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Cuchulainn
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### Re: American options with two free boundaries

Just saw this thread late. It turns out the double exercise boundary happens naturally for American options under negative interest rates. And you may use the integral formulation to solve the exercise boundary, using similar techniques as in the single boundary case (see Andersen an Lake for example). A paper of mine on this has been accepted for publication in JCF (not sure when it will be actually published with the backlog).
Don't suppose you have the public domain preprint? is it for calls I suppose? I suppose puts are never optimal?
Are you using LSM or PDE approach

// 20-30 years ago the interest rate was [12,15] percent.
"Compatibility means deliberately repeating other people's mistakes."
David Wheeler

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

Cuchulainn
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Joined: July 16th, 2004, 7:38 am
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### 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?
If we now take $r < 0$ and apply this formal solution we see there is no solution for puts and no exercise points.
"Compatibility means deliberately repeating other people's mistakes."
David Wheeler

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