Part 3: Joint Probability

Suppose that ( X,Y) is the outcome of a certain experiment and suppose that it is known that ( X,Y) must occur in some region S in the plane. Then S is called the sample space for the experiment and X and Y are called random variables.

Moreover, a function p( x,y) is called a joint probability density of the experiment if the small probability DP that ( X,Y) is in a small region in S containing (x,y) is given by

DP = p( x,y) DA

where DA is the area of the small region. Like a mass density, a joint probability density is essentially the probability per unit area of the experiment.

Thus, if R is a region inside of S and if the x and y-axes are both partitioned into h-fine partitions, then the probability P( (X,Y)   in  R) that ( X,Y) is in R is approximated by the Riemann sum

P( ( X,Y)   in  R) » å j  å k  DPjk = å j  å k  p( xj,yk) DAjk

The actual probability of ( X,Y) being in R is the limit of the approximations as h approaches 0:

P( ( X,Y)   in  R) = lim h®0  å j  å k  p( xj,yk) DAjk =  p(x,y) dA

As a result, the density p( x,y) must be non-negative and since there is a 100% chance of ( X,Y) being in the sample space S, we require also that

p( x,y) dA = 1

(1)

EXAMPLE 5    A point ( X,Y) is chosen at random from the unit square with probability density

p( x,y) = ì í î 1     if  0 £ x £ 1  and  0 £ y £ 1 0 otherwise

What is the probability that the point is inside the rectangle R given by [ 0.1,0.6] ×[ 0.3,0.8] ?

       

Solution: Since the outcome ( X,Y) must be in the unit square, the unit square is the sample space S for our experiment. Moreover, notice that

p( x,y)dA = ó õ 1 0  ó õ 1 0  1dydx = ó õ 1 0  dx = 1

thus showing that p( x,y) does satisfy (1) and is a probability density. Indeed, in this experiment, the probability density can be considered the probability of choosing the point (x,y) ``at random'' from the sample space S.  Moreover, the probability that a random selection of ( X,Y) from S will be in R is

P[ ( X,Y)   in  R] = p( x,y) dA = ó õ 0.6 0.1  ó õ 0.8 0.3  1dydx = 0.25

Thus, there is a 25% chance that ( X,Y) will be in R, which is about a 1 in 4 chance.

       

Suppose that the random variable X of a given event has a probability density of p₁( x) and suppose that the random variable Y of an additional event has a probability density of p₂( y) . Then X and Y are said to be independent random variables if their joint density function is

p( x,y) = p1( x) p2( y)

That is, two events are independent if their joint density function is the product of the density functions of the random variables of the individual events.      

EXAMPLE 6    At a certain restaurant, customers must wait an average of 10 minutes for a table. From the time they are seated until they have finished their meal requires an additional 30 minutes, on average. What is the probability that a customer will spend less than an hour at the restaurant, assuming that waiting for a table and completing the meal are independent events?       

Solution: Waiting times are often modeled by exponential probability densities. Indeed, if X and Y are the random variables for ``waiting for a table'' and ``completing the meal,'' respectively, then their probability density functions are respectively

p1( x) = ì ï í ï î 0 if x < 0  1 10 e-x/10 if x ³ 0             p2( y) = ì ï í ï î 0 if y < 0  1 30 e-y/30 if y ³ 0

Since the events are independent, the joint probability for the two events is

p( x,y) = p1( x) p2( y) = ì ï í ï î 0 if x < 0  or  y < 0  1 300 e-x/10e-y/30 if x,y ³ 0

Our goal is to determine the probability that the combined time X+Y is under 60 minutes. That is, we want to know the probability associated with the region R in the 1st quadrant with upper bound x+y = 60:

As a result, the probability that ( X,Y) is in R is

P( X+Y £ 60) = P[ ( X,Y)   in  R] = p( x,y) dA =  1 300 e-x/10e-y/30dA

Since x+y = 60 is the same as y = 60-x, we have a type I integral:

P( X+Y £ 60) =  1 300 ó õ 60 0  ó õ 60-x 0  e-x/10e-y/30dydx = 0.7982

Thus, there is a 79.8% chance that a customer will spend less than an hour at the restaurant.