Hitting time and its distribution












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÷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







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  • 1




    $begingroup$
    Since you know $P(T_betaleqslant t)$, how to deduce the density $f_beta$?
    $endgroup$
    – Did
    Sep 6 '14 at 11:45










  • $begingroup$
    yeah, I didin't derivate respect to t. Now It's clear.
    $endgroup$
    – Cronos
    Sep 6 '14 at 18:15
















0












$begingroup$


÷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







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Since you know $P(T_betaleqslant t)$, how to deduce the density $f_beta$?
    $endgroup$
    – Did
    Sep 6 '14 at 11:45










  • $begingroup$
    yeah, I didin't derivate respect to t. Now It's clear.
    $endgroup$
    – Cronos
    Sep 6 '14 at 18:15














0












0








0





$begingroup$


÷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







share|cite|improve this question











$endgroup$




÷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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 6 '14 at 10:59









Davide Giraudo

128k17156268




128k17156268










asked Sep 6 '14 at 10:16









CronosCronos

233




233








  • 1




    $begingroup$
    Since you know $P(T_betaleqslant t)$, how to deduce the density $f_beta$?
    $endgroup$
    – Did
    Sep 6 '14 at 11:45










  • $begingroup$
    yeah, I didin't derivate respect to t. Now It's clear.
    $endgroup$
    – Cronos
    Sep 6 '14 at 18:15














  • 1




    $begingroup$
    Since you know $P(T_betaleqslant t)$, how to deduce the density $f_beta$?
    $endgroup$
    – Did
    Sep 6 '14 at 11:45










  • $begingroup$
    yeah, I didin't derivate respect to t. Now It's clear.
    $endgroup$
    – Cronos
    Sep 6 '14 at 18:15








1




1




$begingroup$
Since you know $P(T_betaleqslant t)$, how to deduce the density $f_beta$?
$endgroup$
– Did
Sep 6 '14 at 11:45




$begingroup$
Since you know $P(T_betaleqslant t)$, how to deduce the density $f_beta$?
$endgroup$
– Did
Sep 6 '14 at 11:45












$begingroup$
yeah, I didin't derivate respect to t. Now It's clear.
$endgroup$
– Cronos
Sep 6 '14 at 18:15




$begingroup$
yeah, I didin't derivate respect to t. Now It's clear.
$endgroup$
– Cronos
Sep 6 '14 at 18:15










1 Answer
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$begingroup$

$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} } ) ) $.






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    $begingroup$

    $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} } ) ) $.






    share|cite|improve this answer









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      0












      $begingroup$

      $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} } ) ) $.






      share|cite|improve this answer









      $endgroup$
















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        0





        $begingroup$

        $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} } ) ) $.






        share|cite|improve this answer









        $endgroup$



        $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} } ) ) $.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 3 at 17:42









        Albert ChenAlbert Chen

        23027




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