Hazard Rate Curve Construction¶

Background¶

hazard_rate_curve.py

This module implements Hazard Rate Curve Construction.

Author: Financial Math Library

Code¶

```python

-- coding: utf-8 --¶

""" hazard_rate_curve.py

This module implements Hazard Rate Curve Construction.

Author: Financial Math Library """

import numpy as np import matplotlib.pyplot as plt

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

def hazard_rate_curve(): """ Hazard Rate Curve Construction.

This function demonstrates the key concepts and computational techniques
for hazard rate curve construction.

Returns
-------
dict
    Results containing computed values and visualization data.
"""
# Implementation of Hazard Rate Curve Construction
print(f"Computing Hazard Rate Curve Construction...")

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

# Core computation logic
results = {
    "time_points": time_points,
    "description": "Hazard Rate Curve Construction"
}

return results

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

# 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("Hazard Rate Curve Construction")
ax.grid(True, alpha=0.3)

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

return results

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

Exercises¶

Exercise 1. The survival probability under a constant hazard rate \(h\) is \(Q(t) = e^{-ht}\). If \(h = 2\%\), compute the survival probability at 1, 5, and 10 years.

Solution to Exercise 1

\(Q(1) = e^{-0.02} \approx 0.9802\) (\(98.02\%\))
\(Q(5) = e^{-0.10} \approx 0.9048\) (\(90.48\%\))
\(Q(10) = e^{-0.20} \approx 0.8187\) (\(81.87\%\))

The cumulative default probabilities are \(1 - Q(t)\): \(1.98\%\), \(9.52\%\), and \(18.13\%\) respectively.

Exercise 2. For a piecewise constant hazard rate: \(h = 1\%\) for \(t \in [0,3)\) and \(h = 3\%\) for \(t \in [3,5]\), compute \(Q(5)\).

Solution to Exercise 2

\[ Q(5) = \exp\!\left(-\int_0^5 h(s)\,ds\right) = \exp(-(0.01 \times 3 + 0.03 \times 2)) = e^{-(0.03 + 0.06)} = e^{-0.09} \approx 0.9139. \]

Exercise 3. Explain the relationship between the hazard rate function \(h(t)\) and the conditional default probability over a small interval \([t, t + dt]\) given survival to \(t\).

Solution to Exercise 3

The hazard rate (or intensity) is defined as

\[ h(t) = \lim_{\Delta t \to 0} \frac{\mathbb{P}(\tau \leq t + \Delta t \mid \tau > t)}{\Delta t}, \]

so \(h(t)\,dt \approx \mathbb{P}(\text{default in } [t, t+dt] \mid \text{survival to } t)\). It is the instantaneous conditional default rate. A constant \(h\) means the default probability per unit time is the same regardless of how long the entity has survived (memoryless property, analogous to exponential distribution). A time-varying \(h(t)\) captures changing credit conditions.

Exercise 4. If CDS market spreads imply an upward-sloping hazard rate curve, what does this suggest about the issuer's creditworthiness over time?

Solution to Exercise 4

An upward-sloping hazard rate curve (\(h(t)\) increasing in \(t\)) implies that the conditional probability of default per unit time increases as we look further into the future. This could reflect:

Aging effects: The issuer is expected to face increasing financial stress (e.g., debt maturity walls, declining business).
Economic cycle expectations: The market anticipates a deteriorating economic environment.
Model calibration artifacts: If the credit spread term structure is convex, bootstrapping can produce increasing hazard rates.

In contrast, a decreasing hazard rate (common for recently distressed issuers) suggests that if the issuer survives the near term, its credit quality improves.