Why Log-Price Drift Under GBM Is Reduced by Half the Variance

Summary

The document asks why the drift of the logarithm of an asset price differs from the asset’s drift under geometric Brownian motion. Under the risk-neutral measure, the price has drift equal to the risk-free rate, while applying Itô’s lemma gives log price a drift reduced by half the variance rate. The questioner suggests that continuous compounding may explain the adjustment.

The answer gives an expectation-based intuition: the exponential of a normally distributed shock has a mean above one, with the excess determined by half the variance times time. The correction to log drift offsets that exponential growth effect, so the asset itself retains the intended expected growth rate. This is a brief intuition rather than a derivation; it does not expand on Itô’s lemma or distinguish expected log return from log expected price in detail.

Key ideas

• Under risk-neutral GBM, the asset-price drift is the risk-free rate.
• Applying Itô’s lemma reduces the drift of log price by half the variance rate.
• Exponentiating a normal shock raises its expected value above one.
• The log-drift correction offsets that effect to preserve the intended expected growth of the asset.

Tags

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


# 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?

## Answer by siou0107 (score 3, accepted)

https://quant.stackexchange.com/a/50612

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).

This is pretty well explained in the chapter on BS model from Hull’s book Options, futures and other derivatives!

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.