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I am facing a bit weird issue here. I am going through Shreeve book on stochastic calculus and faced the following theorem, while proving $dWdt=0$.

$\sum_{j=0}^{n-1}(W(t_{j+1})-W(t_j))(t_{j+1}-t_j)$

$\le$ $\max_{0 \le k \lt n-1} |(W(t_{j+1})-W(t_j))| \cdot \sum_{j=0}^{n-1}(t_{j+1}-t_j)$

Now he argued that the because $W$ is continuous , the first term in the above equation goes to zero as the time partition goes to infinity. Now the doubt I am having is that,

$(W(t_{j+1})-W(t_j))=y \cdot \sqrt {t_{j+1}-t_j}$, where $y$ is a standard normal random variable. Then the maximum of this term can be anything from $-\infty$ to $\infty$. Now you may argue that as time partition goes to infinity, the time difference goes to zero (so the square root term goes to zero), but even in this case it can be $0 \cdot \infty$ scenario as the random variable has positive probability, which would again be undefined.

Then how the author argued that the maximum of difference in Brownian terms is zero! Please help.

Best Regards,

saz
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user3001408
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    Maybe this example can help: Consider the function $f(x)=e^x\sin(x)$. It is true that the jumps $|f(t_{i+1})-f(t_i)|$ for this function can be arbitrarily large, even if $|t_{i+1}-t_i|$ is small. But if we look at jumps for $t_i$'s in a fixed interval all the jumps can be made arbitrarily small. This is because a continuous function in a compact interval is uniformly continuous. – Nathanson Mar 21 '15 at 13:43

2 Answers2

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  1. Mind that for each fixed $\omega \in \Omega$, the mapping $ t\mapsto W_t(\omega)$ is continuous and therefore $$\sup_{s,t \in [0,T]} |W(t,\omega)-W(s,\omega)|$$ is always finite. Now, the author uses that the mapping is even uniformly continuous on compact intervals and therefore $$\sup_{s,t \in [0,T],|s-t| \leq \delta} |W(t,\omega)-W(s,\omega)| \to 0 \qquad \text{as} \, \, \delta \to 0.$$
  2. You mentioned the identity $$W(t_{j+1}) - W(t_j) = Y \sqrt{t_{j+1}-t_j}.$$ Mind that this equality holds in distribution. Since we are interested in the behavior for each fixed $\omega$ (i.e. in the path behavior of Brownian motion), this identity is of no use. (A similar problem: $W_t = \sqrt{t} W_1$ (in distribution) does not imply $W_t = \sqrt{t} W_1$ (almost surely).)
saz
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  • pretty clear explanation! Thanks! – user3001408 Mar 21 '15 at 14:38
  • hello @saz could you please answer my second question: http://math.stackexchange.com/questions/1200176/stcochastic-integral-and-ito-isometry --- as I am self studying the subject, I would greatly appreciate your kind help! – user3001408 Mar 22 '15 at 13:41
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The idea here is that you first fix an $\omega$ such that $t\mapsto W_t(\omega)$ is continuous on a compact interval hence uniformly continuous in $t$.

I think what's confusing you is that the convergence is not uniform in $\omega$, and this is indeed the case, but this is not what the writer is claiming.

Tyr Curtis
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  • Basically, what you mean here is that given $\omega_T$, and any $\omega_t \subset \omega_T$, then the equality happens, for t<T! – user3001408 Mar 21 '15 at 14:31
  • @user3001408 Sorry, I don't understand what you mean. What's $\omega_N$? – Tyr Curtis Mar 21 '15 at 14:33
  • Just cleared from saz answer. You also wanted to say the samething. Thanks a lot man for the effort! – user3001408 Mar 21 '15 at 14:41
  • hey @TyrCurtis could you please throw some light on my other question: http://math.stackexchange.com/questions/1200176/stcochastic-integral-and-ito-isometry – user3001408 Mar 22 '15 at 17:03