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Kolmogorov tightness criterion says that if $X_N$ is a sequence of continuous process with $X_N(0)=0$ and $E[[X_N(t)-X_N(s)|^p]\leq C_p |t-s|^{1+\beta}$ then for all $\gamma\in (0,\beta/p)$ we have that the laws of $X_N$ are tight on the Holder space $C^\gamma$.

There are some easy examples where the assumption is not necessary. First of all, if $X_N(t)=Z \psi(t)$ for all $N$ where $Z$ is a random variable with no moments and $\psi$ is as smooth as you want, then $P(\|X_N\|_\gamma>K)\to 0$ uniformly in $N$ as $K\to\infty$ and we have tightness on any Holder space.

I am wondering how we get around the lack of moments to prove tightness. For example, suppose that for all pairs $(s,t)$ we have $\lim_{K\to\infty} P(\{|X_N(t)-X_N(s)|>K|t-s|^\gamma\})=0$ uniformly in $N$, then can we conclude tightness on $C^{\gamma'}$ for any $\gamma'<\gamma$?

Generally how you prove KTC is through Garsia-Rodemich-Rumsey inequality but it is hard to handle without moments.

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    $\begingroup$ if you have some specific process $X_{N}$ in mind, we can try to figure it out together. That way we can avoid constructing some counterexample. $\endgroup$ Commented Jul 5, 2023 at 1:24
  • $\begingroup$ I don't think the condition that $\lim_{K\to\infty} P(|X_N(t)-X_N(s)|>K|t-s|^\gamma)=0$ uniformly in $N$ will be enough. I think you also need some uniformity in $s,t$; cf. the moment condition. $\endgroup$ Commented Jul 5, 2023 at 13:15
  • $\begingroup$ @IosifPinelis Very good point, we should also have this is uniform over all $s,t$. $\endgroup$
    – user479223
    Commented Jul 5, 2023 at 13:22

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Unfortunately, this does not hold, in the sense that you cannot conclude tightness even for any particular $\gamma’ < \gamma$. This very simple modification of my example here is a counterexample (I’ve just replaced the $\gamma$ in the exponent with $\gamma’$):

For any fixed positive integer $n \geq 22$, let $X^n$ be defined as follows - uniformly at random pick a point $p$ in, $[\frac{1}{4}, \frac{3}{4}]$, and define

$$ X^n := \begin{cases} 0 & \text{on } [0, p], \\ n(t - p)^{\gamma’} & \text{on } (p, p + \frac{1}{n^{2/\gamma’}}], \\ \frac{1}{n} & \text{on } (p + \frac{1}{n^{2/\gamma’}}, 1]. \\ \end{cases} $$

Then

$$\lim_{K\to\infty} P(\{|X_n(t)-X_n(s)|>K|t-s|^\gamma\})=0$$

indeed uniformly in $n, s, t$ but $\| X^n \|_{\gamma’} = n$ almost surely, which tends to $\infty$ and so tightness in Holder space is not possible.

I can do the calculation in full if needed, but morally the reason why this counterexample works is this:

The requested limit above can be shown to hold even if $X_n$ were the process that was discontinuous at $p$ with a jump of size $\frac{1}{n}$. Of course we have assumed $X_n$ continuous, but it follows that you could have $X_n$ continuous but make the bump essentially as bad as you want in $\gamma’$ Holder norm and still have that limit hold.

I second the suggestion of Thomas Kojar to try to look at the particular process and see if we can come up with some criterion that fits your purpose.

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