On some cadlaguity moment estimates of stochastic jump processes

TL;DR

This paper derives estimates for the moments of cadlag (right-continuous with left limits) properties of stochastic jump processes, building on Fernique's results on the compactness of their distributions.

Contribution

It introduces new cadlaguity moment estimates for jump processes, extending Fernique's compactness results to provide quantitative bounds.

Findings

Derived cadlaguity moment estimates for jump processes.

Extended Fernique's compactness results to stochastic processes with jumps.

Provided tools for analyzing the regularity of jump processes.

Abstract

Using the results of X. Fernique on the compactness of distributions of cadlag random functions, we derive some cadlaguity moment estimates for stochastic processes with jumps.

Equations

$$\mathbf{E}\left[{\left|X_{t_1}-X_{t_2}\right|}^p\wedge {\left|X_{t_2}-X_{t_3}\right|}^p\right]\le C{\left|t_1-t_3\right|}^{1+r}$$

$$\mathbf{E}\left[{\left|X_{t_1}-X_{t_2}\right|}^p\right]\le C{\left|t_1-t_2\right|}^{1+r}$$

$$\Delta \left(f;(\sigma ,\tau )\right)=\underset{\sigma \le s\le t\le u\le \tau }{\sup }d(f\left(s\right),f\left(t\right))\wedge d(f\left(t\right),f\left(u\right)).$$

$${\left[f\right]}_{\mu }+|f{]}_{\mu }+[f{|}_{\mu }<\infty ,$$

$${\left[f\right]}_{\mu }$$

$$|f{]}_{\mu }$$

$${\left[\left[f\right]\right]}_{\mu ,p}$$

$$||f]{]}_{\mu ,p}$$

$${\left[f\right]}_{\mu }+|f{]}_{\mu }+[f{|}_{\mu }\le C\left([\left[f\right]{]}_{\mu ,p}+||f]{]}_{\mu ,p}+[[f|{|}_{\mu ,p}\right).$$

$$\underset{0\le t\le 1}{\sup }\left|f\left(t\right)\right|\le C\left({\left|f\right|}_{L_p\left(\left[0,1\right]\right)}+[\left[f\right]{]}_{\mu ,p}+||f]{]}_{\mu ,p}+[[f|{|}_{\mu ,p}\right).$$

$${\left[\left[f\right]\right]}_{\mu ,p}^p$$

$$\mathbf{E}\left[\Delta {\left(X;s,t,u\right)}^p\right]$$

$$\mathbf{E}\left[d{\left(X\left(1\right),X\left(t\right)\right)}^p\right]$$

$$\mathbf{E}\left[d{\left(X\left(0\right),X\left(t\right)\right)}^p\right]$$

$$\mathbf{E}\left({\left[\left[X\right]\right]}_{\mu ,p}^p+||X]{]}_{\mu ,p}^p+[[X|{|}_{\mu ,p}^p\right)\le N{C}_0.$$

$$\mathbf{E}\left[\underset{0\le t\le 1}{\sup }{\left|X_t\right|}^p\right]\le N[C_0+\mathbf{E}{\int }_0^1{\left|X_t\right|}^{p}dt].$$

$$\mathbf{E}\left({\left[\left[X\right]\right]}_{\mu ,p}^p\right)\le C_0\int {\int }_{s<u}\frac{{\left|u-s\right|}^{2+r}}{{\left|u-s\right|}^{\mu p+3}}dsdu\le N{C}_0{\int }_0^1{u}^{r-\mu p}du.$$

$$\mathbf{E}\left[{\left|X_t-X_s\right|}^p\wedge {\left|X_t-X_u\right|}^p\right]$$

$$\mathbf{E}\left[{\left|X_0-X_t\right|}^p\right]$$

$$\mathbf{E}\left[{\left|X_1-X_t\right|}^p\right]$$

$$X_t^n=X_{{\pi }_n\left(t\right)},t\in \left[0,1\right],$$

$$\underset{n}{\sup }\mathbf{E}\left[{\left[\left[X^n\right]\right]}_{\mu ,p}^p+||X^n]{]}_{\mu ,p}^p+[[X^n|{|}_{\mu ,p}^p\right]<\infty .$$

