Unit 6, lesson 46

Discrete delta hedging and hedging error

Derives the leading residual of a delta hedge rebalanced on a finite grid, shows how gamma exposure turns realized versus model variance into hedging error, and explains the role of trading frequency and costs.

  • Video 46
  • Black-Scholes formulas, Greeks and hedging

A trader buys a one-year option at its Black-Scholes price, sells the model delta in shares and plans to adjust the hedge once a day. Between two trading dates the share holding stays fixed while the option’s delta moves with the stock, so the hedged position gains or loses money even when the model’s volatility is right. How large is that residual, which sign does it have, and how does it depend on the size of the realized move? This lesson derives the leading term of the one-step residual, shows why it remains random on every finite grid, and relates the accumulated error to a comparison of realized and model variance weighted by gamma.

Key terms

Discrete delta hedging

Definition Fix trading dates 0=t0<t1<⋯<tn=T0 = t_0 < t_1 < \dots < t_n = T with steps hk=tk+1−tkh_k = t_{k+1} - t_k. Let V(t,S)V(t, S) be the Black-Scholes price of a European option computed with a model volatility σm>0\sigma_m > 0 and constant rr and qq, with Greeks Δ=VS\Delta = V_S, Γ=VSS\Gamma = V_{SS} and Θ=Vt\Theta = V_t. A discrete delta hedge of a long option holds the option, is short Δk=VS(tk,Stk)\Delta_k = V_S(t_k, S_{t_k}) shares on [tk,tk+1)[t_k, t_{k+1}), and keeps the balance in the money-market account, so that the option, the shares and this cash are worth zero together at tkt_k, with earlier gains set aside in the same account. The holding Δk\Delta_k is chosen from the information at tkt_k and is not changed inside the step. The one-step residual ΔΠk\Delta\Pi_k is the change in value of this position over [tk,tk+1][t_k, t_{k+1}]. With r=q=0r = q = 0, writing S=StkS = S_{t_k}, h=hkh = h_k and ΔS=ΔSk=Stk+1−Stk\Delta S = \Delta S_k = S_{t_{k+1}} - S_{t_k}, it is

ΔΠk=V(tk+h,S+ΔS)−V(tk,S)−Δk ΔS.\begin{aligned} \Delta\Pi_k &= V(t_k + h, S + \Delta S) \\[0.4em] &\quad - V(t_k, S) - \Delta_k\,\Delta S . \end{aligned}

With rr and qq nonzero it also contains the interest on the cash balance ΔkS−V\Delta_k S - V and the dividends owed on the short shares. For tk<Tt_k < T the price is smooth, and a second-order Taylor expansion with the state frozen at (tk,S)(t_k, S) gives ΔV≈Θh+Δk ΔS+12Γ(ΔS)2\Delta V \approx \Theta h + \Delta_k\,\Delta S + \tfrac12\Gamma(\Delta S)^2. The short shares cancel the delta term, and the carry adds about ((r−q)SΔk−rV)h\bigl((r - q)S\Delta_k - rV\bigr)h. The model pricing PDE Θ+12σm2S2Γ+(r−q)SΔ−rV=0\Theta + \tfrac12\sigma_m^2 S^2\Gamma + (r - q)S\Delta - rV = 0 then leaves

ΔΠk≈12Γ[(ΔS)2−σm2S2h].\begin{aligned} \Delta\Pi_k &\approx \tfrac12\Gamma\bigl[(\Delta S)^2 \\[0.4em] &\qquad - \sigma_m^2 S^2 h\bigr]. \end{aligned}

The omitted terms, such as 16VSSS(ΔS)3\tfrac16 V_{SSS}(\Delta S)^3 and the mixed term ΘS h ΔS\Theta_S\,h\,\Delta S, are of higher order for small steps, but they matter for large moves and near expiry close to the strike, where the derivatives change quickly. A short option with the opposite share position has every sign reversed.

In plain words The hedge removes the first-order exposure to the stock at each trading date and then sits still. Over the step, the option’s curvature earns half gamma times the squared move, while time decay charges half gamma times the squared move the model expects. The residual is the gap between the two.

