**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

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 we get

Since *p + q* = 1 we have

If *n* is made sufficiently large 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:

Rearranging the terms, we have

for values of *c _{k}* which don’t involve

*n, σ*or

*θ*. Since

*μ = np*and σ

^{2}=

*np*(1 –

*p*), the coefficient of the

*θ*term is 0 and the coefficient of the

*θ*term is 1. Thus,

^{2}Since the coefficient of each term in the sum has form

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.

Thank you soooo much