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?

  • 1
    $\begingroup$ Binom/n is assuming that the n bernoullis are perfectly correlated, whereas in the binomial you are assuming they are independent $\endgroup$
    – seanv507
    Commented Sep 25, 2016 at 18:37
  • 2
    $\begingroup$ If $X$ is binomial, is $X/n$ Bernoulli distributed? $\endgroup$
    – Tim
    Commented Sep 25, 2016 at 18:41

1 Answer 1


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:

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


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

Which follows from independence and being identically distributed.


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


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


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