Editing Black–Scholes model (section)

===American options===
The problem of finding the price of an [[American option]] is related to the [[optimal stopping]] problem of finding the time to execute the option. Since the American option can be exercised at any time before the expiration date, the Black–Scholes equation becomes a variational inequality of the form:
:<math>\frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} + rS\frac{\partial V}{\partial S} - rV \leq 0</math><ref>{{cite web|author = André Jaun |url=http://www.lifelong-learners.com/opt/com/SYL/s6node6.php|title=The Black–Scholes equation for American options|access-date=May 5, 2012}}</ref>
together with <math>V(S, t) \geq H(S)</math> where <math>H(S)</math> denotes the payoff at stock price <math>S</math> and the terminal condition: <math>V(S, T) = H(S)</math>.

In general this inequality does not have a closed form solution, though an American call with no dividends is equal to a European call and the Roll–Geske–Whaley method provides a solution for an American call with one dividend;<ref name="Ødegaard">{{cite web|author=Bernt Ødegaard |year=2003|url=http://finance.bi.no/~bernt/gcc_prog/recipes/recipes/node9.html#SECTION00920000000000000000|title=Extending the Black Scholes formula|access-date=May 5, 2012}}</ref><ref name="Chance2">{{cite web|author=Don Chance|year=2008 |url=http://www.bus.lsu.edu/academics/finance/faculty/dchance/Instructional/TN98-01.pdf|title= Closed-Form American Call Option Pricing: Roll-Geske-Whaley|access-date=May 16, 2012}}</ref> see also [[Black's approximation]].

Barone-Adesi and Whaley<ref>{{cite journal|author= Giovanni Barone-Adesi|author2= Robert E Whaley|name-list-style= amp|title=Efficient analytic approximation of American option values|journal=Journal of Finance|volume=42 | issue = 2|date=June 1987|pages=301–20|url=https://ideas.repec.org/a/bla/jfinan/v42y1987i2p301-20.html|doi=10.2307/2328254|jstor= 2328254}}</ref> is a further approximation formula. Here, the stochastic differential equation (which is valid for the value of any derivative) is split into two components: the European option value and the early exercise premium. With some assumptions, a [[quadratic equation]] that approximates the solution for the latter is then obtained. This solution involves [[root-finding algorithms|finding the critical value]], <math>s*</math>, such that one is indifferent between early exercise and holding to maturity.<ref name="Ødegaard2">{{cite web|author=Bernt Ødegaard |year=2003|url=http://finance.bi.no/~bernt/gcc_prog/recipes/recipes/node13.html|title=A quadratic approximation to American prices due to Barone-Adesi and Whaley|access-date=June 25, 2012}}</ref><ref name="Chance3">{{cite web|author=Don Chance|year=2008 |url=http://www.bus.lsu.edu/academics/finance/faculty/dchance/Instructional/TN98-02.pdf|title= Approximation Of American Option Values: Barone-Adesi-Whaley|access-date=June 25, 2012}}</ref>

Bjerksund and Stensland<ref>Petter Bjerksund and Gunnar Stensland, 2002.  [http://brage.bibsys.no/nhh/bitstream/URN:NBN:no-bibsys_brage_22301/1/bjerksund%20petter%200902.pdf Closed Form Valuation of American Options]</ref> provide an approximation based on an exercise strategy corresponding to a trigger price. Here, if the underlying asset price is greater than or equal to the trigger price it is optimal to exercise, and the value must equal <math>S - X</math>, otherwise the option "boils down to: (i) a European [[Barrier option#Types|up-and-out]] call option… and (ii) a rebate that is received at the knock-out date if the option is knocked out prior to the maturity date". The formula is readily modified for the valuation of a put option, using [[put–call parity]]. This approximation is computationally inexpensive and the method is fast, with evidence indicating that the approximation may be more accurate in pricing long dated options than Barone-Adesi and Whaley.<ref>[http://www.global-derivatives.com/index.php?option=com_content&task=view&id=14 American options]</ref>

