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Brownian bridge and first hitting times

  1. Feb 6, 2013 #1

    Letting [itex]W[/itex] be a standard brownian motion, we define the first hitting times
    [itex]T_{a}=inf\{t:W(t)=a\}[/itex] with [itex]a<0[/itex]
    [itex]T_{b}=inf\{t:W(t)=b\}[/itex] with [itex]b>0[/itex]

    The probability of one hitting time being before an other is :

    I'm looking for this probability in the case of a brownian bridge :
    [itex]P\{T_{a}<T_{b} | W(t)=x\}[/itex] with [itex]x<a[/itex]

    Could some one help me please?

    Thx !
  2. jcsd
  3. Feb 6, 2013 #2


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    Your question looks confusing. Your last statement has x < a. Also did you mean W(0) = x? Your T definitions use t also.
  4. Feb 6, 2013 #3
    I did mean what I wrote,

    A standard brownian motion has [itex]W(0)=0[/itex]

    I use [itex]t[/itex] in the definition [itex]T_{a}=inf\{t:W(t)=a\}[/itex]to say the first hitting times are defined as : [itex]T_{a}[/itex] is the smallest time [itex]t[/itex] where the brownian motion does hit the level [itex]a[/itex] i.e. [itex]W(t)=a[/itex]

    In the expression of the probability [itex]P\{T_{a}<T_{b}|W(t)=x\}[/itex], [itex]t[/itex] is the 'time of one observation'.
    If you look for [itex]P\{T_{a}<t\}[/itex], you're looking for the probability for [itex]W[/itex] to hit a level [itex]a[/itex] before [itex]t[/itex] ([itex]t[/itex] being the end of the observation).

    Here, in the expression [itex]P\{T_{a}<T_{b}|W(t)=x\}[/itex], I'm looking for the probability for one hitting time to be before the other, knowing that at the time [itex]t[/itex], the brownian motion will be equal to [itex]x[/itex], i.e. [itex]W(t)=x[/itex]; and thus form a brownian bridge between [itex]0[/itex] at time [itex]0[/itex] ([itex]W(0)=0[/itex]) and [itex]x[/itex] at time [itex]t[/itex] ([itex]W(t)=x[/itex])

    [itex]P\{T_{a}<T_{b}|W(t)=x\}[/itex] could be asked for [itex]x\inℝ[/itex].
    In my case I have [itex]x<a[/itex]. This implies, by continuity of the brownian motion, that [itex]T_{a}[/itex] does exist before [itex]t[/itex]. It might make it simpler, it might not...
    Last edited: Feb 6, 2013
  5. Feb 6, 2013 #4
    I don't know if it's any use but, I've thought of this :

    [itex]P\{T_{a}<T_{b}|W(t)=x\}=\frac{p\{T_{a}<T_{b}\cap W(t)=x\}}{p\{W(t)=x\}}[/itex]

    where capital [itex]P[/itex] are probabilities and small [itex]p[/itex] are probability density functions.

    we know :
    [itex]p\{W(t)=x\}=\frac{1}{\sqrt{2\pi t}}exp( -\frac{x^2}{2t})[/itex]

    we don't know :
    [itex]p\{T_{a}<T_{b}\cap W(t)=x\}[/itex]

    but we know that :
    [itex]\int^{+∞}_{-∞} p\{T_{a}<T_{b}\cap W(t)=x\}dx=\frac{b}{b-a}[/itex]
    Is that right?
  6. Feb 7, 2013 #5


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    I have not studied this problem in any detail, so I don't know if I can be of any further help.
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