Why a Martingale GBM Has a Negative Log-Price Drift
Summary
The document clarifies an apparent contradiction in simulating a geometric Brownian motion under a risk-neutral measure with zero interest rates. For the price process to be a martingale, the drift of its logarithm must offset the convexity introduced by exponentiation. As a result, the simulated log-price has a negative expected change even though the expected price remains at its initial level.
The explanation rests on Jensen’s inequality: the average of exponentials is not the exponential of the average. It also cautions that a sample of simulated paths may not reproduce the theoretical expectation closely, especially with a limited number of runs. Increasing the simulation count and computing a confidence interval can help distinguish sampling variation from a coding or modeling error. The answer gives the central intuition but does not provide a numerical example or a procedure for constructing that interval.
Key ideas
- A zero-rate geometric Brownian price is a martingale when its log drift offsets the diffusion convexity term.
- The expected log-price can fall while the expected price stays unchanged.
- Exponentiation is nonlinear, so the exponential of an average is not the average of exponentials.
- Finite simulations can differ from the theoretical expectation because of sampling variation.
- More paths and a confidence interval can help assess whether a simulated deviation is meaningful.
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# Simulating a GBM with martingale condition - Ito process moving downwards
# Simulating a GBM with martingale condition - Ito process moving downwards
I want to correctly simulate a $\mathcal{Q}$ - martingale $S$, which is a geometric Brownian motion and an exponential of a process $X$, \begin{equation} X_t = X_0 + \mu t + \sigma B_t = X_{t-\Delta t} + \mu \Delta t + \sigma B_{\Delta t}, \end{equation} where $X_0 = 0$ and $B$ is a Brownian motion under $\mathcal{Q}$, such that \begin{equation} S_t = S_0 \exp(X_t) = S_0 \exp(\mu t + \sigma B_t) = S_{t-\Delta t} \exp(\mu \Delta t + \sigma B_{\Delta t}), \end{equation} with $\mu = -\sigma^2/2$ from the martingale condition (no interest rates, or $r=0$).
But when I run many (eg. N=1000) simulations of $(X_t)_{t=\Delta t}^T$ over a one-year time horizon ($T=1$, using the first equation above for simulation) with $\Delta t = 1/250$, the average of $X_T$ is significantly lower than $X_0 = 0$, which implies that also $S_T$ is on average significantly lower than $S_0$. This seems understandable to me since I learnt that the above equation for $S_t$ is the solution of the dynamics $dS/S = \mu dt + \sigma dB_t$, and that, from Ito's lemma applied to the latter, in order for $S$ to be a martingale, the drift $\mu$ of the process $X$ needs to equal $-\sigma^2/2$; thus $X$ should go down on average. However, from the martingale property of $S$, I would expect $S_T$ to be on average on the level of $S_0$. What is wrong? Can anybody write a concise illustration of the concept?
## Answer by AFK (score 1, accepted)
https://quant.stackexchange.com/a/17199
The average of the exponentials is not the exponential of the average. It is always higher due to convexity (Jensen inequality). So there is no contradiction between the average of $X_T$ being negative and the average of $S_T$ being $S_0$.
So the question is: are your results really significantly different from what you would expect? Have you tried increasing your simulation number? Do you know how to compute a confidence interval as a function of N?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.