Let $B(t)$, $t\geq0$, be a standard Brownian motion. I would like to prove that $$\lim_{L\rightarrow\infty} P\left(\sup_{0\leq s\leq t}|B(s)|>L\right)=0,$$ for each $t\geq0$.

In my class notes, I have the following proof: since $\sup_{0\leq s\leq t}|B(s)|$ has the same probability distribution as $|B(t)|$, then $$P\left(\sup_{0\leq s\leq t}|B(s)|>L\right)=P(|B(t)|>L), $$ and by Chebyshev's inequality, that probability is bounded by $L^{-1} E(|B(t)|)$, which tends to $0$ as $L\rightarrow\infty$.

I do not understand why $\sup_{0\leq s\leq t}|B(s)|$ has the same probability distribution as $|B(t)|$. I know that $\sup_{0\leq s\leq t}B(s)$ (with no absolute value) possesses the same probability distribution as $|B(t)|$, but this is not the same statement. Could you please elaborate on this? Thank you!


You are correct in your doubts. $\sup_{s<t}|B_s|$ does not have the same distribution as $|B_t|$, so your notes have a mistake.

One way around this; you can write $$ \sup_{0<s<t}|B_s|=\max\left(\sup_{0<s<t}B_s,\sup_{0<s<t}-B_t\right) $$ and then the event the LHS is more than $L$ is the union of the events that both arguments of the RHS are more than $L$. Therefore, $P(\sup_{0<s<t}|B_s|>L)\le 2P(|B_t|>L)$.

However, there is an even easier proof. For any (finite) random variable $X$, we have $\lim_{L\to\infty}P(X>L)=0$. This follows from continuity of probability, since the events $\{X>L\}$ decrease to the empty set as $L\to\infty$.


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