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5-3 TWO CONTINUOUS RANDOM VARIABLES 163

y

Figure 5-11 The
conditional probability 1500
density function for Y,
given that x 1500, is
nonzero over the solid
0
line. 0 1500 x

0.002y
6 0.001x 0.002y
f 1x2 6 10 e dy 6 10 e a e ` b
6 0.001x
X
0.002 x
x
e 0.002x 0.003x
6 10 e a b 0.003e for x
0
6 0.001x
0.002
This is an exponential distribution with 0.003. Now, for 0 x and x y the conditional
probability density function is

6 0.001x 0.002y
6 10 e
f Y |x 1y2 f XY 1x, y2 f 1x2
x
0.003e 0.003x
0.002e 0.002x 0.002y for 0 x and x y
The conditional probability density function of Y, given that x 1500, is nonzero on the solid
line in Fig. 5-11.
Determine the probability that Y exceeds 2000, given that x 1500. That is, determine
P 1Y
2000 0 x 15002. The conditional probability density function is integrated as follows:

P 1Y
2000|x 15002 Y |1500 1y2 dy 0.002e 0.002115002 0.002y dy
f
2000 2000
e 0.002y
3 e 4
3
0.002e a ` b 0.002e a b 0.368
0.002 0.002
2000
Definition
Let R denote the set of all points in the range of (X, Y) for which X x. The condi-
x
tional mean of Y given X x, denoted as E1Y 0 x2 or 0 , is
Y x

E1Y | x2 y f Y |x 1 y2 dy
R x
2
and the conditional variance of Y given X x, denoted as V 1Y 0 x2 or Y0 x , is
2 2 2
V1Y 0 x2 1 y Y | x 2 f Y | x 1 y2 dy y f Y | x 1y2 dy Y |x (5-20)

R x R x