Martingales, martingale transform, $L_2$ norm and $\textbf{Itô′s isometry}$.

I've asked a question for something similar in my first post but figured that this new question of mine needs a new post.

$$\mathbf{Definition}$$: We have that $$C$$ and $$X$$ are stochastic processes. The process $$(C∘X)$$ is martingale transform, where $$(C∘X)_n:=\sum_{k=1}^n C_k(X_k-X_{k-1})=\sum_{k=1}^nC_kΔΧ_k,$$ when $$n\geq1$$ and $$(C∘X)_0=X_0.$$

$$\mathbf{Theorem}$$: Let $$\mathbf{F}$$ be a history, the process $$X$$ satisfies $$X\in \mathbf{F}$$ and $$C$$ is a predictable process.

1)If in addition $$0\leq C_n(\omega)\leq K$$ and $$X$$ is a supermartingale, then $$Y=(C∘X)$$ is a supermartingale.

2) If in addition $$|C_n(\omega)|\leq K$$ and $$X$$ is a martingale, then $$Y=(C∘X)$$ is a martingale.



So i am trying to understand the $$\mathbf{Itô's~isometry}$$ and find the $$L_2$$ norm of $$Y$$ that way. This is my idea: Let $$\mathcal{X}_0^{2,c}$$ denote the family of all discrete martingales $$X$$ with $$X_0=0$$ such that $$||X||_{\mathcal{X}_0^{2,c}}:=\sqrt{\sup_{n \geq 0}\mathbb{E}\left[X_n^2\right]}<\infty$$. So we have: $$\mathbb{E}\left[(C∘X)_n^2\right]=\sum_{k=1}^{\infty}\mathbb{E}\left[C_k^2(X_{k}-X_{k-1})^2\right].~~~~\textbf{(1)}$$ We know that if $$X$$ is a martingale then $$X^2$$ is a martingale. Given that $$X \in\mathcal{X}_0^{2,c}$$ it follows that $$\mathbb{E}\left[(X_{k}-X_{k-1})^2|\mathcal{F}_{k-1}\right]=\mathbb{E}\left[X_{k}^2-X_{k-1}^2|\mathcal{F}_{k-1}\right].$$ Therefore, the relation $$\textbf{(1)}$$ can be written as $$\mathbb{E}\left[(C∘X)_\infty^2\right]=\mathbb{E} \left[\sum_{k=1}^{\infty}C_k^2\mathbb{E}\left[X_{k}^2-X_{k-1}^2|\mathcal{F}_{k-1}\right]\right]=\mathbb{E}\left[ \int_{0}^{\infty}C_k^2~d⟨X⟩_k\right].~~~~\textbf{(2)}$$

From the last expression $$\textbf{(2)}$$ above, for a progressively-measurable process $$C$$, we define $$||C||_{L^2(X)}=\sqrt{\mathbb{E}\left[ \int_{0}^{\infty}C_k^2~d⟨X⟩_k\right]}.$$

It is not hard to check that the family $$L_2(X)$$ of all progressively-measurable processes for which $$||C||_{L_2(X)} < ∞$$ forms a vector space,and that $$|| · ||_{L_2(X)}$$is a norm there. We also note that $$C$$ is predictable. So from $$\textbf{(2)}$$ we get that $$||Y||_{L_2}=||C∘X||_{\mathcal{X}_0^{2,c}}=||C||_{L_2(X)},~~~\forall~C.$$

Is that right? I tried using $$\mathbf{Itô's~isometry}$$ to find the $$L_2$$ norm of $$Y$$. If it's not right can someone help me find the solution ?

This is mostly right, but it looks like you might be confusing continuous time and discrete time somewhat. The martingale transform you write, $$(C \circ X)_n = \sum_{k=1}^n C_k(X_{k}-X_{k-1})$$, is generally defined for discrete time processes so I'm a little confused about why you mention continuous martingales in your definition of $$\mathcal{X}_0^{2,c}$$. This isn't too big of a problem, but it makes it somewhat confusing when you talk about simple processes since in discrete time all processes are simple processes.
I also don't think I've heard something be called a simple predictable process of another process before. It is not the case that every simple predictable process is in $$L^2(X)$$. However, if we know that $$C$$ is predictable and $$\|C\|_{L^2(X)} < \infty$$ then Ito's isometry applies and $$\|C\|_{L^2(X)} = \|C\circ X \|_{\mathcal X_0^{2,c}}.$$
• Oh my God yeah i thought i had continuous martingales in my head. I'll edit it :D. So $C$ only has to be predictable and $||C||_{L^2(X)}< \infty$. I'll edit that too! Thank you very much for spending time looking at my solution. I really appreciate it ! Jun 10, 2020 at 23:48