Deriving the Piecewise ODE Solution for a Randomized American Put

# ODE Solution in Carr's Randomized American Put


# ODE Solution in Carr's Randomized American Put












In Carr's 1998 paper Randomization and the American Put, he sets up the following ODE for the value of an American put with expiration given by the first jump time of a Poisson process with rate $\lambda = 1/T$ (Eq. 10 in link): $$ \frac{\sigma^2}{2}S^2 P^{(1)}_{SS}(S) + rSP^{(1)}_S(S) - rP^{(1)}(S) = \lambda[P^{(1)}(S) - (K - S)^+], \qquad S > \underline{S}_1 \quad (1) $$ subject to the boundary conditions $$ \lim_{S \, \uparrow \infty} P^{(1)}(S) = 0, \qquad \lim_{S \, \downarrow \underline{S}_1} P^{(1)}(S) = K - \underline{S}_1, \qquad \lim_{S \, \downarrow \underline{S}_1} P^{(1)}_S(S) = -1\;. $$ Here $P^{(1)}(S)$ is the randomized American put value and $\underline{S}_1 < K$ is the (unknown) optimal exercise boundary. Carr gives the following solution, which I am struggling to derive myself: \begin{align*} P^{(1)}(S) = \begin{cases} p^{(1)}(S) + b^{(1)}(S) \qquad & \text{if } S > K \\ KR - S + c^{(1)}(S) + b^{(1)}(S) \qquad & \text{if } S \in (\underline{S}_1, K) \\ K - S \qquad & \text{if } S \leq \underline{S}_1 \end{cases} \end{align*} Here, $p^{(1)}(S)$ is the randomized value of a European put paying $(K-S)^+$ at the first jump: $$ p^{(1)}(S) = \left(\frac{S}{K}\right)^{\gamma - \epsilon} (qKR - \hat{q}K), \qquad S > K $$ where $$ \gamma = \frac{1}{2} - \frac{r}{\sigma^2}, \qquad R = \frac{1}{1 + rT}, \qquad \epsilon = \sqrt{\gamma^2 + \frac{2}{R\sigma^2 T}} $$ and $$ p = \frac{\epsilon - \gamma}{2\epsilon}, \qquad q = 1-p, \qquad \hat{p} = \frac{\epsilon - \gamma + 1}{2\epsilon}, \qquad \hat{q} = 1 - \hat{p} \;. $$ The quantity $b^{(1)}(S)$ is described as the present value of interest received below the critical stock price $\underline{S}_1$ until the first jump, $$ b^{(1)}(S) = \left(\frac{S}{\underline{S}_1}\right)^{\gamma - \epsilon} qKRrT \;. $$ Finally, $c^{(1)}(S)$ is the randomized value of a European call paying $(S - K)^+$ at the first jump time, $$ c^{(1)}(S) = \left(\frac{S}{K}\right)^{\gamma + \epsilon} (\hat{p}K - pKR), \qquad S < K\;. $$

Here's what I did for the case $S \in (\underline{S}_1, K)$, for which I'm not arriving at Carr's solution. Using the change of variables $x \mapsto \log S$, it's easy to show the homogeneous solution to $(1)$ is $$ P^{(1)}_{homo}(S) = c_1 S^{\gamma + \epsilon} + c_2 S^{\gamma - \epsilon} $$ and that the particular solution is $$ P^{(1)}(S) = c_1 S^{\gamma + \epsilon} + c_2 S^{\gamma - \epsilon} + RK - S\;. $$ Using the boundary conditions $\lim_{S \, \downarrow \underline{S}_1} P^{(1)}(S) = K - \underline{S}_1$ and $\lim_{S \, \downarrow \underline{S}_1} P^{(1)}_S(S) = -1$, I get that \begin{align*} c_1 & = \underline{S}_1^{-\gamma - \epsilon}(pK - pKR), \\ c_2 & = \underline{S}_1^{\epsilon - \gamma}qKRrT \end{align*} This is almost the correct solution, if only we had $c_1 = K^{-\gamma - \epsilon}(\hat{p}K - pKR)$. That is, replacing $\underline{S}_1$ with $K$ and the first $p$ with $\hat{p}$. Can anyone spot where I may have made a mistake?