5

Let $\{X_n\}_{n \ge 1}$ be a sequence of $p$-dimensional infinitely divisible (ID) random vectors, and let $\{\mu_n\}_{n \ge 1}$ be their corresponding Lévy measures.

Suppose we have two conditions:

  1. The sequence of random vectors converges weakly to an ID random vector $X$: $$ X_n \Rightarrow X. $$
  2. The second moments of the Lévy measures are uniformly bounded. That is, there exists a constant $C < \infty$ such that: $$\sup_{n \ge 1} \int_{\mathbb{R}^p} |x|^2 \, d\mu_n(x) \le C. $$

My question is: do these two conditions together imply that the function $g(x) = |x|^2$ is uniformly integrable with respect to the sequence of measures $\{\mu_n\}$? Specifically, does it follow that: $$ \lim_{M \to \infty} \left( \sup_{n \ge 1} \int_{\{|x|^2 > M\}} |x|^2 \, d\mu_n(x) \right) = 0? $$

I understand that the uniform bound from condition (2) is the key ingredient for proving the tightness of the sequence of measures $\{|x|^2 d\mu_n\}$. However, I suspect this property can only be established rigorously by using the weak convergence from condition (1) to prevent scenarios where mass escapes to infinity. The formal path to combine these two hypotheses is what is not apparent to me.

Is the proposition true? If so, could you provide a standard reference or a sketch of the standard proof?

Gregório Vieira
  • 95
  • 4

1 Answers1

1

$$\DeclareMathOperator{\real}{\mathbb{R}}$$ $$\DeclareMathOperator{\nat}{\mathbb{N}}$$ $$\DeclareMathOperator{\de}{d}$$ Take as $\{\mu_n\}_{n \in \nat}$ the following: $$\mu_n = \frac{1}{n^2}\delta_{x_n}$$ With: $$\lim_{n \to \infty} |x_n| = +\infty \hspace{1cm} \forall n \in \nat \hspace{0.25cm} |x_n|=n$$ Surely these are Lévy measures and the first condition is satisfied: $$\forall f \in C^0_b(\real^p) \hspace{1cm} \int f \de \mu_n = \frac{f(x_n)}{n^2} \implies \mu_n \stackrel{\ast}{\rightharpoonup} \nu=0 $$ For the second condition we have: $$\forall n \in \nat \hspace{1cm} \int |x|^2 \de \mu_n = 1$$ But $|x|^2$ can't be uniformly integrable with respect to $\{\mu_n\}_{n \in \nat}$ because for any given $M>0$ we have: $$ \sup_{n \in \nat}\left \{\int_{\{|x|^2 > M\}}|x|^2\de \mu_n \right\} \ge 1$$

theBmax
  • 478