Derive an upper bound for the total variation of quadratic covariation [Exercise 1.5.7 in Karatzas and Shreve].

Question

Recall that given two continuous integrable martingale $X$ and $Y$, we can define the quardatic covariation as $$\langle X,Y\rangle=\dfrac{\langle X+Y\rangle-\langle X-Y\rangle}{4},$$ where $\langle \cdot\rangle$ is the quadratic variation defined from the unique increasing natural martingale in Doob's decomposition theorem.

In the book by Karatzas and Shreve, they gave an exercise as follows:

$\langle \cdot,\cdot\rangle$ on the set of square integrable martingales $\mathcal{M}_{2}$ satisfies the following properties: for any members $X,Y,Z$ of $\mathcal{M}_{2}$ and real numbers $\alpha,\beta$, we have:

$(1)$ $\langle \alpha X+\beta Y, Z\rangle=\alpha\langle X,Z\rangle+\beta\langle Y,Z\rangle$;

$(2)$ $\langle X,Y\rangle=\langle Y,X\rangle;$

$(3)$ $|\langle X,Y\rangle|^{2}\leq\langle X\rangle \langle Y\rangle;$

$(4)$ For almost every $\omega\in\Omega$, $$\xi_{t}(\omega)-\xi_{s}(\omega)\leq\dfrac{1}{2}\Big[\langle X\rangle_{t}(\omega)-\langle X\rangle_{s}(\omega)+\langle Y\rangle_{t}(\omega)-\langle Y\rangle_{s}(\omega)\Big]$$ for $0\leq s<t<\infty$, where $\xi_{t}$ denotes the total variation of $\xi:=\langle X,Y\rangle$ on $[0,t]$.

I have proved $(1)-(3)$. The proof of $(1)$ can be read here: Prove that the quadratic covariation is a bilinear form. $(2)$ is immediate. $(3)$ follows from Cauchy-Schwarz since we have proved in $(1)$ and $(2)$ that $\langle \cdot,\cdot\rangle$ is a bilinear symmetric form, and it is clear that it is positive semi-definite.

However, I got stuck in $(4)$. By definition the total variation can be written as $$\xi_{t}(\omega)=\sup_{\mathcal{P}}\sum_{i=0}^{n(p)-1}|\langle X,Y\rangle_{t_{i+1}}-\langle X,Y\rangle_{t_{i}}|$$ where the sup is taken over the collection $\mathcal{P}$ of all partitions $(t_{0},\cdots, t_{n})$ of $[0,t]$.

So, if $\mathcal{P}$ is the partition of $[0,t]$ and $\mathcal{G}$ is the partition of $[0,s]$ for $s<t$, then we have \begin{align*} \xi_{t}(\omega)-\xi_{s}(\omega)&=\sup_{\mathcal{P}}\sum_{i=0}^{n(p)-1}|\langle X,Y\rangle_{t_{i+1}}-\langle X,Y\rangle_{t_{i}}|-\sup_{\mathcal{G}}\sum_{j=0}^{m(g)-1}|\langle X,Y\rangle_{s_{i+1}}-\langle X,Y\rangle_{s_{i}}|\\ &\leq \sup_{\mathcal{P}}\sum_{i=0}^{n(p)-1}|\langle X,Y\rangle_{t_{i+1}}-\langle X,Y\rangle_{t_{i}}|-\sum_{j=0}^{m(g)-1}|\langle X,Y\rangle_{s_{i+1}}-\langle X,Y\rangle_{s_{i}}|\\ &\leq \sup_{\mathcal{P}}\sum_{i=0}^{n(p)-1}|\langle X,Y\rangle_{t_{i+1}}|+|\langle X,Y\rangle_{t_{i}}|-\sum_{j=0}^{m(g)-1}|\langle X,Y\rangle_{s_{i+1}}-\langle X,Y\rangle_{s_{i}}|. \end{align*}

But then I got stuck. What should I do? Thank you!

