[PROBABILITY] Joint probability function for two DISCRETE variables

[libelec]

Homework Statement

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

The Attempt at a Solution

This is what I thought:

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)

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)

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

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

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)

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}

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

Any ideas, please?

Thanks.

 

[libelec]

Anybody?