Differentiation with Ito's Formula

Question

Suppose that $(a,b,c)$ are reals and $B$ is a Brownian motion.

If we have: $$dX_t = (a-bX_t)dt + dB_t $$ and : $$ Z_t = \exp\left(c\int_{0}^{t} X_sdB_s -\frac{c^2}{2}\int_0^tX_s^2ds\right)$$

How to show that: $$ dZ_t = cX_tZ_tdB_t$$

I don't know how to apply Ito's Formula as it's hard to find $\phi$ such as : $$Z_t = \phi(t, B_t)$$

Then I have to show that's it is a local martingale, would it help?

Could you help me?

Thanks a lot.

Answer 1

The idea :

Define the Ito process $Y_t$ as

$$dY_t=-\frac{c^2}{2}X_tdt+cX_tdB_t$$

Apply the Ito's lemma to $Z_t=e^{Y_t}$ wrt $Y_t$

$$dZ_t=e^{Y_t}dY_t+\frac{1}{2}e^{Y_t}d<Y_t,Y_t>$$ $$dZ_t=Z_tdY_t+\frac{1}{2}Z_td<Y_t,Y_t>$$ We have $$d<Y_t,Y_t>=c^2X_t^2dt$$

Thus, $$dZ_t=Z_t\left(-\frac{c^2}{2}X_tdt+cX_tdB_t\right)+\frac{1}{2}Z_tc^2X_t^2dt$$

Finally, $$dZ_t=Z_tcX_tdB_t$$

The process has no drift, therefore it is a local martingale

Answer 2

Let $Y_t$ satisfy $$dY_t=cX_t dB_t-\frac{c^2}{2}X_t^2dt.$$ Let $f(x)=\exp(x)$. Then $$df(Y_t)=f'(Y_t)dY_t+\frac{1}{2}f''(Y_t)d[Y]_t.$$ Noting that $f(Y_t)=Z_t$ gives $$dZ_t=Z_tdY_t+\frac{1}{2}Z_td[Y]_t.$$ Note that $d[Y]_t=c^2 X_t^2 dt$, so $$dZ_t=Z_t\left(cX_t dB_t-\frac{c^2}{2}X_t^2dt+\frac{c^2}{2} X_t^2 dt\right).$$ Your result follows.