Reflexion principle equivalence of statements
$begingroup$
$B_t$ is a Brownian motion, $S_t$ is defined as
$$ S_t := sup_{0 leq s leq t} B_s $$
I want to show that
$$ P(S_tgeq b , B_t leq a) = P(B_t geq 2b-a) $$
for $a leq b$ implies
$$ P(S_t geq a) = 2 P (B_t geq a).$$
I have tried posing $b=a$, however this leads to
$$ P(S_tgeq a) = P(S_tgeq a , B_t leq a) = P(B_t geq a) $$
Which is different from the second statement.
One of those four must be wrong:
- My one-line conclusion
- The first statement (page 2 of https://www.ceremade.dauphine.fr/~mischler/Enseignements/ProcContM1/PrincipeReflexion.pdf)
- The second statement (page 2 of https://ocw.mit.edu/courses/sloan-school-of-management/15-070j-advanced-stochastic-processes-fall-2013/lecture-notes/MIT15_070JF13_Lec7.pdf)
- Mathematics (we are still unsure whether mathematics is consistent, right?).
probability-theory stochastic-processes brownian-motion
asked Jan 2 at 11:11
ThePunisherThePunisher
605616
$endgroup$
add a comment |
$begingroup$
$B_t$ is a Brownian motion, $S_t$ is defined as
$$ S_t := sup_{0 leq s leq t} B_s $$
I want to show that
$$ P(S_tgeq b , B_t leq a) = P(B_t geq 2b-a) $$
for $a leq b$ implies
$$ P(S_t geq a) = 2 P (B_t geq a).$$
I have tried posing $b=a$, however this leads to
$$ P(S_tgeq a) = P(S_tgeq a , B_t leq a) = P(B_t geq a) $$
Which is different from the second statement.
One of those four must be wrong:
- My one-line conclusion
- The first statement (page 2 of https://www.ceremade.dauphine.fr/~mischler/Enseignements/ProcContM1/PrincipeReflexion.pdf)
- The second statement (page 2 of https://ocw.mit.edu/courses/sloan-school-of-management/15-070j-advanced-stochastic-processes-fall-2013/lecture-notes/MIT15_070JF13_Lec7.pdf)
- Mathematics (we are still unsure whether mathematics is consistent, right?).
probability-theory stochastic-processes brownian-motion
asked Jan 2 at 11:11
ThePunisherThePunisher
605616
$endgroup$
add a comment |
$begingroup$
$B_t$ is a Brownian motion, $S_t$ is defined as
$$ S_t := sup_{0 leq s leq t} B_s $$
I want to show that
$$ P(S_tgeq b , B_t leq a) = P(B_t geq 2b-a) $$
for $a leq b$ implies
$$ P(S_t geq a) = 2 P (B_t geq a).$$
I have tried posing $b=a$, however this leads to
$$ P(S_tgeq a) = P(S_tgeq a , B_t leq a) = P(B_t geq a) $$
Which is different from the second statement.
One of those four must be wrong:
- My one-line conclusion
- The first statement (page 2 of https://www.ceremade.dauphine.fr/~mischler/Enseignements/ProcContM1/PrincipeReflexion.pdf)
- The second statement (page 2 of https://ocw.mit.edu/courses/sloan-school-of-management/15-070j-advanced-stochastic-processes-fall-2013/lecture-notes/MIT15_070JF13_Lec7.pdf)
- Mathematics (we are still unsure whether mathematics is consistent, right?).
probability-theory stochastic-processes brownian-motion
asked Jan 2 at 11:11
ThePunisherThePunisher
605616
$endgroup$
$B_t$ is a Brownian motion, $S_t$ is defined as
$$ S_t := sup_{0 leq s leq t} B_s $$
I want to show that
$$ P(S_tgeq b , B_t leq a) = P(B_t geq 2b-a) $$
for $a leq b$ implies
$$ P(S_t geq a) = 2 P (B_t geq a).$$
I have tried posing $b=a$, however this leads to
$$ P(S_tgeq a) = P(S_tgeq a , B_t leq a) = P(B_t geq a) $$
Which is different from the second statement.
