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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?

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You may want to check with Prof. Ward Whitt's webpage. He has hundreds of papers on queueing theory and have addressed Levy processes in many of his publications. Try to find something in there.

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