Quadratic Variation of Brownian motion indexed by cadlag increasing function

Question

Let $(B_t)_{t \geq 0}$ a Brownian motion on $\mathbb R^n$ and $\ell_t \in \mathbb S$, where $\mathbb S$ is the space of all increasing càdlàg function from $(0,\infty)$ to $(0,\infty)$ with $\lim_{s\downarrow 0} = 0$. Then the quadratic variation of $(B_{\ell_t})_{t\geq 0}$ is given by $$[B_{\ell}]_t = \ell_t - \sum_{0<s\leq t}\Delta \ell_s + \sum_{0<s\leq t} | \Delta B_{\ell_s} |^2.$$

Heuristically, this is very clear: the square bracket can be written as the sum of a continuous and discontinuous part. Since the cts. part $[B_\ell]_t^c = l_t$ we get the first summand + the jumps which is the third summand. Finally, since $\ell_t$ is càdlàg, we have to subtracte the jumps $\Delta \ell_s$.

My thoughts so far: By Definition, the quadratic variation of a semimartingale $X$ is given by $$[X] = X^2 - \int X_- \ \mathrm dX.$$ We have to calculate the stochastic integral for $X_t := B_{\ell_t}$. Let $\Pi = \{0 \leq s_1^k \leq ... \leq s_{n(k)}^k \leq t\}$ be a partition of the interval $[0,T]$. Then $$\sum_i B_{\ell(s_j)} (B_{\ell(s_{j+1})} - B_{\ell(s_j)}) \overset{ucp}\longrightarrow \int_0^t B_{\ell_{t-}} \ \mathrm d B_{\ell_t}.$$

If there are reasonable sets A und B, then using the elementary equality $b^2 - a^2 - (b-a)^2 = 2a (b-a)$, we find \begin{align*} 2 \sum_i B_{\ell(s_j)} (B_{\ell(s_{j+1})} - B_{\ell(s_j)}) &= \sum_i (B_{\ell(s_{j+1})}^2 - B_{\ell(s_j)}^2) - \sum_i (B_{\ell(s_{j+1})} - \sum B_{\ell(s_j)})^2\\ &= \underbrace{ \sum_{i,B} (B_{\ell(s_{j+1})}^2 - B_{\ell(s_j)}^2) }_{= \ B_{\ell_t}} - \sum_{i,A} (B_{\ell(s_{j+1})}^2 - B_{\ell(s_j))^2 }\\ &\qquad - \Bigg( \underbrace{ \sum_{i,B} (B_{\ell(s_{j+1})} - B_{\ell(s_j)})^2 }_{\overset{ucp}{\longrightarrow} \ \ell_t} - \sum_{i,A} (B_{\ell(s_{j+1})} - B_{\ell(s_j)} )^2 \Bigg)\\ &\overset{k\to \infty}\longrightarrow B_{\ell_t}^2 - \sum_{0<s\leq t} (B_{\ell_s} - B_{\ell_{s-}} )^2 - (\ell_t - \sum_{0<s\leq t} (\ell_s - \ell_{s-})) \\ &= B_{\ell_t}^2 - \ell_t - \sum_{0<s\leq t} \Delta(B^2)_s + \sum_{0<s\leq t} \Delta \ell_s. \end{align*}

Is this decomposition of the sums possible? My idea was something like $$J(\epsilon) := \{s \in [0,t] : |\Delta X_s| > \epsilon\}.$$ Since $s \mapsto X_s$ is càdlàg, $J$ is a.s. finite, each $\epsilon > 0$ and $\sum_{s \in J(\epsilon)} \overset{\epsilon \to 0}\longrightarrow \sum_{0<s\leq t}$ a.s.

Is there an easier way? Thanks.

Answer 1

I think it is not obvious that the third and fourth term (in your calculation) converge to their prospective limits; however, the idea of your proof is basically correct.

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Fix $t>0$. By definition,

$$[B_{\ell}]_t = \mathbb{P}-\lim_{|\Pi| \to 0} \sum_{j=1}^n (B_{\ell_{t_j}}-B_{\ell_{t_{j-1}}})^2;$$

here $\Pi = \{0=t_0<\ldots<t_n = t\}$ denotes a partition of the interval $[0,t]$ with mesh size $|\Pi| := \max_j |t_j-t_{j-1}|$. For fixed $\epsilon>0$ we denote by $S = S(\epsilon) := \{s_1<s_2<\ldots<s_m\}$ the jump times of $\ell$ up to time $t$ with jump height exceeding $\epsilon$. (Since $\ell$ is càdlàg, there are only finitely many such jumps.) It follows directly from the continuity of the sample paths of Brownian motion and the fact that $|S|<\infty$ that

\begin{equation} \sum_{j: (t_{j-1},t_j] \cap S \neq \emptyset} (B_{\ell_{t_j}}-B_{\ell_{t_{j-1}}})^2 \xrightarrow[]{|\Pi| \to 0} \sum_{s \leq t, \Delta \ell_s \geq \epsilon} \Delta B_{\ell_s}^2 \tag{1} \end{equation}

almost surely. Using exactly the same argumentation as for the "standard" quadratic variation of Brownian motion (that is, $\ell(t)=t$), we find

