Timeline for Why the distribution of M(t) is the same as X(t)?
Current License: CC BY-SA 4.0
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| Jun 30, 2021 at 8:18 | comment | added | Mateusz Kwaśnicki | @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. | |
| Jun 21, 2021 at 0:49 | comment | added | Will Sawin | @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? | |
| May 27, 2021 at 10:25 | review | Close votes | |||
| Jun 24, 2021 at 3:05 | |||||
| May 27, 2021 at 10:00 | history | edited | Luis Yanka Annalisc | CC BY-SA 4.0 |
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| May 27, 2021 at 7:08 | comment | added | Mateusz Kwaśnicki | $X(t)$ 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!). | |
| May 27, 2021 at 2:20 | history | asked | Luis Yanka Annalisc | CC BY-SA 4.0 |