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Let $(X_t)_{t \geq 0}$ a Lévy process and $\varepsilon>0$. Is there anything known about the asymptotics of the probability

$$\mathbb{P}(|X_t| > \varepsilon)$$

as $t \to 0$? Obviously, by the stochastic continuity, this probability converges to $0$ - but how fast? I tried to apply Markov's inequality (assuming that the corresponding moment exists); then I get

$$\mathbb{P}(|X_t|>\varepsilon) \leq \frac{1}{\varepsilon} \cdot \mathbb{E}(|X_t|)$$

Unfortunately, I'm not aware of an estimate of the form

$$\mathbb{E}(|X_t|) \leq C \cdot f(t)$$

for some constant $C>0$ and a function $f$ (which should converge to $0$ as $t \to 0$).

The background is the following: When I tried to answer this question about Lévy processes, I was able to prove that it suffices to show that

$$m \cdot \mathbb{P}(|X_{1/m}|>\varepsilon)^2$$

converges to $0$ as $m \to \infty$. And that's where I'm stuck...

Thanks for any suggestions & hints!

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saz
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2 Answers2

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Lemma Let $(X_t)_{t \geq 0}$ be a one-dimensional Lévy process and $\varepsilon>0$. Then $$\limsup_{t \to 0} \frac{1}{t} \cdot \mathbb{P}(|X_{t}|>\varepsilon) \leq C $$ for some constant $C>0$.

Proof: Denote by $\psi$ the characteristic exponent of $(X_t)_{t \geq 0}$. By the truncation inequality,

$$\DeclareMathOperator \re {Re} \DeclareMathOperator \im {Im}\mathbb{P}(|X_{t}|>\varepsilon) \leq 7\varepsilon \cdot \int_0^{1/\varepsilon} (1-\re e^{-t \cdot \psi(y)}) \, dy \tag{1}$$

Recall that for any $u \geq 0$, $v \in \mathbb{R}$, we have

$$|1-e^{-(u+\imath \, v)}| = \left| \int_0^{u+\imath \, v} e^-z \, dz \right| \leq |u+\imath \, v|.$$

Since $\text{Re} \, \psi(y) \geq 0$ for any $y \in \mathbb{R}$, this shows that

$$\re(1-e^{-t \psi(y)}) \leq |1-e^{-t \psi(y)}| \leq t |\psi(y)|.$$

Since $y \mapsto \psi(y)$ is continuous, we obtain

$$|1-\re e^{-t \cdot \psi(y)}| \leq C \cdot t, \qquad 0 \leq y \leq 1/\varepsilon$$

for some constant $C>0$. Plugging this estimate in $(1)$ finishes the proof.

Remark Using the $d$-dimensional version of the truncation inequality $(1)$, it's not difficult to show the corresponding statement for any $d$-dimensional Lévy process. If $X$ has finite moments of second order, then the statement follows rather easily from Markov's inequality. Moreover, the example of the Poisson process shows that one cannot expect a faster convergence.

saz
  • 120,083
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Denote by $\psi$ the characteristic exponent of the Lévy process $(X_t)_{t \geq 0}$. Then it follows that

$$\psi(\xi) = \lim_{n \to \infty} \frac{1- e^{-\frac{1}{n} \psi(\xi)}}{\frac{1}{n}} = \lim_{n \to \infty} n \int (1-e^{\imath \, y \xi}) \, d\mu_n(y)$$

where $X_{1/n} \sim \mu_n$. Integrating both sides from $-1$ to $1$ yields

$$\infty > \int_{-1}^1 \psi(\xi) \, d\xi = \lim_{n \to \infty} n \int \left(1- \frac{\sin y}{y} \right) \, d\mu_n(y).$$

Now as

$$1- \frac{\sin y}{y} \geq c > 0$$

for any $|y| \geq \delta$, we conclude that

$$\limsup_{n \to \infty} n \int_{B(0,\delta)^c} \mu_n(dy) < \infty.$$

By definition, this equivalent to

$$\limsup_{n \to \infty} n \cdot \mathbb{P}(|X_{1/n}| \geq \delta) < \infty.$$

saz
  • 120,083