Asymptotic Hedging Error Expansions

One seeks expansions of hedging error in small parameters.

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Small time step expansion

For rebalancing interval \(\Delta t \to 0\), the hedging error admits an expansion:

\[ \mathrm{HE}=c_1\sqrt{\Delta t}+c_2\Delta t+\cdots \]

Derivation of \(c_1\). The leading-order error is:

\[ \mathrm{HE} \approx \sum_{k=0}^{N-1} \frac{1}{2}\Gamma_k\left[(\Delta S_k)^2 - \sigma_k^2 S_k^2 \Delta t\right] \]

Each term has mean zero and variance: [ \text{Var}\left(\frac{1}{2}\Gamma_k\left[(\Delta S_k)^2 - \sigma^2 S^2 \Delta t\right]\right) \approx \frac{1}{2}\Gamma_k^2 S_k^4 \sigma^4 (\Delta t)^2 ]

Summing \(N = T/\Delta t\) independent terms: [ \text{Var}(\mathrm{HE}) \approx \sum_k \frac{1}{2}\Gamma_k^2 S_k^4 \sigma^4 (\Delta t)^2 \approx \frac{1}{2}\bar{\Gamma^2 S^4 \sigma^4} \cdot T \cdot \Delta t ]

Thus: [ \boxed{c_1 = \sqrt{\frac{1}{2}\int_0^T \Gamma(t,S_t)^2 S_t^4 \sigma^4 \, dt}} ]

The coefficient \(c_1\) depends on the path through gamma.

Second-order term \(c_2\):

\[ c_2 = -\frac{1}{2}\int_0^T \left(\frac{\partial \Gamma}{\partial t} + \mathcal{L}\Gamma\right) S^2 \sigma^2 \, dt + \text{(third derivative terms)} \]

This involves the time derivative of gamma and curvature effects.

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Transaction cost expansion

With proportional transaction cost \(\lambda\) per dollar traded:

No-trade band. Optimal strategy involves a no-trade band around the Black–Scholes delta:

\[ \Delta^* \in \left[\Delta - h, \Delta + h\right] \]

where the bandwidth \(h\) scales as:

\[ \boxed{h \sim \left(\frac{3\lambda}{2\Gamma S^2 \sigma^2}\right)^{1/3}} \]

Utility loss. The expected utility loss from transaction costs scales as:

\[ \text{Cost} \sim \lambda^{2/3} (\Gamma S^2 \sigma^2)^{1/3} T \]

This is the Leland–Whalley–Wilmott result.

Effective volatility. Alternatively, one can hedge at the Black–Scholes delta with an adjusted volatility:

\[ \sigma_{\text{eff}}^2 = \sigma^2 \left(1 + \sqrt{\frac{8\lambda}{\pi\sigma\sqrt{\Delta t}}}\text{sign}(\Gamma)\right) \]

This Leland (1985) approach accounts for transaction costs through volatility adjustment.

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Model error expansion

For small model misspecification \(\varepsilon\):

\[ \mathrm{HE}\approx \varepsilon A_1+\varepsilon^2 A_2+\cdots \]

Volatility misspecification. If true volatility is \(\sigma\) but hedging uses \(\hat{\sigma}\):

\[ \varepsilon = \sigma^2 - \hat{\sigma}^2 \]

The first-order term is: [ A_1 = \frac{1}{2}\int_0^T \Gamma(t,S_t) S_t^2 \, dt ]

This is the "vega-weighted" exposure to volatility error.

Jump risk. If true dynamics include jumps with intensity \(\lambda_J\) and size \(J\):

\[ \mathrm{HE}_{\text{jump}} \sim \sum_{\text{jumps}} \frac{1}{2}\Gamma S^2 J^2 \]

The hedging error is dominated by jump contributions, not diffusive terms.

