MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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).

share|cite|improve this question
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
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

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.
share|cite|improve this answer
@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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.