# Hedge error - Willmot and Ahmad

I'm currently reading the paper: Willmot and Ahmad: Which free lunch would you like today, Sir? Delta Heding, volatility arbitrage. In case 1: They delta hedge with the actual volatility, by going long in the option and shorting delta. There is one part of their derivation of the guarenteed profit that's confusing my quiet a bit. The specific step is: $$dV^i-dV^a -r(V^i-V^a)dt = e^{rt}d(e^{-rt}(V^i- V^a))$$ Can anybody explain this part of the equation?

• Is your question why the equality holds in the equation you wrote?
– user34971
Commented Mar 16, 2021 at 11:27
• Yeah, I don't quite understand how the two sides equal each other. Commented Mar 16, 2021 at 11:33
• So I'm assuming that we are discounting the value $$dV^i$$ and $$dV^a$$. However, I can't get from the left side to the right side. Commented Mar 16, 2021 at 11:35

\begin{align} d \left(e^{-rt} \left(V^i - V^a \right)\right) &= \left(d e^{-rt} \right) \left(V^i-V^a \right) + e^{-rt} d(V^i - V^a)\\ &= (-e^{-rt} r dt) (V^i - V^a) + e^{-rt} (dV^i - dV^a) \\ &= e^{-rt} [ -r (V^i - V^a)dt + (dV^i - dV^a) ] \end{align}
So multiplying everything by $$e^{rt}$$ gives the result.