**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 *f _{m,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:

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!

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