# Is the distribution of a univariate Brownian motion (Wiener process) at a fixed time $t$, necessarily normal?

According to the wikipedia article on Wiener processes, the unconditional probability density function of a Wiener process follows a normal distribution with mean $0$ and variance $t$.

I think this means that the probability of observing $x$ at time $t$ is given by the normal distribution (with parameters as above).

My question:****I am wondering where normality, as well as the variance, $t$ comes from? I think of a Wiener process (or brownian motion) as the limit of a random walk. The random walk is given by iid mean $0$ and variance $\sigma^2$ random variables, so I would think that we don't know the distribution of the Wiener process at time $t$, exactly?

I believe that normality is perhaps because of something like a law of large numbers (because my naiveté perhaps lies in thinking only the random variable at time $t$, matters for the realization at $t$, but the realization at $t$ depends on the previous realizations), or perhaps Donsker's Theorem (according to the wiki). I am not too familiar with DOnskers theorem though. Donsker's theorem seems to require each random variable to have variance $1$, though... what if it has variance $\sigma^2$? Do we just normalize this out (and subsequently carriage it around)?

• If you view Brownian motion $W$ as a limit of random walks then the gaussianity of each $W_t$ follows from the arch classical central limit theorem (and there is no need to wake up Donsker for that). – Did Nov 23 '16 at 9:46
• Ah, because we have $iid$. Thank you. – user106860 Nov 23 '16 at 16:14

## 1 Answer

Viewing brownian motion as the limit of random walks, we can use a CLT to establish that the distribution of each $W_t$ is gaussian. Also, we should be able to get the variance from the CLT as well.