$$\mathbf{E}\left[{\int }_0^1\frac{{\left|X_0^n-X_t^n\right|}^p}{t^{\mu p+1}}dt\right]$$

$$\mathbf{E}\left[\int \frac{{\left|X_1^n-X_t^n\right|}^p}{{\left|1-t\right|}^{\mu p+1}}dt\right]\le C,n\ge 1.$$

$$\mathbf{E}\int \int {\int }_{s<t<u}\frac{{\left|X_t^n-X_s^n\right|}^p\wedge {\left|X_t^n-X_u^n\right|}^p}{{\left|u-s\right|}^{\mu p+3}}dsdtdu$$

$$\sum _{i=0}^{2^n-3}\sum _{j=i+1}^{2^n-2}\sum _{k=j+1}^{2^n-1}{\int }_{\frac{k}{2^n}}^{\frac{k+1}{2^n}}{\int }_{\frac{j}{2^n}}^{\frac{j+1}{2^n}}{\int }_{\frac{i}{2^n}}^{\frac{i+1}{2^n}}\frac{{\left(\frac{k}{2^n}-\frac{i}{2^n}\right)}^{1+r}}{{\left(u-s\right)}^{\mu p+3}}dsdtdu$$

$$\mathbf{E}\int \int {\int }_{s<t<u}\frac{{\left|X_t^n-X_s^n\right|}^p\wedge {\left|X_t^n-X_u^n\right|}^p}{{\left|u-s\right|}^{\mu p+3}}dsdtdu\le C,n\ge 1.$$

$$N\left(f;\left(\sigma ,\tau \right)\right)=\underset{\sigma <\theta \le \tau }{\inf }\underset{s\in [\sigma ,\theta ),u\in \left[\theta ,\tau \right]}{\sup }[d\left(f(\sigma ),f(s)\right)\vee d\left(f(u),f\left(\tau \right)\right)].$$

$$N\left(f;\eta \right)=\underset{\sigma <\tau \le \sigma +\eta }{\sup }N\left(f;\left(\sigma ,\tau \right)\right)=\underset{0<\tau -\sigma \le \eta }{\sup }N\left(f;\left(\sigma ,\tau \right)\right).$$

$$\frac{1}{2}N\left(f;\left(\sigma ,\tau \right)\right)\le \Delta \left(f;\left(\sigma ,\tau \right)\right)\le 2N\left(f;\left(\sigma ,\tau \right)\right).$$

$$\Delta \left(f;\left(0,1\right)\right)=\underset{0\le s\le t\le u\le 1}{\sup }d(f\left(s\right),f\left(t\right))\wedge d(f\left(t\right),f\left(u\right))\le 2N\left(f;\left(0,1\right)\right).$$

$$N\left(f;\left(s,t\right)\right)\le 2N\left(f;\left(\sigma ,\tau \right)\right),\Delta \left(f;(s,t)\right)\le \Delta \left(f;(\sigma ,\tau )\right).$$

$${\left[f\right]}_{~\mu }:=\underset{\eta >0}{\sup }\frac{N\left(f;\eta \right)}{{\eta }^{\mu }},f\in D^{\mu }\left(\left[0,1\right]\right).$$

Full text

1. Introduction

As suggested by Kolmogorov, it was proved in [2] (1956) that if $X_{t},~{}0\leq t\leq 1,$ is a separable real valued process (see [5]) such that

[TABLE]

with $r>0,p>0,$ and $C$ independent of $t$, then $X$ has no discontinuities of the second kind with probability 1. If

