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suppose that X(t) is a s.p. on T with $EX(t)^2<+\infty$. we give two kinds of continuity of X(t).

  1. X(t) is continuous a.s.
  2. X(t) is m.s. continuous, i.e. $\lim\limits_{\triangle t \rightarrow 0}E(X(t+\triangle t)-X(t))^2=0$.

Then, what's the relationship between these two kinds of continuity.

2 Answers 2

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I don't know if there is a clear relation between both concepts.

For example if you take the Brownian Motion it satisfies 1 and 2 but if you take a Poisson process then it only satisfies 2 (although it satisfies a weaker form of condition 1 which is continuity in probability).

The question is what do you want to do with those processes ?

Regards

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    @Jim : I think that it is true that 1 implies 2 under existence of second moment and an additionnal domination assumption. Take $t_n \to t$ then $Y_n=(X_{t_n}-X_t)^2$ is positive r.v., if the sequence $X_{t_n}$ is bounded by an $L^1$ variable $Z$ (and so $Y_n$) then dominated convergence theorem shows implication 1=>2 under the condition that domination is true for all such sequence converging to $t$. So to exhibit a counter-example you should find a process such that this cannot be done but I'm not sure how to do that or if it is possible, also maybe the condition might be weakened. Regards2011-10-20
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For an example which is a.s. but not m.s. continuous, take your time interval to be $[0, \infty]$, and let $X_t$ be a standard one-dimensional Brownian motion started at 0 and stopped the first time it reaches 1. (That is, if $B_t$ is a standard Brownian motion, take $T = \inf\{t > 0 : B_t = 1\}$ and $X_t = B_{t \wedge T}$.) Since Brownian motion is recurrent, we have $X_t \to 1$ a.s. as $t \to \infty$, and so by setting $X_{\infty} = 1$ we get an a.s. continuous stochastic process on $[0,\infty]$.

However, $X_t$ is not m.s. continuous. If it were, then by Cauchy-Schwarz we would have $E[X_t] \to E[X_\infty] = 1$ as $t \to \infty$. But $X_t$ is a martingale and so $E[X_t] = 0$ for all $t \in [0,\infty)$.

If you don't like using $[0,\infty]$ as your time interval, then apply a time change: let $Y_t = \begin{cases} X_{t/(1-t)}, & t < 1 \\ 1, & t \ge 1.\end{cases}$ Now $Y_t$ is a.s. continuous but $E[Y_t] = 0$ for $t < 1$, $E[Y_t] = 1$ for $t \ge 1$. Note that $Y_t$ is a standard example of a local martingale which is not a martingale.