What is the distribution of minimum of Brownian motion on arbitrary interval? The Next CEO of Stack OverflowConsidering Brownian bridge as conditioned Brownian motionTime scaling of Brownian motionBrownian motion starts fresh variantBrownian Motion and stochastic integration on the complete real lineExistence of the Brownian Motion using the Kolmogorov extension theoremFinding a probability measure such that the time-shifted Brownian motion is also a Brownian motionCan a time-shifted Brownian motion be also a Brownian motionMaximum process of Brownian motiongeometric Brownian motion with drift closedJoint distribution maximum Brownian motion and prior times
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What is the distribution of minimum of Brownian motion on arbitrary interval?
The Next CEO of Stack OverflowConsidering Brownian bridge as conditioned Brownian motionTime scaling of Brownian motionBrownian motion starts fresh variantBrownian Motion and stochastic integration on the complete real lineExistence of the Brownian Motion using the Kolmogorov extension theoremFinding a probability measure such that the time-shifted Brownian motion is also a Brownian motionCan a time-shifted Brownian motion be also a Brownian motionMaximum process of Brownian motiongeometric Brownian motion with drift closedJoint distribution maximum Brownian motion and prior times
$begingroup$
We know that $P(min_0 leq sleq t B_t leq x)=2P(B_tleq x)$. This can be found in any standard stochastic calculus textbook.
However I am curious about instead of the interval $[0,t]$ if we consider an arbitrary interval $[a,b]$. What will the distribution be then?
By law of total probability we have $P(min_a leq sleq b B_t leq x)=P(min_a leq sleq b B_t leq x|B(a)geq x)$ then using Bayes' theorem we have that this is $P(min_a leq sleq b B_t leq x,B(a)geq x)/P(B(a)geq x)$. Then here I am stuck.
Is there another way?
probability stochastic-processes stochastic-calculus brownian-motion
asked Mar 27 at 19:36
user658409user658409
82
New contributor
user658409 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
$begingroup$
We know that $P(min_0 leq sleq t B_t leq x)=2P(B_tleq x)$. This can be found in any standard stochastic calculus textbook.
However I am curious about instead of the interval $[0,t]$ if we consider an arbitrary interval $[a,b]$. What will the distribution be then?
By law of total probability we have $P(min_a leq sleq b B_t leq x)=P(min_a leq sleq b B_t leq x|B(a)geq x)$ then using Bayes' theorem we have that this is $P(min_a leq sleq b B_t leq x,B(a)geq x)/P(B(a)geq x)$. Then here I am stuck.
Is there another way?
probability stochastic-processes stochastic-calculus brownian-motion
asked Mar 27 at 19:36
user658409user658409
82
New contributor
user658409 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
We know that $P(min_0 leq sleq t B_t leq x)=2P(B_tleq x)$. This can be found in any standard stochastic calculus textbook.
However I am curious about instead of the interval $[0,t]$ if we consider an arbitrary interval $[a,b]$. What will the distribution be then?
By law of total probability we have $P(min_a leq sleq b B_t leq x)=P(min_a leq sleq b B_t leq x|B(a)geq x)$ then using Bayes' theorem we have that this is $P(min_a leq sleq b B_t leq x,B(a)geq x)/P(B(a)geq x)$. Then here I am stuck.
Is there another way?
probability stochastic-processes stochastic-calculus brownian-motion
asked Mar 27 at 19:36
user658409user658409
82
New contributor
user658409 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
We know that $P(min_0 leq sleq t B_t leq x)=2P(B_tleq x)$. This can be found in any standard stochastic calculus textbook.
However I am curious about instead of the interval $[0,t]$ if we consider an arbitrary interval $[a,b]$. What will the distribution be then?
By law of total probability we have $P(min_a leq sleq b B_t leq x)=P(min_a leq sleq b B_t leq x|B(a)geq x)$ then using Bayes' theorem we have that this is $P(min_a leq sleq b B_t leq x,B(a)geq x)/P(B(a)geq x)$. Then here I am stuck.
Is there another way?
probability stochastic-processes stochastic-calculus brownian-motion
probability stochastic-processes stochastic-calculus brownian-motion
asked Mar 27 at 19:36
user658409user658409
82
New contributor
user658409 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 27 at 19:36
user658409user658409
82
New contributor
user658409 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 27 at 19:42
user658409
asked Mar 27 at 19:36
user658409user658409
82
New contributor
user658409 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 27 at 19:36
user658409user658409
82
asked Mar 27 at 19:36
user658409user658409
82
82
New contributor
user658409 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user658409 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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user658409 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
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1 Answer
1
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oldest
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$begingroup$
Let $W_a,b:=min_ale sle bB_s$. Then
beginalign
mathsfP(W_a,ble w)&=mathsfE[mathsfP(W_a,b-B_ale w-B_amid B_a)]\
&=int_mathbbR mathsfP(W_0,b-ale w-x)phi(x;0,a),dx,
endalign
where $phi(x;mu,sigma^2)=(2pisigma^2)^-1/2exp-(x-mu)^2/2sigma^2$.
answered Mar 27 at 19:49
d.k.o.d.k.o.
10.5k630
$endgroup$
$begingroup$
Hi I was just wondering if you could expand this out more afterward? I write this down and try to simplify but there's not much simplifications. Thanks for your time.
$endgroup$
– user658409
Mar 28 at 1:24
$begingroup$
@user658409 It is an integral involving normal cdf. I am not aware of any simplification of this expression; though, it can be computed numerically.
$endgroup$
– d.k.o.
2 days ago
$begingroup$
Thank you for your time.
