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 Ygiven that x1500.
The conditional probability density function for Ywas determined in Example 5-17.
Because fY 1500 (y) is nonzero for y 1500,
Integrate by parts as follows:
With the constant 0.002e^3 reapplied
5-3.4 Independence
The definition of independence for continuous random variables is similar to the definition for
discrete random variables. If fXY(x, y)fX(x) fY(y) for all xand y, Xand Yare independent.
Independence implies that fXY(x, y)fX(x) fY(y) for all xand y. If we find one pair of xand y
in which the equality fails, Xand Yare not independent.
E 1 Y^0 x 15002 2000
1500
0.002
e^3
e^3
1 0.002 21 0.002 2
e^3
0.002
120002
1500
0.002
e^3 a
e0.002y
1 0.002 21 0.002 2
`
1500
b
(^)
1500
ye0.002y dyy
e0.002y
0.002
`
1500
(^)
1500
a
e0.002y
0.002
b dy
E 1 Y x 15002
1500
y 1 0.002e0.002^115002 0.002y 2 dy0.002e^3
1500
ye0.002y dy
For continuous random variables Xand Y, if any one of the following properties is
true, the others are also true, and Xand Yare said to be independent.
(1)
(2)
(3)
(4) for any sets Aand Bin the range
of Xand Y, respectively. (5-21)
P 1 XA, YB 2 P 1 XA 2 P 1 YB 2
fX 0 y 1 x 2 fX 1 x 2 for all x and y with fY 1 y 2
0
fY 0 x 1 y 2 fY 1 y 2 for all x and y with fX 1 x 2
0
fXY 1 x, y 2 fX 1 x 2 fY 1 y 2 for all x and y
Definition
EXAMPLE 5-19 For the joint distribution of times in Example 5-15, the
Marginal distribution of Ywas determined in Example 5-16.
Conditional distribution of Ygiven Xxwas 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
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