How is $nq-1 = \sum_{m=0}^n (m-1){n \choose m} p^{n-m}q^m$

Question

$$nq-1 = \sum_{m=0}^n (m-1){n \choose m} p^{n-m}q^m$$ I found this formula in my facility planning book. The math part of it I don't understand. I tried taking it apart like this $\sum_{m=0}^n (m){n \choose m}p^{n-m}q^m-\sum_{m=0}^n {n \choose m}*p^{n-m}q^m$ and calculate each term of the summation and it doesn't work. Btw we have $\ p+q=1$.

I appreciate it if anyone can help me.

Answer 1

$$\begin{align*} \sum_{m=0}^n(m-1)\binom{n}mp^{n-m}q^m&=\sum_{m=0}^nm\binom{n}mp^{n-m}q^m-\sum_{m=0}^n\binom{n}mp^{n-m}q^m\\ &\overset{*}=\sum_{m=0}^nn\binom{n-1}{m-1}p^{n-m}q^m-(p+q)^n\\ &=n\sum_{m=0}^{n-1}\binom{n-1}mp^{n-(m+1)}q^{m+1}-1\\ &=nq\sum_{m=0}^{n-1}\binom{n-1}mp^{(n-1)-m}q^m-1\\ &\overset{*}=nq(p+q)^{n-1}-1\\ &=nq-1 \end{align*}$$

The starred steps use the binomial theorem.

Answer 2

Let $X \sim \text{Bin}(n,q)$. Then: $$ \mathrm{E}[X] = \sum_{m=0}^n m{n \choose m}q^mp^{n-m} $$ Binomial theorem asserts that: $$ \sum_{m=0}^n {n \choose m}q^mp^{n-m} = (p + q)^n = 1 $$ We know that for a binomial random variable, $\mathrm{E}[X] = nq$. Therefore: $$ \sum_{m=0}^n (m-1){n \choose m}q^mp^{n-m} = \sum_{m=0}^n m{n \choose m}q^mp^{n-m} - \sum_{m=0}^n {n \choose m}q^mp^{n-m} = \mathrm{E}[X] - 1 = nq-1 $$

Answer 3

The second sum is $1$ by the binomial theorem. For the first sum, note that \begin{align} \sum_{m=0}^n m{n \choose m}p^{n-m}q^m &=\sum_{m=1}^n m{n \choose m}p^{n-m}q^m\\ &=\sum_{m=1}^n n{n-1 \choose m-1}p^{n-m}q^m\\ &=n q\sum_{m=1}^n {n-1 \choose m-1}p^{n-m}q^{m-1}\\ &=n q\sum_{m=0}^{n-1} {n-1 \choose m}p^{n-1-m}q^m\\ &=n q (p+q)^{n-1}\\ &=n q \end{align}