[TABLE]

is assumed instead of of (1.1), then $X$ paths are Hölder continuous (Kolmogorov, 1934). It can be shown (e.g. [8], [11]), that under (1.2), the Hölder continuity is a consequence of the well-known Sobolev embedding theorem. In that case, $X$ Hölder norm moment estimates can be derived. In this note, we estimate the moments of time supremum and càdlàg Hölder coefficient of $X$ in terms of some integrated time differences of $X$ from which, using assumption (1.1), we can derive the classical claim about the existence of a càdlàg modification of $X$. On the other hand, the estimate obtained could be helpful in the construction of the solutions to SPDEs driven by jump processes when the method of characteristics with a time reversal is used (see [4]). Some different type moment estimates were derived in [10] by imposing assumptions on the cumulative distribution function of the quantities introduced in [3] (see [7], Section 4 of Chapter III, as well).

Our note is organized as follows. In Section 2, we introduce some notation and state the main claim. Some auxiliary results are presented in Section 3, and the main theorem is proved in Section 4.

2. Notation and main result

Let $E$ be a Polish space with distance $d$ and $D\left(\left[0,1\right],E\right)$ be the standard space of $E$-valued càdlàg functions on $\left[0,1\right].$ For $0\leq s\leq t\leq u\leq 1$, denote $\Delta\left(f;s,t,u\right)=d(f\left(s\right),f\left(t\right))\wedge d(f\left(t\right),f\left(u\right)).$ For $\sigma<\tau$, let us introduce the standard modulus of càdlàguity

[TABLE]

For $\mu\in\left(0,1\right)$, we define $\mu$-Hölder càdlàg function space $D^{\mu}\left([0,1],E\right)$ to be the set of of all $f\in D(\left[0,1\right],E)$ such that

[TABLE]

where

[TABLE]

For $\mu\in\left(0,1\right),p>1,f\in D\left(\left[0,1\right],E\right),$ let

[TABLE]

Our main result is the following estimate.

Theorem 1**.**

Let $p>1,~{}\mu\in(0,1)$. There is $C>0$ such that for any $f\in D^{\mu}([0,1],E)$

[TABLE]

Moreover, if $E=\mathbf{R}^{d},d\left(x,y\right)=\left|x-y\right|,x,y\in\mathbf{R}^{d}$, then

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Remark 1**.**

Obviously, for any $E$-valued measurable function $f$ on $\left[0,1\right],$

[TABLE]

Corollary 1**.**

Let $\left(\Omega,\mathcal{F},\mathbf{P}\right)$ be a probability space and $X:\left[0,1\right]\times\Omega\rightarrow E$ be a measurable function, $p>1,r>0$. Assume that

[TABLE]

for some $C_{0}>0$. Then for each $\mu\in\left(0,1\right),\mu<r/p,$ there is a constant $N=N\left(\mu,r,p\right)$ so that

[TABLE]

If $E=\mathbf{R}^{k},d\left(x,y\right)=\left|x-y\right|,x,y\in\mathbf{R}^{k}$, and  $X_{\cdot}\in D^{\mu}\left(\left[0.1\right],E\right)$ a.s. with $\mu\in\left(0,1\right),\mu<r/p,$ then, in addition,

[TABLE]

Proof.

Let $r-\mu p>0$. Then

[TABLE]

Similarly, the other terms can be estimated.

If $E=\mathbf{R}^{k},d\left(x,y\right)=\left|x-y\right|,x,y\in\mathbf{R}^{k}$, and $X_{\cdot}\in D^{\mu}\left(\left[0.1\right],E\right)$ a.s. with $\mu\in\left(0,1\right),\mu<r/p,$ then the last estimate obviously follows by Theorem 1 and (2.1).

Corollary 2**.**

Let $X_{t},t\in\left[0,1\right],$ be a real valued and stochastically continuous process. Assume that

[TABLE]

for all $0\leq s<t<u\leq 1,$ and some $r>0,p>1$. Then $X$ has a càdlàg modification.

Proof.

