Brownian and Geometric Brownian Motion Distributions Under Measure Changes
Summary
Brownian motion with drift, X_t = μt + σW_t, is normally distributed with mean μt and variance σ²t. Exponentiating a normally distributed variable gives a lognormal distribution, so the stated exponential process has lognormal levels; its logarithm has mean μt and variance σ²t. The response also gives the conventional geometric Brownian motion solution, which includes the drift adjustment −σ²/2 and an initial value. This distinction matters because the exponential process as written is not the usual solution to an SDE with level drift μ.
Under a change to an equivalent measure, the response says the drift changes while volatility remains the same, shifting the distribution accordingly. This is a concise account rather than a full derivation: the precise drift adjustment depends on the chosen measure and assumptions, and the explanation does not specify them. The probability measure alone does not imply equal probabilities for outcomes; these continuous distributions are characterized by their parameters.
Key ideas
- Brownian motion with drift is normal with mean μt and variance σ²t.
- Exponentiating that process gives lognormal levels.
- The conventional geometric Brownian motion solution includes a −σ²/2 drift adjustment in its exponent.
- An equivalent measure change can alter drift while preserving volatility under the usual change-of-measure setup.
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# Theoretical distribution of (geometric) Brownian motion (with drift)
# Theoretical distribution of (geometric) Brownian motion (with drift)
I am working on a simulation study which focuses on both the Brownian motion with drift (1) and the geometric Brownian motion (2). I denote them by $X_t$.
What are the theoretical distributions of these processes under the measure P (thus equal probabilities of $\frac{1}{2}$)?
I am getting confused. I know that the processes are of the type:
$(1) X_t = \mu t + \sigma W_t, \\ (2) X_t = \exp (\mu t + \sigma W_t). $
where $W_t \sim N(0,t)$.
However, then what is the theoretical distribution of $X_t$? Are they simply the normal distribution and log-normal distribution? My thought was that it follows a normal distribution with mean $\mu t$ and variance $\sigma^2 t$. The geometric BM would then follow a log-normal distribution with the same parameters. Is that correct?
Furthermore, how do these distributions change under a different (equivalent) measure Q?
Thank you!
## Answer by Magic is in the chain (score 2)
https://quant.stackexchange.com/a/42111
Yes you got it right: the first is normal and the second is log normal as its log is normal.
The GBM solution is $X_t=X_0 e^{\left( \mu-\frac{\sigma^2}{2}\right)t+\sigma W_t}=e^{\ln X_0 +\left( \mu-\frac{\sigma^2}{2}\right)t+\sigma W_t}$. As the exponent is $ N \left [ \ln X_0 +\left( \mu-\frac{\sigma^2}{2}\right)t, \sigma^2 t\right]$, $X_t$ is log normal with the same parameters.
When you change the measure, the drift will change whilst the vol will remain unchanged. The pdf will then be scaled to account for this change in drift.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.