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Hello,

Suppose $(X_{n}(t))_{n\geq 1}$ is a sequence of real valued stochastic processes, and $T>0$ a fixed number.

Do we have the following implication ?

$\displaystyle{ \lim_{n \to \infty} \sup_{t\in[0,T]}} \mathbb{E}[|X_n(t)|] =0$ implies $\displaystyle{ \lim_{n \to \infty} \mathbb{E}[\sup_{t\in[0,T]}}|X_n(t)|] =0$

If not, what are the weakest conditions on $X_n(t)$ such that the above implication is true ?


Edit 2 : is the implication true if \begin{equation} \mathbb{E}\left[\displaystyle{\sup_{n>0}}\ |X_n(t+h)-X_n(t)|\right]\leq c(h) \end{equation}

with $\displaystyle{\lim_{h\to 0}}\ c(h)=0$


Edit 1 : is the implication true if \begin{equation} \displaystyle{\sup_{n>0}}\ \mathbb{E}\left[|X_n(t+h)-X_n(t)|\right]\leq c(h) \end{equation} with $\displaystyle{\lim_{h\to 0}}\ c(h)=0$. Proven false by Jeff Schenker (cf below).

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The implication is false: Let $X_n(t)$ be a stochastic process on $[0,1]$ with $X_n(t)=1$ for an interval $[(i-1)/2^n,i/2^n]$ where $i$ is chosen uniformly at random from the set $\{1,\ldots,2^n\}$. Then for any fixed $t$, $X_n(t)$ is 1 with probability $2^{-n}$ so that the left side is $2^{-n}$. On the other hand, the right side is 1. –  Anthony Quas Sep 28 '11 at 11:57
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Hi I wonder if for some (good) local martingales the implication can be proved using BDG inequlities. Regards –  The Bridge Sep 28 '11 at 12:05
    
@Anthony Quas : Thank you for this nice counter-example. Do you think the implication is still false if one assumes that $X_n(t)$ is continuous on $[0,T]$, with a modulus of continuity independent of $n$? –  user16215 Sep 28 '11 at 12:59
    
I don't know if it can helps, but I imagine your assumption with $c(h)$ implies that the family $\{X_n\}$ is tight for the uniform convergence topology, meaning the law of $X_n$ converges weakly for this topology to the law of some random continuous process $X$. –  kaleidoscop Sep 28 '11 at 14:07

1 Answer 1

This is false even with your edit. Here is a counter example with $T=1$.

Let $j$ be a random integer chosen uniformly from $\{0,\ldots,n-1\}$. Let $X_n(t)$ be a piecewise linear function on $[0,1]$ as follows:

  1. $X_n(t)=0$ if $t\not \in J_n$ where $J_n=[\frac{j}{n},\frac{j+1}{n}]$.
  2. If $t\in J_n$ then the graph of $X_n(t)$ has a "tent shape": it vanishes at each endpoint and increases linearly with slope $2n$ as we move toward the midpoint so that it takes value $1$ at the midpoint.

The resulting function $X_n(t)$ is piecewise linear on $[0,1]$, bounded by $1$ and the slope of any linear segment is bounded by $2n$. Then

  1. For each $t$, $\mathbb{E}[|X_n(t)|]\le \frac{1}{n}$ since $X_n(t)\neq 0$ only with probability $\frac{1}{n}$ and $0\le X_n(t)\le 1$.
  2. $\mathbb{E}[\sup_t |X_n(t)|]=1$ since $\sup_t |X_n(t)|=1$ for every outcome.
  3. $ \mathbb{E}[|X_n(t+h)-X_n(t)|]\le C h $ where $C$ is a constant independent of $n$, since $|X_n(t+h)-X_n(t)|\neq 0$ only with probability less than $\frac{c}{n}$ and is bounded by $2 n h$ for every outcome.
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@Jeff Schenker: thank you very much for your counter-example. I guess one needs to control the continuity of $X_n(t)$ in a stronger sense ? –  user16215 Sep 28 '11 at 15:57

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