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!


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.

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  • $\begingroup$ Thank you! And are the parameters correct too, then? Does it mean that the parameters change for every $t$ ? $\endgroup$ – Emily Oct 8 '18 at 20:09
  • $\begingroup$ Yes both mean and variance scale with t. t for example could be your simulation time steps which could be just one long step for European if you use the exact distribution. $\endgroup$ – Magic is in the chain Oct 8 '18 at 20:32
  • $\begingroup$ Your second equation should have have the $-\frac{\sigma^2}{2}t $ term, though I assumed you only left it out to simplify. $\endgroup$ – Magic is in the chain Oct 9 '18 at 7:05
  • $\begingroup$ Still not clear? $\endgroup$ – Magic is in the chain Oct 26 '18 at 10:09

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