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I'm struggling to understand the proof as to why the stochastic exponential is a continuous non-negative local martingale. My notes say the following:

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where 3.6 is $Z_t = 1 + \int^t_0 Z_s dX_s$.

I don't understand why showing that 3.7 holds is enough to show that $\mathcal{E}(M)_t$ is a non-negative continuous martingale. I feel like its related to the Ito Isometry because:

$$ \mathbb{E}\left[\int^T_0 e^{2M_t - \langle M\rangle_t} \mathrm{d}\langle M\rangle_t\right] = \mathbb{E}\left[\int^T_0 \left(e^{M_t - \frac{1}{2}\langle M\rangle_t}\right)^2 \mathrm{d}\langle M\rangle_t\right] = \mathbb{E}\left[\left(\int^T_0 e^{M_t - \frac{1}{2}\langle M\rangle_t} \mathrm{d}M_t\right)^2\right] $$

We know that an Ito integral against a martingale is also a martingale. So $\int^T_0 e^{M_t - \frac{1}{2}\langle M\rangle_t} \mathrm{d}M_t$ is a Martingale. Then showing that the expectation of this integral squared is finite just shows that the integral is a square integrable martingale. But how does this imply that the integrand is itself a martingale?

Thanks

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1 Answer 1

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In general, an Ito integral is a local martingale and it is a true martingale provided that the integrand is in $L^2$. This is precisely what the remark is telling you, as saying that the integrand is in $L^2$ is precisely: $$E\left[\left( \int_0^T \mathcal{E}(M)_s d M \right)^2\right] = E\left[ \int_0^T \mathcal{E}(M)_s^2 d \langle M\rangle_s \right] = E\int_0^T \exp \left(2M_s - \langle M \rangle_s \right)d \langle M\rangle_s < \infty $$ The first equality is indeed Ito's isometry. Notice that $\mathcal{E}(M)$ is a fixed point for the stochastic integral operator, akin to how $e^x$ is a fixed point for classical integration/differentiation, so that is why you see it as the integrand and the result of the integral as well.

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