac{xe^{-(x-\mu)/s}} {s\left(1+e^{-(x-\mu)/s}\right)^2}} \!

    slightly simplified:

    E(X) = e^{\mu/s}/s \int_{-\infty}^{\infty} {\frac{xe^{-x/s}} {\left(1+e^{-(x-\mu)/s}\right)^2}} \!

    Various substitutions I've tried don't seem to work, u={1+e^{-(x-\mu)/s}}, etc haven't worked to well for me.
    Since  \frac{e^{-A}}{\left( 1 + e^{-A}\right)^2} = \frac{1}{4} \, \text{sech}^2 A it follows that \frac{e^{-(x-\mu)/s}} {s \left( 1 + e^{-(x-\mu)/s} \right)^2} = \frac{1}{4s} \, \text{sech}^2 \left( \frac{x - \mu}{2s}\right).


    Therefore: E(X) = \int_{-\infty}^{+\infty} \frac{x}{4s} \, \text{sech}^2 \left( \frac{x - \mu}{2s}\right) \, dx.


    Substitute w = \frac{x - \mu}{2s} and split the integral up: E(X) = s \int_{-\infty}^{+\infty} w \, \text{sech}^2 w \, dw + \frac{\mu}{2}  \int_{-\infty}^{+\infty} \text{sech}^2 w \, dw .


    The first integral is equal to zero since the integrand is an odd function of w. The second integral is simple to perpetrate and is equal to 2.

    Then E(X) = \mu.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: February 6th 2011, 02:01 PM
  2. Expected value of probability distribution
    Posted in the Statistics Forum
    Replies: 8
    Last Post: June 5th 2010, 04:50 AM
  3. Logistic distribution
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: October 24th 2009, 05:18 PM
  4. Replies: 1
    Last Post: October 5th 2009, 02:45 PM
  5. Expected value of the MLE of a binomial distribution
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: September 24th 2009, 04:57 AM

Search Tags


/mathhelpforum @mathhelpforum