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Let $(X_t, Y_t)$ be a two-dimensional $(\mathscr{F}_t)$-Brownian motion started from 0. We set, for every $t \geq 0$ $$\mathscr{A}_t = \int_0^t X_s dY_s - \int_0^t Y_s dX_s$$ (Levy's area).

Let $f$ be a twice continuously differentiable function on $\mathbb{R}_+$. Give the canonical decomposition of the semimartingales $$Z_t = \cos(\lambda \mathscr{A}_t)$$ $$W_t = -\frac{f'(t)}{2} (X_t^2 + Y_t^2) + f(t)$$ Verify that $\langle Z, W \rangle_t = 0$.

In previous parts I showed that $\langle \mathscr{A}, \mathscr{A} \rangle_t = \int_0^t (X_s^2 + Y_s^2) ds$ and $E[\exp(i \lambda \mathscr{A}_t)] = E[\cos(\lambda \mathscr{A}_t)]$. I think Girsinov's theorem might be helpful since it gives a way to get the canonical decomposition of semimartingales but overall I feel stuck.

This is from Le Gall's book Brownian Motion, Martingales, and Stochastic Calculus, which I'm self-studying in a reading group.


UPDATE: I think I figured out the solution for $Z_t$; apply Itô's formula to get

$$\cos(\lambda \mathscr{A}_t) = -\lambda \int_0^t \sin (\lambda \mathscr{A}_s) d\mathscr{A}_s + 1 - \frac{\lambda^2}{2} \int_0^t \cos(\lambda \mathscr{A}_s)(X_s^2 + Y_s^2) ds$$

$-\lambda \int_0^t \sin (\lambda \mathscr{A}_s) d\mathscr{A}_s$ is the martingale while $1 - \frac{\lambda^2}{2} \int_0^t \cos(\lambda \mathscr{A}_s)(X_s^2 + Y_s^2) ds$ is the finite variation process.

Not sure how to get $W_t$ broken down but its form strongly resembles something related to Itô's formula.

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  • $\begingroup$ The canonical decomposition means that the semimartingale $X_t$ can be written as a sum of local martingale and a predictable finite variation process. Correct? $\endgroup$ – Harto Saarinen Feb 23 '18 at 15:00
  • $\begingroup$ I don't think the finite variation process need be deterministic (or predictable if you prefer). It can be random. $\endgroup$ – cgmil Feb 23 '18 at 15:17
  • $\begingroup$ Predictable doesnt mean same as deterministic. See google.fi/amp/s/almostsure.wordpress.com/2016/11/22/… It quarantees that the decomposition is unique if I remember correctly. $\endgroup$ – Harto Saarinen Feb 23 '18 at 23:21
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I managed to solve this on my own.

Applying Itô's formula to the function $F(a) = \cos(\lambda a)$ gives the decomposition of $Z_t$:

$$\cos(\lambda \mathscr{A}_t) = -\lambda \int_0^t \sin(\lambda \mathscr{A}_s) d\mathscr{A}_s + 1 - \frac{\lambda^2}{2} \int_0^t \cos(\lambda \mathscr{A}_s) (X_s^2 + Y_s^2) ds$$

The local martingale part is $-\lambda \int_0^t \sin(\lambda \mathscr{A}_s) d\mathscr{A}_s$ and the finite variation part is $1 - \frac{\lambda^2}{2} \int_0^t \cos(\lambda \mathscr{A}_s) (X_s^2 + Y_s^2) ds$.

For $W_t$, apply Itô's formula to the process $G(X_t, Y_t, t)$, where $G(x, y, t) = (x^2 + y^2) f'(t)$. After some work Itô's formula will reveal that

$$W_t = -\left(\int_0^t X_s f'(s)dX_s + \int_0^t Y_s f'(s) dY_s\right) + f(0) - \frac{1}{2}\int_0^t (X_s^2 + Y_s^2) f''(s)ds$$

The local martingale part is $-\left(\int_0^t X_s f'(s)dX_s + \int_0^t Y_s f'(s) dY_s\right)$ and the finite variation part is $f(0) - \frac{1}{2}\int_0^t (X_s^2 + Y_s^2) f''(s)ds$.

To show that $\langle W, Z \rangle_t = 0$, first show that $\langle \mathscr{A}, X \rangle_t = -\int_0^t Y_s ds$ and $\langle \mathscr{A}, Y \rangle_t = \int_0^t X_s ds$, and use the fact that the bracket "commutes" with stochastic integrals, and worry only about the martingale parts of the processes; after some work the result will appear.

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