Hitting time and its distribution
÷I'm reading an italian book about casual process (Probabilità e modelli aleatori of Enzo Orsingher). At pag 105 there's the probability of the stopping time $T_beta$.
$$P{T_beta leq t}=P{max_{0leq sleq t} B(s) geq beta}=$$
$$=2P{B(t) geq beta} = sqrt{frac{2}{pi}} int_{frac{beta}{sqrt{t}}}^{infty} exp{-frac{w^2}{2}}dw$$
Because the book used the last result :
$$P{T_beta leq t}=P{max_{0leq sleq t} B(s) geq beta}= 2P{B(t) geq beta}$$
Now I don't undersatnd why $T_beta$ has the inverse gaussian density
$$f_{beta}(t)= frac{|beta|}{sqrt{2pi t^3}}expleft{-frac{beta^2}{2t}right}$$
for $t >0$.
Where did it find?
inverse brownian-motion stopping-times
edited Sep 6 '14 at 10:59
Davide Giraudo
125k16150260
asked Sep 6 '14 at 10:16
Cronos
233
add a comment |
÷I'm reading an italian book about casual process (Probabilità e modelli aleatori of Enzo Orsingher). At pag 105 there's the probability of the stopping time $T_beta$.
$$P{T_beta leq t}=P{max_{0leq sleq t} B(s) geq beta}=$$
$$=2P{B(t) geq beta} = sqrt{frac{2}{pi}} int_{frac{beta}{sqrt{t}}}^{infty} exp{-frac{w^2}{2}}dw$$
Because the book used the last result :
$$P{T_beta leq t}=P{max_{0leq sleq t} B(s) geq beta}= 2P{B(t) geq beta}$$
Now I don't undersatnd why $T_beta$ has the inverse gaussian density
$$f_{beta}(t)= frac{|beta|}{sqrt{2pi t^3}}expleft{-frac{beta^2}{2t}right}$$
for $t >0$.
Where did it find?
inverse brownian-motion stopping-times
edited Sep 6 '14 at 10:59
Davide Giraudo
125k16150260
asked Sep 6 '14 at 10:16
Cronos
233
1
Since you know $P(T_betaleqslant t)$, how to deduce the density $f_beta$?
– Did
Sep 6 '14 at 11:45
yeah, I didin't derivate respect to t. Now It's clear.
– Cronos
Sep 6 '14 at 18:15
add a comment |
÷I'm reading an italian book about casual process (Probabilità e modelli aleatori of Enzo Orsingher). At pag 105 there's the probability of the stopping time $T_beta$.
$$P{T_beta leq t}=P{max_{0leq sleq t} B(s) geq beta}=$$
$$=2P{B(t) geq beta} = sqrt{frac{2}{pi}} int_{frac{beta}{sqrt{t}}}^{infty} exp{-frac{w^2}{2}}dw$$
Because the book used the last result :
$$P{T_beta leq t}=P{max_{0leq sleq t} B(s) geq beta}= 2P{B(t) geq beta}$$
Now I don't undersatnd why $T_beta$ has the inverse gaussian density
$$f_{beta}(t)= frac{|beta|}{sqrt{2pi t^3}}expleft{-frac{beta^2}{2t}right}$$
for $t >0$.
Where did it find?
inverse brownian-motion stopping-times
edited Sep 6 '14 at 10:59
Davide Giraudo
125k16150260
asked Sep 6 '14 at 10:16
Cronos
233
÷I'm reading an italian book about casual process (Probabilità e modelli aleatori of Enzo Orsingher). At pag 105 there's the probability of the stopping time $T_beta$.
$$P{T_beta leq t}=P{max_{0leq sleq t} B(s) geq beta}=$$
$$=2P{B(t) geq beta} = sqrt{frac{2}{pi}} int_{frac{beta}{sqrt{t}}}^{infty} exp{-frac{w^2}{2}}dw$$
Because the book used the last result :
$$P{T_beta leq t}=P{max_{0leq sleq t} B(s) geq beta}= 2P{B(t) geq beta}$$
Now I don't undersatnd why $T_beta$ has the inverse gaussian density
$$f_{beta}(t)= frac{|beta|}{sqrt{2pi t^3}}expleft{-frac{beta^2}{2t}right}$$
for $t >0$.
