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This seems like a simple question, but I am stumped. I know the proofs for quadratic variation and cross variation, etc..., but for some reason can't understand why the following doesn't make sense to do: \begin{equation} [Z,Z]([0,t])=\lim_{\delta_n\to 0}\sum_{i=1}^{n}(\,Z(t_i)-Z(t_{i-1})\,)^2 \leq \lim_{\delta_n\to 0}\sum_{i=1}^{n}(\max_i [Z(t_i)-Z(t_{i-1})]\,)^2 = 0 \\\implies \lim_{\delta_n\to 0}\sum_{i=1}^{n}(\,Z(t_i)-Z(t_{i-1})\,)^2 = 0 \end{equation}

I am unsure why this reasoning can't hold, and it's making the proofs for QVs of other processes confusing. It seems like I have passed the limit through the summation and then assumed the product of the maxes is $0$, so if those don't hold I am wondering why not. It seems like passing a limit that is essentially $\lim_{n \to \infty}$ through the sum doesn't make sense to do, so would that be the mistake here? I was thinking of it in the sense of: \begin{equation} \lim_{n \to \infty} \sum_{i=1}^nx_i(n) = \lim_{n \to \infty} \sum_{i=1}^n\lim_{n \to \infty}x_i(n) \end{equation} or \begin{equation} \lim_{n \to \infty} \sum_{i=1}^nx_i(n) = \sum_{i=1}^{\infty}\lim_{n \to \infty}x_i(n) \end{equation} Thanks for the help!

Slade
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    How are you concluding the sum of max’s Is 0? – Alex R. May 15 '19 at 05:53
  • Since the Brownian motion is (uniformly) continuous so as the partition goes to $0$, the max of the increment goes to $0$. I am unsure if that makes sense – Slade May 15 '19 at 05:55
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    But you are summing asymptotically infinitely many such increments. – Alex R. May 15 '19 at 06:19
  • I'm a total beginner at this stuff so I can't really tell where the flaw in my logic is, unless it's relating to passing the limit through the summation (is that what you're saying?). I am trying to understand why it's not justified. For example, like here, https://math.stackexchange.com/questions/23057/under-what-condition-we-can-interchange-order-of-a-limit-and-a-summation – Slade May 15 '19 at 06:25
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    Also have a look here : https://math.stackexchange.com/questions/92938/quadratic-variation-of-brownian-motion – TheBridge May 15 '19 at 13:54

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Note that the $n$ in the upper bound of the sum depends on the partition and hence depends on your $\delta_n$. In fact $n \to \infty$ as $\delta_n \to 0$ (to chop up $[0,t]$ in to blocks of size $\delta_n$ you need more and more blocks as $\delta_n$ gets smaller). This means that you can't pass the limit through the sum because the upper bound of the sum depends on the thing you take the limit with respect to.

What is happening is that whilst $(\max_i [Z(t_i)-Z(t_{i-1})]\,)^2$ gets small when $\delta_n$ gets small, the number of such terms contributing to the sum gets larger and these two effects push the opposite way which prevents necessary convergence to $0$.

Rhys Steele
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  • Thanks for the help! How about in the case where it's written just with the upper limit of the sum as infinity, and the limit operator just would apply to the brownian motion terms? Is there a reason why writing the summation that way would be invalid? Then it looks like taking the limit inside could be justified using something like DCT – Slade May 15 '19 at 07:33
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    What would the terms in your infinite sum even be? At the moment you have for each $n$ a partition consisting of $n$ points with mesh size $\delta_n$. What are you putting for the infinitely many other terms? – Rhys Steele May 15 '19 at 07:42
  • Oh wow. That makes a lot of sense. I think that's what I was missing this whole time. Your answers are super helpful. Thanks! – Slade May 15 '19 at 07:45