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We know about Kolmogorov Criterion for the tightness of a stochastic process $X_n(t)$

1.The sequence $(X_{n}(0))_{n\geq0}$ is tight.

2.There exist constants $\gamma\geq0$,$\alpha>1$, $K>0$ and an integer $n_0$ such that $$E(|X_{n}(t_{2})-X_{n}(t_{1})|^{\gamma})\leq K|t_{2}-t_{1}|^{\alpha}, \forall n \geq n_0$$ for all $t_{1},t_{2}$.

My first question: what should the $n_0$ depend? Could it depend on the $t_{1}$ and $t_{2}$?

My second question: Is there any other criterion for tightness with the parameter $\alpha=1$ for the version of the moment condition?

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  • $\begingroup$ What space is your stochastic process in? If it's say C[0,1], then the result holds for all n and all t1,t2 in [0,1] $\endgroup$
    – Alex R.
    Commented Sep 17, 2010 at 20:15
  • $\begingroup$ Yes it is in space C[0,1], thanks for your help $\endgroup$
    – syh2010
    Commented Sep 20, 2010 at 6:54

1 Answer 1

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$n_0$ must be independent of $t_1$ and $t_2$, of course. If it's not, the processes might be even discontinuous. For instance, $X_n$ is a Poisson process with parameter $1/n$. Then $$E(|X_n(t_1)-X_n(t_2)|^2)\le |t_1-t_2|^2$$ for all $n>|t_1-t_2|^{-1}$ (for all $n\ge 1$ if $t_1=t_2$).

And the same answer works for the second question: when $\alpha=1$, the processes need not to be continuous. In some special cases, where you have higher moments controlled by a lower one polynomially, it may help (e.g. in the Gaussian case $\gamma=2$ and $\alpha=1$ is enough).

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