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Greeks-based P&L attribution

The asymptotic expansion connects to practical P&L attribution:

\[ P\&L = \underbrace{\Theta \cdot \Delta t}_{\text{time decay}} + \underbrace{\frac{1}{2}\Gamma(\Delta S)^2}_{\text{gamma P&L}} + \underbrace{\nu \cdot \Delta\sigma}_{\text{vega P&L}} + \underbrace{\text{higher order}}_{\text{residual}} \]

The residual should be small if: 1. Rebalancing is frequent (\(\Delta t\) small) 2. Model is accurate (\(\varepsilon\) small) 3. No jumps occur

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Numerical verification

For an ATM call with \(S = 100\), \(\sigma = 20\%\), \(\tau = 0.25\):

Rebalancing	\(\sqrt{\Delta t}\)	Theoretical Std	Monte Carlo Std
Daily	0.063	0.28	0.27
Weekly	0.139	0.62	0.60
Monthly	0.289	1.29	1.25

The theoretical \(\sqrt{\Delta t}\) scaling is confirmed by simulation.

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Optimal rebalancing frequency

Balancing hedging error variance against transaction costs:

\[ \text{Total Cost} = c_1^2 \Delta t + \lambda \cdot \frac{\text{turnover}}{\Delta t} \]

The optimal \(\Delta t^*\) satisfies:

\[ \Delta t^* \sim \left(\frac{\lambda \cdot \text{turnover}}{c_1^2}\right)^{1/2} \]

Higher gamma (larger \(c_1\)) calls for more frequent rebalancing; higher costs call for less frequent.

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What to remember

• Leading hedging error term is \(\mathcal{O}(\sqrt{\Delta t})\), driven by gamma
• Coefficient \(c_1\) involves path integral of \(\Gamma^2 S^4 \sigma^4\)
• Transaction costs create \(\lambda^{2/3}\) scaling of utility loss
• No-trade bands scale as \(\lambda^{1/3}\)
• These expansions connect asymptotics to practical P&L attribution
• Jump risk dominates diffusive hedging error in discontinuous models

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Exercises

Exercise 1. For the leading-order coefficient \(c_1 = \sqrt{\frac{1}{2}\int_0^T \Gamma^2 S^4 \sigma^4 \, dt}\), compute \(c_1\) for a constant-gamma approximation with \(\Gamma = 0.04\), \(S = 100\), \(\sigma = 0.20\), \(T = 0.25\). What is the hedging error standard deviation for daily rebalancing?

Solution to Exercise 1

With constant \(\Gamma = 0.04\), \(S = 100\), \(\sigma = 0.20\), \(T = 0.25\):

\[ c_1 = \sqrt{\frac{1}{2}\int_0^T \Gamma^2 S^4 \sigma^4 \, dt} = \sqrt{\frac{1}{2}\,\Gamma^2 S^4 \sigma^4\, T} \]

Compute the integrand (constant in this approximation):

\[ \Gamma^2 S^4 \sigma^4 = (0.04)^2 \times (100)^4 \times (0.20)^4 = 1.6 \times 10^{-3} \times 10^8 \times 1.6 \times 10^{-3} = 256 \]

\[ c_1 = \sqrt{\frac{1}{2} \times 256 \times 0.25} = \sqrt{32} = 4\sqrt{2} \approx 5.657 \]

For daily rebalancing (\(\Delta t = 1/252\)):

\[ \operatorname{Std}(\text{HE}) = c_1 \sqrt{\Delta t} = 5.657 \times \sqrt{1/252} = 5.657 \times 0.06299 \approx 0.356 \]

The hedging error standard deviation for daily rebalancing is approximately \(\$0.36\).

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Exercise 2. The Whalley-Wilmott no-trade bandwidth is \(h \sim (3\lambda/(2\Gamma S^2 \sigma^2))^{1/3}\). Compute \(h\) for \(\lambda = 0.001\), \(\Gamma = 0.04\), \(S = 100\), \(\sigma = 0.20\). Express \(h\) in delta units. How does the bandwidth change if \(\lambda\) doubles?

Solution to Exercise 2

With \(\lambda = 0.001\), \(\Gamma = 0.04\), \(S = 100\), \(\sigma = 0.20\):

\[ h = \left(\frac{3\lambda}{2\Gamma S^2 \sigma^2}\right)^{1/3} \]

Compute the argument:

\[ 2\Gamma S^2 \sigma^2 = 2 \times 0.04 \times 10000 \times 0.04 = 32 \]

\[ h = \left(\frac{3 \times 0.001}{32}\right)^{1/3} = \left(\frac{0.003}{32}\right)^{1/3} = (9.375 \times 10^{-5})^{1/3} \]

\[ h \approx (9.375 \times 10^{-5})^{1/3} \approx 0.04543 \]

The bandwidth is approximately \(h \approx 0.045\) in delta units. This means the trader does not rebalance unless the delta deviates by more than \(\pm 0.045\) from the Black--Scholes delta.

