Black Scholes Formula¶

Background¶

Analytical BS formulas and option pricing. UNCHANGED API - improved internal implementation using utility functions.

Code¶

```python

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black_scholes/black_scholes_formula.py¶

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""" Analytical BS formulas and option pricing. UNCHANGED API - improved internal implementation using utility functions. """

from .black_scholes_base import BlackScholesBase from .black_scholes_utils import bs_call_price, bs_put_price

class BlackScholesFormula(BlackScholesBase): """ Analytical Black-Scholes formula implementation. API UNCHANGED - delegates to utility functions for consistency. """

def price(self):
    """
    Calculate call and put option prices.
    UNCHANGED API - now delegates to utility functions.
    """
    return (
        bs_call_price(self.S0, self.K, self.T, self.r, self.sigma, self.q),
        bs_put_price(self.S0, self.K, self.T, self.r, self.sigma, self.q)
    )

if name == "main": pass ```

Exercises¶

Exercise 1. Derive the Black-Scholes call price formula starting from the risk-neutral expectation $C = e^{-rT}E^Q[\max(S_T - K, 0)]$.

Solution to Exercise 1

Under the risk-neutral measure, $S_T = S_0 \exp\!\bigl((r - q - \frac{1}{2}\sigma^2)T + \sigma\sqrt{T}\,Z\bigr)$ where $Z \sim \mathcal{N}(0,1)$.

\[ C = e^{-rT}\int_{-\infty}^{\infty} \max(S_T(z) - K, 0)\,\phi(z)\,dz \]

The payoff is positive when $S_T > K$, i.e., when $z > -d_2$ where $d_2 = [\ln(S_0/K) + (r - q - \frac{1}{2}\sigma^2)T]/(\sigma\sqrt{T})$. Splitting and evaluating the integrals:

\[ C = S_0 e^{-qT}\mathcal{N}(d_1) - K e^{-rT}\mathcal{N}(d_2) \]

where $d_1 = d_2 + \sigma\sqrt{T}$ and $\Phi$ is the standard normal CDF.

Exercise 2. Compute the call and put prices for $S_0 = 100$, $K = 105$, $T = 0.5$, $r = 0.03$, $\sigma = 0.25$, $q = 0$.

Solution to Exercise 2

\[ d_1 = \frac{\ln(100/105) + (0.03 + 0.03125) \times 0.5}{0.25\sqrt{0.5}} = \frac{-0.04879 + 0.03063}{0.17678} = \frac{-0.01816}{0.17678} = -0.1027 \]

\[ d_2 = -0.1027 - 0.17678 = -0.2795 \]

From tables: $\Phi(-0.1027) \approx 0.4591$, $\Phi(-0.2795) \approx 0.3899$.

\[ C = 100 \times 0.4591 - 105 \times e^{-0.015} \times 0.3899 \approx 45.91 - 40.32 = \$5.59 \]

By put-call parity: $P = C - S_0 + Ke^{-rT} = 5.59 - 100 + 103.44 = \$9.03$.

Exercise 3. Show that the price() method delegates to utility functions bs_call_price and bs_put_price. What is the software engineering benefit of this delegation pattern?

Solution to Exercise 3

The price() method simply returns (bs_call_price(...), bs_put_price(...)), passing through all parameters. This delegation pattern has several benefits:

Reusability: The utility functions can be called independently without creating a class instance (useful for vectorized computations or quick calculations).
Testability: Pure functions are easier to unit test than class methods.
Consistency: All pricing code (analytical, MC, numerical) can share the same utility functions, ensuring identical $d_1, d_2$ calculations.
Maintainability: Bug fixes in the utility function automatically propagate to all callers.

Exercise 4. Verify put-call parity $C - P = S_0 e^{-qT} - K e^{-rT}$ algebraically using the Black-Scholes formulas.

Solution to Exercise 4

\[ C - P = [S_0 e^{-qT}\mathcal{N}(d_1) - Ke^{-rT}\mathcal{N}(d_2)] - [Ke^{-rT}\mathcal{N}(-d_2) - S_0 e^{-qT}\mathcal{N}(-d_1)] \]

\[ = S_0 e^{-qT}[\mathcal{N}(d_1) + \mathcal{N}(-d_1)] - Ke^{-rT}[\mathcal{N}(d_2) + \mathcal{N}(-d_2)] \]

Since $\Phi(x) + \Phi(-x) = 1$ for all $x$:

\[ C - P = S_0 e^{-qT} - Ke^{-rT} \]

This confirms put-call parity holds for the Black-Scholes formulas with dividends.