Geometric Poisson process

Question

• $\lambda >0$
• $ \sigma > 1$
• $n_t$ si a Poisson process
• $S_t = S_0 \exp( N_t \log( \sigma +1) - \lambda \sigma t ) = S_t e^{- \lambda \sigma t } ( \sigma +1 )^{N_t} $
• $ M_t= N_t - \lambda t$ is the compensated Poisson measure

We want to prove the result slide 14 :

$$S_t= S(0) + \int_0^t S(u_{-} ) dM_u$$

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• $X_t=N_t \log( \sigma +1) - \lambda \sigma t$
• $f(x)=S_0 e^x$
• We apply Ito formula with jumps.

\begin{align*} f(X_t)&= f(X_0)+ \int_{0}^t f'(X_u) dX^{c} + \sum_{0<u \leq t} [ f(X_u) - f({X_u}^-) ] \\ f(X_t)&= S_0 - \int_{0}^t \lambda s S_s ds + \sum_{0<u \leq t} [ f(X_u) - {X_u}^- ] \\ f(X_u) - f({X_u}^-)&= S_{u^{-}} \left( e^{ \ln (\sigma +1)}-1 \right) \mathbb{1}_{ \Delta N_u \ne 0} \\ \sum_{0<u \leq t} [ f(X_u) - {X_u}^- ] &= \int_0^t S_{u-} dN_u \\ \int_{0}^t \lambda s S_s ds &= \int_{0}^t \lambda u S_{u ^{-} } ds \\ S_t&= S_0 + \int_0^t S_{u-} dM_u \\ \end{align*}

With respect to Lebesgue measure $\int S_{ u^{-} }du = \int S_u du $

references :

• stochastic integral
• quadratic variation