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?


1 Answer 1


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? $\endgroup$
    – qarabala
    Commented May 21, 2020 at 6:49

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