Ito formula for jump-Diffusion

Suppose $X_t$ is a jump-diffusion process with evolution given by $$X_t=X_0+\int_0^t a_s\,ds +\int_0^t b_s \, dWs + \sum_{i=1}^{N_t} \Delta X_i$$ where $a_t$ is the drift term

$b_t$ is volatility term

And $\Delta X_i$ corresponds to jump $i$ in the stock price. Then cont and tankov state that for a function $f(X_t,t)$ $$(1) \space\space\space df(X_t,t)=\frac{\partial f(X_t,t)}{\partial t}dt+a_t\frac{\partial f(X_t,t)}{\partial x} \, dt+\frac{b_t^2\sigma^2}{2} \frac{\partial f(X_t,t)}{\partial x^2} \, dt+\\ b_t\frac{\partial f(X_t,t)}{\partial x} \, dW_t+\left(f(X_{t-}+\Delta X_t)-f(X_{t-})\right).$$ Using this theorem for $f(\cdot)=\ln(\cdot)$ and $S_t$ described by the stochastic diff. eq. (below) $$\color{red} {\frac{dS_t}{S_t}=\mu dt +\sigma dW_t+dj}$$ we get $$(2) \space\space\space d\ln S_t =\frac{\partial \ln S_t}{\partial t}\,dt+\mu S_t \frac{\partial \ln S_t}{\partial S_t} \, dt+\sigma S_t\frac{\partial \ln S_t}{\partial S_t} \, dW_t+(\ln j_tS_t-\ln S_t) \\ \mu S_t\frac{1}{S_t} \, dt + \dfrac{\sigma^2S_t^2}{2} \left(\frac{1}{-S_t^2}\right) \, dt + \sigma S_t\frac{1}{S_t} \, dW_t + (\ln j_t+\ln S_t-\ln S_t)=\\ \left(\mu-\frac{1}{2}\right) \, dt+\sigma \,dW_t+\ln j_t$$ this is solved to give $$\ln S_t=\ln s_0 + \left(\mu-\dfrac{1}{2}\right)t+\sigma W_t + \sum_{i=1}^{N_t} \ln j_t$$ now my question is :how he apply ito differential to (1) by $f(x)=\ln x$ to obtain (2)...I am sorry to ask this question, but I can't use Ito formula to obtain (2).

The problem here is that you are using the Ito formula for Lévy processes – your process is not a Lévy process, but a semimartingale that is driven by a Lévy process (read the 8.3.4 – "Ito Formula for Seminartingales" in the Tankov book). For example, at a point $t$ where there is a jump $j_t$, your process doesn't satisfy $$S_t = S_{t-} + j_t$$ but rather $$S_t = j_tS_{t-}$$ More precisely, you process is a Dolean-Dades exponential of a Lévy process.

So you have to use Ito for semimartingales (proposition 8.19 in Tankov), i.e.

\begin{align} f(t, X_t) - f(0, X_0) &= \int^t_0 \frac{\partial f}{\partial s}(s, X_s) ds + \int_0^t\frac{\partial f}{\partial x}(s, X_s) dX_s \\ &+ \frac{1}{2} \int^t_0 \frac{\partial^2 f}{\partial x^2}(s, X_s) d[X, X]_s^c \\ &+\sum_{s \leq t, \Delta X_s \neq 0} [f(s, X_s) - f(s, X_{s-}) - \Delta X_s\frac{\partial f}{\partial x}(s, X_s)] \end{align} where $[X, X]_s^c$ is the quadratic variation of the continuous part of $X$.

