Chapter 18 Deviation from the Mean628
An independence condition is necessary. If we ignored independence, then we
would conclude that VarŒRCRçDVarŒRçCVarŒRç. However, by Theorem 18.4.4,
the left side is equal to 4 VarŒRç, whereas the right side is 2 VarŒRç. This implies that
VarŒRçD 0 , which, by the Lemma above, essentially only holds ifRis constant.
The proof of Theorem 18.4.7 carries over straightforwardly to the sum of any
finite number of variables. So we have:
Theorem 18.4.8.[Pairwise Independent Additivity of Variance] IfR 1 ;R 2 ;:::;Rn
arepairwiseindependent random variables, then
VarŒR 1 CR 2 CCRnçDVarŒR 1 çCVarŒR 2 çCCVarŒRnç: (18.14)
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 Lemma 18.4.2, so by linearity of variance, we have
Lemma(Variance of the Binomial Distribution).IfJhas the.n;p/-binomial dis-
tribution, then
VarŒJçDnVarŒIkçDnp.1 p/: (18.15)
18.5 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 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.
18.5.1 A Voter Poll
How did polling give an advance estimate of the fraction of the Massachusetts
voters who favored Scott Brown over his Democratic opponent?
Suppose at some time before the election thatpwas the fraction of voters favor-
ing Scott Brown. We want to estimate this unknown fractionp. Suppose we have
some random process —say throwing darts at voter registration lists —which will
select each voter with equal probability. We can define a Bernoulli variable,K, by
the rule thatKD 1 if the random voter most prefers Brown, andKD 0 otherwise.
