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Thread: Combining two dependant discrete random variables

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    Combining two dependant discrete random variables




    Hi,
    I’m looking for a way to combine two discrete random variables (which I have as probability distributions). The combination should be the product (or other operation) of the two variables.
    This would be easy if they were independent, but they’re not. There is a known correlation between the variables.

    Question: how to combine two discrete random variables with correlation?
    Given: The marginal probabilities of the two variables & a correlation function
    Result: either the individual probabilities in a probability table or the complete probability distribution of the combination.

    Simple example:
    Variables A and B are the distributions:
    PA(a=1, 4) = [0.75, 0.25]
    PB(b=4, 8, 10) = [0.25, 0.25, 0.5]

    Their joint probability function is shown in their joint probability table and joint value table:
    P B=4 8 10
    A=1 ? ? ? 0.75
    4 ? ? ? 0.25
    0.25 0.25 0.5 1

    value B=4 8 10
    A=1 4 8 10
    4 16 32 40

    (tables are clearer in attached file)


    The correlation between the two variables is: b = 10 – 2/3*a

    P(A*B)(4, 8, 10, 16, 32, 40) = ?
    Last edited by simcc; 12-16-2011 at 03:45 AM. Reason: attachment

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    Re: Combining two dependant discrete random variables


    The correlation between the two variables is: b = 10 – 2/3*a
    I am not sure what the expression is on the RHS, but I assume that you are given the correlation coefficient of the two random variables to be a fixed constant \rho \in (-1, 1)

    Note that, as you displayed in a talbe, the number of support points of the joint probability mass function = 2 \times 3 = 6 (6 unknowns)

    Also, from the table you have (2 - 1)(3 - 1) = 2 degrees of freedom, i.e. you need to have 2 conditions to solve the joint. The four known conditions come from the fact that the sum of row/columns are equal to the marginal, and one of the five equations are redundant.

    From the definition of correlation,

    E[XY] = \rho SD[X]SD[Y] + E[X]E[Y]

    so you have the 5th equation. But you lack the 6th one to uniquely solve the table.

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