Expected Shortfall Estimation¶

Background¶

expected_shortfall_estimation.py

This module implements Expected Shortfall Estimation.

Author: Financial Math Library

Code¶

```python

-- coding: utf-8 --¶

""" expected_shortfall_estimation.py

This module implements Expected Shortfall Estimation.

Author: Financial Math Library """

import numpy as np import matplotlib.pyplot as plt

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

def expected_shortfall_estimation(): """ Expected Shortfall Estimation.

This function demonstrates the key concepts and computational techniques
for expected shortfall estimation.

Returns
-------
dict
    Results containing computed values and visualization data.
"""
# Implementation of Expected Shortfall Estimation
print(f"Computing Expected Shortfall Estimation...")

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

# Core computation logic
results = {
    "time_points": time_points,
    "description": "Expected Shortfall Estimation"
}

return results

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

# 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("Expected Shortfall Estimation")
ax.grid(True, alpha=0.3)

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

return results

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

Exercises¶

Exercise 1. Expected Shortfall (ES) at confidence level $\alpha$ is defined as $\text{ES}_\alpha = \mathbb{E}[L \mid L > \text{VaR}_\alpha]$. If the loss distribution is normal with $\mu = 0$ and $\sigma = \$1M$, compute $\text{ES}_{95\%}$.

Solution to Exercise 1

For a standard normal, $\text{VaR}_{95\%} = \sigma \times \Phi^{-1}(0.95) = 1{,}000{,}000 \times 1.645 = \$1{,}645{,}000$.

The ES for a normal distribution is:

\[ \text{ES}_\alpha = \sigma \times \frac{\phi(\Phi^{-1}(\alpha))}{1 - \alpha} = 1{,}000{,}000 \times \frac{\phi(1.645)}{0.05}. \]

Since $\phi(1.645) = 0.1031$: $\text{ES}_{95\%} = 1{,}000{,}000 \times 0.1031/0.05 = \$2{,}063{,}000$.

Exercise 2. Explain why ES is considered a more coherent risk measure than VaR, specifically regarding the subadditivity property.

Solution to Exercise 2

VaR can violate subadditivity: $\text{VaR}(A + B) > \text{VaR}(A) + \text{VaR}(B)$ is possible, meaning diversification appears to increase risk. This is counterintuitive and makes VaR unsuitable for risk aggregation. ES is always subadditive: $\text{ES}(A + B) \leq \text{ES}(A) + \text{ES}(B)$, meaning diversification always reduces (or maintains) risk as measured by ES. This is because ES averages the tail losses rather than looking at a single quantile, capturing the shape of the tail distribution.

Exercise 3. Describe how to estimate ES from historical return data using the empirical distribution.

Solution to Exercise 3

Given $n$ historical losses $L_1, \ldots, L_n$ sorted in descending order:

Compute $k = \lfloor n(1-\alpha) \rfloor$ (the number of losses exceeding VaR).
The empirical ES is the average of the $k$ largest losses:

\[ \widehat{\text{ES}}_\alpha = \frac{1}{k}\sum_{i=1}^{k} L_{(i)}, \]

where $L_{(1)} \geq L_{(2)} \geq \cdots$. For $n = 1000$ and $\alpha = 95\%$, $k = 50$, so ES is the average of the 50 largest losses.

Exercise 4. If historical daily losses (in $M) in the worst 5 days out of 100 are $\{3.2, 2.8, 2.5, 2.1, 1.9\}$, compute $\text{ES}_{95\%}$ and $\text{VaR}_{95\%}$.

Solution to Exercise 4

$\text{VaR}_{95\%} = \$1.9M$ (the 5th worst loss, the $95$th percentile).

$\text{ES}_{95\%} = \frac{3.2 + 2.8 + 2.5 + 2.1 + 1.9}{5} = \frac{12.5}{5} = \$2.5M$.

ES exceeds VaR by $\$0.6M$, reflecting the severity of losses beyond the VaR threshold.