Mathematical Finance Models

Example: Pricing a European Call Option

Given the following parameters:

• Current stock price \( S_0 = \$100 \).
• Strike price \( K = \$105 \).
• Time to expiration \( T = 1 \) year.
• Risk-free rate \( r = 5\% \).
• Volatility \( \sigma = 20\% \).

Calculate the price of a European call option.

Step 1: Compute \( d_1 \) and \( d_2 \).

Step 2: Compute \( N(d_1) \) and \( N(d_2) \).

Using standard normal distribution tables or a calculator:

Step 3: Plug into the Black-Scholes formula.

The price of the European call option is approximately \$8.02.

Example: Pricing a European Put Option

Using the same parameters as above, calculate the price of a European put option.

Step 1: Compute \( N(-d_1) \) and \( N(-d_2) \).

Step 2: Plug into the Black-Scholes formula.

The price of the European put option is approximately \$8.00.

Note: The call and put prices satisfy put-call parity: \( C - P = S_0 - K e^{-rT} \).

Example: Calculating the Greeks

Using the same parameters as the previous examples (\( S_0 = 100 \), \( K = 105 \), \( T = 1 \), \( r = 0.05 \), \( \sigma = 0.2 \)), calculate the delta and gamma of the call option.

Step 1: Recall \( d_1 \) and \( N(d_1) \).

The probability density function \( N'(d_1) \) is:

Step 2: Calculate Delta.

This means the call option price increases by approximately \$0.5422 for every \$1 increase in the stock price.

Step 3: Calculate Gamma.

This means the delta of the call option increases by approximately 0.0198 for every \$1 increase in the stock price.

Example: Verifying the Black-Scholes Formula for a European Call Option

The Black-Scholes formula for the price of a European call option is:

where:

and \( N(\cdot) \) is the cumulative distribution function of the standard normal distribution.

Verification: We verify that this formula satisfies the Black-Scholes PDE. Let \( V(t, S) = C(S, t) \). Compute the partial derivatives:

1. First Partial Derivative with Respect to \( S \):

   \[ \frac{\partial C}{\partial S} = N(d_1) + S \frac{\partial N(d_1)}{\partial S} - K e^{-r(T-t)} \frac{\partial N(d_2)}{\partial S} \]

   Using the chain rule and the fact that \( \frac{\partial d_1}{\partial S} = \frac{\partial d_2}{\partial S} = \frac{1}{S \sigma \sqrt{T - t}} \), we get:

   \[ \frac{\partial C}{\partial S} = N(d_1) \]

2. Second Partial Derivative with Respect to \( S \):

   \[ \frac{\partial^2 C}{\partial S^2} = \frac{\partial N(d_1)}{\partial S} = \frac{1}{S \sigma \sqrt{T - t}} N'(d_1) \]

   where \( N'(d_1) = \frac{1}{\sqrt{2 \pi}} e^{-d_1^2 / 2} \).

3. Partial Derivative with Respect to \( t \):

   \[ \frac{\partial C}{\partial t} = S \frac{\partial N(d_1)}{\partial t} - K e^{-r(T-t)} \left( -r N(d_2) + \frac{\partial N(d_2)}{\partial t} \right) \]

   Using the chain rule and simplifying, we find:

   \[ \frac{\partial C}{\partial t} = - \frac{S \sigma}{2 \sqrt{T - t}} N'(d_1) - r K e^{-r(T-t)} N(d_2) \]

4. Substitute into the Black-Scholes PDE:

   Substitute \( \frac{\partial C}{\partial t} \), \( \frac{\partial C}{\partial S} \), and \( \frac{\partial^2 C}{\partial S^2} \) into the Black-Scholes PDE:

   \[ \frac{\partial C}{\partial t} + \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 C}{\partial S^2} + r S \frac{\partial C}{\partial S} - r C = 0 \]

   After substitution and simplification, the equation holds true, verifying that the Black-Scholes formula satisfies the PDE.

1. Construct a replicating portfolio: Assume the derivative can be replicated by a self-financing portfolio of the underlying asset \( S_t \) and a risk-free bond \( B_t = e^{rt} \). Let \( \Delta_t \) be the number of units of \( S_t \) held at time \( t \). The value of the portfolio is: \[ V_t = \Delta_t S_t + \left( V_t - \Delta_t S_t \right) e^{r(T-t)}. \]
2. Self-financing condition: The change in the portfolio value is due to changes in \( S_t \) and \( B_t \): \[ dV_t = \Delta_t dS_t + r \left( V_t - \Delta_t S_t \right) dt. \]
3. Discounted portfolio is a martingale: Under the risk-neutral measure \( \mathbb{Q} \), the discounted asset price \( \tilde{S}_t = e^{-rt} S_t \) is a martingale. The discounted portfolio value \( \tilde{V}_t = e^{-rt} V_t \) must also be a martingale (no arbitrage): \[ \tilde{V}_t = \mathbb{E}^\mathbb{Q} \left[ \tilde{V}_T \mid \mathcal{F}_t \right]. \]
4. Substitute the payoff: At maturity \( T \), \( V_T = \Phi(S_T) \), so: \[ \tilde{V}_T = e^{-rT} \Phi(S_T). \] Thus: \[ \tilde{V}_t = \mathbb{E}^\mathbb{Q} \left[ e^{-rT} \Phi(S_T) \mid \mathcal{F}_t \right]. \] Multiplying both sides by \( e^{rt} \) gives the risk-neutral valuation formula: \[ V_t = \mathbb{E}^\mathbb{Q} \left[ e^{-r(T-t)} \Phi(S_T) \mid \mathcal{F}_t \right]. \]

