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Thread: probability-discrete randon variables, pmfs

  1. #1
    Junior Member
    Nov 2008

    Thumbs down probability-discrete randon variables, pmfs

    I've never even covered this stuff before guys :S The below numbers where in a table but I couldnt recreate it on here , hope you clever people can figure it out

    Discrete random variables X and Y have joint probability mass function:

    Y=0 & X=0 =>0.2
    Y=1 & X=0 =>0.2
    Y=2 & X=0 =>0.1
    Y=0 & X=1 =>0.2
    Y=1 & X=1 =>0.3
    Y=2 & X=1 =>0.0

    (a) Find the marginal means and variances of both X and Y .
    (b) Find the conditional mean of Y given X = 1.
    (c) Find the correlation between X and Y
    Last edited by mitch_nufc; Feb 22nd 2009 at 02:24 AM.
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  2. #2
    MHF Contributor matheagle's Avatar
    Feb 2009
    To get the marginal distribution of just one rv, you need to sum over the other one.
    For example P(Y=0)=P(Y=0 and X=0) + P(Y=0 and X=1)=.4
    You are summing over all values of x
    P(Y=1)=P(Y=1 and X=0) + P(Y=1 and X=1)=.5
    Since these probabilities sum to one
    we have .1 left over for P(Y=2)
    The last probability, Y=3 & X=1 =>0 should be erased.
    Anything that has probably zero shouldn't be listed.
    From here you should be able to get Y's mean and variance.
    As for the conditional mean of Y when X=1,
    you need the distribution of Y when X=1.
    You only need these two
    Y=0 & X=1 =>0.2
    Y=1 & X=1 =>0.3
    P(Y=0|X=1)=P(Y=0and X=1)/P(X=1)=.2/.5=.4
    Thus P(Y=1|X=1)=1-P(Y=0|X=1)=.6
    So, E(Y|X=1)=(0)(.4)+(1)(.6)=.6
    The correlation between two rvs is the covariance divided by the st deviations
    AND by the Cauchy-Schwarz inequality it has to be between -1 and 1.
    Last edited by matheagle; Feb 22nd 2009 at 05:17 PM.
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