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.

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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 ?


1 Answer 1


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 $\|C\|_{L^2(X)} = \|C\circ X \|_{\mathcal X_0^{2,c}}.$

  • $\begingroup$ 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 ! $\endgroup$
    – Luck-e
    Jun 10, 2020 at 23:48

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