In the case of the classic Geometric Brownian motion $$dS_t = \mu S_t dt + \sigma S_tdW_t$$ we solve it as $$ S_t = S_0 \exp\left[ \left(\mu - \frac{\sigma^2}{2}\right)t + \sigma dW_t\right] $$ and simulate $S_{t_{i+1}} = S(t_i) \exp\left[ \left(\mu - \frac{\sigma^2}{2}\right)(t_{i+1}-t_{i}) + \sigma dW_t\right]$ with $W_t = \sqrt{t_{i+1}-t_i}Z_{i+1}$.

However, I am working with the slightly different version $$dS_t = \mu S_t dt + \sigma S_t^{\beta/2} dW_t$$ When I solve it using the Ito's Lemma, I get $$S_t = S_0 \exp\left[ \left(\mu - \frac{\sigma^2}{2} S_t^{\beta-2}\right)t + \sigma S^{\beta/2 - 1}_t dW_t\right]$$ and have no idea how to simulate it using normal distribution, since $S_t$ is sitting inside. Is it possible to sample from this process?


No need to use ito's lemma. You can simulate your process directly from the equation: $$dS_t = \mu S_t dt + \sigma S_t^{\beta/2} dW_t$$ which means that: $$S_{t_{i+1}}=S_{t_i}+\mu S_{t_i}(t_{i+1}-t_i)+\sigma S_{t_i}^{\beta/2}\sqrt{t_{i+1}-t_i}Z_{i}$$ where $Z_i$ is a realization of normal distribution with mean 0 and variance equal to 1.

  • $\begingroup$ How do we know that the joint distribution of the samle is the same as the joint distribution of the stochastic process? $\endgroup$ – Cebiş Mellim May 21 '20 at 6:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.