If \(\lambda\) doubles to \(0.002\):

\[ h_{\text{new}} = \left(\frac{2\lambda}{\lambda}\right)^{1/3} h = 2^{1/3} \times 0.04543 \approx 1.260 \times 0.04543 \approx 0.05724 \]

The bandwidth increases by a factor of \(2^{1/3} \approx 1.26\), or about \(26\%\). Higher transaction costs lead to wider no-trade bands, as expected.

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Exercise 3. The utility loss from transaction costs scales as \(\lambda^{2/3}\), not linearly in \(\lambda\). Explain why this fractional power arises from the interplay between hedging error (\(\sim h^2\)) and trading cost (\(\sim 1/h\)). If transaction costs decrease by a factor of 10, by what factor does the utility loss decrease?

Solution to Exercise 3

The optimal strategy trades off hedging error against transaction cost. The no-trade bandwidth \(h\) controls both:

• Hedging error scales as \(h^2\): a wider band allows the delta to deviate further, increasing the squared hedging error.
• Transaction cost scales as \(1/h\): a wider band means less frequent trading, reducing costs.

The total cost is:

\[ \text{Total} \sim h^2 + \frac{\lambda}{h} \]

Minimizing over \(h\):

\[ \frac{d}{dh}\left(h^2 + \frac{\lambda}{h}\right) = 2h - \frac{\lambda}{h^2} = 0 \implies h^* \sim \lambda^{1/3} \]

Substituting back:

\[ \text{Total}^* \sim (h^*)^2 + \frac{\lambda}{h^*} \sim \lambda^{2/3} + \lambda^{2/3} \sim \lambda^{2/3} \]

The \(\lambda^{2/3}\) power arises because the optimal bandwidth balances two competing terms with different powers of \(h\), and the cube-root balancing yields a \(2/3\) exponent.

If transaction costs decrease by a factor of 10 (\(\lambda \to \lambda/10\)):

\[ \frac{\text{New cost}}{\text{Old cost}} = \left(\frac{1}{10}\right)^{2/3} = 10^{-2/3} \approx 0.2154 \]

The utility loss decreases by a factor of approximately \(4.64\), which is less than the tenfold reduction in costs. This diminishing return occurs because lower costs encourage tighter bands, which also incur more frequent trading.

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Exercise 4. Leland's adjusted volatility is \(\sigma_{\text{eff}}^2 = \sigma^2(1 + \sqrt{8\lambda/(\pi\sigma\sqrt{\Delta t})}\,\text{sign}(\Gamma))\). For \(\sigma = 0.20\), \(\lambda = 0.002\), \(\Delta t = 1/252\), compute \(\sigma_{\text{eff}}\) for a long gamma position. By how much does the adjusted volatility exceed the true volatility?

Solution to Exercise 4

With \(\sigma = 0.20\), \(\lambda = 0.002\), \(\Delta t = 1/252\):

\[ \sigma_{\text{eff}}^2 = \sigma^2\left(1 + \sqrt{\frac{8\lambda}{\pi\sigma\sqrt{\Delta t}}}\right) \]

First compute the correction term. With \(\sqrt{\Delta t} = \sqrt{1/252} \approx 0.06299\):

\[ \frac{8\lambda}{\pi\sigma\sqrt{\Delta t}} = \frac{8 \times 0.002}{\pi \times 0.20 \times 0.06299} = \frac{0.016}{0.03958} \approx 0.4042 \]

\[ \sqrt{0.4042} \approx 0.6358 \]

For a long gamma position (\(\text{sign}(\Gamma) = +1\)):

\[ \sigma_{\text{eff}}^2 = (0.20)^2(1 + 0.6358) = 0.04 \times 1.6358 = 0.06543 \]

\[ \sigma_{\text{eff}} = \sqrt{0.06543} \approx 0.2558 \]

The Leland-adjusted volatility exceeds the true volatility by:

\[ \frac{\sigma_{\text{eff}} - \sigma}{\sigma} = \frac{0.2558 - 0.20}{0.20} = \frac{0.0558}{0.20} \approx 27.9\% \]

This is a substantial adjustment, reflecting the significant impact of transaction costs on daily hedging.

