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Unfortunately I've trouble to see the following: If you work with stochastic process $X$ you often want to approximate this in the following sense, define:

$$ X^{(n)}(s,\omega) = X\left(\frac{(k+1)t}{2^n},\omega\right)\cdot \mathbf1\left\{s\in \left(\frac{kt}{2^n},\frac{(k+1)t}{2^n}\right]\right\} $$

for $t>0,n\ge 1, k=0,1,\dots, 2^n-1$ and $\mathbf1$ is the charateristic function of a set.

Now suppose $X$ has continuous trajectories. I don't see why

$$ \sum_{k=0}^{2^n-1} X\left(\frac{(k+1)t}{2^n},\omega\right)\cdot \mathbf1\left\{s\in \left(\frac{kt}{2^n},\frac{(k+1)t}{2^n}\right]\right\} \to X_s(\omega)$$

for all $(s,\omega) \in [0,t]\times\Omega$ as $n\to \infty$. Even more it's enough that the paths of $X$ are right continuous. I'm very thankful for your answer. This question comes up while reading Brownian Motion and Stochastic Calculus (Karatszas, Shreve).


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Usually these results are stated with an a.s. (almost sure) precision rather than "for all $\omega$" – Ilya Feb 29 '12 at 11:28
up vote 2 down vote accepted

Well, this problem is rather from analysis since the only probabilistic fact is used here is that for a.a. $\omega\in \Omega$ the function $s\mapsto X(s,\omega)$ is right-continuous in $s$. For the simplicity of notation, let us put the function $f:[0,t]\to \mathbb R$ to be right-continuous: $$ \lim\limits_{h\downarrow 0}f(s+h) = f(s)\quad\text{ for all }s\in[0,t) $$ and define $$ \begin{cases} t_{k,n} &= \frac{kt}{2^n}, \\ \\ f_{k,n} &= f\left(t_{k+1,n}\right), \\ \\ I_{k,n}(s) &= 1_{(t_{k,n},t_{k+1,n}]}(s) \end{cases} $$ for $n\in\mathbb N $ and $k = 0,1,2,\dots,2^{n}-1$. Let us show that for any $s\in [0,t)$ we have $$ \lim\limits_{n\to\infty}\sum\limits_{k=0}^{2^n-1}f_{k,n}I_{k,n}(s) = f(s). $$ Fix $s$ and let $\hat k(n) = \{k:I_{k,n}(s) = 1\}$ be the index of the interval of the partition where $s$ falls. Then: $$ \lim\limits_{n\to\infty}\sum\limits_{k=0}^{2^n-1}f_{k,n}I_{k,n}(s) = f_{\hat k(n),n} = f\left(t_{\hat k(n)+1}\right) $$ and $s<t_{\hat k(n)+1}$ toghether with the fact that $t_{\hat k(n)+1}-s\leq 2^{-n}t$ imply that $t_{\hat k(n)+1}\downarrow s$ with $n\to\infty$. As a result, $\lim\limits_{n\to\infty}f\left(t_{\hat k(n)+1}\right) = f(s)$ by right-continuiuty. Finally, if $s =t$ then $$ \lim\limits_{n\to\infty}\sum\limits_{k=0}^{2^n-1}f_{k,n}I_{k,n}(t) = f(t) $$ and there is nothing to prove.

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@ Ilya: Thanks a lot for your help! – user20869 Feb 29 '12 at 12:19
@hulik: you're welcome. In fact, the same argument could be applied to any sequence of partitions maximal length of the interval of which converges to zero. You can also tune this approach and realize the approximation for the left-continuous processes. – Ilya Feb 29 '12 at 12:40
@ Ilya: If I want to approximate a left-continuous process, then I would do the following: $I_{n,k}(s)=\mathbf1_{[t_{k,n},t_{k+1,n})}(s)$, $t_{k,n}$ as above and $f_{n,k}=f(t_{k,n})$ is this right? – user20869 Feb 29 '12 at 12:50
@hulik: it is, you've got it – Ilya Feb 29 '12 at 12:53
@ Ilya: Thank you again for your patience – user20869 Feb 29 '12 at 12:59

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