9.3 The P&L of a delta-hedged option

9.3.1The gamma–theta P&L#

Proof

The cash account is $B_t=X_t-V^i_t+\Delta^i_tS_t$. Self-financing means $dX_t=dV^i_t-\Delta^i_t\,dS_t+rB_t\,dt$. By Itô's lemma, with $d\langle S\rangle_t=\sigma_t^2S_t^2dt$,

$$dV^i_t=\Theta^i_t\,dt+\Delta^i_t\,dS_t+\tfrac12\Gamma^i_t\sigma_t^2S_t^2\,dt.$$

Substituting, the $dS$ terms cancel:

$$dX_t=\left(\Theta^i_t+\tfrac12\Gamma^i_t\sigma_t^2S_t^2\right)dt+r\left(X_t-V^i_t+\Delta^i_tS_t\right)dt.$$

The Black–Scholes PDE at $\sigma_i$ reads $\Theta^i+\tfrac12\sigma_i^2S^2\Gamma^i+rS\Delta^i-rV^i=0$, so $\Theta^i-rV^i+rS\Delta^i=-\tfrac12\sigma_i^2S^2\Gamma^i$. Hence $dX_t-rX_tdt=\tfrac12\Gamma^i_tS_t^2(\sigma_t^2-\sigma_i^2)dt$, which is (9.10) after multiplying by $e^{-rt}$. Integrate from 0 to $T$ with $X_0=0$; at $T$ the option is worth its payoff, so $X_T$ is the realised P&L.∎

The drift $\mu$ does not appear. A hedged option is not a directional bet. It is a bet on $\sigma_t^2-\sigma_i^2$.

Proposition 9.3 is the continuous-time, exact version of M08 (8.28), which gave the local approximation from the theta–gamma relation (8.27). The proof above adds the financing terms and the distinction between the vol used to price and hedge ($\sigma_i$) and the vol that is realised ($\sigma_t$).

9.3.2Daily P&L and path dependence#

Discrete form Hedging once a day, over a step $\Delta t$ with return $\Delta S/S$, a second-order Taylor expansion of $V^i$ gives (M08 (8.28) with $\sigma=\sigma_i$)

Read it as a contest: the option gains $\tfrac12\Gamma(\Delta S)^2$ from the move and loses theta, which at the money is approximately $\tfrac12\Gamma S^2\sigma_i^2\Delta t$ (9.8). The breakeven daily move is $|\Delta S/S|=\sigma_i\sqrt{\Delta t}$. At $\sigma_i=14\%$ and one calendar day, that is $0.14/\sqrt{365}=0.733\%$. A long hedged option wins on any day the underlying moves more than 0.733% either way and loses on the others. The total is

Path dependence (9.11) weights each instant's variance gap by $\Gamma^iS^2$, the dollar gamma. For a fixed realised variance, the P&L is large when big moves happen while spot is near the strike and gamma is high, and small when they happen far from the strike. Two paths with identical realised volatility can give opposite P&L signs. Sinclair (Volatility Trading, ch. on hedging) shows simulated distributions of this effect.

9.3.3Expected P&L under GBM#

Proof

(a) Take expectations in (9.11) and swap $E$ and $\int$ (Fubini; the integrand is bounded on $[0,T-\epsilon]$ and integrable near $T$). We need $E[\Gamma^i(t,S_t)S_t^2]$. The identity $S\varphi(d_1)=Ke^{-r\tau}\varphi(d_2)$ (M08) gives

$$\Gamma^iS^2=\frac{S\varphi(d_1)}{\sigma_i\sqrt\tau}=\frac{Ke^{-r\tau}\varphi(d_2)}{\sigma_i\sqrt\tau},\qquad\tau=T-t.$$

Under GBM, $\ln S_t=\ln S_0+(\mu-\tfrac12\sigma_r^2)t+\sigma_r\sqrt t\,Z$, so $d_2=\alpha+\beta Z$ with $\alpha=[\ln(S_0/K)+(\mu-\tfrac12\sigma_r^2)t+(r-\tfrac12\sigma_i^2)\tau]/(\sigma_i\sqrt\tau)$ and $\beta=\sigma_r\sqrt t/(\sigma_i\sqrt\tau)$. The Gaussian convolution identity $E[\varphi(\alpha+\beta Z)]=\varphi\big(\alpha/\sqrt{1+\beta^2}\big)/\sqrt{1+\beta^2}$ (complete the square) gives, with $\sigma_i\sqrt\tau\sqrt{1+\beta^2}=\sqrt{\sigma_i^2\tau+\sigma_r^2t}=\sqrt{v(t)}$ and $\alpha/\sqrt{1+\beta^2}=a(t)$,

$$E\left[\Gamma^iS_t^2\right]=\frac{Ke^{-r\tau}\varphi(a(t))}{\sqrt{v(t)}}.$$

Multiply by $\tfrac12e^{-rt}(\sigma_r^2-\sigma_i^2)$ and integrate; $e^{-rt}e^{-r\tau}=e^{-rT}$.