Example Take a constructed hedged long option with S=100S = 100, Γ=0.02\Gamma = 0.02 per currency unit, σm=0.2\sigma_m = 0.2 and r=q=0r = q = 0, rebalanced daily with h=1/252h = 1/252 year. The model charge is

σm2S2h=0.04⋅104252≈1.587302\begin{aligned} \sigma_m^2 S^2 h &= \frac{0.04 \cdot 10^4}{252} \\[0.4em] &\approx 1.587302 \end{aligned}

in squared currency units. A move of ΔS=2\Delta S = 2 gives the leading residual 0.01⋅(4−1.587302)≈0.0241270.01 \cdot (4 - 1.587302) \approx 0.024127, a move of ΔS=1\Delta S = 1 gives 0.01⋅(1−1.587302)≈−0.0058730.01 \cdot (1 - 1.587302) \approx -0.005873, and no move gives −0.015873-0.015873.

Gamma exposure

Definition The gamma exposure of a position is its second derivative in the stock price, Γ=VSS\Gamma = V_{SS} for one long option and −Γ-\Gamma for one short option, in units of shares per currency unit. The delta hedge has zero gamma, so the gamma of the hedged position is that of the option. In the leading residual it multiplies the excess of the squared move over the model charge, and in terms of the return ΔS/S\Delta S/S

ΔΠk≈12ΓS2[(ΔSS)2−σm2h],\begin{aligned} \Delta\Pi_k &\approx \tfrac12\Gamma S^2\Bigl[\Bigl(\frac{\Delta S}{S}\Bigr)^2 \\[0.4em] &\qquad - \sigma_m^2 h\Bigr], \end{aligned}

where the cash gamma 12ΓS2\tfrac12\Gamma S^2 is in currency per unit of squared return. Gamma also sets the size of the next trade, since the new delta is Δk+1≈Δk+Γ ΔS\Delta_{k+1} \approx \Delta_k + \Gamma\,\Delta S to first order. For a long call or put, Γ>0\Gamma > 0, so after a rise the hedger shorts more shares at the higher price and after a fall buys shares back at the lower price.

In plain words Gamma says how much the hedge ratio moves when the stock moves. A long gamma position profits from large moves and pays for this through time decay, and a short gamma position collects the time decay and loses on large moves.

Example In the constructed example the cash gamma is 12⋅0.02⋅1002=100\tfrac12 \cdot 0.02 \cdot 100^2 = 100, so a squared daily return of 0.00010.0001, a one percent move, contributes 0.010.01, against the model charge 100⋅0.04/252≈0.015873100 \cdot 0.04/252 \approx 0.015873. For the at-the-money Black-Scholes call with S=K=100S = K = 100, σm=0.2\sigma_m = 0.2, τ=1\tau = 1 and r=q=0r = q = 0, where Γ≈0.019848\Gamma \approx 0.019848, a one-day rise to 102102 raises the delta by about 0.0390450.039045 and a fall to 9898 lowers it by about 0.0403920.040392, close to Γ⋅2≈0.039695\Gamma \cdot 2 \approx 0.039695.

Realized variance

Definition On the grid, the realized variance of the stock over [0,T][0, T] is the sum of squared returns

RV⁡=∑k=0n−1(ΔSkStk)2,\operatorname{RV} = \sum_{k=0}^{n-1} \Bigl(\frac{\Delta S_k}{S_{t_k}}\Bigr)^2 ,

or the same sum with log returns ln⁡(Stk+1/Stk)\ln(S_{t_{k+1}}/S_{t_k}), and RV⁡/T\operatorname{RV}/T is its annualized version. Suppose that under P\mathbb{P} the stock follows dSt=μSt dt+σaSt dWtdS_t = \mu S_t\,dt + \sigma_a S_t\,dW_t with an actual volatility σa>0\sigma_a > 0 and constant μ\mu. Then, given Ftk\mathcal{F}_{t_k} and with S=StkS = S_{t_k}, the squared move over a step of length hh satisfies

