# 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.

• I assume that you don't want to calculate the covariance of $(S_T-K)^+$ with $S_t$ but rather of $E[(S_T-K)^+|F_t]$ ... – Ric Jul 6 '15 at 15:05
• @Richard: I don't have small $t$ at all. I'd like to compute the covariance as of now of the expiration values. Let me add some motivation. – Ulysses Jul 6 '15 at 15:17

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

$$C \cdot \omega = \Delta \cdot S \cdot \sigma_S$$

• I see, essentially this looks similar to just straightforward computation through the log-normal density, and is a bit brute-force. Regarding you second ('instantaneous') approach, how do we get this equation through Ito Lemma? I mean, which differential do we start with? I know we get $$\mathrm dC = \sigma_tS_t\mathrm dW_t$$ but not sure how to get your equality. – Ulysses Jul 7 '15 at 7:25