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170 CHAPTER 5 JOINT PROBABILITY DISTRIBUTIONS

The concept of independence can be extended to multiple continuous random variables.

Definition
Continuous random variables X , X , p , X p are independent if and only if
1
2
1x , x p , x 2 f 1x 2f 1x 2 p f 1x 2 for all x , x , p , x (5-26)
f X 1 X 2 p X p 1 2 p X 1 1 X 2 2 X p p 1 2 p

Similar to the result for only two random variables, independence implies that Equation 5-26
holds for all x , x , p , x . If we find one point for which the equality fails, X , X , p , X p are
2
2
1
p
1
not independent. It is left as an exercise to show that if X , X , p , X p are independent,
2
1
P1X A , X A , p , X A 2 P1X A 2P1X A 2 p P1X A 2
2
1
2
2
p
p
1
p
2
1
p
1
for any regions A , A , p , A p in the range of X , X , p , X , respectively.
1
2
p
1
2
EXAMPLE 5-25 In Chapter 3, we showed that a negative binomial random variable with parameters p and r
can be represented as a sum of r geometric random variables X , X , p , X . Each geometric
2
1
r
random variable represents the additional trials required to obtain the next success. Because
the trials in a binomial experiment are independent, X , X , p , X r are independent random
2
1
variables.
EXAMPLE 5-26 Suppose X , X , and X 3 represent the thickness in micrometers of a substrate, an active layer,
2
1
and a coating layer of a chemical product. Assume that X , X , and X 3 are independent
2
1
and normally distributed with 10000, 1000, 80, 250, 20, and
1
2
1
3
2
4, respectively. The specifications for the thickness of the substrate, active layer, and
3
coating layer are 9200 x 10800, 950 x 1050, and 75 x 85, respectively.
3
1
2
What proportion of chemical products meets all thickness specifications? Which one of the
three thicknesses has the least probability of meeting specifications?
The requested probability is P19200 X 10800, 950 X 1050, 75 X 85.
2
3
1
Because the random variables are independent,
P19200 X 10800, 950 X 1050, 75 X 852
3
1
2
2
P19200 X 1 108002P1950 X 10502P175 X 852
3
After standardizing, the above equals
P1 3.2 Z 3.22P1 2.5 Z 2.52P1 1.25 Z 1.252
where Z is a standard normal random variable. From the table of the standard normal distri-
bution, the above equals
10.99862210.98758210.788702 0.7778
The thickness of the coating layer has the least probability of meeting specifications.
Consequently, a priority should be to reduce variability in this part of the process.