E[(ΔS)2∣Ftk]=σa2S2h+O(h2),Var⁡((ΔS)2∣Ftk)≈2σa4S4h2,\begin{aligned} &\mathbb{E}\bigl[(\Delta S)^2 \mid \mathcal{F}_{t_k}\bigr] \\[0.4em] &\quad = \sigma_a^2 S^2 h + O(h^2), \\[0.4em] &\operatorname{Var}\bigl((\Delta S)^2 \mid \mathcal{F}_{t_k}\bigr) \\[0.4em] &\quad \approx 2\sigma_a^4 S^4 h^2 , \end{aligned}

where the exact mean is S2(e(2μ+σa2)h−2eμh+1)S^2\bigl(e^{(2\mu + \sigma_a^2)h} - 2e^{\mu h} + 1\bigr) and the drift enters only at order h2h^2. The variance comes from ΔS≈σaSh Z\Delta S \approx \sigma_a S\sqrt{h}\,Z with Var⁡(Z2)=2\operatorname{Var}(Z^2) = 2. As the mesh tends to zero, the realized variance tends in probability to the quadratic variation σa2T\sigma_a^2 T of ln⁡S\ln S, and for an adapted volatility process σt\sigma_t to ∫0Tσt2 dt\int_0^T \sigma_t^2\,dt, as in Quadratic variation and stochastic limits.

In plain words Realized variance adds up what the squared moves actually were. One squared move is a noisy reading, with a standard deviation larger than its mean, but many small ones add up to the variance rate times the elapsed time.

Example Five constructed daily returns of 0.010.01, −0.02-0.02, 0.0050.005, 0.0150.015 and −0.01-0.01 have squared sum 0.000850.00085. With h=1/252h = 1/252 this is an annualized variance of 0.00085/(5/252)≈0.042840.00085/(5/252) \approx 0.04284, a realized volatility of about 0.20700.2070, against the model variance 0.04⋅5/252≈0.0007940.04 \cdot 5/252 \approx 0.000794 for σm=0.2\sigma_m = 0.2. At a volatility of 0.20.2 one squared daily return has mean 0.0001590.000159 and standard deviation about 0.0002240.000224, so a single day says little about the variance rate.

Hedging error

Definition The hedging error ETE_T of the strategy is the value at TT of the position that buys the option at its model price V(0,S0)V(0, S_0) at time 00, holds the delta hedge and the cash account until TT, and receives the payoff. It equals the residuals accrued to TT,

ET=∑k=0n−1er(T−tk+1) ΔΠk,E_T = \sum_{k=0}^{n-1} e^{r(T - t_{k+1})}\,\Delta\Pi_k ,

with ET=0E_T = 0 for exact replication, and an option writer who sells at the model price and holds the opposite hedge has the error −ET-E_T. With r=q=0r = q = 0 it is ET=(ST−K)+−V(0,S0)−∑kΔk ΔSkE_T = (S_T - K)^{+} - V(0, S_0) - \sum_k \Delta_k\,\Delta S_k for a call. Under the diffusion of the previous card, the frozen-state approximations give for one step

E[ΔΠk∣Ftk]≈12ΓS2(σa2−σm2)h,Var⁡(ΔΠk∣Ftk)≈12Γ2σa4S4h2.\begin{aligned} &\mathbb{E}[\Delta\Pi_k \mid \mathcal{F}_{t_k}] \\[0.4em] &\quad \approx \tfrac12\Gamma S^2(\sigma_a^2 - \sigma_m^2)h, \\[0.4em] &\operatorname{Var}(\Delta\Pi_k \mid \mathcal{F}_{t_k}) \\[0.4em] &\quad \approx \tfrac12\Gamma^2\sigma_a^4 S^4 h^2 . \end{aligned}

A price jump JJ inside a step leaves the exact residual

ΔΠjump=V(t,S+J)−V(t,S)−VS(t,S) J,\begin{aligned} \Delta\Pi_{\mathrm{jump}} &= V(t, S + J) \\[0.4em] &\quad - V(t, S) - V_S(t, S)\,J , \end{aligned}

which is nonnegative for a long call or put because VV is convex in SS, and which can be far from 12ΓJ2\tfrac12\Gamma J^2 for a large jump. A single stock position cannot offset jumps of every size, and a random volatility leaves a vega exposure. These are separate sources of error from the squared-move residual of a diffusion.

In plain words The hedging error is what is left at maturity after the hedge has done its job, positive when the long-option hedger ends with a gain. On a finite grid it is a random quantity even in a correct model, and its average depends on whether the stock moved more or less than the model assumed.

