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Why Implied and Actual Option Values Match at Expiration

Article Quant Q&A · Author: user1627466
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Summary

The document explains an integration step used in an option hedging argument. The quantity being integrated is a full differential of the discounted difference between option values calculated using implied and actual volatility. By the fundamental theorem for differentials, its integral over time is the endpoint value minus the starting value.

At expiration, the option payoff is determined by the underlying price and no longer depends on which volatility was used to value it. The implied-volatility and actual-volatility option values therefore coincide at that endpoint, leaving the initial difference with the opposite sign. This resolves why the terminal contribution vanishes; the explanation is algebraic and depends on both valuations representing the same option payoff at maturity. It does not develop the broader hedge construction or discuss transaction costs and other market frictions.

Key ideas

  • The integral of a full differential equals the difference between its endpoint values.
  • The discounted difference between the two option valuations must be evaluated at both the initial time and expiration.
  • At expiration, the option payoff is independent of the volatility input used for valuation.
  • The two volatility-based values therefore coincide at maturity, eliminating the terminal difference.

Tags

OptionsVolatilityDerivatives pricing
Full text
# Hedging with actual volatility: problem understanding the math behind the result


# Hedging with actual volatility: problem understanding the math behind the result












From this paper. page 3

We get that the total profit at expiration is the difference in value between the price of the option with actual volatility and the one with implied volatility.

I have tried solving the integral but I don't get the same result. I especially don't understand why the incremental value in both options cancel each other out.

$$e^{r\cdot t_0} \int_{t_0}^T d\left(e^{-r\cdot t}(V^i - V^a)\right) = V^a - V^i$$

I used the expression of the differential to solve the integral in the hope of getting the same result, but I get stuck with terms in $dV^i$ and $dV^a$

I get

$$ \left(e^{-r\cdot T} - e^{-r.\cdot t_0}\right)\left[ (dV^a - dV^i)/r - (V^a - V^i) \right]$$

## Answer by Olaf (score 6, accepted)

https://quant.stackexchange.com/a/7602

The integration is over a full differential, meaning we can write:

$$ \int_{t_i}^T df(t) = f(T) - f(t_i)$$

Now, $V^i$ and $V^a$ represent the 'implied' and 'actual' value of the option, meaning they are time-dependent. This gives:

$$e^{r\cdot t_0} \int_{t_0}^T d\left(e^{-r\cdot t}(V^i - V^a)\right) = e^{-r(T-t_0)} (V^i(T) - V^a(T)) - (V^i(t_i) - V^a(t_i))$$

Next, we use the fact that at the time of expiration the value of the options is completely determined by the stock price, and independent of the volatility. This means: $V^i(T) = V^a(T)$. What remains is:

$$e^{r\cdot t_0} \int_{t_0}^T d\left(e^{-r\cdot t}(V^i - V^a)\right) = V^a(t_i) - V^i(t_i) = V^a - V^i$$

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.