Serving the Quantitative Finance Community

Cuchulainn
Topic Author
Posts: 64427
Joined: July 16th, 2004, 7:38 am
Location: Drosophila melanogaster
Contact:

### American options with two free boundaries

Let's say you are a landowner and you are holding the land with a right to build property on the land or to abandon the land. We model this problem as an American perpetual call option. Let $S$ be the value of the property and $H$ the constant holding costs (e.g. property taxes etc.)

Let $S_{*}$ be the lower critical value of $S$ below which it is optimal to abandon the land.
Let $S^{*}$ be the upper critical value of $S$ at which it is optimal to build the property.

Find explicitly $C(S)$, $S_{*}$  and  $S^{*}$. In a first shot you can take $H = 0$.
"Compatibility means deliberately repeating other people's mistakes."
David Wheeler

http://www.datasimfinancial.com
http://www.datasim.nl

bearish
Posts: 6438
Joined: February 3rd, 2011, 2:19 pm

### Re: American options with two free boundaries

You’re being a little stingy on details. I think we need to know the process for S (I can assume GBM, I suppose, although that doesn’t play really well with a constant non-zero value for H, unless that is a proportional cost), the payoff upon development, and the payoff upon abandonment (0?). And the interest rate better be constant, otherwise the optimal boundaries won’t be.

Cuchulainn
Topic Author
Posts: 64427
Joined: July 16th, 2004, 7:38 am
Location: Drosophila melanogaster
Contact:

### Re: American options with two free boundaries

You’re being a little stingy on details. I think we need to know the process for S (I can assume GBM, I suppose, although that doesn’t play really well with a constant non-zero value for H, unless that is a proportional cost), the payoff upon development, and the payoff upon abandonment (0?). And the interest rate better be constant, otherwise the optimal boundaries won’t be.
You are correct, sorry.

1. S follows a GBM (lognormal) process.
2. r is constant.
3. $C(Z) = 0$, $C(W)=W-X$, $X = strike$
4. As an ODE with 2 free boundaries $H$ is the inhomogeneous term in BS ODE. Probably not a big deal, maybe an offset -H/r.

It's like water-ice-water Stefan problem. The result should subsume the well-known perpetual for 1 free boundary.
"Compatibility means deliberately repeating other people's mistakes."
David Wheeler

http://www.datasimfinancial.com
http://www.datasim.nl

bearish
Posts: 6438
Joined: February 3rd, 2011, 2:19 pm

### Re: American options with two free boundaries

OK, I’ll go for the layup. With H=0, the optimal boundaries are zero and infinity, and the option value equals the initial stock price. There will never be a reason to abandon, thus the zero (effectively irrelevant) lower boundary. The rest of the argument follows from Merton, as long as r>0 (which I should perhaps no longer take for granted). With H>0 it will take a bit of actual work.

Cuchulainn
Topic Author
Posts: 64427
Joined: July 16th, 2004, 7:38 am
Location: Drosophila melanogaster
Contact:

### Re: American options with two free boundaries

A bit of reasoning by analogy is always good but a formula is better. If we scope to the usual perpetual, we use basic calculus to differentiate the solution to maximise the option value before expiry. So, now we have two free boundaries.

So, an explicit solution for

S < $S_*$
$S_*$ <= S <= $S^*$
S > $S^*$

It is not too difficult but how to solve it + assumptions.
"Compatibility means deliberately repeating other people's mistakes."
David Wheeler

http://www.datasimfinancial.com
http://www.datasim.nl

bearish
Posts: 6438
Joined: February 3rd, 2011, 2:19 pm

### Re: American options with two free boundaries

It is “not too difficult” in the sense that we can recycle components from the literature. I would remove the H flow by considering a time zero investment in the riskless asset of H/r, that will generate an interest stream of H dt. Then, upon hitting the lower boundary we get the principal back, so a payment of H/r. Likewise, at the upper boundary we get H/r plus the payment from exercising the option $S^* - X$. Each of the two known and constant payments can then be valued as “rebates” in a perpetual double barrier model. Jamshidian may have been the first to give an explicit solution for this, involving a ratio of hyperbolic sines (but no infinite sum, which one would expect). Optimizing the barrier locations looks like a numerical problem, though. And, when we’re almost done, we must remember to subtract H/r and check that the answer is still positive.

