Marginal Distribution of Brownian Motion

Standard Brownian motion has independent, normally distributed increments starting from 0.

By definition $B_t\sim N(0,t)$ for standard Brownian motion.

1. 1. Recall the defining property

   Standard Brownian motion satisfies $B_t\sim N(0,t)$ for every $t\ge 0$, since $B_t=B_t-B_0$ is a single increment over an interval of length $t$.

2. 2. Substitute t = 2

   Setting $t=2$ gives the marginal distribution.

   $$B_2\sim N(0,2)$$

Answer: N(0,2)