On Ito integral and the choice of partition

Question

If the value of the stochastic integral $$ \int_{t_0}^tg(t',W_{t'})\mathrm{d}W_{t'} = \underset{n\to\infty}{\operatorname{ms-lim}}\sum_{i=1}^ng(\tau_i, W_{\tau_i})(W_{t_i}-W_{t_{i-1}})$$ does depend on the choice of $\;\tau_i:=\alpha\,t_i+(1-\alpha)\,t_{i-1}, \quad \alpha\in[0,1]$, why doesn't the choice of partitioning $[t_0,t]$ matter (and, indeed, this allows choosing equally spaced sub-intervals) ? Here are three different ways in which $[t_0,t]$ can be partitioned

Answer 1

The choice of partition $ t_0<t_1<\ldots<t_n = t$ and the choice of $\tau_i\in[t_i,t_{i+1}]$ are pretty much unrelated matters.

Regarding the latter, in fact, you could pick any (determinstic) $\tau_i$ you like, and by splitting the limit into two sums as Kurt G. did in his answer, you would have $$\underbrace{\lim_{n\to\infty\atop\Pi_n\to 0}\sum_{i=0}^{n-1}\big(g({\boldsymbol\tau}_i,W_{{\boldsymbol\tau}_i})-g(t_i,W_{t_i})\big)(W_{t_{i+1}}-W_{t_i})}_{\text{(A)}}+\underbrace{\lim_{n\to\infty\atop\Pi_n\to 0}\sum_{i=0}^{n-1}g(t_i,W_{t_{i}})(W_{t_{i+1}}-W_{t_i})}_{\text{(B)}}$$ The theory of Itô's calculus guarantees us that the term $(B)$ always (well, provided that $g$ satisfies some regularity conditions) converges to the stochastic integral $\int_{t_0}^t g(s,W_s)dW_s$ as $|\Pi_n|\to 0$, so the only concern is term $(A)$.
For $(A)$, we can approximate $g$ using simple processes, and by using Itô isometry, we can show that the quantity $(A)$ is the limit of a Cauchy sequence in $L^2$ and thus converges to something as the mesh size $|\Pi_n|$ goes to zero. Hence we see that for any choice of $\tau_i := \alpha t_i + (1-\alpha)t_{i+1}$, you can always define a "new integral" $I^\alpha_{[t_0,t]}(f)$ as the corresponding limit and you will always end up with a well-defined object.
As you mentioned however, the choice of $\alpha$ does change the result, so if you want $I^\alpha$ to have nice properties, then you need to pick $\alpha$ accordingly : for instance, for the stochastic integral to be a martingale you would have to pick $\alpha\equiv 1$, while if you want it to satisfy the usual chain rule from calculus you would have to pick $\alpha\equiv \frac{1}{2}$.

So you ask

Why doesn't the choice of partitioning $[t_0,t]$ matter ?

Well, as you can see in the above argument, the choice of partition doesn't matter because the convergence holds for any partition whose mesh size $|\Pi_n|:=\sup_{0\le i\le n-1} |t_{i+1}-t_i|$ goes to zero. So you are free to take any partition you like, as long as the distance between two subsequents points converges to $0$ uniformly, the same result will hold, always (in most practical cases we pick a uniformly spaced grid because it is the easiest to work with, but there is no formal need to).
The proof is essentially an application of Itô isometry and Cauchy-Schwarz, and can be found in Proposition 14.16 of Schilling's Brownian Motion: An Introduction to Stochastic Processes. Intuitively, because you require the mesh size to uniformly converge to zero, you don't have any room to create any local bad behaviors, so under that constraint your sum will always converge to the same object.

Answer 2

It is because $W$ has quadratic variation $t\,:$

\begin{align} &\lim_{n\to\infty\atop\Pi_n\to 0}=\sum_{i=1}^nW_{t_{i+1}}(W_{t_{i+1}}-W_{t_i})\\ &=\underbrace{\lim_{n\to\infty\atop\Pi_n\to 0}\sum_{i=1}^n(W_{t_{i+1}}-W_{t_i})(W_{t_{i+1}}-W_{t_i})}_{\textstyle=\langle W\rangle_t=t}+\underbrace{\lim_{n\to\infty\atop\Pi_n\to 0}\sum_{i=1}^nW_{t_{i}}(W_{t_{i+1}}-W_{t_i})}_{\textstyle =\int_0^tW_s\,dW_s\text{ (Ito)}}\, \end{align} where $\Pi_n=\max\limits_{i=1,...,n}|t_{i+1}-t_i|\,.$

We see from this that the limit of the left hand side cannot be equal to the Ito integral.

Related.