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Risk-Neutral GBM Discretization: Asset and Log-Price Drifts

Article Quant Q&A · Author: Egodym

Summary

The document clarifies which drift belongs in a geometric Brownian motion discretization under the physical measure and under the risk-neutral measure. For the asset-price stochastic differential equation, the risk-neutral drift is the risk-free rate. The drift adjusted by half the variance applies after transforming the process to log prices, rather than to the asset-price equation itself.

The answers show the measure change using Girsanov's theorem, then apply Ito's lemma to derive the log-price process. This yields a log increment with drift equal to the risk-free rate less half the variance, and a normally distributed Brownian increment. The distinction helps avoid mixing an asset-price Euler step with a log-price update. The question's displayed Euler formula appears to contain a duplicated volatility factor, and the responses do not compare discretization error or discuss extensions such as dividends or stochastic rates.

Key ideas

  • Under the risk-neutral measure, the GBM asset-price SDE uses the risk-free rate as its drift.
  • The log-price process has drift equal to the risk-free rate minus half the variance.
  • The risk-neutral Brownian motion can be obtained through a Girsanov measure change.
  • Asset-price and log-price discretizations use different drift expressions.

Tags

Full text
# How to discretize a GBM under P- and Q-measures?


# How to discretize a GBM under P- and Q-measures?












Under the P-measure, a geometric Brownian motion can be specified using the following SDE:

$$dS_t=\mu S_tdt+\sigma S_tdW_t^P$$

and its Euler discretization is

$$S_{t+\Delta t}=S_t + \mu S_t \Delta t + \sigma S_t \sigma \sqrt{\Delta t}Zt$$

Under the Q-measure, should the drift $\mu$ be substituted by $r$ or by $r-\frac{\sigma^2}{2}$?

## Answer by SmallChess (score 1, accepted)

https://quant.stackexchange.com/a/18964

$r-\frac{\sigma^2}{2}$ for the drift only applies to the log-returns. The Euler discretisation simply discretises the SDE directly. You'd use the risk-free rate for you drift under the risk-neutral measure for your question.

For your reference:

Please read the wikipedia for more details.

## Answer by user16891 (score -1)

https://quant.stackexchange.com/a/18968

### Girsanov'Theorem

let $\theta_t$ be an adapted procee such that the solution of SDE $$dL_t=-L_t\, \theta_t \,dW_t , \, L_0=1$$ is a Martingale.We set $Q{{|}_{\mathcal{F}_t}}=L_t\,P{{|}_{\mathcal{F}_t}}$,then $$W_{t}^{Q}=W_{t}^{P}+\int_{0}^{t} \theta_s\,ds$$ is a standard wiener process under Q measure.

### Result

Now we assume $\{S_t\}_{t\geq0}$ be a Geometric Brownian Motion. let $\theta=\frac{\mu-r}{\sigma}$ and $dL_t=-L_t\, \theta \,dW_t$. Then

$$W_{t}^{Q}=W_{t}^{P}+\left(\frac{\mu-r}{\sigma}\right)t$$ is a standard wiener process under Q measure. As a result $$dW_{t}^{Q}=dW_{t}^{P}+\frac{\mu-r}{\sigma}dt$$ and $$ dS_t= \mu S_t dt+\sigma S_t dW_t^P=\mu S_t dt+\sigma \,S_t(dW_t^Q-\frac{\mu-r}{\sigma}dt)=r S_t dt+\sigma S_tdW_t^Q$$

### Euler scheme

First let $x_t=\ln S_t$, by application of Ito's lemma, we have $$d{{x}_{t}}=\left( r-\frac{1}{2}{{\sigma }^{2}} \right)dt+\sigma d{{W}_{t}}$$ and $$x_{t+\Delta t}=x_t+\left( r-\frac{1}{2}{{\sigma }^{2}} \right)\Delta t+\sigma (W_{t+\Delta t}-W_t)$$ but we know ${{W}_{t+\Delta t}}-{{W}_{t}}\sim N(0,\Delta t)$ then $$x_{t+\Delta t}=x_t+\left( r-\frac{1}{2}{{\sigma }^{2}} \right)\Delta t+\sigma \sqrt{\Delta t }Z$$ where $Z$ is a standard normal stochastic Variable.

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