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Given a Lévy Process $X_{t}$ in $\mathbb{R}^{d}$, with $X_{t}^{*}:=\sup_{s\in[0,t]}|X_{s}|$.

I want to show, that for $t>0$ with $E[|X_{t}|]<\infty$ for $t>0$, then $E[X_{t}^{*}]<\infty$.

Attempt:

I know with Etemadi's inequality, it holds that for $a,b>0$ it holds that $$ P[X_{t}^{*}>a+b]\leq \frac{P[|X_{t}|>a]}{P[X_{t}^{*}\leq b/2]} $$ We can choose due to the càdlàg property of $X$ the $b>0$ such that $P[X_{t}^{*}\leq b/2]>0$. That looks like we can use it.

Hope you can help me out.

Michael Hardy
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ziT
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  • What is this Etemadi's Inequality that you mentioned? The inequality that goes by this name as stated in Wikipedia doesn't seem to be the one you used – Evan Aad Jul 12 '16 at 16:50
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    @EvanAad This follows by the proof of Etemadis inequality. I could give the details in a few days, since i am busy. However are you only interested on this mentoined Etemadis inequality or on the proof of the finite supremum? Since one can use another inequality than the stated above to proof it, which is more clear than the above mentoined inequality – ziT Jul 12 '16 at 17:08
  • Thanks. What is the other, clearer inequality? – Evan Aad Jul 12 '16 at 17:48
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    @EvanAad it is the ottivani inequality. Im texting on the fly and gonna give a more explicit answer later. – ziT Jul 12 '16 at 18:03
  • @EvanAad You can use the ineq. of Ottivani-Skorohod. Given $a,b>0$ it holds $P(max_{1\leq k\leq n}|S_k|>a+b)\cdot min_{1\leq k \leq n}P(|S_n-S_k|\leq b)\leq P(|S_n|>a)$. Where $S_k=X_1+\ldots+X_k$ a sequence of i.i.d. rv. Then on $[0,t]$ choose the sequence $t_{m}^{n}:=\frac{mt}{2^{n}}$ for $m\in {1,\ldots,2^{n}}$. Watch $\max_{1\leq m\leq 2^{n}}|X_{t_m^{n}}|$. We have $X_{t_m^{n}}=\sum_{j=1}^{m}X_{t_j^{n}}-X_{t_{j-1}^{n}}$. Note this is a random walk.By the inequality $P(\max_{1\leq m\leq 2^{n}} |X_{t_{m}^{n}}|>a+b)\cdot\min_{1\leq m\leq 2^{n}}P(|X_t-X_{t_{m}^{n}}|\leq b)\leq P(|X_t|>a)$. – ziT Jul 13 '16 at 10:13
  • The conclusion holds now by letting $n\rightarrow \infty$ and you can choose a $b>0$ such that $\frac{1}{c}:=\min_{s\in [0,t]}P(|X_t-X_{s}|\leq b)>0$. Then $P(\sup_{s\in[0,t]}|X_t-X_s|>a+b)\leq P(|X_t|>a)\cdot c$. And the proof now follows by the given answer below. – ziT Jul 13 '16 at 10:37
  • Thanks, ziT. Now that saz has answered my question, I actually find his reply easier to follow. – Evan Aad Jul 16 '16 at 15:54

1 Answers1

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Yeah, applying Etemadi's inequality is a good idea. The following identity, which holds for any non-negative random variable $X$, will also be useful:

$$\mathbb{E}(X) = \int_0^{\infty} \mathbb{P}(X > r) \, dr.$$

Because of the monotonicity of $r \mapsto \mathbb{P}(X > r)$ this implies

$$b \sum_{k=0}^{\infty} \mathbb{P}(X > (k+1)b) \leq \mathbb{E}(X) \leq b \sum_{k=0}^{\infty} \mathbb{P}(X > kb) \tag{1}$$

for any $b>0$.

Now back to your problem: Choose $b>0$ such that $\mathbb{P}(X_t^* \leq b/2)>0$ and set $c:= 1/\mathbb{P}(X_t^* \leq b/2)$. Then, by Etemadi's inequality (for $a=kb$),

$$\mathbb{P}(X_t^* > k b) \leq c \mathbb{P}(|X_t| > (k-1)b)$$

for any $k \in \mathbb{N}$ (see this question for a proof). Summing over $k$ yields

$$\sum_{k=1}^{\infty} \mathbb{P}(X_t^* > kb) \leq c \sum_{k=0}^{\infty} \mathbb{P}(|X_t|>kb). \tag{2}$$

Since, by assumption $X_t \in L^1$, it follows from $(1)$ that

$$\sum_{k=1}^{\infty} \mathbb{P}(X_t^* > kb) \stackrel{(2)}{\leq} c+ c\sum_{k=1}^{\infty} \mathbb{P}(|X_t|>kb) \stackrel{(1)}{\leq} c+\frac{c}{b} \mathbb{E}(|X_t|)<\infty.$$

Using again $(1)$, we conclude

$$\mathbb{E}(X_t^*) \leq b \sum_{k=0}^{\infty} \mathbb{P}(X_t^* >kb) \leq b + b \sum_{k=1}^{\infty} \mathbb{P}(X_t^* > kb)<\infty.$$

Community
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saz
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  • yeah the identity is really usefull. thx – ziT Feb 11 '16 at 19:50
  • @ saz shouldn't be the last line just $\infty >b\sum_{k=0}^{\infty}P(X_{t}^{}>(k+1)b) \geq E(X_{t}^{})$ by (1) and the second last inequality multiplied by b? – ziT Feb 12 '16 at 17:05
  • @ziT Sorry, $(1)$ was the wrong way round. Hopefully, it's correct now. – saz Feb 12 '16 at 21:45
  • @ saz yeah all fine now. except last line should be $\ldots \leq b+b\sum_{k=1}^{\infty} P(X_{t}^{*}>kb)<\infty$. but this tiny typo is not even worth mentoining. you did help me a lot by that. great ideas from you. thanks – ziT Feb 13 '16 at 13:41
  • @ziT You are welcome :). (And thanks for pointing out the typo.) – saz Feb 13 '16 at 14:38
  • What is this Etemadi's Inequality that you mentioned? The inequality that goes by this name as stated in Wikipedia doesn't seem to be the one you used. – Evan Aad Jul 12 '16 at 16:08
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    @EvanAad Well, the inequality which I used is a consequence of Etemadi's inequality (so, yes, by Etemadi's inequality I mean exactly the inequality you mentioned in your previous comment). Do you have Sato's book on Lévy processes by hands? He proves the inequality (which I used) in the first part of the proof of Theorem 25.18. (If you don't have the book, then please open a new question so that I can explain you the proof in more detail.) – saz Jul 12 '16 at 17:02
  • Thanks. As you've suggested, I've opened a new question. – Evan Aad Jul 12 '16 at 18:19
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    @EvanAad I hope you are not in a hurry, because I don't have the time to answer your question right now. I'll come back to you tomorrow, okay? – saz Jul 12 '16 at 18:26
  • Whenever convenient. I am grateful that you even consider answering my question at all. – Evan Aad Jul 12 '16 at 18:36