Hitting Time on a double barrier

[quotes]

Wt stops when it hits a<0 or b>0
Is there a formula to determin P(Wt=a, t<T) and P(Wt=b, t<T)?
Any help would be appreciated

 

[MiloRambaldi]

Wt stops when it hits a<0 or b>0
Is there a formula to determin P(Wt=a, t<T) and P(Wt=b, t<T)?
Any help would be appreciated

Yes, this is a very basic quant interview question.

The formula is trivial, but a proof is a bit nontrivial.

Hint: W is a Martingale.

 

[Adrián Facchi]

The probability is of course zero. I assume that you are interested in barriers, then you may be looking for
( P(Y<H)\ , \qquad Y=min_{t\in[0,T]}[W_t]\ ,\qquad H\in \mathbb{R})
or
( P(Y<H|W_0 = \omega_0))
or even
( P(Y<H|W_T=\omega,W_0 = \omega_0)\ .)
Look at Musiela and Rutkowsy Appendix B4 (pag 579) it may be helpful.

 

[MiloRambaldi]

I didn't read his question carefully. I thought quotes was asking P(Wt=a) where t is the stopping time.

quotes: What are t and T. Is t the stopping time? Are you then asking what P(Wt=a and t < T) for some arbitrary T?

 

[quotes]

Let me clarify the symbols a little bit.
The Process（t,W(t)） stops at
(Ta, a)
(Tb, b)
or (T,W(T))

for the first two occasions, W determines the stop, and for the last, t determines the stop

The question is:

1) Prob(0=<Ta<u<=T)
2) Prob(0=<Tb<u<=T)
3) Prob( a=<W(T)<=u, u∈[a,b])

Of course, differentiate them with the variable u we can generate probability density functions for random variable Ta, Tb & W(T).
I am particularly interested in problem 3, for the first two can be solved by this problem 3 alone.

 

[Alexei]

Would the probability in case 3 be 1? since by definition, T is a stopping time when the BM hits either a or b, hence W(T) is either a or b, given T, P(W(T) hits a or b) = 1.

 

[Alexei]

Probability of W(t), W(0) = 0, hitting a before b is b/(a+b) and a/(a+b) is a probability of hitting b before a; a and b are positive values.

 

[Alexei]

Expected hitting time, E[T] = ab.

 

[quotes]

Would the probability in case 3 be 1? since by definition, T is a stopping time when the BM hits either a or b, hence W(T) is either a or b, given T, P(W(T) hits a or b) = 1.

No, a<W(T)<b and W(t) before W(T) in its path cannot hit either a or b.

T is forced stopping time. You can interpret it as Game Over Time - forget about Wt continuing or stopping at barrier a or b, now at time T it stops at whichever value.

 

[MiloRambaldi]

Let me clarify the symbols a little bit.
The Process（t,W(t)） stops at
(Ta, a)
(Tb, b)
or (T,W(T))

for the first two occasions, W determines the stop, and for the last, t determines the stop

The question is:

1) Prob(0=<Ta<u<=T)
2) Prob(0=<Tb<u<=T)
3) Prob( a=<W(T)<=u, u∈[a,b])

Of course, differentiate them with the variable u we can generate probability density functions for random variable Ta, Tb & W(T).
I am particularly interested in problem 3, for the first two can be solved by this problem 3 alone.

Your notice is confusing (what is u?). Perhaps you are asking the following interesting question?

Let Ta = min{t:W(t)=a} and Tb = min{t:W(t)=b} be the hitting times at a and b.

What is P[Ta <= T & Tb <= T], i.e. what is the probability of hitting both barriers in the time interval [0,T]?

 

[quotes]

Okay, suppose you are observing a ball running along the time horizon before a right bound T.
However it must stays between the line y=a and y=b, once it touches either line, it is finished and no longer runs.
_________y=b__________
O O
0- - - - O- - - - - - - >- - - - - -- -T
O
_________y=a__________

Then we have three senarios:
1) it ends at (t,a) , in which stopping time t is undetermined
2) it ends at (t,b), in which stopping time t is undetermined
3) Luckily it makes to the end and at (T,w), in which the ending position w is undetermined

 

[quotes]

O stands for the ball running. initial spaces are cut off so you may not see the trajectory of O clearly, but the O does a random walk verticly and a straight line movement horizontally.

 

[amir]

Wt stops when it hits a<0 or b>0
Is there a formula to determin P(Wt=a, t<T) and P(Wt=b, t<T)?
Any help would be appreciated

I am not sure but you may find this book useful: Stochastic Optimization in Continuous Time by F.R. Chang
chapter 6: 6 Boundaries and Absorbing Barriers

 

[MiloRambaldi]

Okay, suppose you are observing a ball running along the time horizon before a right bound T.
However it must stays between the line y=a and y=b, once it touches either line, it is finished and no longer runs.
_________y=b__________
O O
0- - - - O- - - - - - - >- - - - - -- -T
O
_________y=a__________

Then we have three senarios:
1) it ends at (t,a) , in which stopping time t is undetermined
2) it ends at (t,b), in which stopping time t is undetermined
3) Luckily it makes to the end and at (T,w), in which the ending position w is undetermined

Okay, its clear now, but its better to use correct mathematical notation/definitions

I defined two stopping times above, the first time BM hits a and b:
Ta = min{t:W(t)=a} Tb = min{t:W(t)=b}

There are three possible mutually exclusive events:

E1 = {Ta < min(Tb, T)} (a is hit before b and before time T)
E2 = {Tb < min(Ta, T)} (b is hit before a and before time T)
E3 = {T <= min(Ta, Tb)} (neither a nor b is hit before time T)

P[E1] + P[E2] + P[E3] = 1

I assume you are asking about P[E1], P[E2] and P[E3]?

P[Ta < T] and P[Tb < T] follow from the reflection principle (a standard quant interview question). Thus P[E3] follows from my question about about the probability of hitting both a and b before time T:

P[E3] = 1 - P[Ta < T or Tb < T] = 1 - P[Ta < T] - P[Tb < T] + P[Ta < T and Tb < T].

I'll post this for the special case b = -a later.

 

[quotes]

I am not sure but you may find this book useful: Stochastic Optimization in Continuous Time by F.R. Chang
chapter 6: 6 Boundaries and Absorbing Barriers

Thank you.

Okay, its clear now, but its better to use correct mathematical notation/definitions

I defined two stopping times above, the first time BM hits a and b:
Ta = min{t:W(t)=a} Tb = min{t:W(t)=b}

There are three possible mutually exclusive events:

E1 = {Ta < min(Tb, T)} (a is hit before b and before time T)
E2 = {Tb < min(Ta, T)} (b is hit before a and before time T)
E3 = {T <= min(Ta, Tb)} (neither a nor b is hit before time T)

P[E1] + P[E2] + P[E3] = 1

I assume you are asking about P[E1], P[E2] and P[E3]?

P[Ta < T] and P[Tb < T] follow from the reflection principle (a standard quant interview question). Thus P[E3] follows from my question about about the probability of hitting both a and b before time T:

P[E3] = 1 - P[Ta < T or Tb < T] = 1 - P[Ta < T] - P[Tb < T] + P[Ta < T and Tb < T].

I'll post this for the special case b = -a later.

Thanks for your notation, but I have seen the formula in infinite sum.

 

[MiloRambaldi]

Thank you.



Thanks for your notation, but I have seen the formula in infinite sum.

I'm not sure probability you mean, but I'll post the question anyways in case anyone else is interested.

 

[quotes]

Here is Milo's solution:

https://www.quantnet.com/forum/threads/double-barrier-ctd.8225/