==== Perpetual put ====
Despite the lack of a general analytical solution for American put options, it is possible to derive such a formula for the case of a perpetual option – meaning that the option never expires (i.e., <math>T\rightarrow \infty</math>).<ref>{{Cite book|title=Heard on the Street: Quantitative Questions from Wall Street Job Interviews|last=Crack|first=Timothy Falcon|publisher=Timothy Crack|year=2015|isbn=978-0-9941182-5-7|edition=16th|pages=159–162}}</ref> In this case, the time decay of the option is equal to zero, which leads to the Black–Scholes PDE becoming an ODE:<math display="block">{1\over{2}}\sigma^{2}S^{2}{d^{2}V\over{dS^{2}}} + (r-q)S{dV\over{dS}} - rV = 0</math>Let <math>S_{-}</math> denote the lower exercise boundary, below which it is optimal to exercise the option. The boundary conditions are:<math display="block">V(S_{-}) = K-S_{-}, \quad {dV\over{dS}}(S_{-}) = -1, \quad V(S) \leq K</math>The solutions to the ODE are a linear combination of any two linearly independent solutions:<math display="block">V(S) = A_{1}S^{\lambda_{1}} + A_{2}S^{\lambda_{2}}</math>For <math>S_{-} \leq S</math>, substitution of this solution into the ODE for <math>i = {1,2}</math> yields:<math display="block">\left[ {1\over{2}}\sigma^{2}\lambda_{i}(\lambda_{i}-1) + (r-q)\lambda_{i} - r \right]S^{\lambda_{i}} = 0</math>Rearranging the terms gives:<math display="block">{1\over{2}}\sigma^{2}\lambda_{i}^{2} + \left(r-q - {1\over{2}} \sigma^{2}\right)\lambda_{i} - r = 0</math>Using the [[quadratic formula]], the solutions for <math>\lambda_{i}</math> are:<math display="block">\begin{aligned}
\lambda_{1} &= {-\left(r-q-{1\over{2}}\sigma^{2} \right ) + \sqrt{\left(r-q-{1\over{2}}\sigma^{2} \right )^{2} + 2\sigma^{2}r}\over{\sigma^{2}}} \\
\lambda_{2} &= {-\left(r-q-{1\over{2}}\sigma^{2} \right ) - \sqrt{\left(r-q-{1\over{2}}\sigma^{2} \right )^{2} + 2\sigma^{2}r}\over{\sigma^{2}}}
\end{aligned}</math>In order to have a finite solution for the perpetual put, since the boundary conditions imply upper and lower finite bounds on the value of the put, it is necessary to set <math>A_{1} = 0</math>, leading to the solution <math>V(S) = A_{2}S^{\lambda_{2}}</math>. From the first boundary condition, it is known that:<math display="block">V(S_{-}) = A_{2}(S_{-})^{\lambda_{2}} = K-S_{-} \implies A_{2} = {K-S_{-}\over{(S_{-})^{\lambda_{2}}}}</math>Therefore, the value of the perpetual put becomes:<math display="block">V(S) = (K-S_{-})\left( {S\over{S_{-}}} \right)^{\lambda_{2}}</math>The second boundary condition yields the location of the lower exercise boundary:<math display="block">{dV\over{dS}}(S_{-}) = \lambda_{2}{K-S_{-}\over{S_{-}}} = -1 \implies S_{-} = {\lambda_{2}K\over{\lambda_{2}-1}}</math>To conclude, for <math display="inline">S \geq S_{-} = {\lambda_{2}K\over{\lambda_{2}-1}}</math>, the perpetual American put option is worth:<math display="block">V(S) = {K\over{1-\lambda_{2}}} \left( {\lambda_{2}-1\over{\lambda_{2}}}\right)^{\lambda_{2}} \left( {S\over{K}} \right)^{\lambda_{2}}</math>