Subordinators and the Arcsine rule

After the general discussion of Levy processes in the previous post, we now discuss a particular class of such processes. The majority of content and notation below is taken from chapters 1-3 of Jean Bertoin’s Saint-Flour notes.

We say X_t is a subordinator if:

  • It is a right-continuous adapted stochastic process, started from 0.
  • It has stationary, independent increments.
  • It is increasing.

Note that the first two conditions are precisely those required for a Levy process. We could also allow the process to take the value \infty, where the hitting time of infinity represents ‘killing’ the subordinator in some sense. If this hitting time is almost surely infinite, we say it is a strict subordinator. There is little to be gained right now from considering anything other than strict subordinators.


  • A compound Poisson process, with finite jump measure supported on [0,\infty). Hereafter we exclude this case, as it is better dealt with in other languages.
  • A so-called stable Levy process, where \Phi(\lambda)=\lambda^\alpha, for some \alpha\in(0,1). (I’ll define \Phi very soon.) Note that checking that the sample paths are increasing requires only that X_1\geq 0 almost surely.
  • The hitting time process for Brownian Motion. Note that this does indeed have jumps as we would need. (This has \Phi(\lambda)=\sqrt{2\lambda}.)


  • In general, we describe Levy processes by their characteristic exponent. As a subordinator takes values in [0,\infty), we can use the Laplace exponent instead:

\mathbb{E}\exp(-\lambda X_t)=:\exp(-t\Phi(\lambda)).

  • We can refine the Levy-Khintchine formula;

\Phi(\lambda)=k+d\lambda+\int_{[0,\infty)}(1-e^{-\lambda x})\Pi(dx),

  • where k is the kill rate (in the non-strict case). Because the process is increasing, it must have bounded variation, and so the quadratic part vanishes, and we have a stronger condition on the Levy measure: \int(1\wedge x)\Pi(dx)<\infty.
  • The expression \bar{\Pi}(x):=k+\Pi((x,\infty)) for the tail of the Levy measure is often more useful in this setting.
  • We can think of this decomposition as the sum of a drift, and a PPP with characteristic measure \Pi+k\delta_\infty. As we said above, we do not want to consider the case that X is a step process, so either d>0 or \Pi((0,\infty))=\infty is enough to ensure this.

Analytic Methods

We give a snapshot of a couple of observations which make these nice to work with. Define the renewal measure U(dx) by:

\int_{[0,\infty)}f(x)U(dx)=\mathbb{E}\left(\int_0^\infty f(X_t)dt\right).

If we want to know the distribution function of this U, it will suffice to consider the indicator function f(x)=1_{X_t\leq x} in the above.

The reason to exclude step processes specifically is to ensure that X has a continuous inverse:

L_x=\sup\{t\geq 0:X_t\leq x\} so U(x)=\mathbb{E}L_x is continuous.

In fact, this renewal measure characterises the subordinator uniquely, as we see by taking the Laplace transform:

\mathcal{L}U(\lambda)=\int_{[0,\infty)}e^{-\lambda x}U(dx)=\mathbb{E}\int e^{-\lambda X_t}dt

=\int \mathbb{E}e^{-\lambda X_t}dt=\int\exp(-t\Phi(\lambda))dt=\frac{1}{\Phi(\lambda)}.

The Arcsine Law

X is Markov, which induces a so-called regenerative property on the range of X, \mathcal{R}. Formally, given s, we do not always have s\in\mathcal{R} (as the process might jump over s), but we can define D_s=\inf\{t>s:t\in\mathcal{R}\}. Then

\{v\geq 0:v+D_s\in\mathcal{R}\}\stackrel{d}{=}\mathcal{R}.

In fact, the converse holds as well. Any random set with this regenerative property is the range of some subordinator. Note that D_s is some kind of dual to X, since it is increasing, and the regenerative property induces some Markovian properties.

In particular, we consider the last passage time g_t=\sup\{s<t:s\in\mathcal{R}\}, in the case of a stable subordinator with \Phi(\lambda)=\lambda^\alpha. Here, \mathcal{R} is self-similar with scaling exponent \alpha. The distribution of \frac{g_t}{t} is thus independent of t. In this situation, we can derive the generalised arcsine rule for the distribution of g_1:

\mathbb{R}(g_1\in ds)=\frac{\sin \alpha\pi}{\pi}s^{\alpha-1}(1-s)^{-\alpha}ds.

The most natural application of this is to the hitting time process of Brownian Motion, which is stable with \alpha=\frac12. Then g_1=S_1-B_1, in the usual notation for the supremum process. Furthermore, we have equality in distribution of the processes (see previous posts on excursion theory and the short aside which follows):

(S_t-B_t)_{t\geq 0}\stackrel{d}{=}(|B_t|)_{t\geq 0}.

So g_1 gives the time of the last zero of BM before time 1, and the arcsine law shows that its distribution is given by:

\mathbb{P}(g_1\leq t)=\frac{2}{\pi}\text{arcsin}\sqrt{t}.

