Where is the mistake in this attempt to find the standard error of an estimator? I'm reading a paper: Bect et al 2012, Sequential design of computer experiments for the estimation of a probability of failure. I'm looking at page 2, but the details of the paper don't matter. What matters is that $\alpha\in(0,1)$, and we have an unbiased estimator $$\alpha_m = \frac1m \sum_{i=1}^m \mathbf 1_{A_i}$$ 
where $\mathbb1_{A_i}$ is an indicator function for the event $A_i$. The authors assert that $$\mathbb E [(\alpha_m-\alpha)^2)]=\frac1m\alpha(1-\alpha)$$
But when I try to work this out for myself, I instead get $\mathbb E [(\alpha_m-\alpha)^2)]= \frac1m \alpha (1-m\alpha)$, which I know is wrong. My reasoning is below. I would appreciate either an explanation for how the authors get their answer, or a diagnosis of what is wrong with my reasoning.
\begin{align*}
\mathbb E [(\alpha_m-\alpha)^2)]&=\mathbb E [\alpha_m^2-2\alpha\alpha_m+\alpha)^2]\\
&= \mathbb E[\alpha_m^2] - \alpha^2
\end{align*}
\begin{align}
\mathbb E[\alpha_m^2] &= \mathbb E \left[ \frac1{m^2} \left(\sum_{i=1}^m \mathbb 1_{A_i}\right) \right]\\
&= \frac 1{m^2} \mathbb E\left[ \sum_{i=1}^m \mathbb1_{A_i} + \sum_{j=1}^{m\cdot a_m-1} j \right]\\
&=\frac 1m\alpha + \frac 1{2m^2}\mathbb E[(m\alpha_m-1)(m\alpha_m)]\\
&=\frac 1m\alpha + \frac1{2}\mathbb E[\alpha_m^2] - \frac1{2m}\alpha\\
&=\frac 1{2m}\alpha + \frac1{2}\mathbb E[\alpha_m^2]\\
\end{align}
which implies that $\mathbb E[\alpha_m^2] = \frac1m \alpha$, which implies that $\mathbb E [(\alpha_m-\alpha)^2)]= \frac1m \alpha (1-m\alpha)$.
 A: Notice that $\sum_{i=1}^m \mathbb{1}_{A_i} \sim Binomial(m, \alpha)$
$$Var\left(\sum_{i=1}^m \mathbb{1}_{A_i}\right)=m\alpha(1-\alpha)$$
\begin{align}\mathbb{E} [(\alpha_m-\alpha)^2]&=Var(\alpha_m)\\&=Var\left(\frac1m\sum_{i=1}^m \mathbb{1}_{A_i}\right) \\
&=\frac1{m^2}m\alpha(1-\alpha)
\\&=\frac1m\alpha(1-\alpha)\end{align}
Alternatively, doing it similar to your way:
\begin{align}\mathbb{E}[\alpha_m^2]&=\mathbb{E}\left[ \frac{1}{m^2}\left( \sum_{i=1}^m \mathbb{1}_{A_i}\right)^2\right] \\
&=\mathbb{E}\left[ \frac{1}{m^2}\left( \sum_{i=1}^m \mathbb{1}_{A_i}+2\sum_{i=1}^m\sum_{j<i}^m\mathbb{1}_{A_i \cap A_j}\right)\right]\\&=\frac1{m^2} \left(m\alpha+2\binom{m}{2}\alpha^2\right)\\
&=\frac1m\left( \alpha +(m-1)\alpha^2\right)\\
&=\frac1m\left(\alpha-\alpha^2 \right)+\alpha^2\end{align}
Hence \begin{align}\mathbb{E} \left( \alpha_m^2\right)-\alpha^2&= \frac1m\alpha(1-\alpha)\end{align}
A: You do omit the square as Siong Thye Goh points out, but I believe that's just a typo, as the next line indicates that you intended to consider all the terms in the product.
Your real mistake is 
$$\mathbb{E}\Bigg\lbrack \frac{1}{m^2}\Big(\sum_{i=1}^{m}\mathbb{1}_{A_i}\Big)^2\Bigg\rbrack = \frac{1}{m^2}\mathbb{E}\Bigg\lbrack\sum_{i=1}^{m}\mathbb{1}_{A_i} + \sum_{j=1}^{ma_m -1}j\Bigg\rbrack$$
because each of the cross terms (the terms in the rightmost sum) occur twice. So:
$$\begin{equation}
\begin{split}
\mathbb{E}\lbrack a_m^2\rbrack = 
\mathbb{E}\Bigg\lbrack \frac{1}{m^2}\Big(\sum_{i=1}^{m}\mathbb{1}_{A_i}\Big)^2\Bigg\rbrack & = \frac{1}{m^2}\mathbb{E}\Bigg\lbrack\sum_{i=1}^{m}\mathbb{1}_{A_i} + 2\sum_{j=1}^{ma_m -1}j\Bigg\rbrack \\
& = \frac{1}{m}a + \frac{1}{m^2}\mathbb{E}\lbrack(ma_m - 1)(ma_m)\rbrack \\
& = \frac{1}{m}a + \mathbb{E}\lbrack a_m^2\rbrack - \frac{1}{m}a \\
& = \mathbb{E}\lbrack a_m^2\rbrack 
\end{split}
\end{equation}$$
A rather disappointing result.