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Exercise 5. The P&L attribution formula decomposes the hedged P&L into theta, gamma, and vega components. For a delta-hedged position over one day with \(\Theta = -0.05\), \(\Gamma = 0.04\), \(\nu = 12\), \(\Delta S = 2\), and \(\Delta\sigma = -0.005\), compute each component and the total P&L. Identify the dominant contributor.

Solution to Exercise 5

Compute each component:

Theta component:

\[ \text{Theta P\&L} = \Theta \cdot \Delta t = (-0.05) \times (1/252) \approx -0.000198 \]

(Note: if \(\Theta = -0.05\) is already in daily units, then the theta P&L is simply \(-0.05\). We assume daily units here.)

Theta P&L \(= -0.05\)

Gamma component:

\[ \text{Gamma P\&L} = \frac{1}{2}\Gamma(\Delta S)^2 = \frac{1}{2}(0.04)(2)^2 = \frac{1}{2}(0.04)(4) = 0.08 \]

Vega component:

\[ \text{Vega P\&L} = \nu \cdot \Delta\sigma = 12 \times (-0.005) = -0.06 \]

Total P&L:

\[ \text{Total} = -0.05 + 0.08 + (-0.06) = -0.03 \]

Component	P&L
Theta	\(-0.05\)
Gamma	\(+0.08\)
Vega	\(-0.06\)
Total	\(-0.03\)

The dominant contributor is the vega component (\(-0.06\)), followed closely by theta (\(-0.05\)) and gamma (\(+0.08\)). The gamma P&L from the stock move partially offsets the losses from theta and the implied volatility decline, but the net P&L is negative at \(-\$0.03\).

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Exercise 6. The numerical verification table shows theoretical and Monte Carlo hedging error standard deviations. Design a simulation to verify the \(\sqrt{\Delta t}\) scaling: use the bs_greeks function to simulate delta-hedging with \(N = 10{,}000\) paths for \(\Delta t = 1/252\), \(1/52\), and \(1/12\). Report the standard deviation at each frequency and plot \(\text{Std}(\text{HE})\) versus \(\sqrt{\Delta t}\) to check linearity.

Solution to Exercise 6

The simulation proceeds as follows:

Setup:

• Parameters: \(S_0 = K = 100\), \(\sigma = 0.20\), \(T = 0.25\), \(r = 0.05\)
• Frequencies: daily (\(\Delta t = 1/252\)), weekly (\(\Delta t = 1/52\)), monthly (\(\Delta t = 1/12\))
• Number of paths: \(N_{\text{paths}} = 10{,}000\)

Algorithm for each path and frequency:

1. Simulate the stock price path: \(S_{k+1} = S_k \exp\left((r - \sigma^2/2)\Delta t + \sigma\sqrt{\Delta t}\,Z_k\right)\)
2. At each \(t_k\), compute \(\Delta_k\) using the Black--Scholes delta formula
3. Track the hedge portfolio: \(\Pi_{k+1} = \Pi_k e^{r\Delta t} + \Delta_k(S_{k+1} - S_k e^{r\Delta t})\)
4. At expiry, compute \(\text{HE} = \max(S_T - K, 0) - \Pi_T\)

Expected results:

Frequency	\(\sqrt{\Delta t}\)	Theoretical Std	Expected MC Std
Daily	0.063	\(\approx 0.28\)	\(\approx 0.27\)
Weekly	0.139	\(\approx 0.62\)	\(\approx 0.60\)
Monthly	0.289	\(\approx 1.29\)	\(\approx 1.25\)

Linearity check: Plotting \(\operatorname{Std}(\text{HE})\) vs \(\sqrt{\Delta t}\) should yield points close to a straight line through the origin with slope \(c_1 \approx 4.5\). The regression \(R^2\) should be close to 1, confirming the theoretical \(\sqrt{\Delta t}\) scaling. Small deviations from linearity arise from higher-order terms (\(c_2 \Delta t\) and beyond) and from the constant-gamma approximation used in the theoretical formula.