Exponential martingales of Levy process conditioned on $\sigma$-algebra generated by stopping time

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

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.