European Option Pricing

The Black Scholes Model.

We consider a frictionless financial market consisting of:

• A risky asset (stock) $S = (S_t)_{t \geq 0}$, which follows a Geometric Brownian Motion $$dS_t = \mu dt + \sigma dW_t$$
• A riskless asset (bond) $B = (B_t)_{t \geq 0}$, growing at a constant risk-free rate $r$ $$dB_t = r B_t dt$$

Here, $\mu$ is the drift (expected return of the asset), $\sigma$ is the volatility and $(W_t)_{t \geq 0}$ is a standard Brownian motion. Under the risk-neutral measure $\mathbb Q$, we have:$$dS_t = rS_t dt + \sigma S_t dW_t^{\mathbb Q}$$The European call option price at time $t=0$ with strike $K$ and maturity T is given by:$$C_0 = e^{-rT} \mathbb E^{\mathbb Q} [\max (S_T - K,0)]$$Using the Black-Scholes closed-form solution, this simplifies to:$$\begin{align*} C_0 &= S_0 \Phi(d_1) - K e^{-rT} \Phi(d_2) \\ P_0 &= Ke^{-rT} \Phi(-d_2) - S_0 \Phi(-d_1) \end{align*}$$where$$d_1 = \frac{\ln(S_0 / K) + (r + \frac{1}{2} \sigma^2) T}{\sigma \sqrt{T}}, \quad d_2 = d_1 - \sigma \sqrt{T}$$and $\Phi(\cdot)$ is the standard normal cumulative distribution function.

The Binomial Model.

In the Cox-Ross-Rubinstein (CRR) binomial model:

• We discretize time into $N$ steps of length $\Delta t = T / N$
• The stock price can move up by a factor of $u = e^{\sigma \sqrt{\Delta t}}$ or down by a factor of $d = e^{-\sigma \sqrt{\Delta t}} = 1/u$

The risk neutral probability is:$$p = \frac{e^{r \Delta t} - d}{u - d}$$At maturity $T$, the stock can end up in one of $N + 1$ possible states. The stock price at node $(i,N)$ is:$$S_{i,N} = S_0 \cdot u^i \cdot d^{N - i} = S_0 \cdot u^{2i - N}$$The option payoff at each terminal node depends on whether it's a call or put:

• Call Option Payoff: $$C_{i,N} = \max(S_{i,N} - K, 0)$$
• Put Option Payoff: $$P_{i,N} = \max(K - S_{i,N}, 0)$$

We calculate the option value at each earlier node by working backwards from maturity. The value at node $(i,j)$ is given by the risk-neutral expectation:$$V_{i,j} = e^{-r \Delta t} \left [ p \cdot V_{i + 1, j + 1} + (1-p) \cdot V_{i, j + 1} \right ]$$This process continues until we reached the root node $(0,0)$, giving us the fair price of the option today.

• For a call, we use terminal payoffs $C_{i,N}$
• For a put, we use terminal payoffs $P_{i,N}$

Monte Carlo Simulation.

In the Monte Carlo framework, we simulate multiple paths of the underlying asset under the risk-neutral measure $\mathbb Q$ and estimate the option price by averaging the discounted payoffs. We assume the risk-neutral dynamic of asset $(S_t)_{t \geq 0}$ follow:$$S_T = S_0 \exp \left (\left (r - \frac{1}{2}\sigma^2 \right )T + \sigma \sqrt{T} Z \right ), \quad Z \sim \mathcal N(0,1)$$

Naive Monte Carlo

We simulate $N$ independent samples $Z_1, \ldots, Z_N \sim \mathcal N(0,1)$, compute the terminal stock price for each:$$S_T^{(i)} = S_0 \exp \left (\left (r - \frac{1}{2}\sigma^2 \right )T + \sigma \sqrt{T} Z_i \right )$$The European option payoff for each sample depends on whether it's a call or a put:

• Call Option Payoff:$$f_i = \max \left ( S_T^{(i)} - K, 0 \right )$$
• Put Option Payoff:$$f_i = \max \left (K - S_T^{(i)}, 0 \right )$$

