PRESENTATION OUTLINE
JOINTLY DISTRIBUTED DISCRETE RANDOM VARIABLES
Definition: The joint probability distribution for X,Y is a probability distribution that gives the probability that each of X,Y...falls in any particular discrete set of values specified for that variable. In case of 2 random variables this is called BIVARIATE DISTRIBUTION.
Concepts and Formulae:
• Let X and Y be discrete random variables.
The joint PMF of X and Y is defined by
p(x,y)=P(X=x,Y =y).
• Let A be any set consisting of pairs (x, y) values. Then,
P [(X, Y ) ∈ A] =
EXAMPLE1
Consider the roll of a fair die and let A = 1 if the number is even (i.e. 2, 4, or 6) and A = 0 otherwise. Furthermore, let B = 1 if the number is prime (i.e. 2, 3, or 5) and B = 0 otherwise.
1 2 3 4 5 6
A 0 1 0 1 0 1
B 0 1 1 0 1 0
Then, the joint distribution of A and B, expressed as a probability mass function, is
P(A=0,B=0) = P{1} = 1/6 ;
P(A=1,B=0) = P{4,6} = 2/6 ;
P(A=0,B=1) = P{3,5} = 2/6 ;
P(A=1,B=1) = P{2} = 1/6
These probabilities necessarily sum to 1, since the probability of some combination of A and B occurring is 1.
EXAMPLE2
A large agency services a no. of customers who have purchased both a homeowners policy and an automobile policy from the agency. For each type of policy, a deductible amount must be specified. For an automobile policy, the choices are $100 and $250 ,whereas for a homeowners policy ,the choices are 0,$100 and $200. Suppose an individual with both types of policy is selected at random from the agency's files.
Let X= the deductible amount on the auto policy and Y= the deductible amount on the homeowners policy. Possible (X,Y) pairs are then (100,0),(100,100),(100,200),(250,0),(250,100),and (250,200); the joint pmf specifies the probability associated with each one of these pairs, with any other pair having the probability zero. Suppose the joint pmf is given in the accompanying joint probability table:
Then p(100,100)=P(X=100 and Y=100) = P($100 deductible on both policies) = .10. The probability P(Y>100) is computed by summing probabilities of all(x,y)pairs for which y>100 :
P(Y>100)= p(100,100)+p(250,100)+p(100,200)+p(250,200)
=.75
