# Binomial and Normal Distributions – Advanced

Property A: The moment generating function for a random variable with distribution B(n, p) is

where q = 1 – p.

Proof: Using the definition of the binomial distribution and the definition of a moment generating function, we have

Observation: You can use the moment generating function to calculate the mean and variance (namely Property 1 of Binomial Distribution).

Theorem 1: If x is a random variable with distribution B(n, p), then for sufficiently large n, the distribution of the variable

where

Proof: By the linear transformation properties of the moment generating function (i.e Property 3 of General Properties of Distributions),

Taking the natural log of both sides, and then expanding the power series of $e^{\theta/\sigma}$ we get

Since p + q = 1 we have

If n is made sufficiently large $\sigma = \sqrt{npq}$ can be made large enough that for any fixed θ the absolute value of the sum above will be less than 1. Let

Thus for sufficiently large n, |z| < 1. The ln term in the previous expression is ln(1+z) where |z|<1, and so we may expand this term as follows:

This means that

Rearranging the terms, we have

for values of ck which don’t involve n, σ or θ. Since μ = np and σ2np(1 – p), the coefficient of the θ term is 0 and the coefficient of the θ2 term is 1. Thus,

Since the coefficient of each term in the sum has form

Thus

and so

But note that by Property 3 of Normal Distribution the moment generating function for a random variable z with distribution N(0, 1) is

The result now follows by Corollary 1 of General Properties of Distributions.

### One Response to Binomial and Normal Distributions – Advanced

1. Isanka says:

Thank you soooo much