# Option price with underlying growth rate distinct from discount rate

Consider a European style option.

The price equation is $$\frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} + rS\frac{\partial V}{\partial S} - rV = 0 \tag1$$ where $$S$$ is the underlying stock price and $$V(t,S)$$ is the option price. This is derived using for example an arbitrage free hedging argument.

Consider a similar equation $$\frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} + rS\frac{\partial V}{\partial S} - r'V = 0. \tag2$$ Note: $$r'\neq r$$. Its solution by the Feynman-Kac formula is $$V(t,S)=\mathbf E^Q\Big[e^{-\int_s^T r'ds} V(T,S(T))\,\big|\,S_t=S\Big]$$ under the measure $$\mathbf Q$$ where $$S(t)$$ is an Ito process $$dS(t) = rdt+\sigma dB$$ with $$B$$ being a Brownian motion.

I suppose you can argue that $$\mathbf Q$$ in Equation (2) is a real world measure. Is there a no arbitrage hedging argument under certain conditions to derive Equation (2)? Perhaps it needs to be under two economies with interest rates $$r$$ and $$r'$$ but somehow unrelated? Maybe some kind of foreign exchange world? Let me know if this question makes sense at all.

• Well you could always define $r \equiv r'+q$ where $q$ is the cost of carry of $S$... and the rest is the usual. Important to ensure $V$ and $S$ are in the same currency in that case though. If that (two-currency setting) is in fact the meaning of your question, then you need to add the dynamics of the FX between the two currencies and it changes the picture a bit. – Ivan Jun 2 '19 at 21:52
• @Ivan: Any requirement on the sign of $q$? – Hans Jun 4 '19 at 3:53