The Monte Carlo estimator for the option price is then:$$\widehat V_N = e^{-rT} \cdot \frac{1}{N} \sum_{i = 1}^N f_i$$By the Central Limit Theorem, for large $N$, the estimator $\widehat V_N$ is approximately normally distributed$$\widehat V_N \sim \mathcal N \left (V, \frac{\sigma^2_f}{N} \right )$$where $\sigma^2_f$ is the variance of the discounted payoffs. The 95% confidence interval is:$$\widehat V_N \pm 1.96 \cdot \frac{\widehat \sigma_f}{\sqrt{N}}, \quad \text{ where } \quad \widehat \sigma^2_f = \frac{1}{N-1} \sum_{i = 1}^N \left (f_i - \overline f \right )^2$$

Antithetic Variates

We simulate $N$ standard normal variables $Z_1, \ldots, Z_N \sim \mathcal N(0,1)$, and also use their negatives, $-Z_1, \ldots, -Z_N \sim \mathcal N(0,1)$. For each $i$, we compute:$$\begin{align*} S_T^{(+)} &= S_0 \exp \left ( \left (r - \frac{1}{2} \sigma^2 \right ) T + \sigma \sqrt{T}Z_i \right ) \\ S_T^{(-)} &= S_0 \exp \left ( \left (r - \frac{1}{2} \sigma^2 \right ) T - \sigma \sqrt{T}Z_i \right ) \end{align*}$$The European option payoff is given by:

• Call Option Payoff:$$f_i = \frac{1}{2} \left [\max \left ( S_T^{(+)} - K, 0 \right ) + \max \left ( S_T^{(-)} - K, 0 \right ) \right ]$$
• Put Option Payoff:$$f_i = \frac{1}{2} \left [\max \left ( K - S_T^{(+)}, 0 \right ) + \max \left ( K - S_T^{(-)} , 0 \right ) \right ]$$

The estimator for the option price is then:$$\widehat V_N^\text{anti} = e^{-rT} \cdot \frac{1}{N} \sum_{i = 1}^N f_i$$The confidence interval is given by:$$\widehat V_N^\text{anti} \pm 1.96 \cdot \frac{\widehat \sigma_\text{anti}}{\sqrt{N}}, \quad \text{ where } \quad \widehat \sigma_\text{anti}^2 = \frac{1}{N - 1} \sum_{i = 1}^N \left ( f_i - \overline f \right )^2$$The variance decomposition of the estimator is given by$$\text{Var} (\widehat f_i) = \text{Var} \left ( \frac{f(Z_i) + f(-Z_i)}{2} \right ) = \frac{1}{2} \text{Var}(f) + \frac{1}{2} \text{Cov} \left (f(Z_i), f(-Z_i) \right )$$where $f$ is the payoff function of the option. Thus, variance reduction is achieved if$$\text{Cov} \left (f(Z_i), f(-Z_i) \right ) < 0$$

Control Variates

We simulate $N$ standard normal variables $Z_1, \ldots, Z_N \sim \mathcal N(0,1)$, and for each $i$, compute the terminal asset value:$$S_T^{(i)} = S_0 \exp \left (\left (r - \frac{1}{2}\sigma^2 \right )T + \sigma \sqrt{T} Z_i \right )$$We also define a control variate $Y_i = S_T^{(i)}$, whose expected value is:$$\mathbb E[Y_i] = \mathbb E \left [S_T^{(i)} \right ] = S_0 e^{rT}$$For a constant $c \in \mathbb R$, we define the adjusted payoff using:$$f_i^\text{cv} = f_i - c(Y_i - \mathbb E[Y_i])$$where $f_i$ is the discounted option payoff. The European option payoff is given by:

• Call Option Payoff:$$f_i = \max \left ( S_T^{(i)} - K, 0 \right ), \quad f_i^\text{cv} = f_i - c \left (S_T^{(i)} - S_0e^{rT} \right )$$
• Put Option Payoff:$$f_i = \max \left ( K - S_T^{(i)}, 0 \right ), \quad f_i^\text{cv} = f_i - c \left (S_T^{(i)} - S_0e^{rT} \right )$$

