probability of a brownian motion being equal to the running maximum

Question

Let $B$ be a standard Brownian motion on $\mathbb{R}$. I would like to show that $$ \mathbb{P} \bigg\{ B_1 = \max_{t \in [0,1]} B_t \bigg\} =0 .$$ I argue that since $\max_{t \in [0,1]} B_t $ has the same distribution as $|B_1|$, $$ \mathbb{P} \bigg\{ B_1 = \max_{t \in [0,1]} B_t \bigg\} = \mathbb{P} \bigg\{ B_1 = |B_1| \bigg\} = \mathbb{P} \bigg\{ B_1 =0 \bigg\} =0 .$$ Is this argument correct (by replacing $\max_{t \in [0,1]} B_t$ by $|B_1|$ simply due to their equality in distribution)?

I am not so convinced about that fact that for any three random variables $X, Y,Z$, if $X$ and $Y$ have the same law, then $\mathbb{P}(X=Z) = \mathbb{P} (Y=Z)$. I just can't prove it rigorously.

Answer 1

The joint distribution of the running maximum $M_1 = \max_{0 \leq t \leq 1}B_t$ and $B_1$ is $$f(m,b) = \frac{2(2m-b)}{\sqrt{2\pi}}\exp(-\frac{(2m-b)^2}{2}), \qquad m\geq 0, b \leq m $$

Hence, $P(M_1 = B_1) = 0$

Note : In your approach, you replaced $M_1$ with $\vert B_1 \vert $ which is not true! They have same distribution but the random variables are different.

Answer 2

$P(B_1 = \max_{t\in[0,1]}B_t) = P(\max_{t\in[0,1]}(B_t-B_1) = 0)= P(\max_{t\in[0,1]}(B_{1-t}-B_1) = 0)$

You can easily check that $(B_{1-t} - B_1)_{t\in [0,1]}$ is itself a Brownian motion, so

$P(\max_{t\in[0,1]}(B_{1-t}-B_1) = 0) = P(\max_{t\in[0,1]}B_t = 0) = P(|B_1| = 0) = 0$

Answer 3

A simpler answer : not only $\max_{t\in[0,1] } B_t$ is distributed like $|B_1|$, but also, $\max_{t\in[0,1] } B_t - B_1$ is distributed like $B_1$. Hence $Pr(\max_{t\in[0,1] } B_t = B_1) = Pr(max_{t\in[0,1] } B_t - B_1 = 0) = Pr(|B_1|=0) = 0$.