Page 112 - Fundamentals of Probability and Statistics for Engineers

Expectations and Moments 95

2
2
2
2
ˆ ‡ ‡ ‡ : 4:41†
Y 1 2 n
Let us verify Result 4.3 for n ˆ 2. The proof for the case of n random
variables follows at once by mathematical induction. Consider
Y ˆ X 1 ‡ X 2 :

We know from Equation (4.38) that
m Y ˆ m 1 ‡ m 2 :

Subtracting m Y from Y , and (m 1 ‡ m 2 ) from (X 1 ‡ X 2 ) yields

Y m Y ˆ X 1 m 1 †‡ X 2 m 2 †

and

2
2
2
ˆ Ef Y m Y † gˆ Ef‰ X 1 m 1 †‡ X 2 m 2 †Š g
Y
2 2
ˆ Ef X 1 m 1 † ‡ 2 X 1 m 1 † X 2 m 2 †‡ X 2 m 2 † g
2 2
ˆ Ef X 1 m 1 † g‡ 2Ef X 1 m 1 † X 2 m 2 †g ‡ Ef X 2 m 2 † g
2
2
ˆ ‡ 2 cov X 1 ; X 2 † ‡ :
1 2
The covariance cov(X 1 , X 2 ) vanishes, since X 1 and X 2 are independent [see
Equation (4.27)], thus the desired result is obtained.
Again, many generalizations are possible. For example, if Z is given by
Equation (4.39), we have, following the second of Equations (4.9),
2
2 2
2 2
ˆ a ‡ ‡ a : 4:42†
Z 1 1 n n
Let us again emphasize that, whereas Equation (4.38) is valid for any set of
random variables X 1 ,... X n , Equation (4.41), pertaining to the variance, holds
only under the independence assumption. Removal of the condition of inde-
pendence would, as seen from the proof, add covariance terms to the right-
hand side of Equation (4.41). It would then have the form
2
2
2
2
ˆ ‡ ‡ ‡ ‡ 2 cov X 1 ; X 2 †‡ 2 cov X 1 ; X 3 †‡ ‡ 2 cov X n 1 ; X n †
Y 1 2 n
n n 1 n
X 2 X X
ˆ ‡ 2 cov X i ; X j † 4:43†
j
jˆ1 iˆ1 jˆ2
i<j

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