Where did it find?
inverse brownian-motion stopping-times
inverse brownian-motion stopping-times
edited Sep 6 '14 at 10:59
Davide Giraudo
125k16150260
asked Sep 6 '14 at 10:16
Cronos
233
edited Sep 6 '14 at 10:59
Davide Giraudo
125k16150260
asked Sep 6 '14 at 10:16
Cronos
233
edited Sep 6 '14 at 10:59
Davide Giraudo
125k16150260
edited Sep 6 '14 at 10:59
Davide Giraudo
125k16150260
edited Sep 6 '14 at 10:59
Davide Giraudo
125k16150260
125k16150260
asked Sep 6 '14 at 10:16
Cronos
233
asked Sep 6 '14 at 10:16
Cronos
233
asked Sep 6 '14 at 10:16
Cronos
233
233
1
Since you know $P(T_betaleqslant t)$, how to deduce the density $f_beta$?
– Did
Sep 6 '14 at 11:45
yeah, I didin't derivate respect to t. Now It's clear.
– Cronos
Sep 6 '14 at 18:15
add a comment |
1
Since you know $P(T_betaleqslant t)$, how to deduce the density $f_beta$?
– Did
Sep 6 '14 at 11:45
yeah, I didin't derivate respect to t. Now It's clear.
– Cronos
Sep 6 '14 at 18:15
1
1
Since you know $P(T_betaleqslant t)$, how to deduce the density $f_beta$?
– Did
Sep 6 '14 at 11:45
Since you know $P(T_betaleqslant t)$, how to deduce the density $f_beta$?
– Did
Sep 6 '14 at 11:45
yeah, I didin't derivate respect to t. Now It's clear.
– Cronos
Sep 6 '14 at 18:15
yeah, I didin't derivate respect to t. Now It's clear.
– Cronos
Sep 6 '14 at 18:15
add a comment |
1 Answer
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$P(B(t) > beta) = P(B(t) > beta | T_beta > t ) cdot P(T_beta > t )
+ P(B(t) > beta | T_beta leq t ) cdot P(T_beta leq t ) \
= 0 + frac{1}{2} P(T_beta leq t ) \
Longrightarrow P(T_beta leq t ) = 2 P(B(t) > beta) $.
$ P(B(t) > beta | T_beta leq t ) = 1/2$ because the Brownian motion is symmetric conditioned on the hitting point (independent increase).
So, $P(T_beta leq t ) = 2 P(B(t) > beta) = 2 - 2 P(B(t) leq beta) \
= 2 - 2 P( frac{ B(t)}{ sqrt{t} } leq frac{ beta }{ sqrt{t} } )
= 2 - 2 Phi (frac{ beta }{ sqrt{t} } ) $
where $Phi$ is the CDF of standard Gaussian distribution.
So $f_{T_beta} (t) = frac{d}{d t}( 2 - 2 Phi (frac{ beta }{ sqrt{t} } ) ) $.
answered Jan 3 at 17:42
Albert Chen
20017
add a comment |
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$P(B(t) > beta) = P(B(t) > beta | T_beta > t ) cdot P(T_beta > t )
+ P(B(t) > beta | T_beta leq t ) cdot P(T_beta leq t ) \
= 0 + frac{1}{2} P(T_beta leq t ) \
Longrightarrow P(T_beta leq t ) = 2 P(B(t) > beta) $.
$ P(B(t) > beta | T_beta leq t ) = 1/2$ because the Brownian motion is symmetric conditioned on the hitting point (independent increase).
So, $P(T_beta leq t ) = 2 P(B(t) > beta) = 2 - 2 P(B(t) leq beta) \
= 2 - 2 P( frac{ B(t)}{ sqrt{t} } leq frac{ beta }{ sqrt{t} } )
= 2 - 2 Phi (frac{ beta }{ sqrt{t} } ) $
where $Phi$ is the CDF of standard Gaussian distribution.
