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For classic GBM stock price model, $$\frac{dS}{S} = \mu \cdot dt + \sigma \cdot dW$$

we have the solution: $$S(t)=S(0)\, \exp\left(\frac{\mu-\sigma^2}{2} t+\sigma W(t)\right).$$

During the calculation, Ito's lemma could be used.

If we replace the dW to Poisson process, it's still possible to solve it.

However, if we change the dW to a quasi-periodic poisson process, how could we solve the PDE?

By quasi-periodic Poisson process I mean:

Let $\tau_i$ as a sequence of i.i.d exponential random variables, all with same mean $\eta$.

Let $N_i$ as a sequence of i.i.d Poisson processes, all with same density $\lambda$

Define a "Resettable Poisson Process Η(t)" as $$H(t) = N_{i-1}(t-\tau_{i-1}),$$ if $\tau_{i-1}<t<\tau_i$ , for the convenience we just define $\tau_0=0$.

How to solve the PDE

$$\frac{dS}{S} = \mu \cdot dt + \rho \cdot dH\, ?$$

The big difference is the quasi-periodic $H(t)$, it's no longer a martingale.....

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No (pde) in here. –  Did Apr 1 '12 at 16:16
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2 Answers 2

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It seems what you have in mind is that $\tau_0=0$ and, for every $n\geqslant1$, $\tau_n=\sigma_1+\cdots+\sigma_n$ with $(\sigma_n)_{n\geqslant1}$ i.i.d. and exponentially distributed (compare with the version in your post).

Anyway, in this setting, there is no Brownian component in the variations of the process $(S_t)_{t\geqslant0}$ whose paths jump at each time $\tau_n$ and when the Poisson process involved jumps, but have finite variation inbetween. For every $\tau_{n-1}\leqslant t\lt\tau_{n}$ and every $n\geqslant1$, $$ \log S_t=\log S_0+\mu t+\rho N_{n-1}(t-\tau_{n-1}). $$ that is, $$ S_t=S_0\cdot\mathrm e^{\mu t}\cdot\sum_{n\geqslant1}\mathrm e^{\rho N_{n-1}(t-\tau_{n-1})}\cdot[\tau_{n-1}\leqslant t\lt\tau_{n}]. $$ In words, $t\mapsto\rho^{-1}(\log S_t-\log S_0-\mu t)$ performs a pure jump process, integer valued, with jumps $n\to n+1$ at rate $\lambda$ for every $n\geqslant0$ and jumps $n\to0$ at rate $1/\eta$ for every $n\geqslant1$.

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thanks for your reply. However, I can't understand your formula, especially the 2nd one, $\Sigma e^{\rho N_{n-1}(t-\tau_{n-1})}$. Sorry I'm stupid, how did you get that? –  athos Apr 1 '12 at 20:02
    
The series is a convenient way to denote $\mathrm e^{\rho N_{n-1}(t-\tau_{n-1})}$ when $t$ is in the interval $[\tau_{n-1},\tau_{n})$ (note that, for each $t$, exactly one term in the sum is not zero). Thus, the second formula is exactly the first one composed by the exponential. –  Did Apr 1 '12 at 20:44
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Although Didier's answer covers the question you have, I think it is worth to add some comments.

Jump processes are used to be thought of as more complicated than a (continuous) Brownian motion, however it's not true when it comes to an integration since there is an crucial difference: for pure jump process it is possible to define an integral pointwise (i.e. for each $\omega\in \Omega$) so the stochasticity does not play a role at all provided some conditions are satisfies (see below). On the other hand, it does not work for Brownian motion which paths are a.s. of unbounded variation and there we have to use the probabilistic structure of paths and defined an integral as $L^2$-limit, which you can informally think of a more weak construction than the pointwise limit.

Suppose, $t\mapsto H_t(\omega)$ for each $\omega\in\Omega$ is a real-valued right-continuous function defined on $t\in [0,\infty)$ s.t. on any finite interval it has finitely many jumps of the finite amplitude and it is constant in between of jumps. Let us use $\tau_k(\omega)$ to be the time of $k$-th jump of trajectory $H_t(\omega)$. Then for each process $a(t)$ you can define an integral $$ \int\limits_0^t a(s-)dH_s(\omega) = \sum\limits_{s\leq t}a(s-)\Delta H_s = \sum\limits_{k: \;\tau_k\leq t}a(\tau_k-)\Delta H_{\tau_k} $$ which is defined for any fixed $\omega\in \Omega$. Here $f(s-) = \lim\limits_{u\uparrow s}f(u)$ is a left-limit of $f$ at the point $s$ and $$ \Delta f(s) = f(s) - f(s-). $$

Because of the pointwise construction, we don't need Ito lemma to solve SDEs which are driven by pure jump stochastic processes as Poisson process, non-homogeneous Poisson process, the process you have described or any other counting process which in any finite interval has finitely many jumps a.s.

Example 1: consider an SDE $$ dS_t = S_{t-}dH_t. $$ The solution can be written recursively: assume that we are given $S_0$. We know that on the interval $[0,\tau_1)$ the function $H_t$ is constant hence $dH_t = 0$ there - so $dS_t = 0$ on this interval and so $$ S_t = S_0\text{ for }t\in [0,\tau_1). $$ At the time $\tau_1$ we have $$ \Delta S_{\tau_1} = S_{\tau_1-}dH_{\tau_1} = S_0\Delta H_{\tau_1} $$ and since $S_{\tau_1} = S_{\tau_1-} + \Delta S_{\tau_1}$ by definition, we have $S_{\tau_1} = S_0(1+\Delta H_{\tau_1})$. By induction we assume that $S_{\tau_k}$ is known and hence as above we obtain: $$ S_t = \begin{cases} S_{\tau_k},&\text{ if }t\in [\tau_k,\tau_{k+1}) \\ S_{\tau_{k}}(1+\Delta H_{\tau_{k+1}}),&\text{ if }t = \tau_{k+1}. \end{cases} $$ As a result, we obtain that the solution can be written in a more compact form as $$ S_t = S_0\cdot 1_{[0,\tau_1)}(t) +S_0\sum\limits_{k\geq 1}1_{[\tau_k,\tau_{k+1})}\prod\limits_{j=1}^k(1+\Delta H_{\tau_j}). $$ Not that here it does not matter which distribution follow the jump times $\tau_k$ since the solution is given pointwise in $\omega\in \Omega$.

Example 2: consider and SDE $$ S_t = f(S_t)dt + S_{t-}dH_t $$ where $f$ is Lipschitz, and suppose that for an ODE $X_t = f(X_t)dt$ with $X_{t_0} = X_0$ the solution is given by the function $X_t = F(t;t_0,X_0)$. As above, we obtain: $$ S_t = F(t;0,S_0)\text{ for }t\in [0,\tau_1) $$ and $S_{\tau_1} = F(\tau_1;0,S_0)(1+\Delta H_{\tau_1})$. Recursively, we obtain $$ S_t = F(t;\tau_1,S_{\tau_1})\text{ for }t\in [\tau_1,\tau_2) $$ and $S_{\tau_2} = F(\tau_1;\tau_1,S_{\tau_1})(1+\Delta H_{\tau_2})$. This method can be e.g. applied to obtain the solution for your problem where $f(S) = \mu \cdot S$.

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