Page 207 - Applied statistics and probability for engineers

Section 5-4/Linear Functions of Random Variables 185

∞ ∞ ∞
∫ ∫ … ∫ + …,
,
+ c 2 x f X X2 1 2 …X p (x , x , , x p1 2 … ) dx dx 2 …dx p
1
−∞ −∞ −∞
∞ ∞ ∞
∫ ∫ … ∫
p
+ c p x f X X 2 …X p (x , x , , x p1 2 … ) dx dx 2 …dx p
1
1
−∞ −∞ −∞
By using Equation 5-10 for each of the terms in this expression, we obtain the following.
Mean of a Linear
Function If Y = c X + c X + …+ c X ,
2
p
1
2
p
1
E Y ( ) = c E X ( ) + c E X ( ) + …+ c E X p ( ) (5-25)
1
p
2
2
1
Furthermore, it is left as an exercise to show the following.
Variance of a Linear
Function If X ,X ,… ,X p are random variables, and Y = c X + c X + …+ c X , then in general,
p
p
2
1
1
1
2
2
2
2
2
V Y ( ) = c V X ( ) + c V X ( ) + …+ c V X p ( ) + 2 ∑ ∑ c c j cov ( X , X j) (5-26)
i
p
i
1
2
2
1
<
i j
If X ,X ,… ,X p are independent,
2
1
2
2
2
V Y ( ) = c V X ( ) + c V X ( ) + …+ c V X p ( ) (5-27)
p
1
2
2
1
Note that the result for the variance in Equation 5-27 requires the random variables to
be independent. To see why the independence is important, consider the following simple
−
example. Let X 1 denote any random variable and dei ne X = X 1 . Clearly, X 1 and X 2 are not
2
(
independent. In fact, ρ XY = −1. Now Y = X + X 2 is 0 with probability 1. Therefore, V Y) = 0
1
regardless of the variances of X 1 and X 2 .
Example 5-30 Negative Binomial Distribution In Chapter 3, we found that if Y is a negative binomial ran-
dom variable with parameters p and r, Y = X + X + … + X , where each X i is a geometric random
r
2
1
2
variable with parameter p, and they are independent. Therefore, E X i ( ) = 1/ p and V X i ( ) = (1 − p) / p . From Equation
5-25, E Y ( ) = r p, and from Equation 5-27, V Y ( ) = r(1 − p)/ p .
2
/
An approach similar to the one applied in Example 5-30 can be used to verify the formulas
for the mean and variance of an Erlang random variable in Chapter 4. An important use of
Equation 5-27 is in error propagation, which is presented in the following example.
Example 5-31 Error Propagation A semiconductor product consists of three layers. Supposing that the

variances in thickness of the irst, second, and third layers are 25, 40, and 30 square nanometers,
respectively, and the layer thicknesses are independent. What is the variance of the thickness of the i nal product?
1 ,
Let X X X 3 , and X be random variables that denote the thicknesses of the respective layers, and the i nal product. Then,
2 ,
X = X + X + X 3
2
1
The variance of X is obtained from Equation 5-27:
V X ( ) = ( ) + ( ) + ( )
V X 3
V X 2
V X 1
= 25 + 40 + 30 = 95 nm 2
/

Consequently, the standard deviation of thickness of the inal product is 95 1 2 = 9 75 nm, and this shows how the
.
variation in each layer is propagated to the i nal product.