Option Payoff Beta and One-Time Minimum-Variance Hedging
Summary
The document asks how to calculate the beta of a call payoff relative to its underlying at expiry under the risk-neutral measure. This covariance ratio is the hedge coefficient that minimizes the variance of the terminal option payoff minus a fixed holding in the underlying, when the hedge is set only once. The question contrasts this terminal minimum-variance hedge with Black–Scholes delta.
The reply says the terminal expectation can be computed from the lognormal distribution and relates a required cross-moment to a power-option payoff integral. It also gives an instantaneous alternative under Black–Scholes: Ito’s lemma links option volatility to delta, underlying price, and underlying volatility. These are distinct horizons and objectives; the instantaneous relationship does not establish that Black–Scholes delta is the optimal hedge for terminal variance. The discussion points toward a calculation but does not provide a fully worked closed-form beta.
Key ideas
- The terminal minimum-variance hedge coefficient is the covariance of the call payoff and underlying divided by underlying variance.
- A one-time terminal hedge can differ from the Black–Scholes delta used for local dynamic hedging.
- The required terminal cross-moment can be expressed using a power-option expectation.
- Under Black–Scholes, Ito’s lemma relates instantaneous option volatility to delta and underlying volatility.
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Full text
# Beta between stock and option
# Beta between stock and option
In Black Scholes model I would like to compute $$ \beta_K = \frac{\mathrm{cov}(C_{K,T},S_T)}{\mathrm{cov}(S_T,S_T)} = \frac{\mathrm{cov}((S_T - K)^+,S_T)}{\mathrm{cov}(S_T,S_T)} $$ with respect to say risk-neutral measure. Here $C_{K,T}$ is the payoff of the call with expiry $T$ and strike $K$ at expiry, and $S_T$ is the price of the underlying at expiry. Of course, this all can be computed just by using the log-normal distribution of $S_T$, however I wondered whether there is some trick to compute it in a faster way.
Motivation: I would like to know which $\beta_K = \Delta$ minimizes the variance of $C - \Delta S$. That is, if you can't hedge dynamically, but only once, would you still choose the BS delta, or whether it is gonna be something else.
## Answer by Brian B (score 1)
https://quant.stackexchange.com/a/18721
For the terminal distributions, I don't have the closed-form solution to hand, but it's computable, since we can price power options (with payoffs like $(S_T^n-K)^+$). You need to find
$$ E[S_T C_{K,T}] = \int_K^\infty x(x-K) \cdot p_{BS}(x) dx \\=-Ke^{(r-q)T} C_{K,T} + \int_K^\infty x^2 \cdot p_{BS}(x) dx $$
The latter formula is just a power-option price, which is a common homework problem whose solution can be found in many places, including this paper.
If an instantaneous treatment is good enough, then you really just need to know the instantaneous volatility $\omega$ of an option $C$ to compute this. If you are willing to believe Black-Scholes, you can apply Ito's Lemma to find that
$$ C \cdot \omega = \Delta \cdot S \cdot \sigma_S $$
Edit: added terminal distribution integrals and comments on instantaneous versus terminalShown 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.