Results 1 to 4 of 4

Math Help - Calculating Probabilities from a Joint Distribution

  1. #1
    Member
    Joined
    Dec 2010
    Posts
    92

    Calculating Probabilities from a Joint Distribution

    I spent a solid 35 minutes writing out a long and extensive post then the forum software decided I was on the page for too long and erased everything. So here it goes again... This is for studying, it was from a past homework assignment.

    I have a joint probability distribution, P(X,Y) that is a 4x4 matrix.

    X
    1 2 3 4
    1 1/4 0 1/8 1/8
    Y 2 1/8 0 1/16 1/16
    3 1/16 0 1/32 1/32
    4 1/16 0 1/32 1/32










    I need to, and cannot, find P(Y|X) and P(X=3|Y).

    Below I use the notation P(x,y) to denote the probability at the coordinates (x,y).

    My attempt for P(Y|X):

    P(Y|X) = Each vector in X..?? (Isn't that just the original table?)
    So,
    P(Y|X) = { P(Y|X=1), P(Y|X=2), P(Y,X=3), P(Y,X=4) }

    (X is across, Y is vertical)
    Y / X 1 2 3 4
    1 1/4 0 1/8 1/8
    2 1/8 0 1/16 1/16
    3 1/16 0 1/32 1/32
    4 1/6 0 1/32 1/32









    This is just the original table, but I'm not sure what else P(Y|X) could mean.


    My attempt for P(X=3|Y):

    I think this wants the probabilities at every point where X=3 and for each Y.

    So,
    P(X=3|Y) = { P(3,1), P(3,2), P(3,3), P(3,4) }
    = { 1/8, 1/16, 1/32, 1/32 }

    Any help is appreciated, thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    3,607
    Thanks
    591

    Re: Calculating Probabilities from a Joint Distribution

    Hey tangibleLime.

    For P(Y|X) you will need to consider each slice of X (i.e. each value of x for X = x) and the distribution will still be a bi-variate distribution with four slices for each X=x. Just do it for a fixed x and then you will have a bi-variate distribution with a conditional distribution for P(Y=y) given a particular value of x observed (so four of these distributions).

    For the second one you need to normalize this distribution by P(Y=y): Recall P(X=x|Y=y) = P(X=x and Y=y)/P(Y=y) so as an example P(X=3|Y=1) = P(X=3 and Y=1)/P(Y=1) = (1/8)/(1/2) = 1/4.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2010
    Posts
    92

    Re: Calculating Probabilities from a Joint Distribution

    So I solve P(Y|X) by putting together essentially a table of the normalized marginals of Y?

    P(Y|X=1) = { P(1,1), P(1,2), P(1,3), P(1,4) } = { 1/4, 1/8, 1/16, 1/16 } => normalizing => { 1/2, 1/4, 1/8, 1/8 }
    P(Y|X=2) = { P(2,1), P(2,2), P(2,3), P(2,4) } = { 0, 0, 0, 0 } => normalizing => { Undefined, Undefined, Undefined, Undefined }
    P(Y|X=3) = { P(3,1), P(3,2), P(3,3), P(3,4) } = { 1/8 + 1/16 + 1/32 + 1/32 } => normalizing => { 1/2, 1/4, 1/8, 1/8 }
    P(Y|X=4) = { P(4,1), P(4,2), P(4,3), P(4,4) } = { 1/8 + 1/16 + 1/32 + 1/32 } => normalizing => { 1/2, 1/4, 1/8, 1/8 }

    Then putting it in a table for easier viewing... (Y is y-axis, X is x-axis)
    1 2 3 4
    1 1/2 Undef 1/2 1/2
    2 1/4 Undef 1/4 1/4
    3 1/8 Undef 1/8 1/8
    4 1/8 Undef 1/8 1/8









    Does this look correct?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    3,607
    Thanks
    591

    Re: Calculating Probabilities from a Joint Distribution

    The undefined part should be 0 since 0*anything is 0 (also probabilistically you have a null event which has probability zero).

    Otherwise though, it looks good (I'm assuming also P(a,b) is P(y=a,x=b)).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Approximate joint probabilities using marginal probabilities
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: June 23rd 2011, 04:04 AM
  2. Joint Probabilities and Probability trees
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: July 2nd 2010, 06:36 PM
  3. Calculating poisson distribution probabilities?
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: October 7th 2009, 07:49 PM
  4. Replies: 0
    Last Post: May 24th 2009, 07:23 AM
  5. Continuous joint probabilities
    Posted in the Advanced Statistics Forum
    Replies: 5
    Last Post: February 26th 2009, 03:08 PM

Search Tags


/mathhelpforum @mathhelpforum