# Show that this is the unique solution of that Stochastic Differential Equation

Reading through a paper, I stumbled across the stochastic differential equation $dS_t = \sigma S_{t-} dX_t$. The claim there was that $S_t = S_0 \exp(\alpha N_t - \beta t)$ should be its unique solution.

The context was the following:

Let $(N_t)_{t \in [0, T]}$ be a standard Poisson process with intensity $\lambda > 0$ on a probability space $(\Omega, \mathcal{F}, P)$. We set $X_t := N_t - \lambda t$, so that $X$ is a martingale; and let $\sigma > 0, \ \alpha := \log (1 + \sigma), \ \beta := \sigma \lambda$.

Now, how can I see that $\ S_t = S_0 \exp(\alpha N_t - \beta t) \$ is the unique solution of the Stochastic Differential Equation $\ d S_t = \sigma S_{t-} d X_t \$ ?

I already know how to show that it is indeed a solution:

First we note that $\Delta S_s = S_0 \exp(\alpha N_{s-}) \cdot (\exp ( \alpha \Delta N_s ) - 1) = S_{s-} \, \sigma \, \Delta N_s$.

Then, we take $f : \mathbb{R} \to \mathbb{R}$ to be the function $S_0 \exp(\cdot)$, and invoke Itō's formula for the $C^2$-function $f$ and the (semi)martingale $X$:

\begin{align*} f(X_t) &= f(0) - \beta \int_0^t f'(X_{s-}) ds \ + \ \sum_{0 < s \leq t} \ (f(X_s) - f(X_{s-}) \ \\ &= S_0 - \beta \int_0^t S_{s-} ds + \sum_{0 < s \leq t} \Delta S_s \\ &= S_0 - \sigma \int_0^t S_{s-} d (\lambda s) + \sigma \, \sum_{0 < s \leq t} S_{s-} \Delta N_s \\ &= S_0 - \sigma \int_0^t S_{s-} d(\lambda s) + \sigma \int_0^t S_{s-} d N_s \\ &= S_0 + \sigma \int_0^t S_{s-} d X_s . \end{align*}

But how can I show that this solution is unique? I guess this can be done somehow via the Itō product formula or so ?!

Hint: consider another solution $Y_t$, so $Z_t = S_t-Y_t$ also solves $dZ_t = \sigma Z_{t-}dX_t$ and $Z_0 = 0$. Prove that $Z_t = 0$ a.s. for all $t$. –  Ilya Nov 28 '11 at 12:52