Property 0: B(n, p) is a valid probability distribution

Proof: the main thing that needs to be proven is that

where f(x) is the pdf of B(n, p). This follows from the well-known Binomial Theorem since

The Binomial Theorem that

can be proven by induction on n.

Property 1

Proof (mean): First we observe

Now

where m = n − 1 and i = k − 1 . But

where fm,p(i) is the pdf for B(m, p), and so we conclude μ = E[x] = np.

Proof (variance):

We begin using the same approach as in the proof of the mean:

Thus,

### 2 Responses to Binomial Distribution – Advanced

1. Robert says:

Two points in the variance proof are not clear. Why does the probability function sum up to one when the parameter C(m,i+1) is used instead of C(m,i) as in:
SUM(i=0,m){C(m,i+1)p^i(1-p)^m-i
Second, Why does the following sum equal to mp when the “i” is part of the multiplicand?
SUM(i=0,m){i*C(m,i+1)p^i(1-p)^m-i}. I am not familiar with this identity. Thanks!

• Charles says:

Robert,
The proofs that were given were not quite right. I have now made a some corrections which should address the issues that you raised. Thanks for bringing these to light.
Charles