Let $X$ be a (cadlag) Levy process with a triplet $(\gamma, \sigma, \nu)$ and it stochastic exponential $\mathcal E$, which is the (cadlag) solution of SDE $d\mathcal E_t=\mathcal E_{t-}dX_t$, $\mathcal E_0=1$. It is known that the explicit formula of $\mathcal E$ is $$\mathcal E_t = e^{X_t - \frac{\sigma^2 t}{2}}\prod_{0\leqslant s \leqslant t}(1+\Delta X_s)e^{-\Delta X_s}$$ (e.g., see Protter, Stochastic Integration and Differential Equations 2005), where the jump $\Delta X_t:=X_t - X_{t-}$.

Assume that $\mathbb F=(\mathcal F_t)_{t\geqslant 0}$ is the augmented natural filtration of $X$, then $\mathbb F$ is right continuous. Since $X$ has strong Markov property, $(X_{\tau +t}-X_\tau)_{t\geqslant 0}$ is independent of $\mathcal F_\tau$ for any stopping time $\tau<\infty$ a.s.

My question is whether the process $(\dfrac{\mathcal E_{t+\tau}}{\mathcal E_\tau})_{t\geqslant 0}$ is independent of $\mathcal F_\tau$? Here we assume that $\Delta X_t \neq -1$ for all $t$ a.s. for making our formula is well-defined.

From the formula above, we only need to check that whether $\prod_{0<s\leqslant t}(1+\Delta X_{\tau +s})e^{-\Delta X_{\tau +s}}$ is independent from $\mathcal F_\tau$? By the strong Markov property of $X$, one can show that $\Delta X_{\tau + s}$ is independent from $\mathcal F_\tau$ for each $s\in (0,t]$. But I'm stuck to prove the same conclusion for the whole product. Does anybody have idea?


1 Answer 1


First of all, the number of jumps $N^{\tau}_t$ that happen between times $\tau$ and $t+\tau,$ the random times $t_i$ at which they happen and the random jumps $\Delta X_{\tau + t_i}$ are jointly independent of $\mathcal{F}_{\tau}$ by the strong Markov property.

Thus for any $A \in \mathcal{F}_{\tau}$ we see that for $Y_{t_i}$ defined according to your problem: $$ \mathbb{E}\bigg[1_A \prod_{i=1}^{N^{\tau}_t} Y_{\tau +t_i} \bigg] = \sum_k\mathbb{E}\bigg[1_A 1_{\{N^{\tau}_{t}=k\}} \prod_{i=1}^{k} Y_{\tau +t_i} \bigg] $$

which now conveniently factorizes due to the assumed inependence.

  • $\begingroup$ Can you explain more details why $N_t^\tau$, $t_i$, $\Delta X_{\tau +t_i}$ are jointly independent of $\mathcal F_\tau$? It is not clear for me. $\endgroup$
    – ntt
    Jan 13, 2018 at 13:31

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