Page 190 - Applied statistics and probability for engineers

168 Chapter 5/Joint Probability Distributions

by property (2) of Equation 5-8. Therefore, c = 1/(volume of region R). Furthermore, by
1 (
⎡
B
property (3) of Equation 5-8, P X , X , … , X p) ∈ ⎤ ⎦
⎣
2
= ∫ ∫ … f X X 2 …X p (x , x , … ) dx dx 2 …dx p = × volume (B > R ) = volumme (B > ) R
∫
c
, x p
1
2
1
B 1 volume ( ) R
When the joint probability density function is constant, the probability that the random vari-
ables assume a value in the region B is just the ratio of the volume of the region B> R to the
volume of the region R for which the probability is positive.
Example 5-15 Probability as a Ratio of Volumes Suppose that the joint probability density function of the
continuous random variables X and Y is constant over the region x + y ≤ 4. Determine the prob-
2
2
ability that X + Y ≤ 1.
2
2
The region that receives positive probability is a circle of radius 2. Therefore, the area of this region is 4π. The area
of the region x + y ≤ 1 is π. Consequently, the requested probability is π ( / 4 π) = 1 4 .
2
2
/
Marginal
Probability Density If the joint probability density function of continuous random variables X , X ,… , X p
Function is f X X 2 … ( x , x ,… p) 1 2
1 X p 1 2 , x , the marginal probability density function of X i is
x , x ,
f X i x i ( ) = ∫ ∫ … ∫ f X X … X p ( 1 2 … , x p) dx dx 2 … dx i−1 dx i+1 … dx p (5-9)
1
1 2
where the integral is over all points in the range of X , X , … , X p for which X i = x i .
2
1
As for two random variables, a probability involving only one random variable, for example,
(

P a < X < b , ) can be determined from the marginal probability distribution of X i or from the
i
joint probability distribution of X , X ,… , X p . That is,
2
1
(
(
∞
∞
P a < X < b) = P −∞ < X < ,… , − ∞ < X i−1 < ,a < X < b ,

i
1
i
−∞ < X i+1 < ,… , − ∞ < X < ∞)
∞
<
p
Furthermore, E X i ( ) and V X i ( ) for i = 1 , , , p can be determined from the marginal prob-
…
2
…
ability distribution of X i or from the joint probability distribution of X , X , , X p1 2 as follows.
Mean and
Variance from Joint ∞ ∞ ∞ ∞
x , x ,…
Distribution E X i ( ) = ∫ ∫ … ∫ x f X X 2 … X p ( 1 2 , x p) dx dx 2 … dx p = ∫ x f X i ( x i ) ddx i
1
i
i
1
−∞ −∞ −∞ −∞
and (5-10)
∞ ∞ ∞ ∞ ∞
x , x ,…
V X i ( ) = ∫ ∫ … ( ∫ x i − μ ) 2 f X X … X p ( 1 2 , x p) dx dx 2 … dx p = ∫ (x i − μ X i ) f X i ( )dx i
2
x i
1
X i
1 2
−∞ −∞ −∞ −∞
Example 5-16 Points that have positive probability in the joint probability distribution of three random variables X , X , X 3
1
2
are shown in Fig. 5-11. Suppose the 10 points are equally likely with probability 0.1 each. The range is the
non-negative integers with x + x 2 + x 3 = 3. The marginal probability distribution of X 2 is found as follows.
1
(
(
(
(
P X 2 = 0) = f X X X 3 0 0) + f X X X 0 0 f X X X 1 0 2) + ( , , 0 1 ) = 0 4.
, ,3) +
+
, ,
, ,
2
1 2
1 2 3 1 2 3 1 2 3 f X X X 3
P (X 2 = ) = f X X X 3 ( , , 1 0 ) + f X X X 3 ( , 0 1, ) + f X X X 3 (1 1 1
1
2
,2
, , ) = 0 3 .
1 2
1 2
1 2
(
,
P X = ) =2 f X X X 3 (1 , , ) +2 0 f X X X 3 3 ( 0 2 1) = 0 2
.
,
2
1 2
1 2
P( 2 X = 3) = f X X X ( 0 3 0) = 0.1
,
,
1 2 3
E
Also, (X 2 ) = 0(0.4) +1(0.3) + 2(0.2) + 3(0.1) =1