Let $X=(X_t)_{t\geq 0}$ be a Levy process in a probability space $(\Omega,\mathscr{F},\mathbb{P})$ endowed with (the natural enlargement of) the filtration generated by $X$ and such that $\cup\mathscr{F}_t$ generates $\mathscr{F}$. Let $\psi$ denote the Laplace exponent of $X$, i.e., $\psi(\beta)= \log \mathbb{E}[e^{\beta X_1}]$ whenever the expectation is finite and for each $t$, let $\mathcal{E}_t(\beta)$ be the exponential martingale of $X$ so that $\mathcal{E}_t(\beta)=e^{\beta X_t-\psi(\beta)t}$. Finally, let $\mathbb{P}^\beta$ be the probability measure induced by $\mathcal{E}(\beta)$ on $(\Omega,\mathscr{F})$.

I'm trying to prove that if $\tau$ is a stopping time and $\mathscr{F}_\tau$ is the $\sigma$-algebra associated with it, then on the set $\{\tau<\infty \}$, the Radon-Nikodym derivative of $\mathbb{P}^\beta\mid_\mathscr{F_\tau}$ with respect to $\mathbb{P}\mid_\mathscr{F_\tau}$ is given by $\mathcal{E}_\tau(\beta)$.

The book that I am following gives the following proof:

For $A\in\mathscr{F}_\tau$, then $A\cap \{\tau\leq t\}\in\mathscr{F}_t$, so

$$ \mathbb{P}^\beta[A\cap \{\tau\leq t\}]=\mathbb{E}[\mathcal{E}_t(\beta)1_{A\cap \{\tau\leq t\} }]=\mathbb{E}\left[\mathbb{E}[\mathcal{E}_t(\beta)1_{A\cap \{\tau\leq t\} }\mid \mathscr{F}_\tau]\right]=\mathbb{E}[\mathcal{E}_\tau(\beta)1_{A\cap \{\tau\leq t\} }] $$ where in the third equality the strong Markov property is used together with the martingale property of $\mathcal{E}(\beta)$. One then concludes using monotone convergence.

I don't understand how the strong Markov property and the martingale property are being used in the equation above. I know that one can, for example, do $$ \mathbb{E}[\mathcal{E}_t(\beta)1_{A\cap \{\tau\leq t\} }\mid \mathscr{F}_\tau] = 1_{A\cap \{\tau\leq t\} }e^{\beta X_\tau-\psi(\beta)t}\mathbb{E}[e^{\beta (X_{(t-\tau)+\tau}-X_\tau)}1_{\{\tau\leq t\} }\mid \mathscr{F}_\tau] $$ and I guess one can conclude from the strong Markov property that $\mathbb{E}[e^{\beta (X_{(t-\tau)+\tau}-X_\tau)}1_{\{\tau\leq t\} }\mid \mathscr{F}_\tau] =e^{\psi(\beta)(t-\tau)}$, but I don't see how this last step is true. I know that, $(X_{\tau+t}-X_\tau)_{t\geq 0}$ is a Levy process which has the same distribution as $X$, but in this case $t$ is a fixed time and not random like $t-\tau$. Moreover, I don't know how one can just "ignore" the indicator function, which in a way seems to be necessary to make sense of the indexing of $X_{t-\tau}$.

Any help would be appreciated. Thanks in advance


1 Answer 1


Is ${\cal E}_t(\beta)$ not a strong Markov process whenever $X_t$ is one ? If so: ${\mathbb E}[{\cal E}_t(\beta)|{\cal F}_\tau]={\cal E}_\tau(\beta)$. The indicator being ${\cal F}_\tau$-measurable goes in for free into this equation.


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