Is $X = \max_{s \in [0,1]} B_s = B_1$, where $B_s \sim \mathcal{N}(0,s)$ is a normal random variable with variance $s$ i.e. $\{B_s\}$ is a standard brownian motion process. I'm not sure if this is true or not.
Edit: My original attempt was wrong so I removed it.
My guess is that you're talking about Brownian motion here. No, it is not true that $\mathbb E[\max_{s \in [0,1]} B_s] = 0$. In fact, $\max_{s \in [0,1]} B_s > \max(0, B_1)$ almost surely.
Let $M_t = \sup_{0\leqslant s\leqslant t} B_s$. Recall that $\mathbb P(M_1\geqslant a) =2\mathbb P(B_1\geqslant a)$ by the reflection principle, and since for any $a>0$, $0<\mathbb P(B_1\geqslant a)<1$, it is not true that $M_1\stackrel{\mathrm d}=B_1$. In fact, for any $t>0$, we have $$M_t\stackrel{\mathrm d}=|B_t|\stackrel{\mathrm d}=M_t-B_t.$$ For the first equality, for $T>0$ we have \begin{align} \mathbb P(M_t\geqslant T) &= \mathbb P(M_t\geqslant T, B_t>T) +\mathbb P(M_t\geqslant T, B_t\leqslant T)\\ &=\mathbb P(B_t>T) + \mathbb P(B_T\geqslant T(2-T))\\ &=\mathbb P(|B_t|\geqslant T). \end{align} For the second equality, for $a>0$ we have \begin{align} \mathbb P(M_t-B_t\geqslant a) &= \mathbb P(B_t\leqslant M_t-a)\\ &= \int_0^\infty\int_{-\infty}^{m-a}-2\varphi'(2x-y)\ \mathsf dy\ \mathsf dx\\ &= \int_0^\infty 2\varphi(x+a)\ \mathsf dx\\ &= 2\int_a^\infty \varphi(x)\ \mathsf dx\\ &= \mathbb P(|B_t|\geqslant a), \end{align} where $\varphi(x) = \frac1{\sqrt{2\pi}} e^{-\frac12 x^2}$ is the density of the standard normal.
