Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to show that:

$\sum_{m=0}^n (m-np)^2 {n \choose m} p^m q^{n-m} = npq$

share|cite|improve this question
do you really want to show it this way or can you use the binomial random variable? – Jean-Sébastien Dec 8 '12 at 18:11
up vote 4 down vote accepted

Here is a way to see it without having to expand anything in the sum. Consider a random variable $X$ following a binomial distribution with parameters $n,p$. By definition, it is the sum of $n$ i.i.d. Bernoulli$(p)$ random variable, who have expectation $p$ and variance $pq$. So since $$ X=X_1+X_2+\cdots X_n, $$ it's expectation is the sum of the expectation of all $X_m$, so $E[X]=\sum_{m=0}^nE[X_m]=np$. Since the $X_m's$ are independant, the variance of their sum is the sum of their variance so $$ Var[X]=npq, $$ $(q=1-p)$. By definition of the variance, you also have $$ Var[X]=\sum_{k=0}^{n}(m-np)^2\binom{n}{m}p^{m}q^{n-m}. $$ thus, both are equal.

share|cite|improve this answer

$(m-np)^2\binom n m=\{m(m-1)+(1-2np)m+n^2p^2\}\binom n m=n(n-1)\binom{n-2}{m-2}+(1-2np)n\binom {n-1}{m-1}+n^2p^2\binom n m$

as $m\binom n m=m\frac{n!}{m!(n-m)!}=mn\frac{(n-1)!}{m\cdot(m-1)!\{(n-1)-(m-1)\}!}=n\binom{n-1}{m-1}$ for $m\ge 1$

and $m(m-1)\binom n m=m(m-1)\frac{n!}{m!(n-m)!}$ $=m(m-1)n(n-1)\frac{(n-1)!}{m(m-1)\cdot(m-2)!\{(n-2)-(m-2)\}!}=n(n-1)\binom{n-2}{m-2}$ for $m\ge 2$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.