Any convergence results other than to a mean?

I am wondering whether there can be a statistic that does not converge to its mean but does converge to some other quantities such as the mode.

Here convergence is convergence in probability (or a.s. convergence is of course ok) w.r.t. sample size, and I assume that statistic depends on some i.i.d. samples. Of course I am excluding trivial cases where the statistics for example converges to zero.

If the statistic takes a form of sample mean, in many cases (i.e. if its mean exists) the law of large numbers takes place. Also I regard quantities like sample quantile converge to their means (which are population quantiles). Therefore I am looking for a convergence result that involves quantities other than the mean.

• You're looking for a sequence $T_n$ of statistics that almost surely converges to a constant $c$ but such that $$c = \mathbb{E}\left[\lim_{n \rightarrow \infty} T_n(X_1, \dots, X_n)\right] \not = \lim_{n \rightarrow \infty}\mathbb{E}\left[T_n(X_1, \dots, X_n)\right)]$$ When $T_n(X_1,\dots, X_n)$ is absolutely bounded by an integrable random variable, then by the dominated convergence theorem you have that the two limits are equal. This covers most practical cases, and so I doubt there are many interesting examples. – Olivier Feb 3 '18 at 1:24
• Thank you for the clear explanation. As long as the dominated convergence theorem takes place, it already ensures the convergence to (the limit of) the mean. I see. That makes an enough reason to think it is hard to come up with a good counter-example. – diadochos Feb 3 '18 at 1:48

Let me give you a real-world example. Consider the case of a bivariate normal variable $S=(s_t,s_{t+1})$. To make it simple, let us also transform $S$ to $S'$ so that it was centered around $(s_t^*,s^*_{t+1})$ and now is centered around $(0,0)$ by transformation. This puts the problem in the errors space centered around $(0,0)$. Now let's define a new variable as $r_t=\frac{s_{t+1}'}{s_t'}$ Since $S'$ is bivariate normal, this requires $r_t$ to follow a Cauchy distribution. Now let us assume that $r_t$ only has support on the non-negative real numbers. Now one has the truncated Cauchy distribution for $r_t$ and let's untransform $S'$ to $S$ by noting that $\mu_r=\frac{S_{t+1}^*}{S_t^*}$, so we now have a density for $r_t$ of $$\left[\frac{\pi}{2}+\tan^{-1}\left(\frac{\mu_r}{\gamma}\right)\right]^{-1}\frac{\gamma}{\gamma^2+(r_t-\mu_r)^2}.$$
Of course, I left out a lot of steps. $\mu_r$ is located at the mode and not the median and of course, no mean exists for this density. Now as to practical application. If $S$ is the set of stock prices for going concerns, then $r_t$ is the instantiation of the reward for investing. Nicely, this also matches the data closely in the real world. The same would be true for anything that experiences exponential growth, such as tumors.
For a variety of reasons, you are left with just a Bayesian solution for this and if $\gamma$ were either marginalized out or was known, then $\hat{\mu_r}$ does converge nicely to $\mu_r$ Do note though that $\bar{r}$ does not converge to any value.