Page 186 - Applied statistics and probability for engineers

164 Chapter 5/Joint Probability Distributions

x = Number of Bars of Signal Strength
FIGURE 5-9 y = Response time
(nearest second) 1 2 3
Conditional
4 0.750 0.400 0.091
probability 3 0.100 0.400 0.091
distributions of Y 2 0.100 0.120 0.364
given X = , f Y xu ( ) in 1 0.050 0.080 0.454
y
x
Example 5-7. Total 1 1 1
Conditional Mean and
Variance The conditional mean of Y given X = x, denoted as E Y x) or μ Y x u is,
u (
u (
E Y x) ∫ y f Y xu ( y) (5-6)
5
y
u (
and the conditional variance of Y given X = x, denoted as V Y x) or σ Y x , is
2
u
u (
2
V Y x) ( ∫ y − μ Y x) f Y xu ( y) ∫ y f Y xu ( y) − μ 2 Y x

2
5
5
u
u
y y
Example 5-8 Conditional Mean And Variance For the random variables that denote times in Example 5-2,
determine the conditional mean for Y given that x = 1500.
The conditional probability density function for Y was determined in Example 5-6. Because f Yu1500 ( y) is nonzero for
y > 1500,
(
u (
E Y X 51500) 5 ∞ ∫ y 0 002 e 0 002 1500) 2 0 002 y ) dy 5 0 002 e 3 ∞ ∫ ye 2 20 002y dy
(
.
.
.
.
.
1500 1500
Integrate by parts as follows:
∞ e 20 002 y ∞ ∞ ⎛ e 20 002 y ⎞ 1500 ⎛ e 2 . y ∞ ⎞ ⎞
.
.
0 002
∫ ye 20 002 y dy 5 y 2 ∫ ⎜ ⎟ ⎟ dy 5 e 2 ⎜ ) ⎟
3
2
.
.
.
0 002 2 .
.
1500 20 002 1500 1500 ⎝ 20 002⎠ 0 002 ⎝ ( ⎜ 2 . )( 0 002 1500⎠ ⎟
1500 23 e 23 e 23
5 e 1 = (2000 )
0 002 (0 002 )(0 002 ) 0 002
.
.
.
.
3
With the constant 0 002. e reapplied,
(
E Y X51500) 52000
Practical Interpretation: If the connect time is 1500 ms, then the expected time to be authorized is 2000 ms.
Example 5-9 For the discrete random variables in Example 5-1, the conditional mean of Y given X = 1 is obtained
from the conditional distribution in Fig. 5-9:
( μ ( ( ( (
.
.
.
.
.
E Y u1)5 Yu1 5 1 0 05)1 2 0 1)1 3 0 1)1 4 0 75)5 3 55
The conditional mean is interpreted as the expected response time given that one bar of signal is present. The conditional
variance of Y given X = 1 is
3 55 0 1 + (
)
V Y u1) ( 12 3 55 0 05 ( 2. ) . 1 2 3 55 0 1 ( 3 − . ) 2 . 4 − . 2 2 . 0 748
(
2
2
.
+
3 55) 0 75 =
.
.
5