The Levy-Khintchine Formula

Because of a string of coincidences involving my choice of courses for Part III and various lecturers’ choices about course content, I didn’t learn what a Levy process until a few weeks’ ago. Trying to get my head around the Levy-Khintchine formula took a little while, so the following is what I would have liked to have been able to find back then.

A Levy process is an adapted stochastic process started from 0 at time zero, and with stationary, independent increments. This is reminiscent, indeed a generalisation, of the definition of Brownian motion. In that case, we were able to give a concrete description of the distribution of X_1. For a general Levy process, we have


So the distribution of X_1 is infinitely divisible, that is, can be expressed as the distribution of the sum n iid random variables for all n. Viewing this definition in terms of convolutions of distributions may be more helpful, especially as we will subsequently consider characteristic functions. If this is the first time you have seen this property, note that it is not a universal property. For example, it is not clear how to write a U[0,1] random variable as a convolution of two iid RVs. Note that exactly the same argument suffices to show that the distribution of X_t is infinitely divisible.

It will be most convenient to work with the characteristic functions

\mathbb{E}\exp(i\langle \lambda,X_t\rangle).

By stationarity of increments, we can show that this is equal to

\exp(-\Psi(\lambda)t)\quad\text{where}\quad \mathbb{E}\exp(i\langle \lambda,X_1\rangle)=:\exp(-\Psi(\lambda)).

This function \Psi(\lambda) is called the characteristic exponent. The argument resembles that used for Cauchy’s functional equations, by dealing first with the rationals using stationarity of increments, then lifting to the reals by the (right-)continuity of

t\mapsto \mathbb{E}\exp(i\langle \lambda,X_t\rangle).

As ever, \Psi(\lambda) uniquely determines the distribution of X_1, and so it also uniquely determines the distribution of Levy process. The only condition on \Psi is that it be the characteristic function of an infinitely divisible distribution. This condition is given explicitly by the Levy-Khintchine formula.


\Psi(\lambda) is the characteristic function of an infinitely divisible distribution iff

\Psi(\lambda)=i\langle a,\lambda\rangle +\frac12 Q(\lambda)+\int_{\mathbb{R}^d}(1-e^{i\langle \lambda,x\rangle}+i\langle \lambda,x\rangle 1_{|x|<1})\Pi(dx).

for a\in\mathbb{R}^d, Q a quadratic form on \mathbb{R}^d, and \Pi a so-called Levy measure satisfying \int (1\wedge |x|^2)\Pi(dx)<\infty.

This looks a bit arbitrary, so first let’s explain what each of these terms ‘means’.

  • i\langle a,\lambda\rangle comes from a drift of -a. Note that a deterministic linear function is a (not especially interesting) Levy process.
  • \frac12Q(\lambda) comes from a Brownian part \sqrt{Q}B_t.

The rest corresponds to the jump part of the process. Note that a Poisson process is an example of a Levy process, hence why we might consider thinking about jumps in the first place. The reason why there is an indicator function floating around is that we have to think about two regimes separately, namely large and small jumps. Jumps of size bounded below cannot happen too often as otherwise the process might explode off to infinity in finite time with positive probability. On the other hand, infinitesimally small jumps can happen very often (say on a dense set) so long as everything is controlled to prevent an explosion on the macroscopic scale.

There is no canonical choice for where the divide between these regimes happens, but conventionally this is taken to be at |x|=1. The restriction on the Levy measure near 0 ensures that the sum of the squares all jumps up some finite time converges absolutely.

  • \Pi\cdot 1_{|x|\geq 1} gives the intensity of a standard compound Poisson process. The jumps are well-spaced, and so it is a relatively simple calculation to see that the characteristic function is

\int_{\mathbb{R}^d}(1-e^{i\langle \lambda,x\rangle})1_{|x|\geq 1}\Pi(dx).

The intensity \Pi\cdot 1_{|x|<1} gives infinitely many hits in finite time, so if the expectation of this measure is not 0, we explode immediately. We compensate by drifting away from this at rate


To make this more rigorous, we should really consider 1_{\epsilon<|x|<1} then take a limit, but this at least explains where all the terms come from. Linearity allows us to interchange integrals and inner products, to get the term

\int_{\mathbb{R}^d}(1-e^{-i\langle \lambda,x\rangle}+i\langle\lambda,x\rangle 1_{|x|<1})\Pi(dx).

If the process has bounded variation, then we must have Q=0, and also

\int (1\wedge |x|)\Pi(dx)<\infty,

that is, not too many jumps on an |x| scale. In this case, then this drift component is well-defined and linear \lambda, so can be incorporated with the drift term at the beginning of the Levy-Khintchine expression. If not, then there are some \lambda for which it does not exist.

There are some other things to be said about Levy processes, including

  • Stable Levy processes, where \Psi(k\lambda)=k^\alpha \Psi(\lambda), which induces the rescaling-invariance property: k^{-1/\alpha}X_{kt}\stackrel{d}{=}X. The distribution of each X_t is then also a stable distribution.
  • Resolvents, where instead of working with the process itself, we work with the distribution of the process at a random exponential time.

CLT and Stable Distributions

One of the questions I posed at the end of the previous post about the Central Limit Theorem was this: what is special about the normal distribution?

