# 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 $\mathbb E [(\alpha_m-\alpha)^2)]= \frac1m \alpha (1-m\alpha)$.

## 2 Answers

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}

• Thank you -- however, how do we have that $\sum_{i=1}^m \mathbf 1_{A_i}\sim \mathrm{Bin}(m,a)$? The $A_i$ in question are neither independent nor identically distributed. – Ceph Jun 17 '17 at 0:02
• $A_i$ refers to $\{ f(X_i) > u \}$ where $X_i$ are independent random variables with distribution $P_X$. They are indeed independent and identically distributed isn't it? – Siong Thye Goh Jun 17 '17 at 0:15
• @SiongThyeGoh is correct, and in fact, without the assumption that the $A_i$ are iid, the result does not hold. – David Foley Jun 18 '17 at 18:11
• Ah - I missed that the $X_i$ are iid, I was thinking of them as being the points of a LHC design space for some reason. I guess the details of the paper were important after all! Thanks. – Ceph Jun 19 '17 at 12:34

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.

• Yikes, you are correct. Thank you. – Ceph Jun 17 '17 at 0:04
• Thanks for pointing out that that is just a typo, I stopped reading too early. – Siong Thye Goh Jun 17 '17 at 0:25