# Kolmogorov tightness criterion for stochastic processes

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!

• The body of a question should not depend on the title ("the criterion stated above"). Commented Jun 2, 2020 at 22:55
• 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.) Commented Jun 2, 2020 at 22:56

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 $$C([0,1])$$.

• 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? Commented May 13, 2020 at 14:26
• (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 $C^\gamma$. Commented May 13, 2020 at 14:31
• (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 $\|\cdot\|_\gamma^\ast = \|\cdot\|_\infty + [\cdot]_\gamma$. Commented May 13, 2020 at 14:37
• Thank you very much, I think I got it! Commented May 13, 2020 at 14:54
• 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)? Commented May 13, 2020 at 15:02