probability - Expected value of sum squared is sum of expected value squared?

Consider the following expression:

$X \sim BIN(1, p)$

$Var(\bar{X})=Var(\frac{\sum_i{X_i}}{n}) = \frac{1}{n^2} Var(\sum_i X_i) = \frac{1}{n^2} \left( \sum_i E(X_i^2) - ( \sum_i E(X_i) )^2 \right)$

$= \frac{1}{n^2} \left( \sum_i (pq+p^2) - (np)^2 \right) = \frac{1}{n^2} (npq+np^2 -n^2p^2) = \frac{np(1-p)}{n}$

Now I know that the C.R.L.B. is equal to this: $Var(T) \geq \frac{p(1-p)}{n}$

This means that I must get the same expression to show that $\bar{X}$ is the UMVUE.

Clearly my expression has a $n$ which shouldn't be there. I think my mistake is in this part:

$\ \ E(\sum_i X_i)^2 = \left( \sum_i E(X_i) \right)^2 = (np)^2 = n^2p^2$

So I think it should be:

$\ \ E(\sum_i X_i)^2 = \sum_i \left( E(X_i)^2 \right) = np^2$

Because that does yield the correct answer.

So is this indeed the mistake or did I make another mistake?
And I would also greatly appreciate it if someone could explain why this is wrong because I don't get the reasoning behind it.

Thanks in advance.

Answer

$E(\sum_i X_i)^2$ is ambiguous: it could be $\left(E(\sum_i X_i)\right)^2$ or $E\left((\sum_i X_i)^2\right)$

For a binomial distribution (the sum of $n$ independent Bernoulli random variables), the former is $n^2p^2$ (the square of the mean), while the latter is $np(1-p) + n^2p^2$ (the variance plus the square of the mean).

In your expression, you could have

$Var(\bar{X})=Var(\frac{\sum_i{X_i}}{n}) = \frac{1}{n^2} Var(\sum_i X_i) = \frac{1}{n^2} \sum_i Var( X_i) = \frac{np(1-p)}{n^2}= \frac{p(1-p)}{n}$

or

$Var(\bar{X})=Var(\frac{\sum_i{X_i}}{n}) = \frac{1}{n^2} Var(\sum_i X_i) = \frac{1}{n^2} \left( E\left((\sum_i X_i)^2\right) - \left(E(\sum_i X_i)\right)^2 \right)$