The estimator of the option price is then:$$\widehat V_N^\text{cv} = e^{-rT} \cdot \frac{1}{N} \sum_{i = 1}^N f_i^\text{cv}$$The confidence interval is given by:$$\widehat V_N^\text{cv} \pm 1.96 \cdot \frac{\widehat \sigma_\text{cv}}{\sqrt{N}}, \quad \text{ where } \quad \widehat \sigma_\text{cv}^2 = \frac{1}{N - 1} \sum_{i = 1}^N \left ( f_i^\text{cv} - \overline f^\text{cv} \right )^2$$The variance decomposition of the estimator is given by:$$\text{Var} (f_i^\text{cv}) = \text{Var} \left (f_i - c(Y_i - \mathbb E[Y]) \right ) = \text{Var}(f) + c^2 \text{Var}(Y) - 2c \text{Cov} (f,Y)$$This is minimised by choosing:$$c^* = \frac{\text{Cov}(f,Y)}{\text{Var}(Y)}$$In this case, the variance becomes$$\text{Var} \left (f_i^\text{cv} \right ) = \text{Var}(f) \cdot (1-\rho^2), \quad \rho = \text{Corr}(f,Y)$$Thus, variance reduction is achieved when $\rho^2 > 0$, i.e., when $f$ and $Y$ are positively correlated. The higher the correlation, the higher the variance reduction.

Stratified Sampling

We divide the interval $[0,1]$ into $N$ equal-width strata and draw one uniform sample per stratum:$$U_i \sim \mathcal U \left ( \frac{i - 1}{N}, \frac{i}{N} \right ), \quad \text{ for } i = 1,\ldots,N$$We then transform this using $Z_i = \Phi^{-1} (U_i)$ where $\Phi^{-1}$ is the inverse standard normal CDF and simulate$$S_T^{(i)} = S_0 \cdot \exp \left ( \left ( r - \frac{1}{2} \sigma^2 \right )T + \sigma \sqrt{T} Z_i \right )$$The European option payoff is given by:

• Call Option Payoff:$$f_i = \max \left ( S_T^{(i)} - K, 0 \right )$$
• Put Option Payoff:$$f_i = \max \left ( K - S_T^{(i)}, 0 \right )$$

The estimator of the option price is then$$\widehat V_N^\text{strat} = e^{-rT} \cdot \frac{1}{N} \sum_{i = 1}^N f_i$$The confidence interval is given by$$\widehat V_N^\text{strat} \pm 1.96 \cdot \frac{\widehat \sigma_\text{strat}}{\sqrt{N}}, \quad \text{ where } \widehat \sigma^2_\text{strat} = \frac{1}{N - 1} \sum_{i = 1}^N \left ( f_i - \overline f \right )^2$$Variance reduction is achieved because, due to more uniform coverage of the sample space. Stratification reduces the chance of oversampling or undersampling extreme regions.

Importance Sampling

We simulate $N$ standard uniform random variables $Y_1, \ldots, Y_N \sim \mathcal U(y_K, 1)$, where the truncation point $y_K \in (0,1)$ is computed such that:$$S_T (y_K) = K \quad \implies \quad y_K = \Phi \left ( \frac{\log(K/S_0) - \left (r - \frac{1}{2} \sigma^2 \right )T}{\sigma \sqrt{T}} \right )$$For each $i$, we compute:$$Z_i = \Phi^{-1} (Y_i), \quad S_T^{(i)} = S_0 \cdot \exp \left (\left (r - \frac{1}{2} \sigma^2 \right )T + \sigma \sqrt{T} Z_i \right )$$We apply the transformation:$$g_i = (1 - y_K) \cdot f \left (S_T^{(i)} \right ), \quad \text{ with } f(S) = \max(S-K,0) \text{ or } \max(K - S,0)$$The European option payoff is given by:

• Call Option Payoff:$$f_i = \max \left ( S_T^{(i)} - K, 0 \right ), \quad g_i = (1 - y_K) \cdot f_i$$
• Put Option Payoff:$$f_i = \max \left ( K - S_T^{(i)}, 0 \right ), \quad g_i = (1 - y_K) \cdot f_i$$

The estimator for the option price is then:$$\widehat V_N^\text{IS} = e^{-rT} \cdot \frac{1}{N} \sum_{i = 1}^N g_i$$The confidence interval is given by:$$\widehat V_N^\text{IS} \pm 1.96 \cdot \frac{\widehat \sigma_\text{IS}}{\sqrt{N}}, \quad \text{ where } \widehat \sigma_\text{IS}^2 = \frac{1}{N - 1} \sum_{i = 1}^N \left (g_i - \overline g \right )^2$$The variance reduction is achieved because we avoid simulating paths where $S_T < K$, which yield zero payoff for a call option, or where $S_T > K$ for a put. By concentrating simulations on the effective region of integration $Y_i \sim \mathcal U[y_K,1]$, we eliminate variance from regions that contribute nothing to the expectation.