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A common quant interview question: Let (\tau_a = \min\{t:W(t)=a\}) be the first hitting time of Brownian motion (W(t)) at the level (a). Find (P[\tau_a < T]), in other words probability that BM hits the barrier (a) in time (T).
Here is a solution in case anyone is interested. If you have a better one, please share...
Let (B) be Brownian motion and (B_a) be its reflection at (a): (B_a=B) if (a) has not yet been hit and (B_a=2a-B) after (a) has been hit, which is also Brownian motion by the reflection principle.
The two main observation are:
(1) These 2 events are the same
(\{\tau_a)
(2) This can be continued ad infinitum with reflections at (\pm3a, \pm5a, \pm7a), ...
The next step is: (\{B_a\text{ hits }3a\text{ before time }T\}\land\{B_{-a}\text{ hits }-3a\text{ before time }T\}=)
(\{(B_a)_{3a}\text{ hits }5a\text{ before time }T\}\lor\{(B_{-a})_{-3a}\text{ hits }-5a\text{ before time }T\}),
where for example, ((B_a)_{3a}) is BM reflected at (a) and then reflected again at (3a).
The first observation is almost immediate, while the second one is less obvious.
For a formal solution, define stochastic processes recursively:
(U_0=B) and (U_{k+1}=(U_{k})_{(2k+1)a}),
(D_0=B) and (D_{k+1}=(D_{k})_{-(2k+1)a}).
For example, (U_1) is (B) reflected at (a), (D_{1}) is (B) reflected at (-a), (U_2) is (U_1) reflected at (3a), etc ...
These are all Brownian motions by the reflection principle.
Let (E_{k}:=E_k(T)) be the event that (U_{k}) hits ((2k+1)a) before time (T), and
let (F_{k}:=F_k(T)) be the event that (D_{k}) hits (-(2k+1)a) before time (T).
The above observations can now be expressed as:
Claim 1. (E_k\land F_k=E_{k+1}\lor F_{k+1}).
From the usual application of the reflection principle we have:
Claim 2. (P[E_k]=P[F_k]=2N\bl(-(2k+1)a\br)).
The answer follows from claims 1 and 2. The event that BM hits both barriers is (E_0\land F_0), and the probability is:
(P[E_0\land F_0]=P[E_1\lor F_1]
=P[E_1]+P[F_1]-P[E_1\land F_1]=P[E_1]+P[F_1]-P[E_2\lor F_2]
)
(
=P[E_1]+P[F_1]-P[E_2]-P[ F_2] +P[E_2\land F_2]
)
(
=\cdots=4\sum_{k=1}^\infty (-1)^{k+1}N\bl(-(2k+1)a\))
(=4\bl[N(-3a)-N(-5a)+N(-7a)-\cdots] ).
To complete the solution:
Proof of claim 1. Notice that for all (t\le T),
(*) (E_k(t)\land F_k(t)\to U_{k+1}(t)-D_{k+1}(t)=4(k+1)a).
This is a simple induction using the definition of reflection. For example, for the base case (k=0), after both (a) and (-a) have been hit, (U_1(t)-D_1(t)=[2a-B(t)]-[-2a-B(t)]=4a).
(E_k\land F_k\to E_{k+1}\lor F_{k+1}) is now easy to see.
E.g. if (U_k) hits ((2k+1)a) before (D_k) hits the opposite barrier,
then when (D_k(t)=-(2k+1)a), (D_{k+1}(t)=D_k(t)) and
(*) gives (U_{k+1}(t)=4(k+1)a-(2k+1)a=\bl[2(k+1)+1\br]a)
which means that the event (E_{k+1}) has occurred.
The other direction (E_{k+1}\lor F_{k+1}\to E_k\land F_k) follows by induction on (k).
The base case (k=0) is clear.
For (k\ge1) suppose for example that (E_{k+1}) has occurred, with (U_{k+1}(t)=(2k+3)a).
Then so has (E_k(t)), and then by the induction hypothesis, (D_{k-1}(t)\land E_{k-1}(t)) has occurred.
Applying (*) we get
(D_k(t)=U_k(t)-4ka=\bl[2(2k+1)a-U_{k+1}(t)\br]-4ka=-(2k+1)a),
which mean that (D_k(t)) has occurred. This shows that (E_{k+1}\to E_k\land F_k),
and we reach the conclusion by symmetry.
