Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

One of the definitions of stochastic continuity is: $\forall a > 0$ and $\forall s \geq 0$ $$\lim_{t\rightarrow s}\;P(|X(t)-X(s)|>a) = 0.$$ What does it mean intuitively? I know that it implies the jumps but am not sure how and why.

share|improve this question
    
First try to intuitively understand stochastic convergence: en.wikipedia.org/wiki/… –  Rasmus Oct 11 '10 at 13:36
add comment

1 Answer

$a$ is a kind of 'tolerance' parameter on $X$. Imagine an adversary of yours setting this parameter at some level, as small as they wish. They also set an upper bound, say $\epsilon$, on the proportion of possible sample paths that may violate the tolerance condition.

You on the other hand are allowed to choose a time window around the time $s$.

Continuity then requires that regardless of how tight the two tolerances are set by your adversary, you can always choose a time window narrow enough that within that time window the values of $X$ remain within a distance $a$ of $X(s)$ for no less than $(1-\epsilon)$ proportion of the possible sample paths.

The tricky part here is that you are not required to commit to the sample paths which will violate the tolerance. Depending on the choice of $a$ and $\epsilon$ you may allow different sample paths to 'fail'. All that is required is that the proportion of paths that fail be less than $\epsilon$. This makes your task easier. Therefore this type of continuity is know as weak continuity. On the other hand we could have required that almost all sample paths be continuous in the ordinary deterministic sense:

$$P\left(\lim_{t\to s} X(t)=X(s)\right)=1$$

which would have been a stronger condition to fulfil.

As Rasmus points out, these different notions of continuity are a direct counterpart of the more fundamental notions of convergence of random variables.

['Proportion' above should be interpreted in terms of probability mass.]

share|improve this answer
    
Very clearly said!--much better than most of the attempts I have seen to explain this in the statistical literature. –  whuber Oct 11 '10 at 15:14
add comment

Your Answer

 
discard

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.