Consider a sequence of uncorrelated variables $(X_n)$. If these variables are bounded in $L^2$ and have common mean $\mu$, a well-known variant of the law of large numbers states that $$ \frac{1}{n}\sum_{i=1}^n X_i \stackrel{\mathrm{a.s.}}\longrightarrow \mu. $$ This allows for estimation of parameters when observations are not i.i.d., a situation occurring for example for non-uniformly spaced observations of continuous-time processes.
My question is about the variance analogue of the above. Again, assume that $(X_n)$ is a sequence of uncorrelated variables. Assume that the variables have common variance $\sigma^2$. We do not assume that the variables have common mean. Is there an asymptotic result similar to the above yielding $\sigma^2$ as an almost sure limit?
I would imagine something like $$ \frac{1}{n}\sum_{i=1}^n (X_i - \bar{X}_n)^2 \stackrel{\mathrm{a.s.}}\longrightarrow \sigma^2, $$ where $\bar{X}_n$ is the average of $X_1,\ldots,X_n$, but this result is not obvious to me. Assume boundedness in $L^p$ for any $p\ge 1$ necessary.
For completeness, here is an explicit motivation for the above example. Assume that $S_t = \exp(X_t)$ for some Levy process $X$ and $t\ge0$. Here, $S$ denotes an asset price, and this is then an instance of the exponential Levy market model. In this model, continuous returns of period $\Delta$ are given by $\log S_{t+\Delta} / S_t$ and have mean $\Delta \alpha$ and variance $\Delta \beta$ for some $\alpha$ and $\beta$. Making observations $S_{t_1},\ldots,S_{t_n}$, I'm interested in estimating $\beta$. Here, it holds that the variables $$ \frac{1}{\sqrt{t_i-t_{i-1}}}\log\frac{S_{t_i}}{S_{t_{i-1}}} $$ are uncorrelated (in fact, independent) and have the same variance $\beta$. If something like the above almost sure result for convergence to the variance holds, I would then be able to use the above transformed variables to estimate $\beta$.
Consider the example where $E[X_i] = +m>0$ when $i$ is even and $E[X_i]=-m$ when $i$ is odd, where $\mathrm {Var}(X_i)=\sigma^2$ and the fourth moment is bounded
Then $\bar{X}_n=\frac{1}{n}\sum_{i=1}^n X_i \stackrel{\mathrm{a.s.}}\longrightarrow 0$ while $\frac{1}{n}\sum_{i=1}^n (X_i - \bar{X}_n)^2 \stackrel{\mathrm{a.s.}}\longrightarrow \sigma^2 +m^2 \not=\sigma^2$ .