Discounting the Difference Between Two Option Values

Summary

The document explains a differential identity used in a derivation of hedging profit. The expression relates the difference in changes of two option values, adjusted for the risk-free rate, to the change in their discounted value difference. The answer applies the product rule to the discounted quantity: differentiating the exponential discount factor contributes a negative rate term, while differentiating the value difference contributes the two option-value changes.

After factoring out the discount factor, multiplying both sides by its reciprocal yields the stated identity. This is an algebraic step for a constant risk-free rate and is presented to clarify a derivation in a paper on delta hedging and volatility arbitrage. The response does not assess the broader hedging strategy, assumptions about the option values, or whether the paper’s profit result follows; it only explains the discounting transformation.

Key ideas

• Differentiating a discounted value difference requires the product rule.
• The derivative of the exponential discount factor contributes a negative risk-free-rate term.
• Factoring out the discount factor gives the rate-adjusted difference in option-value changes.
• The explanation addresses an algebraic step and does not validate the broader arbitrage argument.

Tags

# Hedge error - Willmot and Ahmad


# 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?

## Answer by user34971 (score 3, accepted)

https://quant.stackexchange.com/a/61772

\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.

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