Calculation of quadratic covariation of stopped processes

I am stuck in computing the quadratic covariation of the following two processes:

Let $0< y <r$ and let $(B_t)$ be a Brownian motion started at $y$.

Let $T_0 = \inf \{ t \geq 0 : B_t = 0 \}$ and let $T_r = \inf \{ t \geq 0 : B_t = r \}$. Also, set $Z_t= B_{T_0 \wedge T_r \wedge t}$. How can we prove that

$$d \langle Z, B \rangle_t = \mathbf{1}_{ \{t \leq T_r\} } \frac{Z_t}{B_t} dt \quad \quad ?$$

• Do you know Itô's formula?
– saz
Mar 14 '15 at 20:02
• Yes. But do you mean Ito product rule?
– erik
Mar 14 '15 at 20:03
• @saz Frankly speaking, I haven't seen Ito's formula being applied to a covariation process.
– erik
Mar 14 '15 at 20:04
• I don't want you to apply Itô's formula to the covariation process, but to the process $(Z_t,B_t)$. Note that $A_t := \langle Z,B \rangle_t$ is the unique (increasing suitable measurable) process such that $Z_t B_t -A_t$ is a martingale.
– saz
Mar 14 '15 at 20:06
• @saz But I have only learnt Ito's formula being applied to a real-valued process (i.e. $F(X_1, \ldots, X_n)$, where $X_1, \ldots, X_n$ are semimartingales and $F: \mathbb{R}^n \rightarrow \mathbb{R}$ is a function.).
– erik
Mar 14 '15 at 20:09

Lemma: Let $f,g \in L^2([0,T] \times \mathbb{P})$ be progressively measurable. Then for all $t \leq T$ $$\langle \int_0^{\cdot} f(s) \, dB_s, \int_0^{\cdot} g(s) \, dB_s \rangle_t = \int_0^t f(s) g(s) \, ds.$$

Proof: We know that the claim holds true for $f=g$. Now, using $$xy = \frac{1}{4} ((x+y)^2-(x-y)^2),$$ we get \begin{align*} &\quad \int_0^t f(s) \, dB_s \cdot \int_0^t g(s) \, dB_s - \int_0^t f(s) g(s)\\ &= \frac{1}{4} \left( \int_0^t (f+g)^2 \, dB_s - \int_0^t (f+g)^2(s) \, ds \right) - \frac{1}{4} \left( \int_0^t (f-g)^2 \, dB_s - \int_0^t (f-g)^2(s) \, ds \right). \end{align*} Since both terms on the right-hand side are martingales, the left-hand side is also a martingale and this finishes the proof.

Hint: Find suitable $f$, $g$ and apply the above lemma.

Solution: $f=1$, $g(s)= 1_{\{s \leq T_0 \wedge T_R\}}$, then \begin{align*} \langle Z,B \rangle_t &\stackrel{\text{Lemma}}{=} \int_0^t 1_{\{s \leq T_0 \wedge T_R\}} \, ds = \int_0^t 1_{\{s \leq T_r\}} \underbrace{1_{\{s \leq T_0\}}}_{\frac{Z_t}{B_t}} , ds. \end{align*}

• I am still confused why $T_0$ is not present in the above expression on the right hand side.
– erik
Mar 15 '15 at 11:27
• @erik Actually, it is present on the right-hand side (mind that $Z_{T_0}=0$ if $T_0<T_r$). What do you get if you apply the above lemma?
– saz
Mar 15 '15 at 12:13
• Are you suggesting to plug $f$ as $B$ and $g$ as $Z$? But it doesn't seem to work? I really don't know how to get the indicator function.
– erik
Mar 15 '15 at 12:17
• @erik No, that's not what I want you to do. Find $f$, $g$ such that $$B_t = \int_0^t f(s) \, dB_s$$ and $$Z_t = \int_0^t g(s) \, dB_s.$$
– saz
Mar 15 '15 at 12:19
• so $f=1$ and $g= \mathbf{1}_{ \{T \leq T_0 \wedge T_R\} }$?
– erik
Mar 15 '15 at 12:22