4.4. THE ABSOLUTE EXCESS OF HEADS OVER TAILS 149
the difference between the number of heads and tails, an excess of heads if
positive, and a “negative excess”, i.e. a deficit, if negative. Rather than the
clumsy extended phrase “the number of heads exceeds the number of tails or
the number of tails exceeds the number of heads” we can say “the absolute
excess of heads|Tn|.” The value Tn also represents the net “winnings”,
positive or negative, of a gambler in a fair coin flip game.
Corollary 5.Under the assumption thatYi= +1with probability 1 / 2 and
Yi=− 1 with probability 1 / 2 , andTn=
∑n
i=1Yi, then for an integersP[|Tn|> s]≈P[
|Z|≥(s+ 1/2)/√
n]
whereZis a standard normal random variable with mean 0 and variance 1.
Proof. Note thatμ=E[Yi] = 0 andσ^2 = Var [Yi] = 1.
P[|Tn|> s] = 1−P[−s≤Tn≤s]
= 1−P[−s− 1 / 2 ≤Tn≤s+ 1/2]
= 1−P[
(−s− 1 /2)/√
n≤Tn/√
n≤(s+ 1/2)/√
n]
≈ 1 −P
[
(−s− 1 /2)/√
n≤Z≤(s+ 1/2)/√
n]
=P
[
|Z|≥(s+ 1/2)/√
n]
The crucial step occurs at the approximation, and uses the Central Limit
Theorem. More precise statements of the Central Limit Theorem such as the
Berry-Esseen inequality can turn the approximation into a inequality.
If we takesto be fixed we now have the answer to our first question:
The probability of an absolute excess of heads over tails greater than a fixed
amount in a fair game of durationnapproaches 1 asnincreases.
The Central Limit Theorem in the form of the half-integer correction
above provides an alternative proof of the Weak Law of Large Numbers for
the specific case of the binomial random variableTn. In fact,
P
[∣
∣
∣
∣
Tn
n∣
∣
∣
∣>
]
≈P
[
|Z|≥(n+ 1/2)/√
n]
=P
[
|Z|≥
√
n+ (1/2)/√
n