First Passage Time of a Brownian Motion plus drift

[quotes]

Shreve 7.2.2

Under Probability P, W(T) is a standard Brownian Motion
while X(T)=W(T)+aT is a Brownian Motion plus drift.
M(T)=Max{X(s); 0<s<T}

\(P \{M(T)\leq m\} = N(\frac{m-aT}{\sqrt{T}})-e^{2am}N(\frac{-m-aT}{\sqrt{T}}),m>0\)

Can we differentiate the above to get first passage time density?

\(\frac{\partial}{\partial T}P\{M(T)\leq m\}\)

I corrected X(T) by M(T)

 

[MiloRambaldi]

Shreve 7.2.2

Under Probability P, W(T) is a standard Brownian Motion
while X(T)=W(T)+aT is a Brownian Motion plus drift.
M(T)=Max{X(s); 0<s<T}

\(P \{M(T)\leq m\} = N(\frac{m-aT}{\sqrt{T}})-e^{2am}N(\frac{-m-aT}{\sqrt{T}}),m>0\)

Can we differentiate the above to get first passage time density?

\(\frac{\partial}{\partial T}P\{M(T)\leq m\}\)

I corrected X(T) by M(T)

No. If \(\tau_a=\min\{t:X(t)=a\}\) is the first passage time of the process \(X\) to \(a\), then the first passage time density is:
\(\frac{\partial}{\partial T}\mathrm{P}[\tau_a<T]\).

In the case where \(X=W\) is Brownian motion and \(a >0\), \(\mathrm{P}[\tau_a<T]=2\cdot\mathrm{P}[W(T)<-a]\), and the first passage time density is similar to what you wrote. So you asked the wrong question
The right question is:

Let \(X(t)=W(t)+mt\) be Brownian motion with a constant drift.
What is \(\mathrm{P}[\tau_a < T] \)?

 

[quotes]

Shreve 7.2.2

Under Probability P, W(T) is a standard Brownian Motion
while X(T)=W(T)+aT is a Brownian Motion plus drift.
M(T)=Max{X(s); 0<s<T}

\(P \{X(T)\leq m\} = N(\frac{m-aT}{\sqrt{T}})-e^{2am}N(\frac{-m-aT}{\sqrt{T}}),m>0\)

Can we differentiate the above to get first passage time density?

\(\frac{\partial}{\partial T}P\{X(T)\leq m\}\)

Sorry for my previous mistake, X(T) shall be M(T) in the equations.

\(P \{M(T)\leq m\} = N(\frac{m-aT}{\sqrt{T}})-e^{2am}N(\frac{-m-aT}{\sqrt{T}}),m>0\)
\(\frac{\partial}{\partial T}P\{M(T)\leq m\}\)

 

[quotes]

The event

{M(T+h)<=m} contains the event {M(T)<=m} where h>0

So {M(T)<=m}-{M(T+h)<=m}={M(T)<=m, X(T+r)>m}

where 0<r<=m

That is where I got the clue.

 

[quotes]

If we differentiate the Probability of event {M(T)<=m, X(T+r)>m}, or that X(T) breaches the upper barrier m in time interval (T,T+r] , with respect to time increment r, we get the hitting time distribution of X(T):
\[\frac{ -\partial P}{\partial T}=\frac{m}{T\sqrt{2\pi T}}e^{-\frac{(m-aT)^2}{2T}}\]

 

[MiloRambaldi]

The event

{M(T+h)<=m} contains the event {M(T)<=m} where h>0

So {M(T)<=m}-{M(T+h)<=m}={M(T)<=m, X(T+r)>m}

where 0<r<=m

That is where I got the clue.

The intuition is correct, but this is straightforward: The first passage time is the random variable \(\tau_m=\min\{t:X(t)=m\}\).

By definition, its cdf is \(P[\tau_m \le T]\), which is equal to \(P[M(T)\le m]\).

Whenever the cdf of a random variable is differentiable, the pdf (i.e. the probability density function) is the derivative of the cdf.

Thus the density function of \(\tau_m\) is \(\frac{\partial}{\partial T}P[\tau_m\le T]=\frac{\partial}{\partial T}P[M(T)\le m]\).

From the formula in Shreve I get the following for the density:

\(\left[-\frac m{2\sqrt{T^3}} -\frac\alpha{2\sqrt T}\right] N'\large(\frac{m-\alpha T}{\sqrt T}\right) -e^{2\alpha m}\left[\frac m{2\sqrt{T^3}} -\frac\alpha{2\sqrt T}\right] N'\large(\frac{-m-\alpha T}{\sqrt T}\right)\).

 

[quotes]

From the formula in Shreve I get the following for the density:

\(\left[-\frac m{2\sqrt{T^3}} -\frac\alpha{2\sqrt T}\right] N'\large(\frac{m-\alpha T}{\sqrt T}\right) -e^{2\alpha m}\left[\frac m{2\sqrt{T^3}} -\frac\alpha{2\sqrt T}\right] N'\large(\frac{-m-\alpha T}{\sqrt T}\right)\).

Simplify

 

[quotes]

See, if you differentiate P{M(T)<=m} with respect to T, you just get negative values.

