Chapter 19 Deviation from the Mean800
sinceRandSare independent:
ExŒ.RCS/^2 çDExŒR^2 C2RSCS^2 ç
DExŒR^2 çC 2 ExŒRSçCExŒS^2 ç
DExŒR^2 çC 2 ExŒRçExŒSçCExŒS^2 ç (by (19.13))
DExŒR^2 çC 2 0 0 CExŒS^2 ç
DExŒR^2 çCExŒS^2 ç
It’s easy to see that additivity of variance does not generally hold for variables
that are not independent. For example, ifRDS, then equation (19.11) becomes
VarŒRCRçDVarŒRçCVarŒRç. By the Square Multiple Rule, Theorem 19.3.4, this
holds iff 4 VarŒRçD 2 VarŒRç, which implies that VarŒRçD 0. So equation (19.11)
fails whenRDSandRhas nonzero variance.
The proof of Theorem 19.3.7 carries over to the sum of any finite number of
variables. So we have:
Theorem 19.3.8.[Pairwise Independent Additivity of Variance] IfR 1 ;R 2 ;:::;Rn
arepairwiseindependent random variables, then
VarŒR 1 CR 2 CCRnçDVarŒR 1 çCVarŒR 2 çCCVarŒRnç: (19.14)
Now we have a simple way of computing the variance of a variable,J, that has
an.n;p/-binomial distribution. We know thatJ D
Pn
kD 1 Ikwhere theIkare
mutually independent indicator variables with PrŒIkD1çDp. The variance of
eachIkisp.1 p/by Corollary 19.3.2, so by linearity of variance, we have
Lemma 19.3.9(Variance of the Binomial Distribution).IfJhas the.n;p/-binomial
distribution, then
VarŒJçDnVarŒIkçDnp.1 p/: (19.15)
19.4 Estimation by Random Sampling
Democratic politicians were astonished in 2010 when their early polls of sample
voters showed Republican Scott Brown was favored by a majority of voters and so
would win the special election to fill the Senate seat that the late Democrat Teddy
Kennedy had occupied for over 40 years. Based on their poll results, they mounted
an intense, but ultimately unsuccessful, effort to save the seat for their party.
