Results 1 to 2 of 2

Thread: Find E[X] and E[E[X|Y]]

  1. #1
    Super Member
    Mar 2006

    Find E[X] and E[E[X|Y]]

    Let X and Y be random variables with $\displaystyle p(x,y)=e^ {-y} $ for $\displaystyle 0<x<y<+ \infty $

    Find $\displaystyle E[X] $ and $\displaystyle E[E[X|Y]] $

    My solution:

    a) First of all, $\displaystyle p(x) = e^{-x} $, so I have $\displaystyle E[X] = \int ^ y _0 xe^{-x}dx = -ye^{-y}-e^{-y}+1 $

    b) Here, $\displaystyle E[X|Y] = \frac {Y}{2} $ and $\displaystyle p(y)=ye^{-y} $, so I have $\displaystyle E[E[X|Y]] = \int E[X|Y=y]p(y)dy = \int ^ \infty _x \frac {y}{2}(ye^{-y})dy= \frac {1}{2} e^{-x}(x^2+2x+2) $

    But according to the law of total expectation, they should be the same thing, so what am I doing wrong? Thank you!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor matheagle's Avatar
    Feb 2009

    Re: Find E[X] and E[E[X|Y]]

    part of this was asked in another question here

    (a) Yes, $\displaystyle f_X(x)=e^{-x}$
    BUT that is the marginal, hence the range is x>0
    you no longer want x<y.
    Plus the E(X) is a constant, which cannot have Y's in it.
    E(X)=1, this is an exponential

    (b) Yes, $\displaystyle E(X|Y)=y/2$

    and $\displaystyle f_Y(y)=ye^{-y}$, y>0 a Gamma RV

    So $\displaystyle E_Y(E(X|Y)) =\int_0^{\infty}(y/2)ye^{-y}dy=(1/2)\int_0^{\infty}y^2e^{-y}dy=(1/2)\Gamma(3)=1.$
    Last edited by matheagle; Sep 5th 2011 at 07:08 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: Mar 22nd 2011, 04:57 PM
  2. Replies: 2
    Last Post: Jul 5th 2010, 08:48 PM
  3. Replies: 1
    Last Post: Feb 17th 2010, 03:58 PM
  4. Replies: 0
    Last Post: Jun 16th 2009, 12:43 PM
  5. Replies: 2
    Last Post: Apr 6th 2009, 08:57 PM

Search Tags

/mathhelpforum @mathhelpforum