Central Limit Theorem for Lévy Process

Question

I am reading a book, which uses the Central Limit Theorem of Lévy Processes $X_t$ without mentioning the exact theorem. Due to the infinite divisible property I can write $X_t$ as a sum of $N$ iid random variables $X^i$ $$ X_t=\sum_{i=1}^N X^i_{t/N} $$ The Problem is, that i want $t\rightarrow \infty$ but for the CLT i have to keep my sequence of my equidistant iid random variables fixed (like $t/N$ fixed). But they do change, as $t\rightarrow \infty$. The book now just says, that with the central limit theorem for Lévy Processes it holds for $t\rightarrow \infty$ \begin{align} \frac{X_t-\overbrace{tE[X_{1}]}^{=E[X_t]}}{\sqrt{t}}\rightarrow \mathcal{N}(0,\operatorname{Var}[X_1])\\ \sqrt{t} \left(\frac{X_t}{t}-E[X_1])\rightarrow \mathcal{N}(0,\operatorname{Var}[X_1]\right) \end{align} I can't find any proofs, lectures or literature about it. Can you help me out?

Answer 1

Without any additional assumptions on the Lévy process $(X_t)_{t \geq 0}$, a central limit theorem does not hold true.

Let $(X_t)_{t \geq 0}$ be a (one-dimensional) Lévy process with Lévy triplet $(b,\sigma^2,\nu)$. Define

$$T(x) := \nu((x,\infty)) + \nu((-\infty,-x))$$

and

$$U(x) := \sigma^2+2 \int_0^x y T(y) \, dy$$

for $x>0$. There is the following statement by Doney and Maller:

1. Suppose that $T(x)>0$ for all $x>0$. Then there exist deterministic functions $a(t),b(t)>0$ such that $$\frac{X_t-a(t)}{b(t)} \stackrel{t \to \infty}{\to} N(0,1) \tag{1}$$ if, and only if, $$\frac{U(x)}{x^2 T(x)} \stackrel{x \to \infty}{\to} \infty.$$
2. Suppose that $T(x)=0$ for all $x>0$ (i.e. the Lévy measure $\nu$ is symmetric). Then $(1)$ holds if, and only if, $\sigma^2>0$. In this case, $a(t) = t \mathbb{E}(X_1)$ and $b(t) = \sigma \sqrt{t}$ is admissible.

In dimension $d>1$ there are CLT-results for Lévy processes by Grabchak.

References:

• R.A. Doney and R.A. Maller.: Stability and Attraction to Normality for Lévy processes at Zero and at Infinity. Journal of Theoretical Probability, Vol. 15, July 2002.
• Michael Grabchack: A note on the multivariate CLT and convergence of Lévy processes at Long and Short Times. ArXiV.

Answer 2

Answer 1 is a bit of an overkill. The question seems to be about the case where the second moment is finite and not something as general as Doney and Maller.

Below are two different simple proofs. Since $\frac{X_t-t EX_1}{\sqrt{\text{Var}(X_1)}}$ is a L'evy process with zero mean and variance $t$, assume w.l.o.g. (in both proofs) that $EX_1=0$ and $\text{Var}(X_1)=1$.

First proof:

By stationary and independent increments and the CLT, $\frac{X_{\lfloor t\rfloor}}{\lfloor t\rfloor}=\frac{1}{{\lfloor t\rfloor}}\sum_{i=1}^{\lfloor t\rfloor}(X_i-X_{i-1})$ converges in distribution to $N(0,1)$.

Also, $\frac{{\lfloor t\rfloor}}{t}\to 1$ and $\text{Var}\left(\frac{X_t-X_{\lfloor t\rfloor}}{\sqrt{t}}\right)=\frac{t-\sqrt{t}}{t}\le \frac{1}{t}\to 0$, so that by Chebyshev's inequality $\frac{X_t-X_{\lfloor t\rfloor}}{\sqrt{t}}$ converges in probability to zero. Therefore, by Slutzky's theorem, $$\frac{X_t}{\sqrt{t}}=\frac{X_{\lfloor t\rfloor}}{\sqrt{\lfloor t\rfloor}} \sqrt{\frac{\lfloor t\rfloor}{t}}+\frac{X_t-X_{\lfloor t\rfloor}}{t}$$ converges in distribution to $N(0,1)$.

Second proof:

$Ee^{i s X_t}=e^{\varphi(s) t}$ where in general $$\varphi(s)=ics-\frac{\sigma^2}{2}s^2+\int_{\mathbb{R}}\left(e^{isx}-1-isx1_{\{|x|\le 1\}}\right)dx$$ and $\nu$ is the Lévy measure satisfying $\int x^2\wedge 1\,\nu(dx)<\infty$ and $\nu(\{0\})=0$.

When $\nu=0$ then $X_t/\sqrt{t}\sim N(0,1)$ for every $t$, so that there is nothing to prove. Therefore, assume otherwise.

When $EX_1$ exists and is finite, then the exponent has this representation: $$\varphi(s)=ics-\frac{\sigma^2}{2}s^2+\int_{\mathbb{R}}\left(e^{isx}-1-isx\right)\nu(dx)$$ and in particular, when it is zero, then $c=0$. Since $b\equiv\int_{\mathbb{R}}x^2\nu(dx)\le \sigma^2+b=\text{Var}(X_1)=1<\infty$, then $\mu(dx)=x^2\nu(dx)/b$ is a probability measure and we can write $$\varphi(s)=-\frac{s^2}{2}\left(\sigma^2+bE\frac{e^{isY}-1-sY}{(sY)^2/2}\right)$$ where $Y$ is distributed $\mu$ (hence $P(Y=0)=0$).

Since $$\frac{e^{ix}-1-ix}{x^2/2}=-\left(\frac{1-\cos x}{x^2/2}+i\frac{x-\sin x}{x^2/2}\right)$$ is bounded and converges to one, it follows, by the bounded convergence theorem, that, as $t\to\infty$, $$\varphi(s/\sqrt{t})t\to -\frac{s^2}{2}(\sigma^2+b)=-\frac{s^2}{2}\,.$$

Therefore, $$Ee^{is X_t/\sqrt{t}}=e^{\varphi(s/\sqrt{t})t}\to e^{-s^2/2}$$ and, by the continuity theorem for characteristic functions, the proof is complete.