The PDE for the American put option price $P(S,\sigma ,r,t)$ is \begin{align*} 0 =& P_t+P_SS(r-\delta)+P_\sigma a(\sigma)+P_r\alpha (r,t) \\ +& \frac{1}{2}P_{SS}S^2\sigma ^2 + \frac{1}{2}P_{\sigma \sigma}b^2(\sigma)+\frac{1}{2}P_{rr}\beta^2(r) \\ +& P_{S\sigma}\sigma Sb(\sigma)\rho _{12}+P_{Sr}\sigma S\beta(\sigma)\rho _{13}+P_{\sigma r}\beta(\sigma)b(\sigma)\rho _{23}-rP \end{align*} that was extracted from stochastic system \begin{align*} dS_t &= (r_t-\delta)S_tdt+\sigma _tS_tdW_t^{(1)} \\ d\sigma _t &=a(\sigma _t)dt+b(\sigma _t)dW^{(2)}_t\\ dr_t &= \alpha(r_t,t)dt+\beta (r_t)dW_t^{(3)} \end{align*} such that $$ dW^{(i)}_tdW^{(j)}_t=\rho_{ij}dt $$

I found this boundary conditions \begin{align} & P(\infty ,\sigma ,r,t)=0 \\ & P(S,\sigma ,r,T)=\max (K-S,0) \\ & P(\bar{S}(T-t),\sigma ,r\,,t\,)=\max (K-\bar{S}(T-t)\,,\,0\,) \\ & {{P}_{S}}(\bar{S}(T-t),\sigma ,r\,,t)=-1 \\ \end{align} Here, $\bar S(T-t)$ is the early exercise price, which depends on the option time-to-maturity $\tau =T-t$.

Now How can I find others boundary conditions ?


1 Answer 1


When $\sigma=0$ , the boundary condition is little more complicated: \begin{align} P_t+(r-\delta)SP_S +\alpha P_r +\beta^2\frac{1}{2} P_{rr}-rP=0 \end{align} When $\sigma\rightarrow\infty$ , we have \begin{align} P(S,\infty,r,t)=0 \end{align} When $r=0$ , then \begin{align} P_t+aP_\sigma+\frac{1}{2}b^2P_{\sigma\sigma}+\sigma S b \rho_{12}P_{S\sigma}=0 \end{align}

  • $\begingroup$ I think the $\sigma\to\infty$ limit is wrong: should be $K$ as for Black-Scholes as long as $a(\sigma)$ doesn't grow too fast. $\endgroup$
    – q.t.f.
    Jun 26, 2015 at 12:47
  • 1
    $\begingroup$ Why??? when $\sigma\rightarrow\infty$ then $\S_T\rightarrow\infty$ $\endgroup$
    – user16651
    Jun 26, 2015 at 15:19

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