Example For the constructed daily step with S=100S = 100, Γ=0.02\Gamma = 0.02, σm=0.2\sigma_m = 0.2 and an actual volatility σa=0.3\sigma_a = 0.3, the expected squared move is 0.09⋅104/252≈3.5714290.09 \cdot 10^4/252 \approx 3.571429 against the model charge 1.5873021.587302, so

E[ΔΠk]≈0.01⋅1.984127≈0.019841,\begin{aligned} \mathbb{E}[\Delta\Pi_k] &\approx 0.01 \cdot 1.984127 \\[0.4em] &\approx 0.019841 , \end{aligned}

with a standard deviation of about 0.0505080.050508 for the same step. A deliberately large constructed move of 1010 gives the leading estimate 0.01⋅(100−1.587302)≈0.9841270.01 \cdot (100 - 1.587302) \approx 0.984127, which is only a second-order estimate. Repricing the at-the-money Black-Scholes call of the gamma exposure card gives an exact residual of 0.9134060.913406 for a rise of 1010 in one day and 1.0088021.008802 for a fall of 1010, while its leading term is 0.9766290.976629 for both.

Trading frequency

Definition The trading frequency is the number of rebalancing dates per unit of time, 1/h1/h on an equally spaced grid. Over a fixed horizon TT with n=T/hn = T/h steps and the state frozen, the one-step noise has standard deviation about (1/2)∣Γ∣σa2S2h(1/\sqrt{2})\lvert\Gamma\rvert\sigma_a^2 S^2 h, and the conditionally uncorrelated steps add up to

sd⁡(ET)≈12∣Γ∣σa2S2Th,E[ET]≈12ΓS2(σa2−σm2)T,\begin{aligned} &\operatorname{sd}(E_T) \\[0.4em] &\quad \approx \tfrac{1}{\sqrt{2}}\lvert\Gamma\rvert\sigma_a^2 S^2\sqrt{Th}, \\[0.4em] &\mathbb{E}[E_T] \\[0.4em] &\quad \approx \tfrac12\Gamma S^2(\sigma_a^2 - \sigma_m^2)T , \end{aligned}

so more frequent trading reduces the noise like h\sqrt{h} and leaves the variance mismatch unchanged. For calls and puts on an equally spaced grid with σa=σm\sigma_a = \sigma_m, the root mean square of the total error shrinks in proportion to h\sqrt{h}, a result stated here without proof. Trading has costs. With a proportional cost κ\kappa per unit of traded value, a rebalance trades about ∣Γ ΔS∣\lvert\Gamma\,\Delta S\rvert shares and costs about κ∣Γ∣S2σa2h/π\kappa\lvert\Gamma\rvert S^2\sigma_a\sqrt{2h/\pi} on average, so the total cost over the horizon grows like 1/h1/\sqrt{h}. Letting h→0h \to 0 with a fixed κ\kappa is therefore a different economic model from frictionless continuous trading, and choosing the trading frequency against costs is a separate optimization problem.

In plain words Trading more often makes each unhedged interval shorter, so the random part of the error averages out better. It does not remove the error from using the wrong volatility, and every extra trade pays costs.

Example With the frozen constructed values Γ=0.02\Gamma = 0.02, S=100S = 100 and σa=σm=0.2\sigma_a = \sigma_m = 0.2 over one year, daily trading gives sd⁡(ET)≈0.3563\operatorname{sd}(E_T) \approx 0.3563, and four trades a day halve it to about 0.17820.1782. A constructed simulation of 2000020000 paths of the at-the-money one-year call with σa=σm=0.2\sigma_a = \sigma_m = 0.2, zero drift and r=q=0r = q = 0 gave standard deviations of about 0.950.95, 0.440.44 and 0.220.22 for 5252, 252252 and 10081008 rebalances, halving with every fourfold increase, and means within sampling error of zero. With a cost of κ=0.001\kappa = 0.001 on every share traded, including the first purchase and the final unwind, the mean costs were about 0.340.34, 0.610.61 and 1.111.11.

