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.

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    $\begingroup$ 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. $\endgroup$ – Ivan Jun 2 '19 at 21:52
  • $\begingroup$ @Ivan: Any requirement on the sign of $q$? $\endgroup$ – Hans Jun 4 '19 at 3:53

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