counterexample for the sufficient condition required for consistency We know that if an estimator, say $\widehat{\theta}$, is an unbiased estimator of $\theta$ and if its variance tends to 0 as n tends to infinity then it is a consistent estimator for $\theta$. But this is a sufficient and not a necessary condition. I am looking for an example of an estimator which is consistent but whose variance does not tend to 0 as n tends to infinity. Any suggestions? Thank you in advance for your time.
 A: If the parameter space has only one element, then it's kinda boring. To make it a little more interesting, let's say parameters space is $\Theta=\{0,1\}$. And the random variable $X$ is normally distributed with mean $\theta$, i.e. $$X \sim N(\theta,1)$$
The natural estimate to consider is $t_n=\frac{\sum x_n}{n}$. It is consistent, but unfortunately also has $V(t_n)\rightarrow 0$.
But if we purturb it a lot, albeit with a small chance, say
$$t_n'=\frac{\sum^{n-1} x_i}{n-1}+\mathbb{1}_{x_n\le\Phi^{-1}(1/n^2)}\times n^{100}$$
then it will still be consistent, as chance of making the error is only $1/n^2$ (for $\theta=0$) or less (for $\theta=1$), but it's variance $V(t_n')$ clearly diverges since when we do make the error - it is big - and will not compensate the probability of the error which is $1/n^2$, at least when $\theta=0$.
That's the basic idea, but it won't probably work for $\theta=1$. But we can cover it by fabricating another example which works for that, and selecting them alternatively... as below.
To tackle $\theta=1$, we can fabricate another $t_n'$ -
$$t_n''=\frac{\sum^{n-1} x_i}{n-1}+\mathbb{1}_{x_n\gt 1+\Phi^{-1}(1-1/n^2)}\times n^{100}$$
To combine, we can create $t_n'''$ by choosing $t_n'$ when $n$ is odd, and $t_n''$ when $n$ is even as the estimate. It will converge in probability to the real $\theta$, but will have infinite variance.
A: Just take an estimator $\hat\theta_n$ which has just two values:
\begin{align*}P(\hat\theta_n = \theta) &= p_n\\P(\hat\theta_n = \delta_n) &= 1-p_n\end{align*}
Now, choose sequences $\delta_n$ and $p_n$ appropriate and you should get a consistent estimator ($p_n \to 0$), whose variance does not converge to zero ($\delta_n \to \infty)$.
