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What is the interpretation of

$$ \lim_{a \to \infty} \sup_n P \Big( \sup_t |X_t^n| \geq a \Big) = 0$$

where $\{X^n_t\}_n$ is a sequence of stochastic processes?

Could we say that the paths $X^n$ almost never "explode"?

harisf
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2 Answers2

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Let $Y_n:=\sup_t|X_t^n|$. Then $\{Y_n\}_{n\ge 1}$ is a sequence of random elements. The condition can be written as $$ \lim_{a\to\infty}\sup_{n\ge 1}\mathsf{P}(Y_n\ge a)=0. $$ This is the definition of the (uniform) tightness of a stochastic process. It is equivalent to saying that for each $\epsilon>0$, there exists $K_{\epsilon}<\infty$ (if each $X_t^n$ is real valued) s.t. $\mathsf{P}(Y_n> K_{\epsilon})<\epsilon$ for each $n\ge 1$, i.e., $$ Y_n=O_p(1). $$

  • When you write "...this is the definition of tightness of a stochastic process" are you referring to the (discrete-time) process $Y_n$ (indexed by $n$)? Or is tightness referring to the each of the processes $X_t^n$ in the sequence indexed by $n$? Do you have a reference for tightness of stochastic processes; I only managed to find definitions for measures. – harisf Oct 07 '21 at 06:54
  • @harisf (1) I'm referring to ${Y_n}$. (2) ${Y_n}$ is tight if the corresponding distributions form a tight family of measures. –  Oct 07 '21 at 07:38
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This condition certainly implies that the paths almost never explode, but is stronger than that. I would interpret this condition as "uniform boundedness in probability": There exists $K \in \mathbb{R}$ such that $|X_t^n| \le K$ for all $n \in \mathbb{N}, t \ge 0$ with high probability.

user6247850
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