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164 CHAPTER 5 JOINT PROBABILITY DISTRIBUTIONS

EXAMPLE 5-18 For the random variables that denote times in Example 5-15, determine the conditional mean
for Y given that x 1500.
The conditional probability density function for Y was determined in Example 5-17.
Because f Y 1500 (y) is nonzero for y
1500,

0.002115002 0.002y 3 0.002y
E1Y x 15002 y 10.002e 2 dy 0.002e ye dy

1500 1500
Integrate by parts as follows:

0.002y

0.002y
0.002y e e
ye dy y ` a b dy
0.002
1500 0.002
1500 1500
1500 e 0.002y
e 3 a ` b
0.002 1 0.00221 0.0022 1500
1500 e 3 e 3
e 3 120002
0.002 10.002210.0022 0.002

3
With the constant 0.002e reapplied
E1Y 0 x 15002 2000

5-3.4 Independence

The definition of independence for continuous random variables is similar to the definition for
discrete random variables. If f (x, y) f (x) f (y) for all x and y, X and Y are independent.
XY
X
Y
Independence implies that f (x, y) f (x) f (y) for all x and y. If we find one pair of x and y
Y
XY
X
in which the equality fails, X and Y are not independent.
Definition
For continuous random variables X and Y, if any one of the following properties is
true, the others are also true, and X and Y are said to be independent.
(1) f 1x, y2 f 1x2 f 1 y2 for all x and y
XY X Y
(2) f Y 0 x 1 y2 f 1y2 for all x and y with f 1x2
0
Y X
(3) f X 0 y 1x2 f X 1x2 for all x and y with f Y 1 y2
0
(4) P1X A, Y B2 P1X A2P1Y B2 for any sets A and B in the range
of X and Y, respectively. (5-21)

EXAMPLE 5-19 For the joint distribution of times in Example 5-15, the
Marginal distribution of Y was determined in Example 5-16.
Conditional distribution of Y given X x was determined in Example 5-17.

Because the marginal and conditional probability densities are not the same for all values of
x, property (2) of Equation 5-20 implies that the random variables are not independent. The