$endgroup$
– user658409
2 days ago
add a comment |
Your Answer
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1 Answer
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active
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1 Answer
1
active
oldest
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active
oldest
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active
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votes
$begingroup$
Let $W_a,b:=min_ale sle bB_s$. Then
beginalign
mathsfP(W_a,ble w)&=mathsfE[mathsfP(W_a,b-B_ale w-B_amid B_a)]\
&=int_mathbbR mathsfP(W_0,b-ale w-x)phi(x;0,a),dx,
endalign
where $phi(x;mu,sigma^2)=(2pisigma^2)^-1/2exp-(x-mu)^2/2sigma^2$.
answered Mar 27 at 19:49
d.k.o.d.k.o.
10.5k630
$endgroup$
$begingroup$
Hi I was just wondering if you could expand this out more afterward? I write this down and try to simplify but there's not much simplifications. Thanks for your time.
$endgroup$
– user658409
Mar 28 at 1:24
$begingroup$
@user658409 It is an integral involving normal cdf. I am not aware of any simplification of this expression; though, it can be computed numerically.
$endgroup$
– d.k.o.
2 days ago
$begingroup$
Thank you for your time.
$endgroup$
– user658409
2 days ago
add a comment |
$begingroup$
Let $W_a,b:=min_ale sle bB_s$. Then
beginalign
mathsfP(W_a,ble w)&=mathsfE[mathsfP(W_a,b-B_ale w-B_amid B_a)]\
&=int_mathbbR mathsfP(W_0,b-ale w-x)phi(x;0,a),dx,
endalign
where $phi(x;mu,sigma^2)=(2pisigma^2)^-1/2exp-(x-mu)^2/2sigma^2$.
answered Mar 27 at 19:49
d.k.o.d.k.o.
10.5k630
$endgroup$
$begingroup$
Hi I was just wondering if you could expand this out more afterward? I write this down and try to simplify but there's not much simplifications. Thanks for your time.
$endgroup$
– user658409
Mar 28 at 1:24
$begingroup$
@user658409 It is an integral involving normal cdf. I am not aware of any simplification of this expression; though, it can be computed numerically.
$endgroup$
– d.k.o.
2 days ago
$begingroup$
Thank you for your time.
$endgroup$
– user658409
2 days ago
add a comment |
$begingroup$
Let $W_a,b:=min_ale sle bB_s$. Then
beginalign
mathsfP(W_a,ble w)&=mathsfE[mathsfP(W_a,b-B_ale w-B_amid B_a)]\
&=int_mathbbR mathsfP(W_0,b-ale w-x)phi(x;0,a),dx,
endalign
where $phi(x;mu,sigma^2)=(2pisigma^2)^-1/2exp-(x-mu)^2/2sigma^2$.
answered Mar 27 at 19:49
d.k.o.d.k.o.
10.5k630
$endgroup$
Let $W_a,b:=min_ale sle bB_s$. Then
beginalign
mathsfP(W_a,ble w)&=mathsfE[mathsfP(W_a,b-B_ale w-B_amid B_a)]\
&=int_mathbbR mathsfP(W_0,b-ale w-x)phi(x;0,a),dx,
endalign
where $phi(x;mu,sigma^2)=(2pisigma^2)^-1/2exp-(x-mu)^2/2sigma^2$.
answered Mar 27 at 19:49
d.k.o.d.k.o.
10.5k630
answered Mar 27 at 19:49
d.k.o.d.k.o.
10.5k630
answered Mar 27 at 19:49
d.k.o.d.k.o.
10.5k630
answered Mar 27 at 19:49
d.k.o.d.k.o.
10.5k630
10.5k630
$begingroup$
Hi I was just wondering if you could expand this out more afterward? I write this down and try to simplify but there's not much simplifications. Thanks for your time.
$endgroup$
– user658409
Mar 28 at 1:24
$begingroup$
@user658409 It is an integral involving normal cdf. I am not aware of any simplification of this expression; though, it can be computed numerically.
$endgroup$
– d.k.o.
2 days ago
$begingroup$
Thank you for your time.
$endgroup$
– user658409
2 days ago
add a comment |
$begingroup$
Hi I was just wondering if you could expand this out more afterward? I write this down and try to simplify but there's not much simplifications. Thanks for your time.
$endgroup$
– user658409
Mar 28 at 1:24
$begingroup$
@user658409 It is an integral involving normal cdf. I am not aware of any simplification of this expression; though, it can be computed numerically.
$endgroup$
– d.k.o.
2 days ago
$begingroup$
Thank you for your time.
$endgroup$
– user658409
2 days ago
$begingroup$
Hi I was just wondering if you could expand this out more afterward? I write this down and try to simplify but there's not much simplifications. Thanks for your time.
$endgroup$
– user658409
Mar 28 at 1:24
$begingroup$
Hi I was just wondering if you could expand this out more afterward? I write this down and try to simplify but there's not much simplifications. Thanks for your time.
$endgroup$
– user658409
Mar 28 at 1:24
$begingroup$
@user658409 It is an integral involving normal cdf. I am not aware of any simplification of this expression; though, it can be computed numerically.
$endgroup$
– d.k.o.
2 days ago
$begingroup$
@user658409 It is an integral involving normal cdf. I am not aware of any simplification of this expression; though, it can be computed numerically.
$endgroup$
– d.k.o.
2 days ago
$begingroup$
Thank you for your time.
$endgroup$
– user658409
2 days ago
$begingroup$
Thank you for your time.
$endgroup$
– user658409
2 days ago
add a comment |
user658409 is a new contributor. Be nice, and check out our Code of Conduct.
user658409 is a new contributor. Be nice, and check out our Code of Conduct.
user658409 is a new contributor. Be nice, and check out our Code of Conduct.
user658409 is a new contributor. Be nice, and check out our Code of Conduct.
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