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

Why for $X\sim B(n,p)$ is $Var(X)=np(1-p)$?

$Var(X)=\sum x_i^2 p_i -(\sum x_i p_i)^2=\sum_{r=0}^n r^2 \binom{n}{r}p^r(1-p)^{n-r}+( \sum_{r=0}^n r \binom{n}{r}p^r(1-p)^{n-r} )^2$

In my short-sightedness, I don't see any viable ways to derive the variance from this.

share|cite|improve this question
up vote 5 down vote accepted

An easier way is to recognize that $X = Y_1 + Y_2 + \cdots Y_n$ where $Y_k$ are independent Bernoulli random variables with parameter $p$. For a Bernoulli random variable $Y_k$, we have $$\text{Var}(Y_k) = p(1-p)$$ Since $Y_k$ are independent, we have that $$\text{Var}(X) = \text{Var}(Y_1) + \text{Var}(Y_2) + \cdots + \text{Var}(Y_n) = np(1-p)$$

To go the direct way, we need to first evaluate couple of summations.

We will evaluate the sums $$\sum_{k = 0}^n k \mathbb{P}(X=k) \text{ and }\sum_{k = 0}^n k^2 \mathbb{P}(X=k)$$ First note that $\mathbb{P}(X=k) = \dbinom{n}k p^k (1-p)^{n-k}$. Hence, $$\sum_{k = 0}^n k \mathbb{P}(X=k) = \sum_{k = 0}^n k \dbinom{n}k p^k (1-p)^{n-k}$$ Note that $$k \dbinom{n}k = \dfrac{n!}{(n-k)! (k-1)!} = n \dbinom{n-1}{k-1}$$ Hence, \begin{align} \sum_{k = 0}^n k \mathbb{P}(X=k) & = \sum_{k = 1}^n n \dbinom{n-1}{k-1} p^k (1-p)^{n-k} = np \sum_{k = 1}^n \dbinom{n-1}{k-1} p^{k-1} (1-p)^{(n-1)-(k-1)}\\ & = np \left( p + (1-p)\right)^{n-1} = np \end{align} Similarly, \begin{align} k^2 \dbinom{n}k & = k \dfrac{n!}{(n-k)! (k-1)!} = n k \dbinom{n-1}{k-1}\\ & = n (k-1) \dbinom{n-1}{k-1} + n \dbinom{n-1}{k-1}\\ & = n(n-1) \dbinom{n-2}{k-2} + n \dbinom{n-1}{k-1} \end{align} Hence, \begin{align} \sum_{k = 0}^n k^2 \mathbb{P}(X=k) & = \sum_{k = 2}^n n(n-1) \dbinom{n-2}{k-2} p^k (1-p)^{n-k} + \sum_{k = 1}^n n \dbinom{n-1}{k-1} p^k (1-p)^{n-k}\\ & = n(n-1)p^2 + n p \end{align} Hence, \begin{align} \text{Var}(X) & = \sum_{k = 0}^n k^2 \mathbb{P}(X=k) - \left(\sum_{k = 0}^n k \mathbb{P}(X=k) \right)^2\\ & = n(n-1)p^2 + n p - (np)^2 = n^2p^2 - np^2 + np - n^2 p^2\\ & = np(1-p) \end{align}

share|cite|improve this answer
Thanks for the answer. Bernoulli trials (at least the LaTex of them) seem much more elegant than brute force, do you know of a good introduction to them? – Alyosha Nov 18 '12 at 20:15
@Alyosha Wiki has good articles on both and – user17762 Nov 18 '12 at 20:31

Compute the expected value of $k$ $$ \begin{align} \mathrm{E}(k) &=\sum_{k=1}^nk\binom{n}{k}p^k(1-p)^{n-k}\\ &=\sum_{k=1}^nnp\binom{n-1}{k-1}p^{k-1}(1-p)^{n-k}\\ &=np(p+(1-p))^{n-1}\\ &=np\tag{1} \end{align} $$ Compute the expected value of $k(k-1)$ $$ \begin{align} \mathrm{E}(k(k-1)) &=\sum_{k=1}^nk(k-1)\binom{n}{k}p^k(1-p)^{n-k}\\ &=\sum_{k=1}^nn(n-1)p^2\binom{n-2}{k-2}p^{k-2}(1-p)^{n-k}\\ &=n(n-1)p^2(p+(1-p))^{n-2}\\ &=n(n-1)p^2\tag{2} \end{align} $$ Add $(1)$ and $(2)$ to get $\mathrm{E}(k^2)$ then subtract the square of $(1)$ to get $$ \begin{align} \mathrm{Var}(k) &=\mathrm{E}(k^2)-\mathrm{E}(k)^2\\ &=n^2p^2-np^2+np-n^2p^2\\ &=np(1-p)\tag{3} \end{align} $$

share|cite|improve this answer

Use a different model. Let $S_n = \sum_{i=1}^n X_i$ where $X_i$ are iid bernoulli random variables with parameter p. Then $S_n$ is Binomial(p,n).

Now $Var(S_n) = \sum_{i=1}^n Var(X_i) = np(1-p)$

share|cite|improve this answer

Replace $1-p$ by $q$, then replace each factor of $r$ by $p\partial/\partial p$.

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.