Concrete mathematics : a foundation for computer science

since (EX) is a constant; hence
8.2 MEAN AND VARIANCE 375
VX = E(X’) - (EX)‘. (8.15)
“The variance is the mean of the square minus the square of the mean.”
For example, the mean of (Xl +X2)’ comes to .98(0M)2 + .02( 100M)2 =
200M’ or to .9801 I(OM)2 + .0198( 100M)’ + .OOOl (200M)2 = 202M2 in the
lottery problem. Subtracting 4M2 (the square of the mean) gives the results
we obtained the hard way.
There’s an even easier formula yet, if we want to calculate V(X+ Y) when
X and Y are independent: We have
E((X+Y)‘) = E(X2 +2XY+Yz)
= E(X’) +2(EX)(EY) + E(Y’),
since we know that E(XY) = (EX) (EY) in the independent case. Therefore
V(X + Y) = E#((X + Y)‘) - (EX + EY)’
= EI:X’) + Z(EX)(EY) + E(Y’)
-- (EX)‘-2(EX)(EY) - (EY)’
= El:X’) - (EX)’ + E(Y’) - (EY)’
= VxtvY. (8.16)
“The variance of a sum of independent random variables is the sum of their
variances.” For example, the variance of the amount we can win with a single
lottery ticket is
E(X:) - (EXl )’ = .99(0M)2 + .Ol(lOOM)’ - (1 M)’ = 99M2 .
Therefore the variance of the total winnings of two lottery tickets in two
separate (independent) lotteries is 2x 99M2 = 198M2. And the corresponding
variance for n independent lottery tickets is n x 99M2.
The variance of the dice-roll sum S drops out of this same formula, since
S = S1 + S2 is the sum of two independent random variables. We have
2
35
6 = ;(12+22+32+42+52+62!- ; = 12
0
when the dice are fair; hence VS = z + g = F. The loaded die has
VSI = ;(2.12+22+32+42+52+2.62)-