# Variance of Bernoulli from Binomial

This has bugging me for days, and I can't find where I'm wrong about the following statement:

Bernoulli variance to Binomial variance:

The variance of a Bernoulli variable is $\operatorname{Var}(Ber) = p(1-p).$

So performing the Bernoulli $n$ times, we have a binomial distribution, with $\operatorname{Var}(Binom) = \operatorname{Var}(\sum_{i=1}^n Ber_i) = \sum_{i=1}^n(\operatorname{Var}(Ber_i)) = np(1-p).$

This is the correct result.

Binomial variance to Bernoulli variance:

However, I can't get this way round right:

$\operatorname{Var}(Binom) = np(1-p)$

$\operatorname{Var}(Binom / n) = (1/n^2) \times \operatorname{Var}(Binom) = p(1-p) / n.$

This is obviously incorrect, but I cannot find out why. Can somebody explain it to me?

• Binom/n is assuming that the n bernoullis are perfectly correlated, whereas in the binomial you are assuming they are independent – seanv507 Sep 25 '16 at 18:37
• If $X$ is binomial, is $X/n$ Bernoulli distributed? – Tim Sep 25 '16 at 18:41

## 1 Answer

A Binomial random variable is the sum of $n$ $i.i.d$ Bernoulli random variables. So, if $X$ is Binomial such that $X=\sum_{i=1}^n Y_i$ it follows that:

$Var(X)=Var(\sum_{i=1}^n Y_i)=n p (1-p)$

However,

$Var(\sum_{i=1}^n Y_i) = \sum_{i=1}^n Var(Y_i) = \sum_{i=1}^n Var(Y_1) = n\times Var(Y_1)$

Which follows from independence and being identically distributed.

Therefore,

$Var(X) = n \times Var(Y_1)$

hence,

$Var(Y_1) = Var(Y_i) = \frac{Var(X)}{n}=\frac{np(1-p)}{n}=p(1-p)$