Why Discrete Delta Hedging Leaves Residual Option Risk
Article Quant Q&A · Author: lkjldfkjhljk
Summary
The document examines why a simulated delta-hedged European call has a nonzero dispersion in its terminal financing cost. The author simulates stock paths with fixed interest and volatility, repeatedly adjusts the stock position using Black–Scholes delta, accrues interest on borrowing, and compares the resulting fee with the discounted option payoff. The reported fee dispersion is much smaller than the payoff dispersion, but it is not zero.
The accepted explanation is that delta hedging offsets first-order price exposure, while higher-order effects remain; with discrete rebalancing, the hedge is not exact along each path. Another response points out that terminal debt varies because the underlying path and the resulting hedge trades vary. The discussion is conceptual and does not fully audit the simulation’s timing conventions or assumptions. Its conclusions should be read within the stated fixed-parameter model, rather than as a general estimate of hedging error under real market conditions.
Key ideas
- Delta hedging targets first-order sensitivity to changes in the underlying price.
- A discrete hedge can leave residual risk from higher-order option exposure and price movement between rebalances.
- Different simulated price paths lead to different hedge trades and terminal financing costs.
- The example reports much lower dispersion in hedged costs than in option payoffs, but not zero dispersion.
- The discussion does not assess the simulation’s timing conventions or model realism.
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Full text
# stdev of delta hedged portfolio for call option !=0. Why?
# stdev of delta hedged portfolio for call option !=0. Why?
I wrote a Monte-Carlo simulation of delta hedging for a european call. R and Sigma are fixed.
I start simulation with zero money and short call option. At each step I borrow money to buy 'delta' of assets. At the end of simulation I look at my debt after option settlement. This is fee I should have been payed for the option.
My question: I've noticed that stddev( my cash after all settlements ) != 0. But, it is still << than stdev of payoff. That means that delta hedging did not hedge me completely. Is it ok? Why?
The following code gives me
required_initial_fee mean = 72.17 stdev = 0.56
discounted_option_payoff mean = 74.34 stdev = 426.68
```
from numpy import log, sqrt, exp, maximum
from math import erf
import pandas as pd
import numpy as np
@np.vectorize
def N_cdf(x):
'Cumulative distribution function for the standard normal distribution'
return (1.0 + erf(x / sqrt(2.0))) / 2.0
def BS_CALL_DELTA(S, K, T, r, sigma):
d1 = ( log(S/K) + (r + 0.5 * sigma**2) * T ) / (sigma * sqrt(T))
return N_cdf(d1)
S0 = 100 # Initial stock price
K = 100 # Strike price
r = 0.05 # Risk-free interest rate
sigma = 0.5 # Volatility
T = 3000 # Time to expiration
N = 100000 # Number of simulations
days_per_year = 250 # Time steps per year
#simulate price path
np.random.seed(2024)
dt = 1.0/days_per_year
# simulate path
simulated_paths = np.zeros((T, N))
simulated_paths[0] = S0
for i in range(0, T-1):
w = np.random.standard_normal(N)
simulated_paths[i+1] = simulated_paths[i] * (1 + r*dt + sigma * np.sqrt(dt)*w)
previous_delta = np.zeros(N)
# Total amount of borrowings is stored in 'debt' variable. Daily continuous interest rate is accrued on this debt
debt = np.zeros(N)
for t_i in range(0, T - 1):
S_i = simulated_paths[t_i]
year_fraction_left_to_maturity = float(T - t_i) / days_per_year
delta = BS_CALL_DELTA(S_i, K, year_fraction_left_to_maturity, r, sigma)
# As Im doing delta hedging, by the end of the period I must own 'delta' amount of shares on the previous step I had 'previous_delta' amount. So I need to buy 'delta-previous_delta' of shares
settled_cash_out = (delta - previous_delta) * S_i
previous_delta = delta
# I lend money to buy stocks. This is my total due
debt = debt * exp(r / days_per_year) + settled_cash_out
# now I need to handle the last day:
S_final = simulated_paths[-1]
option_payoff = maximum.reduce([S_final - K, np.zeros(N)])
discounted_option_payoff = option_payoff * exp( -r * T / days_per_year)
# at the end of the last step I sell all my stocks (I have previous_delta of stocks) and pay payoff
settled_cash_out = -S_final * previous_delta + option_payoff
debt = debt * exp(r / days_per_year) + settled_cash_out
#how much money should I take as a commission to have 0 pnl by the end of the contract, to compensate my spendings on debt
required_initial_fee = debt * exp( -r * T / days_per_year)
print("required_initial_fee", round(required_initial_fee.mean(), 2), round(required_initial_fee.std(), 2))
print("discounted_option_payoff", round(discounted_option_payoff.mean(), 2), round(discounted_option_payoff.std(), 2))
```
## Answer by lkjldfkjhljk (score 0, accepted)
https://quant.stackexchange.com/a/79350
Seems like I got an answer. Delta hedging - is a first order of Tailor series. Neglect of other members creates error. Therefore hedge is not perfect.
## Answer by KaiSqDist (score 0)
https://quant.stackexchange.com/a/79349
Disclaimer - I did not read your code.
According to what I understand, your stdev is the stdev across 100,000 values of total debt at the maturity of those 100,000 simulations (where the underlying stock price evolution is stochastic).
I don't believe there is any reason for the stdev to be zero.
The total debt is a result of borrowing capital to go long on stock to delta hedge (the short European call option) and is continuously readjusted at every single time period to make the total position delta-neutral. You can think of this way:
$$ Underlying \: Spot \rightarrow Delta \rightarrow Hedging \: Stock \: Position \rightarrow Borrowed \: Capital $$
where since the underlying spot is stochastic, the amount of borrowed capital would also follow a random path. Therefore, the total debt at the maturity of the option would definitely be different for each simulation, and the stdev would be non-zero.Shown 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.