Model-Free Bounds Computation¶

Background¶

model_free_bounds.py

This module implements Model-Free Bounds Computation.

Author: Financial Math Library

Code¶

```python

-- coding: utf-8 --¶

""" model_free_bounds.py

This module implements Model-Free Bounds Computation.

Author: Financial Math Library """

import numpy as np import matplotlib.pyplot as plt

======================================================================¶

def model_free_bounds(): """ Model-Free Bounds Computation.

This function demonstrates the key concepts and computational techniques
for model-free bounds computation.

Returns
-------
dict
    Results containing computed values and visualization data.
"""
# Implementation of Model-Free Bounds Computation
print(f"Computing Model-Free Bounds Computation...")

# Create sample data/parameters
n_simulations = 1000
time_points = np.linspace(0, 1, 100)

# Core computation logic
results = {
    "time_points": time_points,
    "description": "Model-Free Bounds Computation"
}

return results

def main(): """Main execution function.""" results = model_free_bounds()

# Create visualization
fig, ax = plt.subplots(figsize=(10, 6))
ax.plot(results["time_points"], "b-", linewidth=2)
ax.set_xlabel("Time")
ax.set_ylabel("Value")
ax.set_title("Model-Free Bounds Computation")
ax.grid(True, alpha=0.3)

plt.tight_layout()
plt.savefig("/tmp/model_free_bounds.png", dpi=150)
print(f"Figure saved to /tmp/model_free_bounds.png")
plt.close()

return results

if name == "main": main() ```

Exercises¶

Exercise 1. The model-free upper bound for a European call option on a non-dividend-paying stock is \(C \leq S_0\). Derive this bound.

Solution to Exercise 1

The call payoff satisfies \(\max(S_T - K, 0) \leq S_T\) for all \(S_T \geq 0\). Taking risk-neutral expectations:

\[ C = e^{-rT}\mathbb{E}^{\mathbb{Q}}[\max(S_T - K, 0)] \leq e^{-rT}\mathbb{E}^{\mathbb{Q}}[S_T] = e^{-rT} \cdot S_0 e^{rT} = S_0. \]

The call price cannot exceed the current stock price, since the call's payoff is always at most \(S_T\) and the stock costs \(S_0\).

Exercise 2. Given call prices at strikes \(K_1 < K_2 < K_3\), derive the butterfly spread inequality as a model-free constraint.

Solution to Exercise 2

A butterfly spread is: long 1 call at \(K_1\), short 2 calls at \(K_2\), long 1 call at \(K_3\), with \(K_2 = (K_1 + K_3)/2\). The payoff \((S - K_1)^+ - 2(S - K_2)^+ + (S - K_3)^+ \geq 0\) for all \(S\). Therefore:

\[ C(K_1) - 2C(K_2) + C(K_3) \geq 0. \]

This convexity constraint must hold model-free. Violation would create an arbitrage opportunity via the butterfly spread.

Exercise 3. Explain how model-free bounds are computed from a set of observed call prices across multiple strikes.

Solution to Exercise 3

Given call prices \(C(K_1), \ldots, C(K_n)\), the model-free bounds for an exotic payoff \(f(S_T)\) are obtained by solving:

\[ \text{Upper:} \quad \sup_{\mu} \int f(s)\,d\mu(s) \quad \text{s.t.} \quad \int (s - K_i)^+ d\mu(s) = C(K_i) \; \forall i. \]

This is a linear programming problem: the decision variable is the distribution \(\mu\), and the constraints ensure consistency with market call prices. The solution identifies the worst-case (or best-case) distribution for the exotic payoff within the set of all distributions matching the observed vanillas.

Exercise 4. If the model-free bounds for a variance swap strike are \([18\%, 22\%]\) while a particular model gives \(20\%\), what does this tell us about model risk?

Solution to Exercise 4

The model-free bounds \([18\%, 22\%]\) represent the range of prices consistent with all available market information (vanilla options). The model price of \(20\%\) lies within this range, so it is arbitrage-free. However, the \(4\%\) width of the bounds indicates significant model uncertainty: different arbitrage-free models can disagree by up to \(4\%\). The model risk (potential pricing error due to model choice) is at most \(2\%\) (half the width). If the bounds were tighter (e.g., \([19.5\%, 20.5\%]\)), model risk would be small and the model choice matters less.