How to show that $Y$(martingale transform) has equal $L_2$ norm to $X$(martingale or supermartingale)?

Question

$\mathbf{Definition}$: 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.

I want to show that $Y$ and $X$ have equal $L_2$ norm.

I've been studying this .pdf http://math.aalto.fi/teaching/stokanal/lecture2.pdf and i trying to understand how i am going to do that. The only thing that i think will help is that the norm of a martingale is a sub-martingale. But i have no idea how to continue.

Answer 1

So the theorem itself is definitely correct, it's just the part about the $L_2$ norms being equal that doesn't work. Here is the proof for part i):

We want to show that $Y$ is a super-martingale, so we need to show $\mathbb{E}[|Y_n|] < \infty$ and $\mathbb{E}[Y_n | \mathcal{F}_{n-1}] \le Y_{n-1}$ for all $n$.

First we will show $\mathbb{E}[|Y_n|] < \infty$ (showing this one first because it makes sure that $\mathbb{E}[Y_n|\mathcal{F}_{n-1}]$ is well-defined). We compute

\begin{align*} \mathbb{E}[|Y_n |]&= \mathbb{E}\left[\left\lvert \sum_{k=1}^n C_k (X_{k}-X_{k-1}) \right\rvert\right] \\ &\le \mathbb{E}\left[ \sum_{k=1}^n |C_k|\cdot |X_{k}-X_{k-1}|\right] \\ &\le \mathbb{E}\left[ \sum_{k=1}^n K|X_{k}-X_{k-1}|\right] \\ &\le \mathbb{E}\left[ \sum_{k=1}^n K (|X_{k}|+|X_{k-1}|) \right] \\ &= \sum_{k=1}^n K (\mathbb{E}[|X_k|] + \mathbb{E}[|X_{k-1}|]) < \infty. \end{align*}

Now we need to show $\mathbb{E}[Y_n | \mathcal{F}_{n-1}] \le Y_{n-1}$. We compute

\begin{align*} \mathbb{E}[Y_n | \mathcal{F}_{n-1}] &= \mathbb{E}[Y_{n-1} + C_n(X_n-X_{n-1}) | \mathcal{F}_{n-1}] & \\ &= Y_{n-1} + \mathbb{E}[C_n(X_n - X_{n-1})|\mathcal F_{n-1}] & \\ &= Y_{n-1} + C_n \mathbb{E}[X_n-X_{n-1}|\mathcal{F}_{n-1}] & \end{align*}

where we were able to pull $C_n$ out of the conditional expectation because $C_n$ is $\mathcal{F}_{n-1}$ measurable. Now since we were told $X$ is a super-martingale, we know $\mathbb{E}[X_n-X_{n-1}|\mathcal{F}_{n-1}] \le 0$, and since $C_n(\omega) \ge 0$ for all $\omega$ this implies $C_n \mathbb{E}[X_n-X_{n-1}|\mathcal{F}_{n-1}] \le 0$. Therefore we conclude $$\mathbb{E}[Y_n | \mathcal{F}_{n-1}] = Y_{n-1} + C_n \mathbb{E}[X_n-X_{n-1}|\mathcal{F}_{n-1}] \le Y_{n-1},$$

so by the definition of a super-martingale, $Y$ is a super-martingale.