Reputation
725
Next privilege 1,000 Rep.
Create new tags
Badges
3 10
Newest
 Yearling
Impact
~10k people reached

Jul
30
comment Show that $\mathbb{P}(\tau_{0}>T)\approx\frac{1}{\sqrt{T}}$ where $\{ B(t) : t\geq 0\}$ is a linear brownian motion started at $B(0)=1$
@bcf: Linear Brownian motion = 1-dimensional Brownian motion.
Jul
30
comment Brownian motion: Strong Markov versus translation invariance
@ByronSchmuland Ouch. That's obvious. Thanks for straightening that out.
Jul
30
accepted Brownian motion: Strong Markov versus translation invariance
Jul
30
comment Brownian motion: Strong Markov versus translation invariance
ARGH. I see. I just got confused by the way it's worded in the book. Thanks.
Jul
30
revised Brownian motion: Strong Markov versus translation invariance
added 438 characters in body
Jul
30
comment Brownian motion: Strong Markov versus translation invariance
Sorry, $T_a = \inf\{t: B_t = a\}$, so essentially the same as $S$.
Jul
29
asked Brownian motion: Strong Markov versus translation invariance
Jul
28
comment Tail field versus germ field of Brownian motion
@Did: Ah, that was dumb of me. Thank you for pointing out that error.
Jul
27
asked Tail field versus germ field of Brownian motion
Jul
26
comment Intuition about Blumenthal's 0-1 law
Dang. This is such a simple yet effective device to explain this idea. Thanks!
Jul
26
accepted Intuition about Blumenthal's 0-1 law
Jul
26
answered Central Limit Theorem and understanding mean for a single object
Jul
26
asked Intuition about Blumenthal's 0-1 law
Jul
26
comment Understanding the Markov property of Brownian motion
awesome. thanks for your help!
Jul
26
accepted Understanding the Markov property of Brownian motion
Jul
24
comment Understanding the Markov property of Brownian motion
@user190080 Durrett. Somewhat frustrating because he often leaves notation unexplained and makes big leaps. But it's possible that I'm just dumb.
Jul
24
comment Understanding the Markov property of Brownian motion
Thanks for this. But 2 follow-up questions: so in 1) I'm still a bit confused. Where is the $\omega$ that the $\mathbb{P}^x$-a.s. refers to? And in 2), so am I correct in saying that the function here is $Y(\omega(\cdot)) = \omega(t)$?
Jul
24
asked Understanding the Markov property of Brownian motion
Jun
28
comment Sum of gamma-ish power series
Perfect. Thanks to the both of you.
Jun
28
asked Sum of gamma-ish power series