Answer 1

Given a partition $\Pi=\{a=t_0<\ldots<t_n=b\}$ of some interval $[a,b]$ set

$$S_{\Pi}(X,Y) := \sum_{t_i \in \Pi} |\langle X,Y \rangle_{t_{i+1}}-\langle X,Y \rangle_{t_i}|.$$

If we add a new point to the partition $\Pi$, then we get a finer partition $\Pi'$ and by the triangle inequality,

$$S_{\Pi'}(X,Y) \geq S_{\Pi}(X,Y).$$

This means that refining the partition makes the variation sum larger.

Claim: $$\xi_t-\xi_s \leq \sup_{\Pi} S_{\Pi}(X,Y)$$ where the supremum is taken over all partitions $\Pi$ of $[s,t]$.

Proof: For fixed $\epsilon>0$ take a partition $P$ of $[0,t]$ and partition $Q$ of $[0,s]$ such that $|\xi_t(\omega)-S_{P}(X,Y)(\omega)| \leq \epsilon$ and $|\xi_s-S_Q(X,Y)(\omega)| \leq \epsilon$. Denote by $P'$ the joint refinement of $P$ and $Q$. By our previous consideration,

\begin{align*} S_P(X,Y)-S_Q(X,Y) &\leq S_{P'}(X,Y) - S_Q(X,Y) \\ &\leq S_{P' \cap [0,s]}(X,Y) + S_{P' \cap [s,t]}(X,Y) - S_Q(X,Y).\end{align*}

Since the partition $P' \cap [0,s]$ is finer than the partition $Q$, we have $$\xi_s(\omega)\leq S_{Q}(X,Y)(\omega)+\epsilon \leq S_{P' \cap [0,s]}(X,Y)(\omega)+\epsilon,$$ and so

\begin{align*} |S_{P' \cap [0,s]}(X,Y)(\omega)-S_Q(X,Y)(\omega)| &\leq |\xi_s(\omega)-S_{P' \cap [0,s]}(X,Y)(\omega)| \\ &\quad + |S_Q(X,Y)(\omega)-\xi_s(\omega)| \\ &\leq 2 \epsilon. \end{align*}

Hence,

$$S_P(X,Y)-S_Q(X,Y) \leq S_{P' \cap [s,t]}(X,Y) + 2 \epsilon \leq \sup_{\Pi} S_{\Pi}(X,Y) +2 \epsilon$$

where the supremum is taken over all partitions $\Pi$ of $[s,t]$. Combining this with the estimate

$$\xi_t(\omega)-\xi_s(\omega) \leq (S_P(X,Y)(\omega)+\epsilon)-S_Q(X,Y)(\omega)$$

this proves the claim.

Claim 2: $$\sup_{\Pi} S_{\Pi}(X,Y) \leq \frac{1}{2} (\langle X \rangle_t-\langle X \rangle_s + \langle Y \rangle_t - \langle Y \rangle_s)$$ where the supremum is taken over all partitions $\Pi$ of $[s,t]$.

It follows from (1) that

$$\langle X,Y \rangle = \frac{1}{4} (\langle X+Y \rangle - \langle X-Y \rangle).$$

In particular,

\begin{align*} |\langle X,Y \rangle_t-\langle X,Y \rangle_s| &\leq \frac{1}{4} \big( |\langle X+Y \rangle_t- \langle X+Y \rangle_s| \big) + \frac{1}{4} \big( |\langle X-Y \rangle_t- \langle X-Y \rangle_s| \big). \end{align*}

Since $\langle X+Y \rangle$ and $\langle X-Y \rangle$ are non-decreasing in time, we can drop the modulus on the right-hand side. Summing over any partition $\Pi$ of the interval $[s,t]$, we get

\begin{align*} S_{\Pi}(X,Y) &\leq \frac{1}{4} (\langle X+Y \rangle_t - \langle X+Y \rangle_s) + \frac{1}{4} (\langle X-Y \rangle_t - \langle X-Y \rangle_s) \\ &\stackrel{(1)}{=} \frac{1}{2} (\langle X \rangle_t - \langle X \rangle_s + \langle Y \rangle_t - \langle Y \rangle_s). \end{align*}