Consistency of a sequence of Bernoullis Anyone got any suggestions? I tried finding an estimator explicitly and showing it had to be consistent, along with showing the consistency of the mle but neither seems to be working. Think I'm just missing a trick somewhere.
Let $X_i\sim\mathrm{Bernoulli}(1/2+\theta\cdot a_i)$ where $\left\{a_i\right\}_{i\geq 1}$ is a sequence 
of known positive constants such that $a_i$ decreases to 0. Show there is a consistent estimator for $\theta$ if and only if $\sum_{i=1}^{\infty}a_i ^2 =\infty$.
 A: Here is an estimator which can consistently estimate $\theta$ and for which the condition $\sum_{i=1}^{\infty}a_i ^2 =\infty$ is relevant.  
We have
$$E(X_i) = \frac 12 + \theta a_i \implies \frac{2E(X_i)-1}{2a_i} = \theta$$
Set 
$$Z_i = \frac{2X_i-1}{2a_i} \implies E(Z_i) = \theta$$
and
$$\text{Var}(Z_i) = \text{Var}(X_i/a_i) = \frac {1}{4a_i^2} -\theta^2$$
Consider the sample mean $\bar Z_n$. Its expected value is always $\theta$, and its variance is
$$\text{Var}(\bar Z_n) = \frac {1}{n^2}\sum_{i=1}^n\frac{1}{4a_i^2} -\frac{\theta^2}{n}$$
Ignoring weird situations, we want this variance to go to zero in order to make certain that $\bar Z_n$ will consistently estimate $\theta$. So we want
$$\frac {1}{n^2}\sum_{i=1}^n\frac{1}{a_i^2} \to 0 \implies \frac {n^2}{\sum_{i=1}^n\frac{1}{a_i^2}} \to \infty$$
Now from the Harmonic-Arithmetic mean inequality , we have that 
$$\frac {n}{\sum_{i=1}^n\frac{1}{a_i^2}}  < \frac 1n \sum_{i=1}^na_i^2 \implies \frac {n^2}{\sum_{i=1}^n\frac{1}{a_i^2}} < \sum_{i=1}^na_i^2$$
Since we want the left side of the inequality to go to infinity, it follows that $\sum_{i=1}^{\infty}a_i ^2 =\infty$ is a necessary condition for this to happen, i.e. a necessary condition for $\bar Z_n$ to consistently estimate $\theta$.

(Follows the original exploratory post). 
We have that 
$$\{a_i\} \to 0 \implies \frac 1 n \sum a_i \to 0 \tag{1}$$
and 
$$\{a_i\} \to 0 \implies \{a_i^2\} \to 0\implies  \frac 1 n \sum a_i^2 \to 0 \tag{2}$$
Now, the OP does not state it but I will assume that the rv's are independent, since the focus is elsewhere. Also, I note that the Bernoulli distribution belongs to the Exponential family and so the sum of observations is a sufficient statistic, it has all the information that the sample can give us. 
More over, Chebychev's Law of Large Numbers applies (mean and variance of each $X_i$ is finite), namely, the sample average of the $X$s converges in probability to the limit of the average of their means, and we have
$$\text{plim}\bar X =  \lim\frac 1n \sum E(X_i) = \frac 12 + \theta \lim\frac1n \sum a_i  = \frac 12 \tag{2}$$
So as sample size tends to infinity, information on $\theta$ is eliminated if we use the sufficient statistic. 
The same we would obtain if we looked at the variance. The variance of $X_i$ is 
$$\text{Var}(X_i) = \left (\frac 12 + \theta a_i\right) \left(1-\frac 12 - \theta a_i\right) = \left (\frac 12 + \theta a_i\right) \left(\frac 12 - \theta a_i\right) = (1/4) - \theta^2 a_i^2$$
and calculating any sample mean will again tend to eliminate the information on $\theta$ as the sample size increases, due to $(2)$.
So the challenge here appears to be that we have to abandon the sufficient statistic.
A: Consider the simpler case where $X_i \sim N(\theta a_i,1)$, then the MLE 
$$\hat{\theta} = \sum_{i=1}^n a_iX_i/\sum_{i=1}^n a_i^2 \sim N(\theta, 1/\sum_{i=1}^n a_i^2).$$ Note that  if $\sum_{i=1}^n a_i^2 $ diverges, then $\hat{\theta}$ is consistent since the variance goes to 0 and it is unbiased; if $\sum_{i=1}^n a_i^2 $ converges, $\hat{\theta}$ is asymptotically normal with constant variance, which implies inconsistency of MLE and there is no other estimator that can achieve consistency in this case. 
Going back to your problem, I think you can show that asymptotic variance of the MLE, by computing the Fisher information, is the same order of $1/\sum_{i=1}^n a_i^2$. If $\sum_{i=1}^n a_i^2 < \infty$, it means the information is not accrued fast enough, MLE will not be consistent and there are no other consistent estimators. If $\sum_{i=1}^n a_i^2 = \infty$, MLE will be consistent.
A: I think I have finally figured out a way to solve this problem. The details are yet to be worked out but I feel if the sum is finite then for two different $\theta_1,\theta_2$ the sequence of likelihood ratios will be mutually contiguous, which would imply there is no consistent estimator. A complete solution will be added when time permits.
