# How To Understand the Drift of ln(S) if S Follows Geometric Brownian Motion

As we know, if an asset S follows geometric Brownian motion, under risk neutral measure, it can be expressed as $$\frac{dS}{S}=rdt+\sigma dW$$, by applying Ito's lemma, $$d(lnS)=(r-0.5*σ^2)dt+σdW(t)$$, for me, the mathematical conversion from $$\frac{dS}{S}$$ to $$d(lnS)$$ makes sense, but I'm trying to make sense intuitively why the drift changes from $$r$$ to $$(r-0.5*σ^2)$$. Here is my understanding (but not 100% sure about it): $$dlnS$$ is continuously compounded rate of stock price, due to continuously compounded feature, it takes into account volatility (or standard deviation), so its real drift should be subtracted by this volatility component, I'm wondering if my understanding is correct?

• This is a result from working with stochastic Ito integrals. It is a result from working with rough objects that is hard to reason if you only know regular smooth calculus. – oliversm Jan 8 '20 at 21:50

Because $$\mathbb{E}\left(e^{\sigma W_t}\right) = e^{\frac{1}{2}\sigma^2T} > 1$$, you need that correction to ensure that your asset grows on average at rate $$\mu$$ (or $$r$$ in the risk-neutral measure).
• $$\mathbb{E}\left(e^{\sigma W_t}\right) = e^{\frac12 \sigma^2T} > 1$$ – Canardini Jan 9 '20 at 17:50