Cuchulainn
Topic Author
Posts: 64427
Joined: July 16th, 2004, 7:38 am
Location: Drosophila melanogaster
Contact:

### Re: American options with two free boundaries

It is “not too difficult” in the sense that we can recycle components from the literature. I would remove the H flow by considering a time zero investment in the riskless asset of H/r, that will generate an interest stream of H dt. Then, upon hitting the lower boundary we get the principal back, so a payment of H/r. Likewise, at the upper boundary we get H/r plus the payment from exercising the option $S^* - X$. Each of the two known and constant payments can then be valued as “rebates” in a perpetual double barrier model. Jamshidian may have been the first to give an explicit solution for this, involving a ratio of hyperbolic sines (but no infinite sum, which one would expect). Optimizing the barrier locations looks like a numerical problem, though. And, when we’re almost done, we must remember to subtract H/r and check that the answer is still positive.
I don't like this approach. Unfortunately, it feels like a fudge. (btw H is like property tax, so seeing it as a kind of rebate is hard to get my head around),
It has commonality with first exit times but the boundaries are unknowns.

Anyways, I thought about it logically and I solved it as a ODE free boundary value problem. We solve 2nd order ODE

$V(S) = AU_{+} + BU_{-}$ where A and B are to be found and so on. In general we have four things to find

. A,B
. $S_*,$$S^*$

These are found from the instantaneous stopping condition for V and the smooth fit for $V'$. We take the derivative but not for optimising stuff but for continuity reasons.

Then afterwards I found the same results from the work of my comrade Shiryaev.
"Compatibility means deliberately repeating other people's mistakes."
David Wheeler

http://www.datasimfinancial.com
http://www.datasim.nl

bearish
Posts: 6438
Joined: February 3rd, 2011, 2:19 pm

### Re: American options with two free boundaries

I know it suffers in your eyes from a total lack of differential equations. But it will give you the exact solution, modulo the optimal location of the exercise boundaries, which should be pretty trivial to solve for numerically - to as high a precision as you care to get. If I can muster the energy I’ll put it together this weekend. Just for fun. I used to do this sort of thing for a living...

Paul
Posts: 11267
Joined: July 20th, 2001, 3:28 pm

### Re: American options with two free boundaries

We see what you are trying to achieve, Cuch. But it would be enormously helpful if you explained what’s what.

What is the independent variable, besides t, the one that presumably evolves randomly? What does it mean and what is your symbol of choice?

What are you abandoning? Just giving away or selling for some amount?

And what do you get if you “exercise”? What’s it then worth and what does it cost?

And the taxes are on the property, that you haven’t built yet? Or are the taxes on whatever you currently own, the land?

We get the concept of two boundaries and smoothness. What I don’t get is everything else!

bearish
Posts: 6438
Joined: February 3rd, 2011, 2:19 pm

### Re: American options with two free boundaries

Right - I was going to say something about that, too. The story only really holds together if the S process is the value of the property once developed, X is the cost of developing it, and H is a tax or some such constant cost that occurs as long as you hold the deed to the land. What we are solving for is then the value of the land inclusive of the right to develop it for this particular purpose, assuming a free walk-away option. It’s an 80’s real option problem, with a very small twist.

Paul
Posts: 11267
Joined: July 20th, 2001, 3:28 pm

### Re: American options with two free boundaries

Exactly. The land and property are all a bit vague at the moment.

DavidJN
Posts: 1781
Joined: July 14th, 2002, 3:00 am

### Re: American options with two free boundaries

"...you are holding the land with a right to build property on the land or to abandon the land. "

By abandon, do you mean sell it? No residual value?

Years ago I spent a fascinating afternoon visiting Goldman Sachs in Calgary, Canada. Goldman is deeply engaged in the integrated energy business and they view the entire business as a chain of options, e.g. sit on an oil property or extract. Once extracted - store or sell it. If selling - sell it for gasoline production or refined products (e.g. plastic), and if selling - direct the output through this pipeline channel or that one. They are plainly very good with real options, how forthcoming they might be is another story, but you might try them.

Cuchulainn
Topic Author
Posts: 64427
Joined: July 16th, 2004, 7:38 am
Location: Drosophila melanogaster
Contact:

### Re: American options with two free boundaries

I'll try to scope the problem.

