= Solution
For the exponential payoff, substituting $U=e^{\theta X}V$ gives $U_X=\theta U$, $U_{XX}=\theta^2U$, and $U_{\sigma X}=\theta e^{\theta X}V_\sigma$. Dividing the backward <partial differential equation> by the nonzero factor $e^{\theta X}$ therefore gives the <exponential payoff transform PDE>
$$
\boxed{V_t+\left(A(\sigma)+\rho\theta\sigma B(\sigma)\right)V_\sigma
+\frac12B(\sigma)^2V_{\sigma\sigma}
+\frac12\theta(\theta-1)\sigma^2V=0.}
$$
The terminal value is
$$
\boxed{V(T,\sigma)=1.}
$$
The correlation changes the first-derivative coefficient, while the original drift of the log price combines with its variance to give $\theta(\theta-1)/2$ rather than $\theta^2/2$.
Back to article page
