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We know that if $M$ is a continuous local martingale then the Dade exponential $\exp(\lambda M_t - \frac{\lambda^2}{2}\langle M,M \rangle_t)$ is also a continuous local martingale for all $\lambda$. The book I am working from leaves as an exercise the following converse:

If $M$ is adapted and continuous, $A$ is an adapted continuous process of finite variation, and $Z_t^\lambda := \exp(\lambda M_t - \frac{\lambda^2}{2}A_t)$ is a local martingale for all $\lambda \in \mathbb{R}$, then $M$ is a local martingale and $\langle M,M \rangle = A$.

Based on a hint in the book, I computed

$$ \left.\frac{\partial}{\partial \lambda} Z_t^{\lambda} \right\rvert_{\lambda=0} = \left.(M_t - \lambda A_t)Z_t^\lambda\right\rvert_{\lambda=0} = M_t $$

so from the definition of differentiability we have

$$ \lim_{\lambda \rightarrow 0} \left( \frac{Z_t^\lambda-1}{\lambda} \right) = M_t $$

for all $t$. Since each $\frac{Z_t^\lambda-1}{\lambda}$ is a local martingale this implies $M_t$ is the limit of local martingales, but I'm not sure if that is enough to show $M_t$ is as well. If this method works, then showing $\langle M,M\rangle = A$ can be done with the same method by taking a second derivative.

Does this work to show $M$ is a local martingale? If not, what is the proper method and way to use the hint?

EDIT: The book is Continuous Martingales and Brownian Motion by Revuz and Yor

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    $\begingroup$ The definition of differentiability implies almost sure convergence, but in order to maintain the respective Markov Properties, you want $L^1$-convergence. Say $X_{n,t}\to X_t$ in $L^1$ for every $t$ and $X_{n,t}$ is an adapted martingale with respect to $\mathcal{F}_t$ for every $n$. Then, for any $A\in \mathcal{F}_s$ and any $s<t,$ we have $\mathbb{E} 1_A X_t=\lim_{n\to \infty} \mathbb{E}1_A X_{n,t}=\lim_{n\to\infty}\mathbb{E}1_A X_{n,s}=\mathbb{E}1_A X_s,$ proving that $(X_t,\mathcal{F}_t)_{t\geq 0}$ is a martingale. $\endgroup$ Jul 23, 2019 at 16:41
  • $\begingroup$ For the sequence of stopping times, could you just use the localizing sequence for $Z$ and take the min with $\tau_n := \inf\{t>0 : |M_t| > n\}$ since we're given $M$ is continuous? $\endgroup$ Jul 23, 2019 at 16:50
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    $\begingroup$ That works for making $M$ bounded, and should allow you to use the Taylor expansion of $\exp$ to get $L^1$-convergence. I guess you want some compatability between the stopping times observing the Markov properties of $Z^{\lambda}$ for different values of $\lambda$ in order for this to work out? $\endgroup$ Jul 23, 2019 at 16:53
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    $\begingroup$ So the problem is that there isn't just one localising sequence for $Z$, right? There's one for every $\lambda$. $\endgroup$ Jul 23, 2019 at 16:54
  • $\begingroup$ Good point. If there's any sort of monotonicity then it works, which we might be able to get by taking subsequences? $\endgroup$ Jul 23, 2019 at 17:04

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I came up with a different way to prove it that doesn't use the derivative or convergence results:

By continuity of $M_t$ and $A_t$, $Z_t > 0$ a.s. so we can take the log to find $M_t = \log(Z_t^1) + \frac 12 A_t$ is a semi-martingale. Therefore $M_t = M_0 + N_t + B_t$ where $N_t$ is a local martingale and $B_t$ is a process of finite variation. Applying Ito's formula to $Z_t$ yields

\begin{align*} dZ_t^\lambda &= \lambda Z_t^\lambda dM_t - \frac{\lambda^2}{2} Z_t^\lambda dA_t + \frac{\lambda^2}{2} Z_t^\lambda d\langle M,M \rangle_t \\ &= \lambda Z_t^\lambda dN_t + Z_t^\lambda \left( \lambda dB_t - \frac{\lambda^2}{2} dA_t + \frac{\lambda^2}{2} d\langle M,M \rangle_t \right) \\ &= \lambda Z_t^\lambda dN_t \end{align*}

because a local martingale must have no drift terms. Since $e^{\lambda N_t - \frac{\lambda^2}{2} \langle N,N \rangle_t}$ is the unique solution to $dX_t = \lambda X_t dN_t$, this gives that $$Z_t^{\lambda} = e^{\lambda N_t - \frac{\lambda^2}{2} \langle N,N \rangle_t} = e^{\lambda M_t - \frac{\lambda^2}{2} A_t} = e^{\lambda N_t + \lambda B_t - \frac{\lambda^2}{2} A_t}, $$ so $B_t = \frac{\lambda}{2}(A_t - \langle N,N \rangle_t)$. Since this holds for all $\lambda \in \mathbb{R}$, we must have that both sides are $0$ so $M_t = N_t$ and $A_t = \langle N,N \rangle_t = \langle M,M \rangle_t$.

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