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 ?!
Thanks for your help! Regards,