Let

[TABLE]

where $\pi_{n}\left(t\right)=k/2^{n}$ if $k/2^{n}\leq s<(k+1)/2^{n}$, $n=1,2,\ldots,k=0,\ldots,2^{n}-1$, and $\pi_{n}\left(1\right)=1-1/2^{n}$. Obviously the sequence $X_{\cdot}^{n}\in D^{\beta}\left(\left[0,1\right]\right)$ a.s. for any $\beta\in\left(0,1\right)$. Let $\mu\in\left(0,1\right),\mu p<r$. It is enough to show that

[TABLE]

Indeed, every $X_{n}$ induces a probability measure $X^{n}\left(\mathbf{P}\right)$ on $D\left(\left[0,1\right]\right)$. The estimate (2.2) implies that the sequence of measures $\left\{X^{n}\left(\mathbf{P}\right),n\geq 1\right\}$ is weakly relatively compact (see [7], [9]). Any weak limit of a weakly converging subsequence $X^{n_{k}}$ has càdlàg paths with probability 1 and, obviously, the same finite-dimensional distributions as $X$. Therefore $X$ has a càdlàg modification according to Lemma 2.24 in [9].

In order to show (2.2), we estimate, using assumptions imposed,

[TABLE]

Similarly,

[TABLE]

In the same vein,

[TABLE]

Note that $r>\mu p$ and for every set of $\{n,i,j,k\}$, $1/2^{n}\leq\left(k-i-1\right)/2^{n}\leq u-s\leq\left(k-i+1\right)/2^{n}$. Hence, $\left(k-i\right)/2^{n}\leq 2\left(u-s\right)$, and

[TABLE]

which shows that

[TABLE]

Thus (2.2) follows, and the statement is proved.

3. Auxiliary results

Following [6], for $0\leq\sigma<\tau\leq 1$, we introduce another modulus of càdlàguity

[TABLE]

Denote for $\eta>0,$

[TABLE]

Clearly, $N\left(f;\eta\right)$ is increasing in $\eta$.

Remark 2**.**

According to Lemma 1.0 in [6],

(a) For any $\sigma<\tau,$

[TABLE]

In particular,

[TABLE]

(b) For $\left(s,t\right)\subseteq\left(\sigma,\tau\right),$

[TABLE]

For $\mu\in\left(0,1\right)$, define

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Remark 3**.**

Obviously,

[TABLE]

and

[TABLE]

also,

[TABLE]

We show that $\left[f\right]_{\mu}$ and $\left[f\right]_{\char 126\relax\mu}$ are equivalent.

Lemma 1**.**

Let $\mu\in(0,1)$. For any $f\in D^{\mu}\left([0,1],E\right)$,

[TABLE]

Proof.

Since for each $\sigma\leq r\leq\tau,\tau\leq\sigma+\eta,$

[TABLE]

it follows that

[TABLE]

and, using Remark 3,

[TABLE]

Hence for any $\eta>0,$ by Remark 2(a),

[TABLE]

and by (3.2),

[TABLE]

The following key estimate was pointed out in [6], Lemma 1.2.4, as an extraction from Theorem 12.5 in [1] (cf. inequality 12.76 in [1]). For the sake of completeness we provide its proof.

Lemma 2**.**

(Lemma 1.2.4 in [6]) For any $f\in D\left(0.1],E\right)$ and every triplet $0\leq\sigma<t<\tau\leq 1$,

[TABLE]

Proof.

Let $0\leq\sigma<t<\tau\leq 1$. By the definition of $N$, for each $\varepsilon\in\left(0,1\right)$, there exist $\sigma<\theta_{1}\leq t<\theta_{2}\leq\tau$ such that

[TABLE]

We split the proof in two cases: $d\left(f\left(\sigma\right),f\left(t\right)\right)\leq d\left(f\left(t\right),f\left(\tau\right)\right)$ and $d\left(f\left(\sigma\right),f\left(t\right)\right)>d\left(f\left(t\right),f\left(\tau\right)\right)$ respectively.

