# 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!

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.
• Thank you! And are the parameters correct too, then? Does it mean that the parameters change for every $t$ ? – Emily Oct 8 '18 at 20:09
• Your second equation should have have the $-\frac{\sigma^2}{2}t$ term, though I assumed you only left it out to simplify. – Magic is in the chain Oct 9 '18 at 7:05