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[PROBABILITY] Joint probability function for two DISCRETE variables

  1. Apr 29, 2010 #1
    1. The problem statement, all variables and given/known data

    X1 and X2 represent the values of two honest dice throws (independent of each other). Find the joint probability function of U and V when:

    a) U = min{X1,X2}, V = X1 + X2

    3. The attempt at a solution

    This is what I thought:

    [tex]P(U = u,V = v) = P(\min \left\{ {{X_1},{X_2}} \right\} = u,{X_1} + {X_2} = v) = P({X_1} + {X_2} = v|\min \left\{ {{X_1},{X_2}} \right\} = u)*P(\min \left\{ {{X_1},{X_2}} \right\} = u)[/tex]

    [tex]P(U = u,V = v) = P({X_1} + u = v)*P(\min \left\{ {{X_1},{X_2}} \right\} = u) = P({X_1} = v - u)*P(\min \left\{ {{X_1},{X_2}} \right\} = u) = \frac{1}{6}*P(\min \left\{ {{X_1},{X_2}} \right\} = u)[/tex]

    Now, my question is, how do I find the probability function of U?

    I can work it out to find the cumulative probability function:

    [tex]P(U \le u) = P(\min \left\{ {{X_1},{X_2}} \right\} \le u) = 1 - P(\min \left\{ {{X_1},{X_2}} \right\} > u) = 1 - P({X_1} > u,{X_2} > u) = 1 - P({X_1} > u)*P({X_2} > u)[/tex]

    [tex]P(U \le u) = 1 - \sum\limits_{{X_1} > u} {\sum\limits_{{X_2} > u} {{p_{{X_1},{X_2}}}} ({X_1},{X_2}) = 1} - \sum\limits_{{X_1} > u} {\sum\limits_{{X_2} > u} {\frac{1}{{36}}} = 1} - \frac{1}{{36}}{\left( {6 - u} \right)^2}[/tex]

    But I don't know how to get the probability function from there (if I can).

    Any ideas, please?

    Thanks.
     
  2. jcsd
  3. Apr 29, 2010 #2
    Anybody?
     
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