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 ?


3 Answers 3


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


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.

  • $\begingroup$ I am not persuaded yet. For instance, I believe that VG process has first 4 moments and is self-decomposable, so each T can be devided to T/N intervals with characteristic function of power T/N. you can increase N enough to get CLT for any length of T so the process should be gaussian but it is VG. What am I missing? $\endgroup$ Mar 21, 2013 at 21:08
  • $\begingroup$ For any Levy process, you can write $X_t \sim \sum_{i=1}^N X^i(t/N)$. This by itself isn't enough. To apply CLT, you need a fixed sequence of random variables: the CLT then says that when you sum $N$ of them and divided by $\sqrt{N}$, that converges weakly to a normal as $N \rightarrow \infty$. BUT, in order to work with a fixed sequence of random variables (namely identical copies of $X_t$, say), you need to have the proper scaling property. The $\sqrt{\cdot}$ in the central limit theorem corresponds exactly to the $2$-exponent particular to Brownian Motion. $\endgroup$
    – quasi
    Mar 21, 2013 at 21:47
  • $\begingroup$ To add some more info if you're interested, if you don't make any assumption of scalability, what you get when you let $N$ tend to infinity is a triangular collection of random variables, in the sense that for each $N$, you have $N$ i.i.d random variables. There is a generalization of the CLT which says that stable distributions (and every Levy process of course has a stable distribution) arise from such arrays. See the [wikipedia page on triangular arrays](en.wikipedia.org/wiki/Infinite_divisibility_(probability) for example. $\endgroup$
    – quasi
    Mar 21, 2013 at 21:54
  • $\begingroup$ What I think the OP is asking is whether in the limit of $t \rightarrow \infty$ the distribution of $X_t$ will converge to Gaussian (assuming finite second moment of the marginal Levy distribution). My understating is that it will by the CLT, no? $\endgroup$
    – Confounded
    May 21, 2020 at 10:46

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:


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:


  • $\begingroup$ Bob, thanks a lot for the edit. My apologies again $\endgroup$
    – Matt Wolf
    Mar 20, 2013 at 6:11
  • $\begingroup$ You're welcome. The edit wasn't any trouble at all. $\endgroup$
    – Bob Jansen
    Mar 20, 2013 at 10:20
  • $\begingroup$ let me reword my question, I am still not convinced in that how central limit theorem fails for all those Levy processes which have not normal distribution ( VG process as an example). I believe CLT needs many i.i.d distributions to be summed. Self-decomposablity says for time T you can break it down to sum of N many T/N distributions. So I say that how sum of these N i.i.d is not normal? $\endgroup$ Mar 21, 2013 at 1:31
  • $\begingroup$ @AmirYousefi, I edited my answer to also include the case when a Levy process is not a Brownian motion. $\endgroup$
    – Matt Wolf
    Mar 21, 2013 at 3:16
  • $\begingroup$ I retracted my up vote because I think compared to my own answer your answer complicates things unnecessarily. I find it difficult to check it is even correct. $\endgroup$
    – Bob Jansen
    Mar 21, 2013 at 10:29

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.

  • $\begingroup$ Right, I have an elementary understanding of these processes so correct me if I am wrong. The only condition for the iid distributions is having finite variance, so for example VG process has finite variance and it is self-decomposable ( so you can divide it to N iid over a period). Why it is not normal? I mean should not it contradict itself? $\endgroup$ Mar 21, 2013 at 1:52
  • $\begingroup$ I like the question very much. The reason must but that it has jumps (because of time jumps due to the Gamma subordinator). But I can not tell you a rigorous reasoning. If there are jumps then it is not Gaussian. I will try to find something. $\endgroup$
    – Richi Wa
    Mar 21, 2013 at 10:15

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