# Mean and variance of a Brownian motion process

Say $X(t),t\geq 0,$ denotes a Brownian motion process with drift parameter $\mu=3$ and variance parameter $\sigma^2 = 9$. If $X(0)=10$ I want to find

1. $E[X(2)]$
2. $Var[X(2)]$
3. $P(X(2)>20)$

My reasoning is the following: Since the Brownian motion has a Gaussian probability density with mean $t\mu$ and variance $t\sigma^2$, I would say that

1. $E[X(2)] = 2\times 3 = 6$
2. $Var[X(2)] = 2\times 9 = 18$
3. $P(X(2)>20) = \int_{20}^\infty{\frac{1}{6\sqrt{2\pi}}e^{-(x-6)^2/36}dx}$

Is my reasoning correct?

• Brownian motion has independent stationary Gaussian increments, where the variance is proportional to the length of the time/index difference. Usually, these things are defined to have X(0) = 0. You have to take X(0) = 10 into account. Yours perhaps can be written as $X_t = 10 + 3 t + 3 Z_t$ where $Z_t$ is the standard Brownian motion. – passerby51 Sep 9 '17 at 15:54
• Can you see how this can be used to find the probability in #3? It's not entirely obvious to me – BillyJean Sep 9 '17 at 16:56
• X(2) will be a Gaussian variable and your integral has the right idea. You just have to get the correct mean and variance of X(2). ($Z_t \sim N(0,t)$.) Your variance, BTW, seems correct, just fix the mean. – passerby51 Sep 9 '17 at 18:08