Case 1**: ** $d\left(f\left(\sigma\right),f\left(t\right)\right)\leq d\left(f\left(t\right),f\left(\tau\right)\right)$, i.e. $\Delta\left(f;\sigma,t,\tau\right)=d\left(f\left(\sigma\right),f\left(t\right)\right)$. Obviously,

[TABLE]

and

[TABLE]

Now,

[TABLE]

and

[TABLE]

Hence

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Using (3.4)-(3.6), we have

[TABLE]

Case 2**: ** $d\left(f\left(\sigma\right),f\left(t\right)\right)>d\left(f\left(t\right),f\left(\tau\right)\right)$, i.e. $\Delta\left(f;\sigma,t,\tau\right)=d\left(f\left(t\right),f\left(\tau\right)\right)$. Obviously,

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and

[TABLE]

Now,

[TABLE]

Therefore, again,

[TABLE]

Since $\varepsilon\in\left(0,1\right)$ is arbitrary,

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For $\mu\in\left(0,1\right),$ define for $f\in D\left(\left[0,1\right],E\right),$

[TABLE]

We will need the following equivalence claim.

Lemma 3**.**

Let $\mu\in\left(0,1\right)$. For any $f\in D^{\mu}\left(\left[0,1\right],E\right),$

[TABLE]

Proof.

Let $K=\left[f\right]_{\char 126\relax\mu}$. According to Remark 3, for any $a>0,$

[TABLE]

By (3.3), for every $\sigma<\tau\leq 1$ such that $\left|\tau-\sigma\right|\leq a,$ we have, by Lemma 2,

[TABLE]

Hence

[TABLE]

and

[TABLE]

Taking $\sup$ in $a>0$ on both sides, we see that

[TABLE]

or

[TABLE]

The statement follows by Lemma 1.

4. Proof of Theorem 1

First we show that there is $C=C\left(\mu,p\right)$ so that with any $\delta\in\left(0,1\right),$

[TABLE]

Assume $0<t\leq 3/4$. Taking $\varepsilon<\frac{1}{4}t$, we have for $\tau^{\prime}\in\left(t-\varepsilon,t\right),\tau^{\prime\prime}\in\left(t,t+\varepsilon\right)$,

[TABLE]

Integrating with respect to $\tau^{\prime},\tau^{\prime\prime}$ over $\bar{Q}=\left[t-\varepsilon,t\right]\times[t,t+\varepsilon],$

[TABLE]

Now

[TABLE]

Taking $\varepsilon=\frac{1}{4}\delta t$ with any $\delta\in\left(0,1\right),$ we have

[TABLE]

By Hölder inequality, $\kappa=\mu+1/p,1/p+1/q=1,$

[TABLE]

Taking $\varepsilon=\frac{1}{4}\delta t$, with any $\delta\in\left(0,1\right),$

[TABLE]

and, the same way,

[TABLE]

The inequality (4.1) follows from (4)-(4.5).

Similarly, with obvious changes, we prove that

[TABLE]

for some $C=C\left(\mu,p\right)$ with any $\delta\in\left(0,1\right).$

Finally, (4.1), (4.6) imply that there is $C=C\left(\mu,p\right)$ so that with any $\delta\in\left(0,1\right),$

[TABLE]

Now,

[TABLE]

Hence, by (4.1), (4.6) and (4), there is $C=C\left(\mu,p\right)$ so that for all $\delta\in\left(0,1\right),t\in\left(0,1\right),$

[TABLE]

Now we estimate $\Delta\left(f;s,t,u\right)$ with $0\leq s<t<u$ $\leq 1$ and $t=\frac{s+u}{2}$.

