# Why is the variance of a binomial distribution not $n^2p(1-p)$?

Using properties of variance, we know that $$Var(aX)$$ is $$a^2Var(X)$$.

A binomial distribution has n many Bernoulli trials, i.e. we can substitute $$Var(X)$$ with $$Var(nB)$$ (where $$X$$ is a binomial variable and $$B$$ a Bernoulli trial) which, using the property described above, gives us $$n^2Var(B)$$.

But, this would give us a result of $$n^2p(1-p)$$ rather than $$np(1-p)$$ which is the supposed variance of a binomial distribution.

Where is the mathematical error in the approach I have taken?

• Taking the sum of $n$ independent Bernoulli trials is not the same as performing a single Bernoulli trial and then multiplying the result by $n$ - the latter approach only has 2 possible outcomes, $0$ and $n$. Commented Mar 8, 2021 at 19:24
• ...and $n^2 p(1-p)$ is the correct variance for that case.
– Ben
Commented Mar 9, 2021 at 5:31

The problem with your solution is at this step. $$Var(X) = Var(nB)$$

Because $$X \neq nB$$

I mean yes, $$X = B_1 + B_2 + ... +B_n$$ because all the $$B_i$$'s are $$0$$ or $$1$$, and X is the number of $$1$$'s in n trails. But you can't call it $$nB$$ because all $$B$$'s don't have the same value. Some of them are $$0$$ and some are $$1$$. Your idea is only true if $$B_1 = B_2 = ... = B_n = B$$.

But if you really want to express binomial distribution in terms of Bernoulli's trails, you can do it like this.

$$X = B_1 + B_2 + ... +B_n$$ $$Var(X) = Var(\sum_{i=1}^n{B_i})$$ You can bring the summation out because by definition of binomial distribution, each trail is independent of the other trails. So it's like they are all independent. $$Var(X) = \sum_{i=1}^n{Var(B_i)}$$ $$Var(X) = \sum_{i=1}^n{pq}$$ Now this you can write as $$npq$$ because all the $$pq$$'s regardless of $$i$$ have the same value $$p(1-p)$$. $$Var(X) = npq$$

It might be worth examining the binomial as a sum of $$n$$ i.i.d. bernoulli trials. Let $$X_i$$ be i.i.d. bernoulli draws. $$Y = \sum_i X_i$$ is then a binomial random variable. The variance of this is

$$\operatorname{Var}(Y) = \operatorname{Var}(\sum_i X_i) = \sum_i \operatorname{Var}(X_i)$$

Where I have used the property that the variances add iff the the covariance is 0 (which is true by assumption). For a given Bernoulli random variable $$\operatorname{Var}(X_i) = p(1-p)$$. Since all the $$X_i$$ are identical,

$$\operatorname{Var}(Y) = \sum_i p(1-p) = n p (1-p)$$