$\sum_{m=0}^N \frac{1}{m+1}{N\choose m}p^m (1-p)^{N-m}=\frac{1-(1-p)^{N+1}}{(N+1)p}$

Question

While reading a journal article, I noticed an equation which looked like the sum of the average binomial coefficient. But, I have no idea how equation was derived.

$$\sum_{m=0}^N \frac{1}{m+1}{N\choose m}p^m (1-p)^{N-m}=\frac{1-(1-p)^{N+1}}{(N+1)p}$$

What should I call this type of equation? Please explain how to prove it.

Answer 1

Call your quantity $f(p,N)$. A brute force method for evaluating it goes like this. First note that in general

$$\sum_{m=0}^N {N \choose m} p^m q^{N-m} = (p+q)^N.$$

If you now look at

$$g(p,q,N):=\sum_{m=0}^N \frac{1}{m+1} {N \choose m} p^{m+1} q^{N-m}$$

(i.e. $pf(p,N)$, but with the $1-p$ replaced by a general $q$) then it follows from the above that

$$\frac{\partial g}{\partial p}=(p+q)^N.$$

Since $g(0,q,N)=0$, you have

$$g(p,q,N)=\int_0^p (r+q)^N dr.$$

You can simplify matters by changing variables to $r+q$ and substituting in $q=1-p$, obtaining

$$f(p,N)=\frac{1}{p} \int_{1-p}^1 r^N dr$$

which is easily evaluated.

Answer 2

$$\sum_{m=0}^N \frac1{m+1}{N\choose m} p^m (1-p)^{N-m}$$ $$=\sum_{m=0}^N \frac1{m+1}\frac{N!}{m! (N-m)!} p^m (1-p)^{N-m}$$ $$=\sum_{m=0}^N \frac{N!}{(m+1)! (N-m)!} p^m (1-p)^{N-m}$$ $$=\frac{1}{(N+1)p}\sum_{m=0}^N \frac{(N+1)!}{(m+1)! (N-m)!} p^{m+1} (1-p)^{N-m}$$ $$=\frac{1}{(N+1)p}(1-(1-p)^{N+1}) $$

Answer 3

The sum is a standard binomial development, to the denominator $m+1$. You can get rid of it by the following transformation:

$$\frac{\displaystyle\binom nm}{m+1}=\frac{n!}{(n-m)!m!(m+1)}=\frac{(n+1)!}{(n+1)(n-m)!(m+1)!}=\frac{\displaystyle\binom{n+1}{m+1}}{n+1}.$$

Now we have a modified binomial sum

$$\sum_{m=0}^n\frac{\displaystyle\binom{n+1}{m+1}}{n+1}p^m(1-p)^{n-m}=\frac1{(n+1)(1-p)}\sum_{m=1}^{n+1}\binom{n+1}{m}p^m(1-p)^{n+1-m}$$

where the first term is missing, thus which equals

$$\frac1{(n+1)(1-p)}(1-(1-p)^{n+1}).$$