(i) Assume $\left|s\right|>\frac{1}{4}\left|u-s\right|$ and $\left|1-u\right|>\frac{1}{4}\left|u-s\right|.$

Let $\varepsilon<\frac{1}{4}\left(u-s\right),s^{\prime}\in\left(s-\varepsilon,s\right),s^{\prime\prime}\in\left(s,s+\varepsilon\right),t^{\prime}\in\left(t-\varepsilon,t\right),t^{\prime\prime}\in\left(t,t+\varepsilon\right),u^{\prime}\in\left(u-\varepsilon,u\right),u^{\prime\prime}\in\left(u,u+\varepsilon\right),$ and

[TABLE]

Let $Q=\left(s-\varepsilon,s\right)\times\left(s,s+\varepsilon\right)\times\left(t-\varepsilon,t\right)\times\left(t,t+\varepsilon\right)\times\left(u-\varepsilon,u\right)\times\left(u,u+\varepsilon\right)$. Then $\left|Q\right|=\varepsilon^{6},$ and

[TABLE]

Let

[TABLE]

and

[TABLE]

Integrating (4.9) over $Q,$

[TABLE]

Now,

[TABLE]

Similarly we estimate the other two terms in $\tilde{A}$ and see that

[TABLE]

Using Hölder inequality, $1/q+1/p=1,p>1,$ with $\kappa=\mu+3/p,$

[TABLE]

Since

[TABLE]

we have

[TABLE]

Similarly estimating the other terms in $\tilde{B}$, we get

[TABLE]

Hence by (4.10) and (4.11), for some $C=C\left(\mu,p\right),$

[TABLE]

if $\left|s\right|>\frac{1}{4}\left|u-s\right|$ and$\left|1-u\right|>\frac{1}{4}\left|u-s\right|$. Taking $\varepsilon=\frac{1}{4}\delta\left(u-s\right)$ with any $\delta\in\left(0,1\right),$ we have

[TABLE]

for some $C=C\left(\mu,p\right)$ if $\left|s\right|>\frac{1}{4}\left|u-s\right|$ and$\left|1-u\right|>\frac{1}{4}\left|u-s\right|$.

(ii) Assume $\left|s\right|\leq\frac{1}{4}\left|u-s\right|$ or $\left|1-u\right|\leq\frac{1}{4}\left|u-s\right|.$

If $s\leq\frac{1}{4}\left|u-s\right|$ (recall $t=\frac{s+u}{2}$), then $s\leq 1/4$ and

[TABLE]

By (4.1), there is $C=c\left(\mu,p\right)$ so that for any $\delta\in\left(0,1\right)\,,s\leq\frac{1}{4}\left|u-s\right|,$

[TABLE]

If $1-u\leq\frac{1}{4}\left|u-s\right|$, then $u\geq\frac{3}{4}$ and $t\geq 1/4$ because

[TABLE]

By (4.6), there is $C=C\left(\mu,p\right)$ so that for any $\delta\in\left(0,1\right)\,,1-u\leq\frac{1}{4}\left|u-s\right|,$ we have

[TABLE]

Hence

[TABLE]

if $\left|s\right|\leq\frac{1}{4}\left|u-s\right|$ or $\left|1-u\right|\leq\frac{1}{4}\left|u-s\right|.$

According to (4.13) and (4.14), there is $C=C\left(\mu,p\right)$ so that

[TABLE]

for any $0\leq s<t<u\leq 1,t=\left(s+u\right)/2.$ Hence for all $\delta\in\left(0,1\right),$

[TABLE]

for some $C=C\left(\mu,p\right)$. Then by Lemma 3 and (4.8), (4.15), there is $C_{1}=C_{1}\left(\mu,p\right)$ so that for all $\delta\in\left(0,1\right)$ we have

[TABLE]

Choosing $\delta$ so that $\delta^{\mu}C_{1}\leq 1/2$ we see that for some $C=C\left(\mu,p\right),$

[TABLE]

If $E=\mathbf{R}^{k},d\left(x,y\right)=\left|x-y\right|,x,y\in\mathbf{R}^{k}$, then we can estimate the supremum of $f$. For each $t,$

[TABLE]

Hence

[TABLE]

and

[TABLE]

The claim of Theorem 1 follows.