Let $X_t$ be a Lévy process which is known to have mean zero and finite variance $t \cdot \sigma^2$, but for which the value of $\sigma^2$ is unknown. How do we estimate $\sigma^2$? One approach would be to fix a number of samples $N$ over a time horizon $[0,T]$, compute the independent increments $Z_j^{(T,N)} = X_{jT/N} - X_{(j-1)T/N}$, and then compute the sample variance $$ Y^{(N)}_T = \sum_{j=1}^N \left(Z_j^{(T,N)}\right)^2 $$ The expectation of $\frac1T Y_T^{(N)}$ is $\sigma^2$, while its variance is $$ \frac{N}{T^2} \text{var}\left(\left(Z_1^{(T,N)}\right)^2\right) = \frac{\sigma^4}{N}\text{kurt}\left(Z_1^{(T,N)}\right) + 2\frac{\sigma^4}{N} $$ Here $\text{kurt}\left(Z_1^{(T,N)}\right)$ is the normalized (excessive) kurtosis, the fourth cumulant divided by the square of the variance.

If $X_t$ is simply Brownian motion then the kurtosis of its increments is zero, so accuracy is encouraged by taking a large number of samples, regardless of the horizon $T$. On the other hand, if $X_t$ is anything other than Brownian motion then its fourth cumulant is positive and scales with $t$, so that its kurtosis is equal to $k/t$ for some $k>0$. In this situation we can write the variance of $\frac1T Y_T^{(N)}$ as $$ \frac{\sigma^4}{N}\text{kurt}\left(Z_1^{(T,N)}\right) + 2\frac{\sigma^4}{N} = \sigma^4\left(\frac{k}{T} + \frac{2}{N}\right) $$ So fixing a time horizon $[0,T]$ and simply taking a large number of samples $N$ is not enough to guarantee a high probability of accuracy.

My question is this: For fixed $T>0$, is it known whether $Y_T^{(N)}$ has a limiting distribution $Y_T$ as $N\to\infty$ for all Lévy processes? Furthermore, if $Y_t$ is treated as a stochastic process, is it known to be a nonnegative Lévy process? Is there a name for the process $Y_t$ associated to the Lévy process $X_t$, assuming it exists?