So $f_{T_beta} (t) = frac{d}{d t}( 2 - 2 Phi (frac{ beta }{ sqrt{t} } ) ) $.
answered Jan 3 at 17:42
Albert Chen
20017
add a comment |
$P(B(t) > beta) = P(B(t) > beta | T_beta > t ) cdot P(T_beta > t )
+ P(B(t) > beta | T_beta leq t ) cdot P(T_beta leq t ) \
= 0 + frac{1}{2} P(T_beta leq t ) \
Longrightarrow P(T_beta leq t ) = 2 P(B(t) > beta) $.
$ P(B(t) > beta | T_beta leq t ) = 1/2$ because the Brownian motion is symmetric conditioned on the hitting point (independent increase).
So, $P(T_beta leq t ) = 2 P(B(t) > beta) = 2 - 2 P(B(t) leq beta) \
= 2 - 2 P( frac{ B(t)}{ sqrt{t} } leq frac{ beta }{ sqrt{t} } )
= 2 - 2 Phi (frac{ beta }{ sqrt{t} } ) $
where $Phi$ is the CDF of standard Gaussian distribution.
So $f_{T_beta} (t) = frac{d}{d t}( 2 - 2 Phi (frac{ beta }{ sqrt{t} } ) ) $.
answered Jan 3 at 17:42
Albert Chen
20017
add a comment |
$P(B(t) > beta) = P(B(t) > beta | T_beta > t ) cdot P(T_beta > t )
+ P(B(t) > beta | T_beta leq t ) cdot P(T_beta leq t ) \
= 0 + frac{1}{2} P(T_beta leq t ) \
Longrightarrow P(T_beta leq t ) = 2 P(B(t) > beta) $.
$ P(B(t) > beta | T_beta leq t ) = 1/2$ because the Brownian motion is symmetric conditioned on the hitting point (independent increase).
So, $P(T_beta leq t ) = 2 P(B(t) > beta) = 2 - 2 P(B(t) leq beta) \
= 2 - 2 P( frac{ B(t)}{ sqrt{t} } leq frac{ beta }{ sqrt{t} } )
= 2 - 2 Phi (frac{ beta }{ sqrt{t} } ) $
where $Phi$ is the CDF of standard Gaussian distribution.
So $f_{T_beta} (t) = frac{d}{d t}( 2 - 2 Phi (frac{ beta }{ sqrt{t} } ) ) $.
answered Jan 3 at 17:42
Albert Chen
20017
$P(B(t) > beta) = P(B(t) > beta | T_beta > t ) cdot P(T_beta > t )
+ P(B(t) > beta | T_beta leq t ) cdot P(T_beta leq t ) \
= 0 + frac{1}{2} P(T_beta leq t ) \
Longrightarrow P(T_beta leq t ) = 2 P(B(t) > beta) $.
$ P(B(t) > beta | T_beta leq t ) = 1/2$ because the Brownian motion is symmetric conditioned on the hitting point (independent increase).
So, $P(T_beta leq t ) = 2 P(B(t) > beta) = 2 - 2 P(B(t) leq beta) \
= 2 - 2 P( frac{ B(t)}{ sqrt{t} } leq frac{ beta }{ sqrt{t} } )
= 2 - 2 Phi (frac{ beta }{ sqrt{t} } ) $
where $Phi$ is the CDF of standard Gaussian distribution.
So $f_{T_beta} (t) = frac{d}{d t}( 2 - 2 Phi (frac{ beta }{ sqrt{t} } ) ) $.
answered Jan 3 at 17:42
Albert Chen
20017
answered Jan 3 at 17:42
Albert Chen
20017
answered Jan 3 at 17:42
Albert Chen
20017
answered Jan 3 at 17:42
Albert Chen
20017
20017
add a comment |
add a comment |
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Since you know $P(T_betaleqslant t)$, how to deduce the density $f_beta$?
– Did
Sep 6 '14 at 11:45
yeah, I didin't derivate respect to t. Now It's clear.
– Cronos
Sep 6 '14 at 18:15