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Brownian bridge and first hitting times 
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#1
Feb613, 02:46 PM

P: 3

Hi,
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] and [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 : [itex]P\{T_{a}<T_{b}\}=\frac{b}{ba}[/itex] 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
Feb613, 03:23 PM

Sci Advisor
P: 6,077

Your question looks confusing. Your last statement has x < a. Also did you mean W(0) = x? Your T definitions use t also.



#3
Feb613, 04:00 PM

P: 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... 


#4
Feb613, 04:53 PM

P: 3

Brownian bridge and first hitting times
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}{ba}[/itex] Is that right? 


#5
Feb713, 06:04 PM

Sci Advisor
P: 6,077

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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