1. infinite time interval
2. American call option; right to build or to abandon the land.
3. H = constant rate of holding costs (we could take H = 0 for starters). H is a bit fuzzy. Do we pay tax on land with no property? I wouldn't put is pat them. Probably stamp duty and sales/appreciation tax etc.  Probably paid at source?

The solution is a fee BVP for an ODE wrt underlying S:

$0 + \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} + r S \frac{\partial V}{\partial S} - r V = H,$

defined in the mushy/continuation? interval $S_* \le S \le S^*$.  When H = 0 Peskir and Shiryaev give an analytical solution. For $H \ne 0$ I'm not so sure (variation of parameters? Likewise, what about finite time intervall? As someone said, is it part;y a real,option?

The special case when $S_* = 0$ is a fixed boundary is well known.
//  When we exercise we get the payoff in the stopping regions(?)

By abandon, do you mean sell it? No residual value?
Not sure, this would be a requirements question?
Last edited by Cuchulainn on January 30th, 2021, 9:47 am, edited 6 times in total.
"Compatibility means deliberately repeating other people's mistakes."
David Wheeler

http://www.datasimfinancial.com
http://www.datasim.nl

Cuchulainn
Topic Author
Posts: 64427
Joined: July 16th, 2004, 7:38 am
Location: Drosophila melanogaster
Contact:

### Re: American options with two free boundaries

I know it suffers in your eyes from a total lack of differential equations. But it will give you the exact solution, modulo the optimal location of the exercise boundaries, which should be pretty trivial to solve for numerically - to as high a precision as you care to get. If I can muster the energy I’ll put it together this weekend. Just for fun. I used to do this sort of thing for a living...
Yes and no. The approach is based on DE but it is good to compare to different approaches.. We look for analytic solution but a numerical solution based on elliptic variational inequality is also possible?
I suppose even binomial method ..??

It was the word 'barrier' that threw me off.
"Compatibility means deliberately repeating other people's mistakes."
David Wheeler

http://www.datasimfinancial.com
http://www.datasim.nl

Paul
Posts: 11267
Joined: July 20th, 2001, 3:28 pm

### Re: American options with two free boundaries

I'll try to scope the problem.

1. infinite time interval
2. American call option; right to build or to abandon the land.
3. H = constant rate of holding costs (we could take H = 0 for starters). H is a bit fuzzy. Do we pay tax on land with no property? I wouldn't put is pat them. Probably stamp duty and sales/appreciation tax etc.  Probably paid at source?

The solution is a fee BVP for an ODE wrt underlying S:

$0 + \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} + r S \frac{\partial V}{\partial S} - r V = H,$

defined in the mushy/continuation? interval $S_* \le S \le S^*$.  When H = 0 Peskir and Shiryaev give an analytical solution. For $H \ne 0$ I'm not so sure (variation of parameters? Likewise, what about finite time intervall? As someone said, is it part;y a real,option?

The special case when $S_* = 0$ is a fixed boundary is well known.
//  When we exercise we get the payoff in the stopping regions(?)

By abandon, do you mean sell it? No residual value?
Not sure, this would be a requirements question?
Still missing the obstacle constraint.

I'm guessing you want
$V \ge \max(S-E,0).$
With a growth rate of $\mu$ instead of $r$ you probably want something like
$V=-\frac{H}{r}+AS^a + BS^b$
where $a$ and $b$ are
$a,b=\frac{1}{\sigma^2}\left( \frac{\sigma^2}{2}-\mu \pm \sqrt{\left( \mu-\frac{\sigma^2}{2} \right)^2+2r \sigma^2} \right).$

$A$, $B$, $S_*$ and $S^*$ given by
$-\frac{H}{r}+AS_*^a+BS_*^b=0,$
$-\frac{H}{r}+AS^{*^a}+BS^{*^b}=S^*-E,$
$AaS_*^{a-1}+BbS_*^{b-1}=0,$
and
$AaS^{*^{a-1}}+BbS^{*^{b-1}}=1.$

Some obvious constraints on size  of $\mu$ and a lot of simplification when $\mu=r$.

When $\mu=r$ we get $a=1$ and $b=-2r/\sigma^2$.

And then when $H=0$ we get the Peskir and Shiryaev solution (drum roll, please)
$V=S.$
There's five minutes typing I'll never get back! And 30 seconds maths.

Given the, ahem, simplicity of (Peskir and Shiryaev's) solution for this problem, perhaps this is not their problem?