Mathematical Methods for Physicists: A concise introduction - Site Map

INTRODUCTION TO PROBABILITY THEORY
Now multiplying both sides by p…1 p† we get
X
mn C m ‰…1 p†p m …1 p† nm ‡ p m‡1 …1 p† nm Šˆnp X nC m p m …1 p† nm :
Combining the two terms on the left hand side, and using Eq. (14.17) in the right
hand side we have
X
mn C m p m …1 p† nm ˆ X mf …m† ˆnp:
…14:19†
Note that the left hand side is just our original expression for m, Eq. (14.18). Thus
we conclude that
m ˆ np
…14:20†
for the binomial distribution.
The variance 2 is given by
2 ˆ X …m m† 2 f …m† ˆX …m np† 2 f …m†;
…14:21†
here we again drop the summation limits for convenience. To evaluate this sum
we ®rst rewrite Eq. (14.21) as
2 ˆ X …m 2 2mnp ‡ n 2 p 2 † f …m†
ˆ X m 2 f …m†2np X mf …m†‡n 2 p 2 X f …m†:
This reduces to, with the help of Eqs. (14.17) and (14.19),
2 ˆ X m 2 f …m†…np† 2 :
…14:22†
To evaluate the ®rst term on the right hand side, we ®rst di€erentiate Eq. (14.19):
X
mn C m ‰mp m1 …1 p† nm …n m†p m …1 p† nm1 Šˆp;
then multiplying by p…1 p† and rearranging terms as before
X
m
2
n C m p m …1 p† nm np X m n C m p m …1 p† nm ˆ np…1 p†:
By using Eq. (14.19) we can simplify the second term on the left hand side and
obtain
X
m
2
n C m p m …1 p† nm ˆ…np† 2 ‡ np…1 p†
or
X
m 2 f …m† ˆnp…1 p ‡ np†:
Inserting this result back into Eq. (14.22), we obtain
2 ˆ np…1 p ‡ np†…np† 2 ˆ np…1 p† ˆnpq;
…14:23†
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