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?