# Euler Scheme for Jump-Diffusion models

Jump-diffusion models (as Merton) have following SDE: $$dS_t=\mu S_tdt+\sigma S_t dW_t+S_tdJ_t$$ where $$J_t=\sum_{i=1}^{N_t}(\xi_i - 1)$$ $$\xi_i$$ - i.i.dn $$N_t$$ - Poisson process

Do we in Euler scheme have sth like this?

### $$S_{t+\Delta t}=S_t+\mu S_t\Delta t+\sigma S_t\Delta W_t+S_t\Delta J_t$$

where $$\Delta J_t = \sum_{i=N_{t}}^{N_{t+\Delta t}}(\xi_i -1)$$ So to calculate $$\Delta J_t$$ we have to smulate random variable from Poisson distribution $$\lambda \Delta t$$ which denotes number of jumps between $$t$$ and $$t+\Delta t$$ and then simulate this number of jumps $$\xi$$, am I right? I know that this SDE has a solution, but I want to compare results. Which number of $$N$$ (time steps) is typically optimal to aproximate solution very well?

• For sufficiently small time steps $\Delta t$, you should be able to simulate the jump probability from a Bernoulli with $p(N_{t\to t+\Delta t}=0)=e^{-\lambda \Delta t}$. Mar 27, 2021 at 12:09
• For larger time steps you can directly compute the bound of the number $N_t$ that you may reasonably expect.Personally, I have always simulated using at most 1 jump, only, with a daily resolution. Mar 27, 2021 at 12:15
• I add my code in Python for Kou model, can you look at this at tell me what is wrong? I calculate call option value from Euler and Exact paths and the exact path works well but Euler path returns very strange result (I use $T=1$ and $N=252$ (time steps) Mar 27, 2021 at 18:12
• Usually, we Euler-discretize not the asset price process but the log asset price process (and recover the price simulation by exponentiation). That way, you should be able to get better results, IMO. Mar 29, 2021 at 9:32
• Math 122 : I prefer your initial Euler scheme on S because it is closest to the stock process, it is more natural as it doesn't resort to any transform Your writing is correct except one thing: epsilon follows a lognormal distribution, not a normal one Often the parameters of epsilon are given like for stocks, ie d epsilon / epsilon = mu_epsilon dt + sigma_epsilon dz with z wiener of epsilon
– Gil
yesterday

Commonly, we employ the Euler scheme for $$\Delta\ln(S_t)$$, not for $$\Delta S_t$$.

Let us specify the jump part as

$$S_{t+}=S_{t}J\Rightarrow dS_t=S_t(J-1)$$ where $$J$$ is a strictly positive random variable. (NB: Under Merton we would have $$\ln(J)\sim N(\mu_J,\sigma_J^2)$$ and $$\mathrm{E}(J)=e^{\mu_J+\frac{1}{2}\sigma_J^2}$$)

And for the solution scheme we arrive at:

\begin{align} \frac{dS_t}{S_t}&=\mu dt + \sigma dW_t+(J-1)dN_t\\ y_t&=\ln{S_t}\\ \Rightarrow dy_t&=\left( \mu-\frac{1}{2}\sigma^2 \right)dt+\sigma dW_t+\left(\ln (S_{t+})-(ln S_t)\right)dN_t\\ &=\left( \mu-\frac{1}{2}\sigma^2 \right)dt+\sigma dW_t+\ln(J) dN_t\\ \Rightarrow y_t&=\left( \mu-\frac{1}{2}\sigma^2 \right)t+\sigma W_t + \sum_{i=i}^{N_t}\ln(J_i)\\ \Rightarrow S_t&=S_0e^{\left( \mu-\frac{1}{2}\sigma^2 \right)t+\sigma W_t} \prod_{i=1}^{N_t}J_i\\ \end{align}

Let's assume that we have the Merton jump diffusion model here. Then the Euler discretization is:

\begin{align} y_t&\leftarrow y_0\\ \epsilon_{1,t} & \sim N(0,\sigma^2)\\ \epsilon_{2,t} & \sim N(\mu_J,\sigma_J^2)\\ N_t&\sim \left\{ \begin{array}{1} 0 & p=e^{-\lambda\Delta t}\\ 1 & 1-p\end{array} \right. \\ y_{t+\Delta t}&\leftarrow y_t+\left(\mu-\frac{1}{2}\sigma^2\right)\Delta t+\sigma\sqrt{\Delta t}\epsilon_1+N_t\epsilon_{2,t} \end{align} and $$S_{t}=e^{y_t}$$ accordingly. And if you simulate under the risk-neutral measure, then of course $$\mu=r_f-\lambda\mathrm{E}^{\mathbb{Q}}(J-1)$$.

HTH?

• Thanks :D Could you explain how the Milstein formula for jump-diffusion processes is derived? here on page 6 is a formula but I don't know how it was derived (and how to use it since we need to have value of a Wiener process in a point of jump) uts.edu.au/sites/default/files/qfr-archive-02/QFR-rp176.pdf Mar 30, 2021 at 8:09
• No sorry, I have not yet delved into that area. You could try your luck over at math SE, though - or you post a separate question on this forum. Mar 30, 2021 at 9:43