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

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.

• What do you mean by solving it ? Commented Aug 24, 2020 at 8:42
• @TheBridge i.e. find a process $S_t$ which satisfies this equation. Commented Aug 24, 2020 at 21:37

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