# Simulation of the geometric Brownian motion under risk-neutral measure

I hope you can help me again. It is clear how to simulate the GBM: $S_{t_{k}}=S_{t_{k}}exp[(\mu-\frac{\sigma^2}{2})\Delta t_{k+1}+\sigma\sqrt{\Delta t_{k+1}}Z]$, where Z is a stand. norm. dis. RV.

By Girsanov define the BM: $\tilde{W}_t=W_t+\frac{\mu-r}{\sigma}t.$ The dynamic under Q is then given by $dS_t=S_trdt+S_t\sigma d\tilde{W}_t$.

But how do I simulate S under the risk-neutral measure Q? My idea is $S_{t_{k}}=S_{t_{k}}exp[(r-\frac{\sigma^2}{2})\Delta t_{k+1}+\sigma\sqrt{\Delta t_{k+1}}Z_1]$. But should I take a new RV $Z_1$?

Thanks for your help!

• I believe you accidentally dropped an Ito term there, shouldn't it be $(r-\frac{\sigma^2}{2})\Delta t_{k+1}$ in the risk-neutral equation? – Alex C Mar 30 '17 at 23:49

I suspect your expression of Gisanov has the wrong sign and should rather read: $$\tilde{W}_t = W_t + \frac{\mu-r}{\sigma} t$$
Equivalently, in differential form $$d\tilde{W}_t = dW_t + \frac{\mu-r}{\sigma} dt \tag{1}$$
Such that the dynamics under the physical measure $$\frac{dS_t}{S_t} = \mu dt + \sigma dW_t$$ becomes, under the risk-neutral measure \begin{align} \frac{dS_t}{S_t} &= \mu dt + \sigma (d\tilde{W}_t - \frac{\mu-r}{\sigma} dt) \tag{see (1)} \\ &= (\mu - \mu + r) dt + \sigma d\tilde{W}_t \\ &= r dt + \sigma d\tilde{W}_t \end{align}
The discretised version of $(1)$ is \begin{align} \Delta \tilde{W}_t &= \Delta W_t + \frac{\mu-r}{\sigma} \Delta \\ &= Z \sqrt{\Delta} t + \frac{\mu-r}{\sigma} \Delta \end{align} with $Z \sim N(0,1)$. So yes if you're looking at the same realisation $\omega \in \Omega$ you should use the same draw from the standard normal $Z$. If you're only looking at statistical properties then it does not matter as long as your draws $Z_i$ are i.i.d.