Infinitely Divisible Distribution and Lévy-Khinchin representation

convergence-divergencelevy-processesprobability theoryvariance

I am studying the book Probability Theory by A. Klenke (3rd Edition). I do not understand the construction of a process from which to obtain the Lévy-Khinchin representation for an infinitely divisible distribution on the real numbers. (pag. 375 of the book).

In particular, I do not understand how the three points below lead to the constraint $\int \text{min}(1,x^2)v(dx)<\infty$ on the so called canonical measure $v$ of the representation.

He starts by defining a real random variable as the sum of independent random variables:

$$X=b+X^N+X_0+\sum_{k=1}^{\infty} (X_k-\alpha_k) $$

where

  • $b\in\mathcal{R}$
  • $X^N=\mathcal{N}_{0,\sigma^2}$ for some $\sigma^2\geq0$
  • $P_{X_k}=\text{CPoi}_{v_k}$, where $v_k$ is the intensity measure of a Compound Poisson Distribution concentrated on $I_k:=(-1/k,-1/(k+1)] \cup [1/(k+1),1/k)$ (convention $1/0=\infty$)
  • $ \alpha_k=E[X_k]=\int x v_k(dx)$

My questions:

  1. For the series to converge is required, as an application of the Kolmogorov's Three Series Theorem, that:
    $$\sum_{k=1}^{\infty} \text{Var}[X_k]<\infty \quad(1)$$
    I know and tried to apply this result to $Z_k=X_k-\alpha_k$: if $Z_k$ are independent r.v. with zero expectation and if $\sum_{k=1}^{\infty} \text{Var}[Z_k]<\infty$ then $\sum_{k=1}^{\infty} Z_k<\infty$ a.s. How is this result related to the Kolmogorov's theorem?

  2. $\text{Var}[X_k]=\int x^2 v_k({dx})$. How is it possible since, if I understood well, the $X_k$ has not zero expectation in general?

  3. Given 2) and letting $v=\sum_{k=0}^{\infty} v_k$, it is said that (1) is equivalent to $\int_{(-1,1)} x^2v(dx)<\infty$ and that, since $v_0$ is always finite, this is equivalent to $\int \text{min}(1,x^2)v(dx)<\infty$. What is the meaning of the last integral, as a restriction on the possible $v$? In particular, I do not understand why the number $1$ shows up in the minimum function and its role in the costruction of the process. Also, is $v_0$ finite because the definition of a Compound Poisson distribution states that (pag. 369)?

Definition: Poisson Compound Distribution
$$\text{CPoi}_v:=e^{-v(\mathcal{R})}\sum_{n=0}^{\infty}\frac{v^{*n}}{n!}$$
where $v$ is a finite measure on the real numbers.

Thanks for the help.

Best Answer

The Lévy measure $\nu$ can be seen as an intensity measure of the jumps of the Lévy process we need to construct. The condition $$\int (1\land x^2)\nu(dx)$$ tells you two things:

  1. The value of $\nu(\{|x|\geq 1\})=\int_{|x|\geq 1} \nu(dx)$ is finite. For the jumps of the Lévy process, this simply means that the number of jumps with jump size at least one is almost-surely finite. We obviously need this condition, since if this number were infinite, the sum of those jumps cannot converge. In your setting, this means $\nu_0$ is finite.

  2. Often times the Lévy measure $\nu$ "blows up" around $0$, meaning $\nu(\{|x|<1\})=\infty$. This means there can be infinitely many "small jumps" with jump size at most $1$. This is where it gets tricky: We want the sum of the jumps $\sum_{k=0}^\infty X_k$ to converge. As you mentioned, Kolmogorovs Three Series Theorem states that it is necessary that $$\sum_{k=1}^\infty \text{Var}(X_k)<\infty$$ where $X_k$ denotes the sum of the jumps with jump size of magnitude between $1/k$ and $1/(k+1)$. As an exercise you might want to show that the amount of such jumps with size between $1/k$ and $1/(k+1)$ is indeed finite, so there is no issue of viewing $X_k$ that way.

If i am not mistaken, you are right in saying that $\text{Var}(X_k)\neq \int x^2\nu_k(dx)$. However, we have $$\text{Var}(X_k)=\mathbb E[X_k^2]-\alpha_k^2 \leq \mathbb E[X_k^2]=\int x^2\nu_k(dx).$$ As a consequence, the convergence of the sum of the $\int x^2 \nu_k(dx)$ is sufficient to show convergence of the sum of $\text{Var}(X_k)$.

Indeed, the choice of $1$ in the condition $\int(1\land x^2)\nu(dx)$ is quite arbitrary. You can choose any other $\epsilon>0$. The point of this condition, however, this that you "control" the number and the magnitude of small jumps, so that their sum is finite, and hence the Lévy process is well defined.

Edit: Additional material about Point processes

Since you have asked, here is a bit more about the connection between the jumps of the Lévy process and the measure $\nu$.

A point process $\mathcal P$ is a "random measure"/"random set function", meaning for a set $A\subset \mathbb R$, $\mathcal P(A)$ is a random variable, i.e. a random ensemble of points in $A$. They are both random in terms of their location in $A$, and random in terms of the number of points in $A$. Now, a Poisson point process is a particular type of point process, where you have a measure $\mu$ on $\mathbb R$, such that the number of points of $\mathcal P$ in $A$ has a Poisson-distribution with parameter $\mu(A)$ (there are additional assumptions, but i won't list them here). In particular, this means that if $\mu(A)<\infty$, the number of points in $A$ will be almost-surely finite.

Now to the Lévy process: We consider a point process $\mathcal P$ on $\mathbb R$ with intensity measure $\nu$. The points of $\mathcal P$ are considered as the "jumps" of the Lévy process in a time interval of length $1$ (we can generalize this for general time intervals, but that's besides the point for now). As mentioned above, the constraint $\int (1\land x^2)\nu(dx)$ ensures that $\nu(\{|x|\geq 1\})<\infty$. Therefore, the number of jumps with jump size greater than 1 will have a $\text{Poi}(\nu(\{|x|\geq 1\}))$-distribution, ensuring that the amount of those will be finite.

Similarly, for any $k$, if we consider the jumps of jump-size between $1/k$ and $1/(k+1)$, the amount of such jumps is equal to $$\mathcal P\bigg(\big\lbrace 1/(k+1)\leq |x|\leq 1/k\big\rbrace\bigg)\sim \text{Poi}(\nu([-1/k,-1/(k+1)]\cup [1/(k+1),1/k]))$$ and hence there will also be an almost-surely finite amount of such jumps. In your example, $X_k$ denotes the sum of those jumps. We know that the sum converges, since it is only a finite sum, i.e. $X_k$ is given by $$X_k=\sum_{i=1}^\tau J_i^{(k)}$$ where the $J_i^{(k)}$ are the points of the Point process $\mathcal P$ in $\lbrace 1/(k+1)\leq |x|\leq 1/k\rbrace$ and $\tau$ is a Poisson-distributed random variable.