How to solve $ \frac{dS(t)}{S(t-)}=\mu dt+\sigma dW(t)+d\left( \sum_{i=1}^{N(t)}\left( V_{i}-1\right) \right) $

Question

Hey I found following Ito formula for jump diffusion process. Let

$$X_{t}=X_{0}+\int_{0}^{t}b_{s}ds+\int_{0}^{t}\sigma_{s}dW_{s}+\sum_{i=1}% ^{N_{t}}\Delta X_{i},$$ where $N_t$ is Poisson process and $W_t$ is standard Wiener process. Then

$$dY_{t}=\frac{\partial f}{\partial t}\left( t,X_{t}\right) dt+b_{t}% \frac{\partial f}{\partial x}\left( t,X_{t}\right) dt+\frac{\sigma_{t}^{2}% }{2}\frac{\partial^{2}f}{\partial x^{2}}\left( t,X_{t}\right) dt+\frac {\partial f}{\partial x}(t,X_{t})\sigma_{t}dW_{t}+\left[ f(X_{t-}+\Delta X_{t})-f(X_{t-})\right] $$

Now I have to solve this SDE: $$ \frac{dS(t)}{S(t-)}=\mu dt+\sigma dW(t)+d\left( \sum_{i=1}^{N(t)}\left( V_{i}-1\right) \right) , $$

Where $V_i$ are i.i.d random variables. $S(t-)$ means left limit and I don't know how to solve it.

Answer 1

Taking note that if $Y_t$ is compound process and $Z_t$ is diffusion without jumps then $d(Y_t.Z_t) = dY_t + dZ_t$

Trying to solve separately

• $dS^1_t= S^1_t(\mu dt +\sigma dW_t)$

• $dS^2_t= S^2_{t-}.dY_t$

and then looking at both solution will lead $S_t=S^1_t.S^2_t$ to be the result.

The first equation I well known and has solution $S^1_t=S^1_0.e^{(\mu-\frac{\sigma^2}{2}).t+\sigma.W_t}$ so I won't go too much in the details (look on the web on geometric Brownian motion).

The second one has solution : $S^2_t=S^2_0.\Pi_{k=1}^{N_t}V_k$

Thank's to proposition 20.14 with $\lambda$ and $\mu_s$ equal to $0$, and $\eta_{T_k}=1-V_k$.

So piecing all this together using $f$ leads to unless mistaken :

$$S_t=S_0.exp((\mu-\frac{\sigma^2}{2}).t+\sigma.W_t).\Pi_{k=1}^{N_t}V_k$$

Regards