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Probably this question has been asked many times before, but still can't put together the puzzle in my head...

Consider a continuous integrand $X$ and a continuous local martingale $Y$, and we try to somehow define the $\int_{0}^{T}X_{t}dY_{t}$ integral. For this, we consider the following partition: $0=t_{0}<t_{1}<\ldots t_{n-1}<t_{n}=T$ and somehow we “obtain” a sequence of a $\left(I_{n}\right)_{n}$ integral approximating sums. Then, we check the limit (under an appropriate convergence) of this sequence and we define the $\int_{0}^{T}X_{t}dY_{t}$ integral as the limit of $\left(I_{n}\right)_{n}$.

My question is why do we stick so strongly to the Itô-integral construction, i.e. why is it so important to evaluate the $X$ integrand at the beginning of each $\left[t_{i-1},t_{i}\right]$ interval and choose $I_{n}$ as

$$I_{n}\dot{=}\sum_{i=1}^{n}X_{t_{i-1}}\left(Y_{t_{i}}-Y_{t_{i-1}}\right)?$$

So far I thought that the importance/ the reason of choosing the evaluating point in the beginning of the $\left[t_{i-1},t_{i}\right]$ interval is because this is the only case when it is always guaranteed that we can achieve some kind of convergence (namely $p$-convergence) in the $\left(I_{n}\right)_{n}$ sequence, then define the integral as

$$\lim_{n}^{p}I_{n}=\int_{0}^{T}X_{t}dY_{t}=\left(X\bullet Y\right)_{T}.$$

I thought we should be happy even for this “not so strong convergence”, because it is surprising that we could “force out” some kind of convergence, although we limited ourselves to always choose the evaluating point at the beginning of $\left[t_{i-1},t_{i}\right]$. However as it turns out this is not the only case...

Reading questions about the Stratonovich-integral (see e.g.: Definitions of the Stratonovich integral and why the "average" definition is arguably correct), I started to think that in many cases the sum will (always?) converge to somewhere and doesn't really matter where to choose the evaluating point because the sum converges anyway (although not to the same value depending on where to choose the evaluating point). Even if we choose the endpoints of the $\left[t_{i-1},t_{i}\right]$ intervals,$\left(I_{n}\right)_{n}$ has a $p$-limit. Do we know any example when the $\left(I_{n}\right)_{n}$ sequence doesn't converge? As far as I see, choosing any $t_{i}^{*}=\lambda t_{i}+\left(1-\lambda\right)t_{i+1}$ point consistently, the integral approximating sum has a $p$-limit. Is it the problem: if we choose $\xi_{i}\in\left[t_{i-1},t_{i}\right]$ points randomly/arbitrary for any $n$, then the $\left(I_{n}\right)_{n}$ sequence doesn't necessary have a $p$-limit? Or the only special thing about the $X\bullet Y$ Itô integral is that it is always “part of the limit” in a sense that

$$I_{n}\overset{p}{\rightarrow}X\bullet Y+something,$$

and we have to prove that this $X\bullet Y$ also exists?

Kapes Mate
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    See the 3rd example in this former question of mine here – Jean Marie Dec 22 '23 at 21:26
  • Ito's construction is strongly tied to the "arrow of time". A classical interpretation is where $X_t$ is the amount of stocks you are holding, and $Y_t$ is the price of the stock per unit in the market. Then, unless you are one of the big hands in the market or an oracle, you have no control over the changes in the price $Y_t$. The choice of the left-endpoint in the construction of Ito's integral models this limitation, meaning that quantities at time $t_{i-1}$, such as $X_{t_{i-1}}$, are what you can only access or control in the market during a short time interval $[t_{i-1},t_i]$. – Sangchul Lee Dec 23 '23 at 03:25
  • For the other question, each choice of $\lambda \in [0, 1]$, with the choice of tags $t_i^*=(1-\lambda)t_{i-1}+\lambda t_{i}$ in each subinterval of a partition, gives rise to a stochastic integral. This includes Ito integral ($\lambda=0$) and Stratonovich integral ($\lambda=\frac{1}{2}$). But I haven't seen any other versions that are practically used in the literature. – Sangchul Lee Dec 23 '23 at 03:34
  • So it is not so important to choose the beginning of the intervals as evaluating points in a sense that the $p$-limit of the approximating sums exists anyway? – Kapes Mate Dec 24 '23 at 15:33

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