Brownian bridge and first hitting times


by Tetef
Tags: bridge, brownian, hitting time, probability
Tetef
Tetef is offline
#1
Feb6-13, 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}{b-a}[/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 !
Phys.Org News Partner Science news on Phys.org
SensaBubble: It's a bubble, but not as we know it (w/ video)
The hemihelix: Scientists discover a new shape using rubber bands (w/ video)
Microbes provide insights into evolution of human language
mathman
mathman is offline
#2
Feb6-13, 03:23 PM
Sci Advisor
P: 5,941
Your question looks confusing. Your last statement has x < a. Also did you mean W(0) = x? Your T definitions use t also.
Tetef
Tetef is offline
#3
Feb6-13, 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...

Tetef
Tetef is offline
#4
Feb6-13, 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}{b-a}[/itex]
Is that right?
mathman
mathman is offline
#5
Feb7-13, 06:04 PM
Sci Advisor
P: 5,941
I have not studied this problem in any detail, so I don't know if I can be of any further help.


Register to reply

Related Discussions
If you flip a coin 100 times how many times will it land on heads? Precalculus Mathematics Homework 18
what is the gravity of a planet 2 times the radius and 3 times the mass of earth Introductory Physics Homework 6
Why are smoke particles used in Brownian motion? What observations of Brownian motion Introductory Physics Homework 1
Proof of Exponential Interarrival times times Set Theory, Logic, Probability, Statistics 2
If you lose 3 times in a row your out, am i hitting all cases? Calculus & Beyond Homework 0