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

• 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. Jul 23, 2019 at 16:41
• 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? Jul 23, 2019 at 16:50
• 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? Jul 23, 2019 at 16:53
• So the problem is that there isn't just one localising sequence for $Z$, right? There's one for every $\lambda$. Jul 23, 2019 at 16:54
• Good point. If there's any sort of monotonicity then it works, which we might be able to get by taking subsequences? Jul 23, 2019 at 17:04

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 $$A_t = \langle N,N \rangle_t = \langle M,M \rangle_t$$.