To simplify, just make a=0 the driftless case:

\(-\frac{m}{2\sqrt{T^3}}N^{'}(m/\sqrt{T})-\frac{m}{2\sqrt{T^3}}N'(-m/\sqrt{T})\)

\(N^{'}(m/\sqrt{T})=N'(-m/\sqrt{T})=\frac{1}{\sqrt{2\pi}}e^{-\frac{m^2}{2T}}\)

And the negation of it corresponds with the density in (3.7.3)

\(\frac{m}{t\sqrt{2\pi t}}e^{-\frac{m^2}{2t}} \)

weird, ha?

 

[quotes]

I understand what went wrong regarding negative sign

These two are not equal but complimentary:
\(P\{\tau _m \le T\}+P\{M(T)<m\}=1\)

 

[MiloRambaldi]

...

And the negation of it corresponds with the density in (3.7.3)

\(\frac{m}{t\sqrt{2\pi t}}e^{-\frac{m^2}{2t}} \)

weird, ha?

Yes, I was very surprised when you pointed out that the expression for density of the first passage time for BM with a drift simplifies to

\(\frac{m}{\sqrt{2\pi t^3}}e^{-\frac{(m-\alpha t)^2}{2t}}.\)

It's not obvious to me why this happens. Maybe Fourier transforms would help?

I was just reading a conference paper on credit modeling where they use this density and they call it the "Bachelier-Levy formula", but I haven't had a chance to look up the references yet.

PS. Personally I avoid Shreve's book because I find the mathematics is obscured by complicated integrals. Mark Joshi's book gives a better account, for example of the derivation of the joint distribution function of ( M(T), X(T) ), in the section on the Reflection Principle.

 

[quotes]

Yes, I was very surprised when you pointed out that the expression for density of the first passage time for BM with a drift simplifies to
\[\frac{m}{\sqrt{2\pi t^3}}e^{-\frac{(m-\alpha t)^2}{2t}}.\]

It's not obvious to me why this happens. Maybe Fourier transforms would help?

It is just your formula given in post 6 simplified and multiplied by (-1)
Try it.

 

[quotes]

PS. Personally I avoid Shreve's book because I find the mathematics is obscured by complicated integrals. Mark Joshi's book gives a better account, for example of the derivation of the joint distribution function of ( M(T), X(T) ), in the section on the Reflection Principle.

Mark Joshi only gaves 2 pages explaining reflection principle in his book Concepts and Practice of Mathematical Finance

The problem with most stochastic finance books is that they only briefly discuss the barrier problem from one or two aspects, thus any book is incomplete treating the various probabilities concerned with barrier. I have worked out some problems myself like the one in this post, but have headache on some others. Yesterday I found tens of books on stochastic process in electronic version, and perhaps one of them will settle the barrier probabilities nicely.

 

[MiloRambaldi]

It is just your formula given in post 6 simplified and multiplied by (-1)
Try it.

Yes, I tried as soon as you said "simplify".

I meant that there must be some mathematical phenomenon explaining why everything "magically" cancels out.

 

[MiloRambaldi]

Mark Joshi only gaves 2 pages explaining reflection principle in his book Concepts and Practice of Mathematical Finance

The problem with most stochastic finance books is that they only briefly discuss the barrier problem from one or two aspects, thus any book is incomplete treating the various probabilities concerned with barrier. I have worked out some problems myself like the one in this post, but have headache on some others. Yesterday I found tens of books on stochastic process in electronic version, and perhaps one of them will settle the barrier probabilities nicely.

Of course it is a matter of personally preference.

But, Joshi's account is shorter because is goes straight to the important points. It shows that there is absolutely no need for all those complex integrals in Shreve.

Have you checked Bjorn's book (I haven't looked at it recently)? Yes, you will probably have to look for research papers, or a straight mathematics book ...

 

[quotes]

I meant that there must be some mathematical phenomenon explaining why everything "magically" cancels out.

\[e^{2\alpha m}N'(d2)=e^{-\frac{-4(m)(\alpha T)}{2T}}N'(d2)=N'(d1)\]

- completion of squares

It is the same way that Shreve used to simplify his complex intgrals.

 

[quotes]

The intuition is correct, but this is straightforward: The first passage time is the random variable \(\tau_m=\min\{t:X(t)=m\}\).

By definition, its cdf is \(P[\tau_m \le T]\), which is equal to \(P[M(T)\le m]\).

Whenever the cdf of a random variable is differentiable, the pdf (i.e. the probability density function) is the derivative of the cdf.

Thus the density function of \(\tau_m\) is \(\frac{\partial}{\partial T}P[\tau_m\le T]=\frac{\partial}{\partial T}P[M(T)\le m]\).

From the formula in Shreve I get the following for the density:

\(\left[-\frac m{2\sqrt{T^3}} -\frac\alpha{2\sqrt T}\right] N'\large(\frac{m-\alpha T}{\sqrt T}\right)
-e^{2\alpha m}\left[\frac m{2\sqrt{T^3}} -\frac\alpha{2\sqrt T}\right] N'\large(\frac{-m-\alpha T}{\sqrt T}\right)\).

\(N'(x)=\frac{e^{-x^2/2}}{\sqrt{2\pi}}\)

\(e^{2\alpha m} N'\large(\frac{-m-\alpha T}{\sqrt T}\right)=\frac{1}{\sqrt{2\pi}}exp\large(2\alpha m-\frac{m^2+\alpha^2 T^2+2\alpha mT}{2T} \right)= N'\large(\frac{m-\alpha T}{\sqrt T}\right)\)

 

[quotes]

And then the four becomes two since the second and fourth term cancels out in the sum.