$$\mathbb{E} \left( \left| \sum_{j: (t_{j-1},t_j] \cap S = \emptyset} (B_{\ell_{t_j}}-B_{\ell_{t_{j-1}}})^2- \sum_{j: (t_{j-1},t_j] \cap S = \emptyset} (\ell_{t_j}-\ell_{t_{j-1}}) \right|^2 \right) \leq C \sup_{j: (t_{j-1},t_j] \cap S = \emptyset} (\ell_{t_{j-1}} - \ell_{t_j})$$ for some constant $C>0$ which does not depend on $\Pi$ and $\epsilon$; the idea is basically to use the scaling property of Brownian motion, for more details see e.g. René Schilling & Lothar Partzsch: Brownian motion - An Introduction to Stochastic Processes, Theorem 9.1. Consequently,

\begin{equation} \limsup_{|\Pi| \to 0} \mathbb{E} \left( \left| \sum_{j: (t_{j-1},t_j] \cap S = \emptyset} (B_{\ell_{t_j}}-B_{\ell_{t_{j-1}}})^2- \sum_{j: (t_{j-1},t_j] \cap S = \emptyset} (\ell_{t_j}-\ell_{t_{j-1}}) \right|^2 \right)\leq C \epsilon. \tag{2} \end{equation}

Moreover, we note that

\begin{equation} \ell_t - \sum_{s \leq t, \Delta \ell_s \geq \epsilon} \Delta \ell_s = \lim_{|\Pi| \to 0} \sum_{j: (t_{j-1},t_j] \cap S = \emptyset} (\ell_{t_j}-\ell_{t_{j-1}}). \tag{3} \end{equation}

Now we are ready to prove the convergence in probability. Fix $\gamma>0$. By the triangle inequality, we have

$$\mathbb{P} \left( \left| \sum_{j} |B_{\ell_{t_j}}-B_{\ell_{t_{j-1}}}|^2 - \left( \sum_{s \leq t} \Delta B_{\ell_s}^2 + \ell_t - \sum_{s \leq t} \Delta \ell_s \right) \right| > \gamma \right) \leq I_1+I_2+I_3+I_4$$

where \begin{align*} I_1 &:= \mathbb{P} \left( \left| \sum_{j: (t_{j-1},t_j] \cap S \neq \emptyset} (B_{\ell_{t_j}}-B_{\ell_{t_{j-1}}})^2- \sum_{s \leq t, \Delta \ell_s \geq \epsilon} \Delta B_{\ell_s}^2 \right|> \frac{\gamma}{4} \right) \\ I_2 &:= \mathbb{P} \left( \left| \sum_{j: (t_{j-1},t_j] \cap S = \emptyset} (B_{\ell_{t_j}}-B_{\ell_{t_{j-1}}})^2- \sum_{j: (t_{j-1},t_j] \cap S = \emptyset} (\ell_{t_j}-\ell_{t_{j-1}}) \right| > \frac{\gamma}{4} \right) \\ I_3 &:= \mathbb{P} \left( \left| \ell_t - \sum_{s \leq t, \Delta \ell_s \geq \epsilon} (\ell_{t_j}-\ell_{t_{j-1}}) - \sum_{j: (t_{j-1},t_j] \cap S = \emptyset} (\ell_{t_j}-\ell_{t_{j-1}}) \right| > \frac{\gamma}{4} \right) \\ I_4 &:= \mathbb{P} \left( \left| \sum_{s \leq t, \Delta \ell_s < \epsilon} \Delta B_{\ell_s}^2 \right| + \left| \sum_{s \leq t, \Delta \ell_s < \epsilon} \ell_s \right|> \frac{\gamma}{4} \right) \end{align*}

We consider the terms separately:

• $I_1$ converges to $0$ as $|\Pi| \to 0$ because of $(1)$ (recall that a.s. convergence implies convergence in probability.)
• Applying Markov's inequality for $p=2$, it follows from $(2)$ that $$\limsup_{|\Pi| \to 0} I_2(|\Pi|,\epsilon) \leq C \epsilon.$$ Letting $\epsilon \to 0$ gives $$\lim_{\epsilon \to 0} \limsup_{|\Pi| \to 0} I_2(|\Pi|,\epsilon)=0.$$
• By $(3)$, $I_3$ converges to $0$ as $|\Pi| \to 0$.
• In order to see that $I_4$ converges to $0$ as $\epsilon \to 0$, we need that $$\sum_{s \leq t} \Delta B_{\ell_s}^2 < \infty$$ for any $t \geq 0$. That's not obvious, but follows e.g. from the well-known fact that $$\sum_{s \leq t} \Delta X_s^2 < \infty$$ for any Lévy process $(X_t)_{t \geq 0}$.

Adding all up gives

$$\lim_{|\Pi| \to 0} \mathbb{P} \left( \left| \sum_{j} |B_{\ell_{t_j}}-B_{\ell_{t_{j-1}}}|^2 - \left( \sum_{s \leq t} \Delta B_{\ell_s}^2 + \ell_t - \sum_{s \leq t} \Delta \ell_s \right) \right| > \gamma \right) = 0.$$

Remark:

• In this answer I use a different definition of quadratic variation; using some elementary calculations it is not difficult to see that the same argumentation applies if we use the definition stated in the above question. If you don't get along with it, don't hesitate to ask.