This formula is less straight-forward to apply, because for example if you insert $S_t$ as $X_t$ and use $\log$ as $f$, then the second term on RHS is $$\int_0^t\frac{\partial f}{\partial x}(s, S_s) dS_s = \int_0^t\frac{1}{S_t}S_t(\mu ds + \sigma dW_s + dj) = \mu t + \sigma W_t + \int^t_0dj$$ How do you interpret the last integral in this situation? It is actually an integral against the jump measure of $S_t$, $J_S(x, t)$. This is pure hand-waving, but for now I just argue that this is the sum of jumps up to time $t$: $$\int^t_0dj = \int_0^t \int_\mathbb R xJ_S(ds, dx) = \sum_{s\leq t, \Delta S_s \neq 0} j_s$$ This part will cancel against the term $- \Delta X_s\frac{\partial f}{\partial x}(s, X_s)$ inside the sum in the Ito for semimartingales so that essentially, you will be left with the usual Ito for Lévy processes, except that the jump part will look like $$f(X_t) - f(X_{t-})$$ instead of $$f(X_{t-} + \Delta X_t) - f(X_{t-})$$ Thus, at a point where there is a jump, you get $$f(S_t) - f(S_{t-}) = f(j_t S_{t-}) - f(S_{t-}) = \ln(j_t) + \ln(S_{t-}) - \ln(S_{t-})$$

As for your solution attempt I'm not sure what you tried to do there – it seems you are trying to derive an Ito formula of your own. I suggest using the original formula with the said modification for the jumps.

Your process corresponds to $$S_t = S_0 + \int^t_0 S_s \mu ds + \int^t_0 \sigma S_s dW_s + \sum S_s j_s$$ from which you can read off $a_t$, $b_t$, and then plug into the Ito formula. For example, $$a_t \frac{\partial f(S_t, t)}{\partial x} dt = \mu S_t \frac{1}{S_t} dt = \mu dt$$

• thanks a million Jun 5 '17 at 0:24
• @Khosrotash I removed that $[X, X]^c = t$. This was wrong, because I forgot you need the quadratic variation of a geometric BM, not regular BM. This doesn't affect the rest of the analysis though.
– user126540
Jun 5 '17 at 6:14
• I am wondering what it will happen if $J_t = 0$? If there is no jumps, the last term related to $\ln J_t$ should disappear. Is that because $S_{t}=S_{t-}(1+J_t)$ instead of $S_t = S_{t-}J_t$? Nov 30 '17 at 20:57

Is my work true ? $$f(x,t)=\ln x \to (It\hat{o} \space formula)\\ df=\frac{\partial f}{\partial t}dt+\frac{\partial f}{\partial x}dx+\frac12\frac{\partial^2 f}{\partial x^2}(dx)^2\\ df=0dt+\frac{1}{ x}dx+\frac12.\frac{-1}{ x^2}.(dx)^2$$ then put $dS_t=\mu S_tdt+\sigma S_tdW_t+S_tdj$ into $df$ $$d(\ln S_t)=0dt+\frac{1}{S_t}dS_t-\frac{1}{2S_t^2}(dS_t)^2=\\ 0dt+\frac{1}{S_t}\color{red} {(\mu S_tdt+\sigma S_tdW_t+S_tdj)}-\frac{1}{2S_t^2}(\color{red} {\mu S_tdt+\sigma S_tdW_t+S_tdj})^2=\\0dt+\mu S_t+\sigma W_t +dj-\frac{1}{2S_t^2}(\mu^2S_t^2dt^2+(\sigma S_tdW_t)^2+(S_tdj)^2+2\mu S_tdt\sigma S_tdW_t+2\sigma S_tdW_tS_tdj+2\mu S_tdtS_tdj)\\=$$ simplify ,and be ware of $$dt^2 \to 0 \\(dW_t)^2 \to dt\\dt.dW_t \to 0\\dW_t.dj \to 0 \\dj.dj\to 0 \\dj.dt \to 0$$ $$\\0dt+\mu S_t+\sigma W_t +dj-\frac{1}{2S_t^2}(\sigma^2 S^2_tdt)$$ so $$dS_t=(\mu-\frac12 \sigma^2)dt+\sigma dW_t+dj$$ vs $$\left(\mu-\frac{1}{2}\right) \, dt+\sigma \,dW_t+\ln j_t$$ I don't now what happen for $dj \to d\ln j$ ? Hint at this point is needed !