Editing Black–Scholes model (section)

==== 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>