Let $(\Omega,\mathcal F, P)$ be a filtered probability space.
Let $n>0$ be a natural number, $T\in \mathbb R$ such that $T>0$, and $\delta=\frac{T}{n}$.
Let $(\sigma^n_i)^n_{i=0}$, $(\tau^n_i)^n_{i=0}$ be two sequences of increasing stopping times in $[0,T]$ with those properties
- $\sigma^n_0=\tau^n_0=0$, and, $\sigma^n_n=\tau^n_n=T$.
- $E[\mid\sigma^n_{i+1}-\sigma^n_{i}\mid]\rightarrow 0$ as $n$ goes to $+\infty$ for every $i=0,...,n-1$.
- $E[\mid\sigma^n_{i}-i\delta\mid]\rightarrow 0$ as $n$ goes to $+\infty$ for every $i=0,...,n$.
- Same property as above for $\tau^n$
Let $(\Theta^n_i)^n_{i=0}$ be a sequence such that,
- $\Theta^n_0=0$
- For $k>0$, $\Theta^{n}_k = \tau^{n}_{w^{X,n}_k}\wedge\sigma^{n}_{w^{Y,n}_k}$, with $w^{X,n}_k := \min\{i\leq k\ :\ \tau^{n}_{i}>\Theta^{n}_{k-1}\}$, and $w^{Y,n}_k := \min\{i\leq k\ :\ \sigma^{n}_{i}>\Theta^{n}_{k-1}\}$.
Roughly speaking $\Theta^n$ represents $\sigma^{n}$, $\tau^{n}$ on the timeline $[0,T]$.
For instance for $n=4$ and $\omega\in\Omega$ such that $\tau^{4}_{0}(\omega)=\sigma^{4}_{0}(\omega)=0<\tau^{4}_{1}<\sigma^{4}_{1}<\sigma^{4}_{2}<\tau^{4}_{2}=\sigma^{4}_{3}<\tau^{4}_{3}<T$
Then $\Theta^{4}_2(\omega)=\sigma^{4}_{1}(\omega)$, $\Theta^{4}_4(\omega)=\sigma^{4}_{3}(\omega)=\tau^{4}_{2}(\omega)$, and, $\Theta^{4}_6(\omega)=T$
Some facts :
We can show that $\Theta^n$ is a sequence of stopping times.
Second $w^{X}_k$ and $w^{Y}_k$ are predictables indexes.
Third $E[\mid\Theta^{n}_{i+1}-\Theta^{n}_{i}\mid]\rightarrow_{n\infty} 0$ for $i=0,..,n-1$ as a direct consequence of 2. and 4.
Fourth $E[\mid\tau^{n}_{i}-\sigma^{n}_{i}\mid]\rightarrow_{n\infty} 0$ for $i=0,..,n-1$ as a direct consequence of 2., 3.+Triangular inequality, and, 4.
My question : For every $i\in\mathbb N$, do we have $E[\mid i-w^{X,n}_{i\wedge n}\mid]\rightarrow_{n\infty} 0$ with $i\wedge n=min(i,n)$ ? If not, do we have at least $E[\mid i-w^{X,n}_{i\wedge n}\mid]=O(1)$ (i.e. bounded for $n$ very large) ? Or impossible to show something, we do not have enough properties ? My goal is to show $E[\mid \Theta^n_i-\Theta^n_{w^{X,n}_{i\wedge n}}\mid]\rightarrow_{n\infty} 0$.
Same questions with $w^{Y,n}$.