Why Lévy Process Increments Do Not Generally Produce a Normal Limit

# Central Limit Theorem and Lévy processes


# Central Limit Theorem and Lévy processes












Lévy processes are self-decomposable and independent on any non-overlapping interval, so how come the distribution of the process at time T,$\phi(T)$, which is the sum of N i.i.d with law $\phi(T/N)$ is not normally distributed ? I cannot find what am I missing here ?

## Answer by quasi (score 4)

https://quant.stackexchange.com/a/7563

I believe your problem is that you're assuming all Lévy processes are stable with exponent $2$. Here is what happens if we try to use your argument: Let $X$ be a Lévy process (that is a martingale, for simplicity). At time $t$, for any $N$, we have $$ X_t \sim\sum_{i=1}^N X^i \left(\frac{t}{N}\right), $$ with each $X^i \left(\frac{t}{N}\right)$ i.i.d. and distributed like $X_{\frac{t}{N}}$. In order to apply the Central Limit Theorem, to deduce the alleged normality of $X_t$, what we would want is $$ X^i \left(\frac{t}{N} \right) \sim \frac{1}{\sqrt{N}} X^i, $$ where now the $X^i$ are a sequence of i.i.d random variables which don't depend on $N$ anymore (i.e. each $X^i \sim X_t$) This is critical, because the CLT only applies to a fixed sequence of random variables. This then would yield $$ X_t \sim \frac{1}{\sqrt{N}} \sum_{i=1}^N X^i, $$ which, as $N \rightarrow \infty$, would fit the template of the Central Limit Theorem (of course up to variance conditions)

This property (that $X_{\frac{t}{N}} \sim \frac{1}{\sqrt{N}} X_t$), however, is not enjoyed by an arbitrary Lévy martingale (although it is satisfied by Brownian Motion). Consequently, if you carried out this argument in general, you would have to use a triangular array of random variables, (see my comment below on these objects) which give rise to stable distributions through a generalization of the CLT.

The more general scaling property for a Lévy process is that $X$ satisfies $$ X_t / t^{1/\alpha} \sim X_1, $$ with $\alpha$ being known as the exponent of the process. This is the basic property of the stable processes, which are a subset of the Lévy processes. The Poisson Process, for example, isn't stable, while the Cauchy process is stable with exponent $1$. A $2$-stable process must be Gaussian

EDIT:

I'd like to add a new argument, which attempts to explain the following fact: If $X$ is a Levy process, then it is Gaussian if and only if it has continuous paths. This is a technical argument, and indeed it basically constitutes half of the Levy decomposition theorem.

Let $\xi_{nj}$ be a triangular null array of independent random variables. This means $\xi_{nj}$ is defined, for $1 \leq j \leq m_n$, for each $n \in \mathbb{N}$, and $\sup_j E\left[ |\xi_{nj}| \wedge 1 \right] \rightarrow 0$ as $n \rightarrow \infty$. This is like a uniform convergence in probability to zero, as $n \rightarrow \infty$. I'm going to cite a Theorem due to Feller and Levy. I found it in "Foundations of Modern Probability" by Kallenberg.

Theorem Let $\xi_{nj}$ be a triangular null array. $\sum_j \xi_{nj}$ converges to a normal r.v. $\xi \sim N(b,c)$ if and only if the following three conditions hold:

$$ \sum_j P(|\xi_{nj}| > \epsilon) \rightarrow 0 \text{ for all } \epsilon > 0, $$ $$ \sum_j E \left[\xi_{nj} ; |\xi_{nj}| \leq 1 \right] \rightarrow b, $$ $$ \sum_j var \left[\xi_{nj} ; |\xi_{nj}| \leq 1 \right] \rightarrow c $$

The proof of this theorem is not easy. Message me if you'd like to get more details. However, the upshot is that, assuming this theorem, we see that path continuity is essentially equivalent to the first condition, with $\xi_{nj}$ representing the $j^{th}$ increment of $X$ with the $n^{th}$ grid. Mean and variance control are the second and third conditions, respectively.

## Answer by Matt Wolf (score 2)

https://quant.stackexchange.com/a/7560

The Lévy theorem states that the conditions that have to be met for $M(t)$ to be a Brownian motion (and hence be normally distributed):

- $M(t)$, for $t>0$, be a martingale relative to some filtration $F(t), t>0$.

- $M(0)= 0$

- $M(t)$ has continuous paths

- $[M,M](t) = t$ for all $t\geq0$;

So, to test each condition you simply differentiate your Lévy process (using Itô Calculus), change to the integrated form and you notice that at $t=0$ the stochastic integral takes the value zero, hence the expectation is always zero (at $t=0$). When you take the expectation of the stochastic integral you can isolate the following moment generating function:

$$\mathbb{E}\left[\mathrm{exp}\left(uM(t)\right)\right] = \mathrm{exp}\left(\frac{1}{2} u^2t\right)$$

which is the MGF for the normal distribution with zero mean and variance $t$. Therefore your $M(t)$ Lévy process follows the same distribution and you just showed that also an $M(t)$ process follows normality.

If your Levy process does not satisfy above conditions but still follows the general definition of a Levy processes as stated here:

http://almostsure.wordpress.com/2010/11/23/levy-processes/

then you can utilize the characteristic functions. Since the characteristic function of a convolution is the product of the characteristic functions of the densities involved, the central limit theorem has yet another restatement: the product of the characteristic functions of a number of density functions becomes close to the characteristic function of the normal density as the number of density functions increases without bound, under the conditions stated above. However, to state this more precisely, an appropriate scaling factor needs to be applied to the argument of the characteristic function.

The following shows how those scaling factors and drift adjustments can be made:

http://www.stats.ox.ac.uk/~winkel/lp1.pdf

## Answer by Richi Wa (score 1)

https://quant.stackexchange.com/a/7566

A very good question. In other words you ask why the central limit theorem does not hold, right? A sum of iid should be somehow normal, right? Looking at the Levy-Kinchin representation we see the Gaussian part, which comes from increments of a continuous process, and the rest from the jumps. So one answer (which is not mathematically rigorous) is the presence of jumps. Another reason is that a Levy process can have infinite moments (also because of the jumps).

If we the process is continuous then it is Gaussian (if and only if). The jumps enrich the model but of course make if much more complicated.