Page 188 - Applied statistics and probability for engineers

166 Chapter 5/Joint Probability Distributions

Rectangular Range for (X , Y )
Let D denote the set of points in two-dimensional space that receive positive probability under
f XY ( x, y). If D is not rectangular, X and Y are not independent because knowledge of X can
restrict the range of values of Y that receive positive probability. If D is rectangular, independ-
ence is possible but not demonstrated. One of the conditions in Equation 5-6 must still be verii ed.
The variables in Example 5-2 are not independent. This can be quickly determined because
the range of (X, Y) shown in Fig. 5-4 is not rectangular. Consequently, knowledge of X changes
the interval of values for Y with nonzero probability.

Example 5-11 Independent Random Variables Suppose that Example 5-2 is modii ed so that the joint
−
−
−
6
probability density function of X and Y is f XY ( x, y) = 2 × 10 exp ( 0.001 x 0.002 y) for x $ 0 and
(

y $ 0. Show that X and Y are independent and determine P X.1000 ,Y ,1000).
Note that the range of positive probability is rectangular so that independence is possible but not yet demonstrated.
The marginal probability density function of X is
f X ( x) ∫ ∞ 310 26 e 20 001 x20 002. y dy 5 0 001. e 20 001 x for x . 0
.
.
5 2
0
The marginal probability density function of Y is
f Y ( y) ∫ ∞ 310 26 e 20 001 x 2 0 002. y dx 5 0 002 e 20 002 y for y . 0
.
.
.
5 2
0
)
Therefore, f XY ( x, y) = f x f y) for all x and y, and X and Y are independent.
(
(
Y
X
To determine the probability requested, property (4) of Equation 5-7 can be applied along with the fact that each
random variable has an exponential distribution. Therefore,
P Y ,1000)
(
(
P X.1000 ,Y ,1000 5 ) P X.1000) (
e ( e )
21
22
.
5 1 2 50 318
Often, based on knowledge of the system under study, random variables are assumed to be
independent. Then probabilities involving both variables can be determined from the marginal
probability distributions. For example, the time to complete a computer search should be
independent of an adult’s height.
Example 5-12 Machined Dimensions Let the random variables X and Y denote the lengths of two dimen-
sions of a machined part, respectively. Assume that X and Y are independent random variables,
2
and further assume that the distribution of X is normal with mean 10.5 millimeters and variance 0.0025 (mm ) and that
2
the distribution of Y is normal with mean 3.2 millimeters and variance 0.0036 (mm ). Determine the probability that
.
.
.
.
10 4 < X < 10 6 and 3 15 < Y < 3 25.
Because X and Y are independent,
(
(
.
, 6 3 15, ,
P 10 4 , X , 10. . Y 3 25 5 ) P 10 4 , X , 10 6) ( . Y 3 25)
.
.
P 3 15, ,
.
.
−
. −
⎛ 1 10 4 10 5 . 10 6 10 5⎞ ⎛ 3 152 3 2 3 252 3.2⎞
.
.
.
.
.
.
5 P ⎜ ⎝ 0 05 , Z , 0 05 ⎟ ⎠ P ⎜ ⎝ 0 06 , Z , . 0 06 ⎟ ⎠
.
.
.
5 P (2 2 , Z , 2) (2P . 0 833, Z , . 0 833)5 . 0 568
where Z denotes a standard normal random variable.
Practical Interpretation: If random variables are independent, probabilities for multiple variables are often much
easier to compute.