Variance, Standard Deviation, and the Law of Averages
537
from Section 8.7.5, E(XI + X 2 ) = A1 + A2. Applying Theorem l(a) of this section with
i = Al + A2 gives
Var(XI + X 2 ) = E((X 1 + X 2 )
2 ) - (A, + A2)2
Regarding (X 1 + X 2 )
2 as the sum of the three random variables X2, 2X 1 X 2 , and X2 allows
us to apply Theorem 3(a) from Section 8.7.5 again:
E((X 1 + X2)2) = E(X2 + 2X 1 X 2 + X2)
= E(X2) + 2E(X 1 X 2 ) + E(X 2 )
Because X 1 and X 2 are independent, Theorem 3 implies that the term 2E(X 1 X 2 ) can be
rewritten as
2E(X 1 X 2 ) = 2E(X 1 ) - E(X 2 ) = 21t2
Replacing E((XI + X 2 )
2 ) by E(X2) +
2 h1h2 + E(X2) in the expression for Var(X 1 +
X 2 ) and simplifying gives
Var(Xi + X 2 ) = E((X 1 + X 2 )
2 ) - (hi + A2)2
= E(X2) + 2/11/2 + E(X2) - (/1 + A2)2
= (X2) + 2tt1A2 + E (X2) _ (A2 + 2g1A2 + /_2)
2
2
2
2
= E(X2) + 2ttlA2 + E(X 2 ) _ _ 1 2 -
2
_2#1I2 _ A2
= E (X2) _-t + E (X2) _-t
= Var(X 1 ) + Var(X 2 ).
Suppose X 1 , X 2 ....
X, are i.i.d. random variables with common expectation
E(Xi) = f and common variance Var(Xi) = o
2 . Consider the random variable
Y (X i +
"'
- + X n)
n
which represents their average. By Theorem 3(c) of Section 8.7.5, we know that
SE(XI +.. + X,)
I1
E(Y) =
= -... , E(Xi)
n
n
i=1
Since the Xi's all have the same mean it, the expectation of their average is
E(Y) = _;
= /1
n
i=1
As for the variance of Y, we know from Theorem 1(b) that
Var(Y)=Var((XI+X2 +'+
Xn))
Var(XI+X21+
+ Xn)
Since the Xi's are i.i.d. random variables,
Var(XI +... + Xn)
Var(X)