Let $ B(t)(t\geq 0) $ be the standard Brownian motion and $ M(t)=\max_{0\leq s\leq t}{B(s)} $. If we define $ X(t)=M(t)-B(t) $ as a new stochastic process, how can I show that $ X(t) $ has the same distribution as $ B(t) $? I have tried to use the property of Brownian motion to calculate $ P(M(t)>x,B(t)>y) $, but I cannot figure out this probability when $ x>y $. Can anyone give some hint or thoughts about it?
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asked May 27, 2021 at 2:20
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4$\begingroup$ $X(t)$ has the same distribution as $M(t)$ or $|B(t)|$ rather than $B(t)$ (note that $X(t) \geqslant 0$, while $B(t)$ can be negative). This is a standard result known as the reflection principle and it follows as a corollary from the strong Markov property (with certain caveats!). $\endgroup$– Mateusz KwaśnickiCommented May 27, 2021 at 7:08
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$\begingroup$ @MateuszKwaśnicki Surely if $X(t)$ has the same distribution as $|B(t)|$, then it doesn't have the same distribution as $M(t)$, because $M(t)$ is nondecreasing while $|B(t)|$ can decrease? $\endgroup$– Will SawinCommented Jun 21, 2021 at 0:49
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1$\begingroup$ @WillSawin: The problem lies in the meaning of the word "distribution", I suppose. For a fixed $t$, the random variables $X(t)$, $|B(t)|$ and $M(t)$ have the same distribution. (That said, of course these are different random variables: no two of them are equal with probability one). Additionally, $X(t)$ and $|B(t)|$ have equal law as stochastic processes: both are reflected Brownian motions, but of course law of $M(t)$ is different. $\endgroup$– Mateusz KwaśnickiCommented Jun 30, 2021 at 8:18
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