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I want to prove that $Z_t$ the Doléans-Dade exponential is a local martingale i.e. that there exists a stopping time $\tau_n$ tending to infinity such that the stopping process $\mathbb{1}_{\tau_n>0}Z_{\tau_n}$ is a martingale. I don't know how to start and how to prove the existence of such stopping time ! any hint or help will be appreciated.. thank you for your time :)

Doléans-Dade exponential: $$ Z_t=e^{-\int_0^t\beta_s dW_s-\frac{1}{2}\int_0^t \beta_s^2 ds} $$

where $W_t$ is a Brownian motion and $\beta_t$ is stochastic process.

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    $\begingroup$ You're never going to prove this by demonstrating an explicit sequence of stopping times, it's a case of finding the right equivalents. I would usually define the exponential as a local martingale satisfying a certain SDE though... $\endgroup$ – John Fernley Jun 22 '16 at 11:52
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    $\begingroup$ ie Ito's formula $\endgroup$ – Chill2Macht Jun 22 '16 at 14:21
  • $\begingroup$ the only SDE I have is $d\xi_t=\beta_t dt+dW_t$ where $(\xi_t,\mathcal{F}_t)$ is an Itô process adapted to the filtration $\mathcal{F}_t$ ; now what is the equivalent of my statement using this SDE ? :) $\endgroup$ – houda Jun 22 '16 at 15:51
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Define $X_t = -\int_0^t\beta_s\,dW_s$ and $Y_t = -\frac{1}{2}\int_0^t\beta_s^2\,ds$. Then $Z_t = e^{X_t+Y_t}$. Even though you did not mention it I am guessing that there is a condition on $(\beta_s)_{s\geq 0}$ that makes $X$ a local martingale. Since Brownian motion has continuous paths, $X$ is continuous as well, which implies $Z$ is continuous as well. Hence, (needless to say) $X$ and $Y$ are both semimartingales. Since $(x,y) \mapsto e^{x+y}$ is twice continuously differentiable in all its arguments, Ito's formula applies, i.e.

$$dZ_t = Z_tdX_t + Z_tdY_t + \frac{1}{2}Z_td[X,X]_t+ \frac{1}{2}Z_td[Y,Y]_t + Z_td[X,Y]_t$$

You can check the following identities yourself. $$dX_t = -\beta_tdW_t$$ $$dY_t = -\frac{1}{2}\beta_t^2dt$$ $$d[X,X]_t = \beta^2_tdt$$ $$d[X,Y]_t = d[Y,Y]_t = 0$$

Substituting these identities into the SDE above, $$dZ_t = -Z_t\beta_tdW_t - Z_t\frac{1}{2}\beta_t^2dt + Z_t\frac{1}{2}\beta^2_tdt$$ Hence, $dZ_t = -Z_t\beta_tdW_t$. I mentioned above that there must be a condition on $\beta$ that makes $X$ a local martingale. The most general condition for this is that $$\int_0^t\beta_s^2\,ds < \infty \quad \text{a.s.} \qquad (\triangle)$$ for every $t < \infty$. We will require the same for $Z$ to be a local martingale, i.e. $$\int_0^tZ_s^2\beta_s^2\,ds < \infty \quad \text{a.s.} \qquad (\square)$$ for every $t < \infty$. Now fix an arbitrary $0 < t < \infty$. Let $\Omega_t$ denote the full measure set where $(\triangle)$ is true. Let $\Omega_c$ be the full measure set where $Z$ is continuous. Fix $\omega \in \Omega_c \cap \Omega_t$. Since $s \mapsto Z_s(\omega)$ is continuous, $s \mapsto 1_{[0,t]}(s)Z_s(\omega)$ is bounded. Then, $$\int_0^tZ_s(\omega)^2\beta_s(\omega)^2\,ds \leq \sup_{s \in [0,t]}Z_s(\omega)^2 \int_0^t\beta_s(\omega)^2\,ds < \infty$$ Since $\Omega_c \cap \Omega_t$ has full measure, $(\square)$ is true. So then, $Z$ is a local martingale.

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