One of those four must be wrong:
- My one-line conclusion
- The first statement (page 2 of https://www.ceremade.dauphine.fr/~mischler/Enseignements/ProcContM1/PrincipeReflexion.pdf)
- The second statement (page 2 of https://ocw.mit.edu/courses/sloan-school-of-management/15-070j-advanced-stochastic-processes-fall-2013/lecture-notes/MIT15_070JF13_Lec7.pdf)
- Mathematics (we are still unsure whether mathematics is consistent, right?).
probability-theory stochastic-processes brownian-motion
probability-theory stochastic-processes brownian-motion
asked Jan 2 at 11:11
ThePunisherThePunisher
605616
asked Jan 2 at 11:11
ThePunisherThePunisher
605616
asked Jan 2 at 11:11
ThePunisherThePunisher
605616
asked Jan 2 at 11:11
ThePunisherThePunisher
605616
asked Jan 2 at 11:11
ThePunisherThePunisher
605616
605616
add a comment |
add a comment |
1 Answer
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$begingroup$
Your (my) one-line conclusion is wrong:
$$ P(S_t geq a) = P(S_t geq a, B_t > a) + P(S_t geq a, B_tleq a).$$
I don't know why on earth you (I) assumed that $P(S_t geq a, B_t > a) = 0$.
Since $B_t > a$ implies $S_t >a$, then
$$ P(S_t geq a, B_t > a) = P(B_t geq a).$$
Hence
$$ P(S_t geq a) = P(S_t geq a, B_t > a) + P(S_t geq a, B_tleq a) = 2 P(B_t geq a). $$
The consistency of mathematics is saved... at least for now...
answered Jan 2 at 11:27
ThePunisherThePunisher
605616
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Your (my) one-line conclusion is wrong:
$$ P(S_t geq a) = P(S_t geq a, B_t > a) + P(S_t geq a, B_tleq a).$$
I don't know why on earth you (I) assumed that $P(S_t geq a, B_t > a) = 0$.
Since $B_t > a$ implies $S_t >a$, then
$$ P(S_t geq a, B_t > a) = P(B_t geq a).$$
Hence
$$ P(S_t geq a) = P(S_t geq a, B_t > a) + P(S_t geq a, B_tleq a) = 2 P(B_t geq a). $$
The consistency of mathematics is saved... at least for now...
answered Jan 2 at 11:27
ThePunisherThePunisher
605616
$endgroup$
add a comment |
$begingroup$
Your (my) one-line conclusion is wrong:
$$ P(S_t geq a) = P(S_t geq a, B_t > a) + P(S_t geq a, B_tleq a).$$
I don't know why on earth you (I) assumed that $P(S_t geq a, B_t > a) = 0$.
Since $B_t > a$ implies $S_t >a$, then
$$ P(S_t geq a, B_t > a) = P(B_t geq a).$$
Hence
$$ P(S_t geq a) = P(S_t geq a, B_t > a) + P(S_t geq a, B_tleq a) = 2 P(B_t geq a). $$
The consistency of mathematics is saved... at least for now...
answered Jan 2 at 11:27
ThePunisherThePunisher
605616
$endgroup$
add a comment |
$begingroup$
Your (my) one-line conclusion is wrong:
$$ P(S_t geq a) = P(S_t geq a, B_t > a) + P(S_t geq a, B_tleq a).$$
I don't know why on earth you (I) assumed that $P(S_t geq a, B_t > a) = 0$.
Since $B_t > a$ implies $S_t >a$, then
$$ P(S_t geq a, B_t > a) = P(B_t geq a).$$
Hence
$$ P(S_t geq a) = P(S_t geq a, B_t > a) + P(S_t geq a, B_tleq a) = 2 P(B_t geq a). $$
The consistency of mathematics is saved... at least for now...
answered Jan 2 at 11:27
ThePunisherThePunisher
605616
$endgroup$
Your (my) one-line conclusion is wrong:
$$ P(S_t geq a) = P(S_t geq a, B_t > a) + P(S_t geq a, B_tleq a).$$
I don't know why on earth you (I) assumed that $P(S_t geq a, B_t > a) = 0$.
Since $B_t > a$ implies $S_t >a$, then
$$ P(S_t geq a, B_t > a) = P(B_t geq a).$$
Hence
$$ P(S_t geq a) = P(S_t geq a, B_t > a) + P(S_t geq a, B_tleq a) = 2 P(B_t geq a). $$
The consistency of mathematics is saved... at least for now...
answered Jan 2 at 11:27
ThePunisherThePunisher
605616
answered Jan 2 at 11:27
ThePunisherThePunisher
605616
answered Jan 2 at 11:27
ThePunisherThePunisher
605616
answered Jan 2 at 11:27
ThePunisherThePunisher
605616
605616
add a comment |
add a comment |
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