Applied Statistics and Probability for Engineers

5-1 TWO DISCRETE RANDOM VARIABLES 149

Rectangular Range for (X, Y)!

If the set of points in two-dimensional space that receive positive probability under

fXY(x,y) does not form a rectangle, Xand Yare not independent because knowledge of X

can restrict the range of values of Ythat receive positive probability. In Example 5-1

knowledge that X3 implies that Ycan equal only 0 or 1. Consequently, the marginal

probability distribution of Ydoes not equal the conditional probability distribution

for X3. Using this idea, we know immediately that the random variables Xand Ywith

joint probability mass function in Fig. 5-1 are not independent. If the set of points in two-

dimensional space that receives positive probability under fXY(x,y) forms a rectangle,

independence is possible but not demonstrated. One of the conditions in Equation 5-7 must

still be verified.

Rather than verifying independence from a joint probability distribution, knowledge of

the random experiment is often used to assume that two random variables are independent.

Then, the joint probability mass function of Xand Yis computed from the product of the

marginal probability mass functions.

EXAMPLE 5-9 In a large shipment of parts, 1% of the parts do not conform to specifications. The supplier

inspects a random sample of 30 parts, and the random variable Xdenotes the number of parts

in the sample that do not conform to specifications. The purchaser inspects another random

sample of 20 parts, and the random variable Ydenotes the number of parts in this sample that

do not conform to specifications. What is the probability that and?

Although the samples are typically selected without replacement, if the shipment is large,

relative to the sample sizes being used, approximate probabilities can be computed by assum-

ing the sampling is with replacement and that Xand Yare independent. With this assumption,

the marginal probability distribution of Xis binomial with n30 and p0.01, and the mar-

ginal probability distribution of Yis binomial with n20 and p0.01.

If independence between Xand Ywere not assumed, the solution would have to proceed

as follows:

However, with independence, property (4) of Equation 5-7 can be used as

P 1 X 1, Y 12 P 1 X 12 P 1 Y 12

fXY 1 0, 0 2 fXY 1 1, 0 2 fXY 1 0, 1 2 fXY 1 1, 1 2

P 1 X0, Y 12 P 1 X1, Y 12

P 1 X 1, Y 12 P 1 X0, Y 02 P 1 X1, Y 02

X 1 Y 1

fY 031 y 2

For discrete random variables Xand Y, if any one of the following properties is true,

the others are also true, and X and Yare independent.

(1)

(2)

(3)

(4) for any sets Aand Bin the range

of Xand Y, respectively. (5-7)

P 1 XA, YB 2 P 1 XA 2 P 1 YB 2

f (^) Xƒy 1 x 2 fX 1 x 2 for all x and y with fY 1 y 2
0
f (^) Yƒx 1 y 2 fY 1 y 2 for all x and y with f X 1 x 2
0

fXY 1 x, y 2 fX 1 x 2 fY 1 y 2 for all x and y

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