# Is $\max_{s \in [0,1]} B_s = B_1$ where $B_s \sim \mathcal{N}(0,s)$

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.

• You have NOT told us the JOINT distribution of the random variables $B_s,\,0\le s\le1. \qquad$ Commented Jan 5, 2020 at 19:48

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.
• You are right, my "proof" is actually completely wrong since the $\max$ is over an uncountable sequence. So can we actually say anything about $\max_{s \in [0,1]}B_s$ apart from the fact that $\mathbb{P}(\max_{s \in [0,1]}B_s \geq c) = 2 \mathbb{P}(B_1 \geq c)$ ? Commented Jan 5, 2020 at 17:44
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.
• How can we prove that $M_t = | B_t |$ in distribution ? I think most of my confusion stemmed from the fact that I was thinking $M_t$ is normally distributed for some reason. Commented Jan 5, 2020 at 19:56