(b) With $\mu=r$ the numerator of $a(t)$ is $\ln(S_0/K)+rT-\tfrac12v(t)$, so $a(t)=d_2(v)$, where $d_2(v)$ is Black–Scholes $d_2$ written in terms of total variance $v=\sigma^2T$. Change variable $v=v(t)$, $dv=(\sigma_r^2-\sigma_i^2)dt$:

$$E[e^{-rT}X_T]=\int_{\sigma_i^2T}^{\sigma_r^2T}\frac{Ke^{-rT}\varphi(d_2(v))}{2\sqrt v}\,dv.$$

The integrand is $\partial BS/\partial v$: from vega $=S\varphi(d_1)\sqrt T$ and $d\sigma/dv=1/(2\sigma T)$, $\partial BS/\partial v=S\varphi(d_1)/(2\sqrt v)=Ke^{-rT}\varphi(d_2)/(2\sqrt v)$. The integral is therefore $BS(\sigma_r)-BS(\sigma_i)$.∎

With $\mu\neq r$ the answer differs, because drift moves spot away from (or toward) the strike and changes the average dollar gamma. In the worked example below, raising $\mu$ from 6% to 15% cuts the seller's expected profit from 84.7 to 82.8 index points per unit. The tests in m09_delta_hedge.py verify (9.14) against Monte Carlo for both drifts.

9.3.4Hedging volatility and robustness#

Proof

Let $V^r$ be the Black–Scholes value at $\sigma_r$. Repeat the proof of Proposition 9.3 with $V^r$ in place of $V^i$ and $\sigma_t\equiv\sigma_r$: the bracket $(\sigma_t^2-\sigma_r^2)$ vanishes, so the self-financing portfolio "long $V^r$, short $\Delta^r$" grows at exactly $r$. The trader's actual book differs from that portfolio only in the initial cash: she paid $V^i_0$ instead of $V^r_0$. So her book at $T$ equals the riskless growth of $V^r_0-V^i_0$, that is $X_T=e^{rT}(V^r_0-V^i_0)$.∎

So there is a trade-off (Carr and Madan 1998; Ahmad and Wilmott 2005; Sinclair, Volatility Trading):

• Hedge at implied: the mark-to-market P&L is smooth day by day (the book is always worth $V^i$ minus hedge), but the final P&L is path-dependent through dollar gamma, and only its expectation is pinned down.
• Hedge at realised: the final P&L is deterministic, $BS(\sigma_r)-BS(\sigma_i)$, but the daily mark-to-market swings, because the hedge is wrong relative to market prices. And you need to know $\sigma_r$, which you never do.

In the code (Exercise m09_delta_hedge; the worked example below, 20,000 GBM paths with simulate_gbm_paths(..., seed=1)), hedging a 30-day Nifty call at realised vol instead of implied vol cut the P&L standard deviation from 34.3 to 19.2 index points with daily-ish (200-step) hedging (34.1–34.5 and 19.1–19.3 over seeds 0–4); at 800 steps the realised-vol hedge's remaining noise was 9.7, half that, consistent with discrete-hedging error of order $1/\sqrt N$ (Bertsimas, Kogan and Lo 2000).

Proof

For the short option, (9.10) with $\sigma_i$ replaced by $\sigma_h$ gives $d(e^{-rt}X_t)=\tfrac12e^{-rt}\Gamma^hS^2(\sigma_h^2-\sigma_t^2)dt$ (Itô's lemma does not need $\sigma_t$ constant). A convex payoff has $\Gamma^h\ge0$ everywhere (the Black–Scholes value of a convex payoff is convex in $S$). Both factors are non-negative.∎

This is why option sellers think in terms of "selling vol above where it will realise". The converse is the danger: one period in which $\sigma_t$ exceeds $\sigma_h$ while gamma is large can erase months of carry. For non-convex payoffs (a call spread), $\Gamma$ changes sign and the result fails.

9.3.5A hedged Nifty call#