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(Geometric Poisson process). Let $N(t)$ be a Poisson process with intensity $\lambda>0$, and let $S(0)>0$ and $\sigma>-1$ be given. Using Theorem 11.2.3 rather than the Itô-Doeblin formula for jump processes, show that $$ S(t)=\exp \{N(t) \log (\sigma+1)-\lambda \sigma t\}=(\sigma+1)^{N(t)} e^{-\lambda \sigma t} $$ is a martingale.

     (Geometric Poisson process). Let $N(t)$ be a Poisson process with intensity $\lambda>0$, and let $S(0)>0$ and $\sigma>-1$ be given. Using Theorem 11.2.3 rather than the Itô-Doeblin formula for jump processes, show that
$$
S(t)=\exp \{N(t) \log (\sigma+1)-\lambda \sigma t\}=(\sigma+1)^{N(t)} e^{-\lambda \sigma t}
$$
is a martingale.

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Stochastic Calculus for Finance II : Continuous-Time Models

Stochastic Calculus for Finance II : Continuous-Time Models

Steven E. Shreve 1st Edition

Chapter 11, Problem 3

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Step 1

Given the Poisson process $N(t)$ with intensity $\lambda > 0$, and constants $S(0) > 0$ and $\sigma > -1$, the process $S(t)$ is defined as: $$ S(t) = \exp \{N(t) \log (\sigma+1) - \lambda \sigma t\} = (\sigma+1)^{N(t)} e^{-\lambda \sigma t}. $$ Show more…

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(Geometric Poisson process). Let $N(t)$ be a Poisson process with intensity $\lambda>0$, and let $S(0)>0$ and $\sigma>-1$ be given. Using Theorem 11.2.3 rather than the Itô-Doeblin formula for jump processes, show that $$ S(t)=\exp \{N(t) \log (\sigma+1)-\lambda \sigma t\}=(\sigma+1)^{N(t)} e^{-\lambda \sigma t} $$ is a martingale.

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