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Repost of the solution \(and fixed typos in last paragraph\):
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 \(\{\tau_a<T \}\land\{\tau_{-a}<T\}=\)
\(\{B_a\text{ hits }3a\text{ before time }T\}\lor\{B_{-a}\text{ hits }-3a\text{ before time }T\}\).
\(2\) 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 \(proved below\).
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\br\)\)
\(=4\bl[N\(-3a\)-N\(-5a\)+N\(-7a\)-\cdots\br] \).
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,
\(E_{k-1}\(t\)\land F_{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 \(F_k\(t\)\) has occurred. This shows that \(E_{k+1}\to E_k\land F_k\),
and we reach the conclusion by symmetry.
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 \(\{\tau_a<T \}\land\{\tau_{-a}<T\}=\)
\(\{B_a\text{ hits }3a\text{ before time }T\}\lor\{B_{-a}\text{ hits }-3a\text{ before time }T\}\).
\(2\) 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 \(proved below\).
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\br\)\)
\(=4\bl[N\(-3a\)-N\(-5a\)+N\(-7a\)-\cdots\br] \).
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,
\(E_{k-1}\(t\)\land F_{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 \(F_k\(t\)\) has occurred. This shows that \(E_{k+1}\to E_k\land F_k\),
and we reach the conclusion by symmetry.
