Results 1 to 3 of 3

Math Help - Conditional Expectation of a joint PMF

  1. #1
    Member
    Joined
    Oct 2009
    Posts
    107
    Thanks
    2

    Conditional Expectation of a joint PMF

    (please refer to attached image)
    The question appears to be simple enough, but i have two queries

    A) does E[X1 X2] mean the same as E[X1 | X2]

    B) If not/so, how exactly do I go about computing this. I've seen a few formulas in my lectures notes for computing conditional expectations for discrete random variables,
    however I find it difficult to understand and apply the notation/procedure.

    Any help is appreciated!


    edit: ok, after some more research, i've found that
    E(X1 X2] simply means The expectations of X1 and X2 multiplied by each other.

    so, what I want to ask now is this.
    is the PMF of X1, given that table:

    X1 | -1 | 0 | 1 |
    px(X1)| 1/3 | 0 | 1/3 |



    And finally, how do i find out if X1 and X2 are independent?
    Attached Thumbnails Attached Thumbnails Conditional Expectation of a joint PMF-conditional-joint-pmf.png  
    Last edited by 99.95; September 19th 2013 at 06:00 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    4,170
    Thanks
    765

    Re: Conditional Expectation of a joint PMF

    Hey 99.95.

    For a) the answer is no. E[X1*X2] is not the same as E[X1|X2]. E[X1X2] = Integral over region for (X1,X2) x1x2*f(x1,x2)dx1dx2 whereas E[X1|X2=x2] = Integral x1*f(x1|X2=x2)dx1.

    X1 and X2 are independent if for all values of X1 and X2 you have the joint distribution to be separable. In other words:

    P(X1 = x1, X2 = x2) = P(X1 = x2)P(X2 = x2) for all x1 and x2.

    To compute conditional expectations you need to find the conditional distribution. To do this use the relationship P(A|B) = P(A and B)/P(B) and then integrate with respect to the mean of A.

    To get you started on the table exercise, note that

    E[X1X2] = -1*-1*P(X1=-1,X2=-1) + 0*-1*P(X1=0,X2=-1) + ... + 1*1*P(X1=1,X2=1).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2009
    Posts
    107
    Thanks
    2

    Re: Conditional Expectation of a joint PMF

    thanks chiro,
    i forgot to update but yes i figured that out. Although I don't understand what I am supposed to integrate over? or is that when i'm given a PDF instead of in tabular form.

    i found E[X1X2] = 0, E[Y1Y2]=0 and E[Z1Z2] = -2/3

    I also just wanted to ask, instead of computing the E[X1X2] and E[Y1Y2] by -1*-1*P(X1=-1,X2=-1)+...+
    is there some way to say, 'it is symmetrical around 0, so E[X1X2] = 0?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Conditional expectation
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: October 10th 2011, 12:20 PM
  2. Replies: 0
    Last Post: July 22nd 2011, 02:39 AM
  3. conditional expectation
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: January 24th 2010, 01:31 AM
  4. Conditional expectation
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: January 6th 2010, 04:34 PM
  5. Expectation & Conditional Expectation
    Posted in the Advanced Statistics Forum
    Replies: 5
    Last Post: February 1st 2009, 11:42 AM

Search Tags


/mathhelpforum @mathhelpforum