Applied Statistics and Probability for Engineers

5-3 TWO CONTINUOUS RANDOM VARIABLES 163

This is an exponential distribution with   0.003. Now, for the conditional

probability density function is

The conditional probability density function of Y, given that x1500, is nonzero on the solid

line in Fig. 5-11.

Determine the probability that Yexceeds 2000, given that x1500. That is, determine

The conditional probability density function is integrated as follows:

0.002e^3 a

e0.002y

0.002

`

2000

b0.002e^3 a

e^4

0.002

b0.368

P 1 Y 2000|x 15002  

2000

fY | 15001 y 2 dy 

2000

0.002e0.002^115002 0.002y dy

P 1 Y 

20000 x 15002.

0.002e0.002x0.002y for 0x and xy

fY |x 1 y 2 fXY 1 x, y 2 fx 1 x 2 

610 ^6 e0.001x0.002y

0.003e0.003x

0 x and xy

 610 ^6 e0.001x a

e0.002x

0.002

b0.003e0.003x for x

0

fX 1 x 2 

x

610 ^6 e0.001x0.002y dy 610 ^6 e0.001x a

e0.002y

0.002

`

x

b

Let Rxdenote the set of all points in the range of (X, Y) for which Xx. The condi-

tional meanof Ygiven Xx, denoted as

and the conditional varianceof Ygiven Xx, denoted as is

V 1 Y 0 x 2  (5-20)

Rx

1 yY | x 22 fY | x 1 y 2 dy

Rx

y^2 fY | x 1 y 2 dy^2 Y |x

V 1 Y 0 x 2 or ^2 Y 0 x,

E 1 Y | x 2 

Rx

y fY |x 1 y 2 dy

E 1 Y 0 x 2 or Y (^0) x, is
Definition
y
0 x
0
1500
1500
Figure 5-11 The
conditional probability
density function forY,
given thatx1500, is
nonzero over the solid
line.
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