2
$\begingroup$

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?

$\endgroup$
  • 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 Sep 25 '16 at 18:37
  • 2
    $\begingroup$ If $X$ is binomial, is $X/n$ Bernoulli distributed? $\endgroup$ – Tim Sep 25 '16 at 18:41
5
$\begingroup$

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,

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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