More precisely, for a large class of variables (those with finite variance) the limit in distribution of S_n after a natural rescaling is distributed as N(0,1). As a starting point for investigating similar results for a more general class of underlying distributions, it is worth considering what properties we might require of a distribution if it is to appear as a limit in distribution of sums of IID RVs, rescaled if necessary.

The property required is that the distribution is stable. In the rest of the post I am going to give an informal precis of the content of the relevant chapter of Feller.

Throughout, we assume a collection of IID RVs, X,X_1,X_2,\ldots, with the initial sums S_n:=X_1+\ldots+X_n. Then we say X is stable in the broad sense if


for some deterministic parameters c_n,\gamma_n for every n. If in fact \gamma_n=0 then we say X is stable in the strict sense. I’m not sure if this division into strict and broad is still widely drawn, but anyway. One interpretation might be that a collection of distributions is stable if they form a non-trivial subspace of the vector space of random variables and also form a subgroup under the operation of adding independent RVs. I’m not sure that this is hugely useful either though. One observation is that if \mathbb{E}X exists and is 0, then so are all the \gamma_ns.

The key result to be shown is that

c_n=n^{1/\alpha} for some 0<\alpha\leq 2.

Relevant though the observation about means is, a more useful one is this. The stability property is retained if we replace the distribution of X with the distribution of X_1-X-2 (independent copies naturally!). The behaviour of c_n is also preserved. Now we can work with an underlying distribution that is symmetric about 0, rather than merely centred. The deduction that \gamma_n=0 still holds now, whether or not X has a mean.

Now we proceed with the proof. All equalities are taken to be in distribution unless otherwise specified. By splitting into two smaller sums, we deduce that


Extending this idea, we have


Note that it is not even obvious yet that the c_ns are increasing. To get a bit more control, we proceed as follows. Set v=m+n, and express


from which we can make the deduction

\mathbb{P}(X>t)\geq \mathbb{P}(X_1>0,X_2>t\frac{c_v}{c_n})=\frac12\mathbb{P}(X_2>t\frac{c_v}{c_n}). (*)

So most importantly, by taking t>>0 in the above, and using that X is symmetric, we can obtain an upper bound

\mathbb{P}(X_2>t\frac{c_v}{c_n})\leq \delta<\frac12,

in fact for any \delta<\frac12 if we take t large enough. But since


(which should in most cases be \frac12), this implies that \frac{c_v}{c_n} cannot be very close to 0. In other words, \frac{c_n}{c_v} is bounded above. This is in fact regularity enough to deduce that c_n=n^{1/\alpha} from the Cauchy-type functional equation (*).

It remains to check that \alpha\leq 2. Note that this equality case \alpha=2 corresponds exactly to the \frac{1}{\sqrt{n}} scaling we saw for the normal distribution, in the context of the CLT. This motivates the proof. If \alpha>2, we will show that the variance of X is finite, so CLT applies. This gives some control over c_n in an n\rightarrow\infty limit, which is plenty to ensure a contradiction.

To show the variance is finite, we use the definition of stable to check that there is a value of t such that

\mathbb{P}(S_n>tc_n)<\frac14\,\forall n.

Now consider the event that the maximum of the X_is is >tc_n and that the sum of the rest is non-negative. This has, by independence, exactly half the probability of the event demanding just that the maximum be bounded below, and furthermore is contained within the event with probability <\frac14 shown above. So if we set


we then have

\frac14>\mathbb{P}(S_n>tc_n)\geq\frac12\mathbb{P}(\max X_i>tc_n)=\frac12[1-(1-\frac{z}{n})^n]

\iff 1-e^{-z(n)}\leq \frac12\text{ for large }n.

So, z(n)=n(1-F(tc_n)) is bounded as n varies. Rescaling suitably, this gives that

x^\alpha(1-R(x))<M\,\forall x,\,\text{for some }M<\infty.

This is exactly what we need to control the variance, as:

\mathbb{E}X^2=\int_0^\infty \mathbb{P}(X^2>t)dt=\int_0^\infty \mathbb{P}(X^2>u^2)2udu

=\int_0^\infty 4u\mathbb{P}(X>u)du\leq \int_0^\infty 1\wedge\frac{4M}{u^{-(\alpha-1)}}du<\infty,

using that X is symmetric and that \alpha>2 for the final equalities. But we know from CLT that if the variance is finite, we must have \alpha=2.

All that remains is to mention how stable distributions fit into the context of limits in distribution of RVs. This is little more than a definition.

We say F is in the domain of attraction of a broadly stable distribution R if

\exists a_n>0,b_n,\quad\text{s.t.}\quad \frac{S_n-b_n}{a_n}\stackrel{d}{\rightarrow}R.

The role of b_n is not hugely important, as a broadly stable distribution is in the domain of attraction of the corresponding strictly stable distribution.

The natural question to ask is: do the domains of attraction of stable distributions (for 0<\alpha\leq 2) partition the space of probability distributions, or is some extra condition required?

Next time I will talk about stable distributions in a more analytic context, and in particular how a discussion of their properties is motivated by the construction of Levy processes.