I know that if $W(t)$ is a Brownian motion on the interval $[0,t]$, then $$\mathbb{E}[W(t)]=0$$ but why $$\mathbb{E}[W(1)]=0?$$ Lets say that $W(1)=a$, then why we do not have $$\mathbb{E}[W(1)]=\mathbb{E}[a]=a?$$
2026-03-28 03:02:28.1774666948
Brownian motion expected value of $W(1)$.
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$W(1)$ is a random variable and you cannot just suppose that it is equal to $a$. In fact, you know its distribution. If $W$ is a standard Brownian motion then $W(t) \sim \mathcal{N}(0,t)$ for every $t$. In particular, $W(1) \sim \mathcal{N}(0,1)$ which is why $\mathbb{E}[W(1)] = 0$.