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I am searching for the criterion stated above and also here: The question about Kolmogorov tightness criterion.

It should state the following: If a sequence of stochastic processes $(X^n)$ fulfills:

$$\mathbb{E}[|X^n_t-X^n_{t'}|^p]\leq C|t-t'|^\alpha$$

then it is tight.

I don't know if my googling skills are just too bad, but I can't find any source for that. Thanks for any advice!

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    $\begingroup$ The body of a question should not depend on the title ("the criterion stated above"). $\endgroup$
    – LSpice
    Jun 2, 2020 at 22:55
  • $\begingroup$ Also, @Glorfindel, I think that blank lines around displayed formulas should be discouraged. Markdown regards them as paragraph breaks, which they aren't semantically. That is, inline $$display$$ inline and inline␤$$display$$␤inline are both fine (and display as probably intended), but inline␤␤$$display$$␤␤inline, while it seems to display OK, is semantically wrong. (I mention this semantic issue because the original edit in that part of the post seems to have been semantic, too.) $\endgroup$
    – LSpice
    Jun 2, 2020 at 22:56

1 Answer 1

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I assume that you want to show tightness in a space such as $C([0,1])$, as was the case in the question you link to. In fact, the approach of this answer will show tightness in $C^\beta([0,1])$ for every $\beta \in (0, \frac{\alpha - 1}{p})$.

Firstly, note that the statement you write cannot be sufficient for tightness since if $X^n$ is a sequence of constant processes then your condition trivially holds. Such a sequence need not be tight. The extra condition in your linked question that $(X_0^n)_{n \geq 1}$ is a tight sequence in $\mathbb{R}$ prevents such counterexamples.

The key point is that from the proof of Kolmogorov's Continuity Criterion one can derive control on Holder norms of your process. For $\gamma \in (0, \frac{\alpha - 1}{p})$, one has the bound $$\mathbb{E}([X^n]_\gamma^p) \leq C(p, \alpha, \gamma) \cdot C$$ where $C(p,\alpha,\gamma)$ is a constant depending only on $p, \alpha$ and $\gamma$ (but is independent of $n$) and $[\cdot]_\gamma$ is the usual $\gamma$-Holder seminorm on $C^\gamma([0,1])$. See this answer for a proof.

Let $\|X\|_\gamma = |X_0| + [X]_\gamma$ denote (a norm equivalent to) the usual $\gamma$-Holder norm. Fix here $\varepsilon > 0$. By tightness of $(X_0^n)$ there is an $M_1$ such that $$\sup_n \mathbb{P}(|X_0^n| > M_1) \leq \varepsilon$$

Also, by Markov's inequality and our above control on the Holder seminorm, we have that for $M_2$ sufficiently large, $$\sup_n \mathbb{P}([X^n]_\gamma > M_2) \lesssim M_2^{-p} \leq \varepsilon.$$

Hence $$\sup_n\mathbb{P}(\|X^n\|_\gamma > M_1 + M_2) \leq \sup_n\mathbb{P}(|X_0^n| > M_1) + \sup_n \mathbb{P}([X^n]_\gamma > M_2) \lesssim \varepsilon.$$ Finally, by compactness of the embedding $C^\gamma([0,1]) \to C([0,1])$ the closed ball of radius $M_1 + M_2$ in $C^\gamma([0,1])$ is relatively compact in $C([0,1])$ so the above inequality yields tightness of your sequence in $C([0,1])$.

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  • $\begingroup$ Hi Rhys, first of all thank you very much for your helpful answer! Let me ask two questions to clarify my understanding of your answer: in the closing line, after the "Hence", you show that the sequence is tight in $C^\gamma([0,1])$, correct? And, why do you need that second norm $||\cdot ||_\gamma$? Why can't you just show tightness by using the usual $\gamma$-norm? $\endgroup$
    – max_muster
    May 13, 2020 at 14:26
  • $\begingroup$ (1/2) No, in the closing line I show that sequence is in a closed ball in $C^\gamma$ with high probability. This doesn't immediately show tightness since closed balls in $C^\gamma$ are not compact. That's why the next part where you use the compact embedding is necessary. Of course, one could replace $\gamma$ with $\gamma _ \varepsilon$ for $\varepsilon$ suitably small for the main body of the proof and then compactly embed $C^{\gamma+\varepsilon} \to C^\gamma$ to get tightness in $C^\gamma$. $\endgroup$ May 13, 2020 at 14:31
  • $\begingroup$ (2/2) For the second question, $[X]_\gamma := \sup_{s \neq t}\frac{|X_s - X_t|}{|s-t|^\gamma}$ does not give control on $\sup_t |X_t|$ so you can't hope to get nice properties of a closed ball for this seminorm under the embedding $C^\gamma \to C[0,1]$. $\|\cdot\|_\gamma$ is equivalent to the usual norm (not seminorm) on $C^\gamma$ and is more convenient to work with since tightness of $X_0^n$ gives us immediate control on the first term in $\|\cdot\|_\gamma$. This would take more a bit more work if we worked with the usual norm $\|\cdot\|_\gamma^\ast = \|\cdot\|_\infty + [\cdot]_\gamma$. $\endgroup$ May 13, 2020 at 14:37
  • $\begingroup$ Thank you very much, I think I got it! $\endgroup$
    – max_muster
    May 13, 2020 at 14:54
  • $\begingroup$ Sorry, one more question: would it simplify the argumentation, if I assume $\sup\limits_{n\geq 0}\sup\limits_{0\leq s\leq 1}\mathbb{E}[|X_s^n|^p]\leq c$ for some constant c (independent from n)? $\endgroup$
    – max_muster
    May 13, 2020 at 15:02

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