Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 39 / 1 / b

For the exponential payoff, substituting $U=e^{\theta X}V$ gives $U_X=\theta U$, $U_{XX}={\theta }^{2}U$, 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 PDEThe terminal value isThe 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 ${\theta }^2/2$.