Gamma-weighted variance comparison

Definition Summing the leading residuals with r=0r = 0 gives

ET≈∑k12ΓkStk2×[(ΔSkStk)2−σm2hk],\begin{aligned} E_T &\approx \sum_{k} \tfrac12\Gamma_k S_{t_k}^2 \\[0.4em] &\quad \times \Bigl[\Bigl(\frac{\Delta S_k}{S_{t_k}}\Bigr)^2 - \sigma_m^2 h_k\Bigr], \end{aligned}

with Γk=Γ(tk,Stk)\Gamma_k = \Gamma(t_k, S_{t_k}), a comparison of each realized squared return with the model variance, weighted by the cash gamma at that date. For continuous hedging the comparison is exact under these assumptions. Trading is frictionless and continuous, under P\mathbb{P} the stock follows dSt=μtSt dt+σtSt dWtdS_t = \mu_t S_t\,dt + \sigma_t S_t\,dW_t with adapted μt\mu_t and σt\sigma_t, the long option is hedged by a short position of VS(t,St)V_S(t, S_t) shares computed from the model with constant σm\sigma_m, rr and qq, cash earns rr, and the dividends owed on the short shares are paid from cash. Itô’s formula for V(t,St)V(t, S_t) uses σt\sigma_t, the model PDE uses σm\sigma_m, and the difference of the two drifts gives

ET=∫0Ter(T−t) 12Γ(t,St)×St2(σt2−σm2) dt,\begin{aligned} E_T &= \int_0^T e^{r(T - t)}\,\tfrac12\Gamma(t, S_t) \\[0.4em] &\qquad \times S_t^2\bigl(\sigma_t^2 - \sigma_m^2\bigr)\,dt , \end{aligned}

for calls and puts, applying Itô on [0,T−ε][0, T - \varepsilon] and letting ε→0\varepsilon \to 0. With Γ>0\Gamma > 0 and σt>σm\sigma_t > \sigma_m at all times the continuous hedge gains on every path, and with σt<σm\sigma_t < \sigma_m it loses. The amount depends on the path through Γ(t,St)St2\Gamma(t, S_t)S_t^2, which is largest near the strike close to expiry. On a finite grid the random squared-shock noise of the previous cards comes on top.

In plain words The hedger effectively buys variance at the model rate and receives the realized rate, in amounts set by the cash gamma at each moment. Realized variance above the model rate helps a long gamma position, and the help counts most where gamma is large. It is not a guaranteed gain from owning an option, because realized variance can fall short and because the weights depend on where the stock goes.

Example For the five constructed daily returns of the realized variance card, with Γ=0.02\Gamma = 0.02 held fixed, σm=0.2\sigma_m = 0.2 and the stock starting at 100100, the leading residuals are about −0.005873-0.005873, 0.0246120.024612, −0.013102-0.013102, 0.0065580.006558 and −0.005987-0.005987, which add to about 0.0062080.006208. In the simulation of the previous card with σa=0.3\sigma_a = 0.3 in place of 0.20.2, the mean error was about 3.963.96 for every trading frequency, with standard deviations of about 2.332.33, 1.931.93 and 1.861.86 for 5252, 252252 and 10081008 rebalances, which do not shrink to zero. With zero drift and r=q=0r = q = 0 the stock gains of the hedge have mean zero on every grid, so the exact mean error is C(σ=0.3)−C(σ=0.2)≈3.957971C(\sigma = 0.3) - C(\sigma = 0.2) \approx 3.957971.

Check yourself

  1. A trader is short an option and delta hedged with shares, so the position gamma is −0.015-0.015 per currency unit. The stock is at 5050, the model volatility is 0.250.25, the actual volatility is 0.200.20 and the hedge is rebalanced weekly, h=1/52h = 1/52. Find the expected leading residual for one week and the leading residual for a weekly move of 33, and explain the signs.
Show the answer

The model charge is σm2S2h=0.0625⋅2500/52≈3.004808\sigma_m^2 S^2 h = 0.0625 \cdot 2500/52 \approx 3.004808, and the expected squared move is σa2S2h=0.04⋅2500/52≈1.923077\sigma_a^2 S^2 h = 0.04 \cdot 2500/52 \approx 1.923077. With 12Γ=−0.0075\tfrac12\Gamma = -0.0075,

E[ΔΠ]≈−0.0075⋅(−1.081731)≈0.008113.\begin{aligned} \mathbb{E}[\Delta\Pi] &\approx -0.0075 \cdot (-1.081731) \\[0.4em] &\approx 0.008113 . \end{aligned}

The short gamma position gains on average because the stock is expected to move less than the model charge assumes. For a move of 33 the residual is −0.0075⋅(9−3.004808)≈−0.044964-0.0075 \cdot (9 - 3.004808) \approx -0.044964, a loss, because the squared move 99 exceeds the model charge. That move is about 2.162.16 weekly standard deviations, since σaSh≈1.386751\sigma_a S\sqrt{h} \approx 1.386751. Both numbers are second-order estimates.

