Merton Structural Model¶

Background¶

merton_structural_model.py

This module implements Merton Structural Model.

Author: Financial Math Library

Code¶

```python

-- coding: utf-8 --¶

""" merton_structural_model.py

This module implements Merton Structural Model.

Author: Financial Math Library """

import numpy as np import matplotlib.pyplot as plt

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

def merton_structural_model(): """ Merton Structural Model.

This function demonstrates the key concepts and computational techniques
for merton structural model.

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

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

# Core computation logic
results = {
    "time_points": time_points,
    "description": "Merton Structural Model"
}

return results

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

# 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("Merton Structural Model")
ax.grid(True, alpha=0.3)

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

return results

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

Exercises¶

Exercise 1. In the Merton structural model, the firm's equity is a call option on the firm's assets with strike equal to the debt face value. If $V_0 = \$100M$, $D = \$80M$, $\sigma_V = 25\%$, $r = 3\%$, and $T = 1$, compute $d_1$ and $d_2$.

Solution to Exercise 1

\[ d_1 = \frac{\ln(V_0/D) + (r + \sigma_V^2/2)T}{\sigma_V\sqrt{T}} = \frac{\ln(100/80) + (0.03 + 0.03125)}{0.25}. \]

\[ d_1 = \frac{0.2231 + 0.06125}{0.25} = \frac{0.2844}{0.25} = 1.1374. \]

\[ d_2 = d_1 - \sigma_V\sqrt{T} = 1.1374 - 0.25 = 0.8874. \]

Exercise 2. Using the values from Exercise 1, compute the probability of default (under the risk-neutral measure) and the credit spread.

Solution to Exercise 2

The risk-neutral default probability is $\mathbb{P}^*(\text{default}) = \mathcal{N}(-d_2) = \Phi(-0.8874) \approx 0.1874 = 18.74\%$.

The debt value is $D_0 = D\,e^{-rT}\,\mathcal{N}(d_2) + V_0\,\mathcal{N}(-d_1) = 80\,e^{-0.03}\Phi(0.8874) + 100\,\Phi(-1.1374)$.

$D_0 = 77.63 \times 0.8126 + 100 \times 0.1277 = 63.09 + 12.77 = \$75.86M$.

The credit spread is $s = -\ln(D_0/(D\,e^{-rT}))/T = -\ln(75.86/77.63) = -\ln(0.9772) = 0.0230 = 2.30\%$.

Exercise 3. Explain the main limitation of the Merton model regarding short-term default prediction.

Solution to Exercise 3

In the Merton model, default can only occur at debt maturity $T$ (when $V_T < D$). The model produces near-zero credit spreads for short maturities because asset values follow a continuous diffusion -- the probability of $V$ dropping below $D$ in a short time is negligible (the firm is currently solvent with $V_0 > D$). This contradicts market observation, where even short-dated CDS spreads for investment-grade firms are nonzero. Extensions like the first-passage model (Black-Cox), which allows default at any time when $V$ hits a barrier, address this by producing positive short-term spreads.

Exercise 4. In the Merton model, how does increasing asset volatility $\sigma_V$ affect equity value and credit spread simultaneously?

Solution to Exercise 4

Since equity is a call option on assets, increasing $\sigma_V$ increases equity value (options benefit from volatility). Simultaneously, the probability of default $\mathcal{N}(-d_2)$ increases because higher volatility makes it more likely that $V_T < D$. This increases the credit spread.

The wealth transfer effect: higher volatility transfers value from debtholders to equityholders. Debtholders face more downside risk, while equityholders benefit from the upside. This is why equity markets may react positively to risk-increasing strategies while credit markets react negatively -- a phenomenon captured by the Merton model.