  1. A colleague says that if the actual volatility equals the model volatility, a daily delta hedge has no error. Using the frozen values Γ=0.02\Gamma = 0.02, S=100S = 100, σa=σm=0.2\sigma_a = \sigma_m = 0.2 and h=1/252h = 1/252, find the standard deviation of one daily residual and of the accumulated error over one year, and the effect of trading four times as often. What would happen to the mean error with σa=0.3\sigma_a = 0.3?
Show the answer

The mean of the leading residual is zero, but its standard deviation is

12 Γσ2S2h=12⋅8252≈0.022448,\begin{aligned} \tfrac{1}{\sqrt{2}}\,\Gamma\sigma^2 S^2 h &= \tfrac{1}{\sqrt{2}} \cdot \frac{8}{252} \\[0.4em] &\approx 0.022448 , \end{aligned}

since Γσ2S2=0.02⋅0.04⋅104=8\Gamma\sigma^2 S^2 = 0.02 \cdot 0.04 \cdot 10^4 = 8. Over 252252 conditionally uncorrelated days the standard deviation grows to about 0.022448⋅252≈0.35630.022448 \cdot \sqrt{252} \approx 0.3563. With four trades a day each residual is four times smaller and there are four times as many, so the total standard deviation is multiplied by 144=12\tfrac14\sqrt{4} = \tfrac12, giving about 0.17820.1782. The colleague is right only about the mean. With σa=0.3\sigma_a = 0.3 the frozen mean error over the year is 12⋅0.02⋅104⋅(0.09−0.04)=5\tfrac12 \cdot 0.02 \cdot 10^4 \cdot (0.09 - 0.04) = 5 at every trading frequency, so more frequent trading would not remove it.

  1. For the at-the-money Black-Scholes call with S=K=100S = K = 100, σm=0.2\sigma_m = 0.2, τ=1\tau = 1 and r=q=0r = q = 0, where Δ≈0.539828\Delta \approx 0.539828 and Γ≈0.019848\Gamma \approx 0.019848, a hedger is long the call and short Δ\Delta shares. Over one day the stock moves by +10+10 or by −10-10. Compare the leading residual with the exact residual, and explain why the two exact values differ.
Show the answer

The leading term is the same for both moves,

12⋅0.019848×(100−1.587302)≈0.976629.\begin{aligned} &\tfrac12 \cdot 0.019848 \\[0.4em] &\quad \times (100 - 1.587302) \\[0.4em] &\quad \approx 0.976629 . \end{aligned}

Repricing gives C(110,τ=1−1/252)−C(100,1)−10Δ≈0.913406C(110, \tau = 1 - 1/252) - C(100, 1) - 10\Delta \approx 0.913406 for the rise and 1.0088021.008802 for the fall. Both exact residuals are positive, because for a move this large the convexity gain far exceeds the one-day time decay Θh≈−3.969525/252≈−0.015752\Theta h \approx -3.969525/252 \approx -0.015752. They differ because the expansion stops at second order. The third derivative here is VSSS=−Γ(1+d1/(στ))/S≈−0.000298V_{SSS} = -\Gamma\bigl(1 + d_1/(\sigma\sqrt{\tau})\bigr)/S \approx -0.000298, since gamma falls as the stock rises above this level. Its term 16VSSS(ΔS)3\tfrac16 V_{SSS}(\Delta S)^3 is about −0.0496-0.0496 for the rise and +0.0496+0.0496 for the fall, which brings the estimates to about 0.9270.927 and 1.0261.026, and higher-order terms account for the rest. A move of 1010 in one day is almost eight daily standard deviations at the model volatility 0.20.2, so the leading term shows the sign but should not be read as the exact hedge result.

How this lesson connects