Consider the Black-Scholes model where the stock price \( S_t \) follows geometric Brownian motion under the real-world measure \( \mathbb{P} \):

1. Apply Girsanov's Theorem: Define a new Brownian motion \( \tilde{W}_t \) under \( \mathbb{Q} \) as: \[ \tilde{W}_t = W_t + \frac{\mu - r}{\sigma} t. \] The Radon-Nikodym derivative is: \[ \frac{d\mathbb{Q}}{d\mathbb{P}} \bigg|_{\mathcal{F}_t} = \exp \left( -\frac{\mu - r}{\sigma} W_t - \frac{1}{2} \left( \frac{\mu - r}{\sigma} \right)^2 t \right). \]
2. Stock price dynamics under \( \mathbb{Q} \): Substitute \( W_t = \tilde{W}_t - \frac{\mu - r}{\sigma} t \) into the SDE for \( S_t \): \[ dS_t = \mu S_t dt + \sigma S_t \left( d\tilde{W}_t - \frac{\mu - r}{\sigma} dt \right) = r S_t dt + \sigma S_t d\tilde{W}_t. \]
3. Discounted stock price is a martingale: The discounted stock price \( \tilde{S}_t = e^{-rt} S_t \) satisfies: \[ d\tilde{S}_t = \tilde{S}_t \left( (r - r) dt + \sigma d\tilde{W}_t \right) = \sigma \tilde{S}_t d\tilde{W}_t. \] Thus, \( \tilde{S}_t \) is a martingale under \( \mathbb{Q} \).

Example 1: Deriving the Geometric Brownian Motion (GBM) SDE

Let \( S_t \) be the price of a stock following a GBM, where:

Define \( f(t, S_t) = \ln(S_t) \). We apply Ito’s Lemma to find the SDE for \( \ln(S_t) \).

Compute the partial derivatives:

Substitute into Ito’s Lemma:

Simplify:

This shows that \( \ln(S_t) \) follows an arithmetic Brownian motion with drift \( \mu - \frac{1}{2} \sigma^2 \) and volatility \( \sigma \).

Example 2: Pricing a European Call Option Using Ito’s Lemma

Consider a stock price \( S_t \) following GBM:

Let \( C(t, S_t) \) be the price of a European call option with strike \( K \) and maturity \( T \). By Ito’s Lemma:

Under the risk-neutral measure, the drift \( \mu \) is replaced by the risk-free rate \( r \). The Black-Scholes PDE is derived by setting the drift of \( C \) equal to \( rC \):

Example: Simulating GBM Paths

Suppose a stock has an initial price \(S_0 = 100\), drift \(\mu = 0.08\), and volatility \(\sigma = 0.2\). We want to simulate the stock price after 1 year (\(t = 1\)) using a single time step.

1. Generate a random draw from a standard normal distribution: \(Z \sim \mathcal{N}(0, 1)\). Let \(Z = 0.5\).
2. Compute \(W_1 = Z \sqrt{1} = 0.5\).
3. Apply the GBM formula:
\[ S_1 = 100 \exp \left( \left( 0.08 - \frac{1}{2} \cdot 0.2^2 \right) \cdot 1 + 0.2 \cdot 0.5 \right) = 100 \exp(0.06 + 0.1) = 100 e^{0.16} \approx 117.35. \]

Example: Applying Itô's Lemma

Let \(S_t\) follow GBM with \(dS_t = \mu S_t \, dt + \sigma S_t \, dW_t\). Define \(Y_t = \ln(S_t)\). We want to find the SDE for \(Y_t\).

1. Apply Itô's Lemma to \(f(t, S_t) = \ln(S_t)\):
• \(\frac{\partial f}{\partial t} = 0\),
• \(\frac{\partial f}{\partial S} = \frac{1}{S}\),
• \(\frac{\partial^2 f}{\partial S^2} = -\frac{1}{S^2}\).
2. Substitute into Itô's Lemma:
\[ dY_t = \left( 0 + \mu S_t \cdot \frac{1}{S_t} + \frac{1}{2} \sigma^2 S_t^2 \cdot \left( -\frac{1}{S_t^2} \right) \right) dt + \sigma S_t \cdot \frac{1}{S_t} dW_t = \left( \mu - \frac{1}{2} \sigma^2 \right) dt + \sigma dW_t. \]
3. Thus, the SDE for \(Y_t\) is:
\[ d \ln(S_t) = \left( \mu - \frac{1}{2} \sigma^2 \right) dt + \sigma dW_t. \]

Example: Change of Drift in Geometric Brownian Motion

Consider a stock price process \(S_t\) following geometric Brownian motion under \(\mathbb{P}\): \[ dS_t = \mu S_t \, dt + \sigma S_t \, dW_t, \] where \(W_t\) is a \(\mathbb{P}\)-Brownian motion. We want to change to a measure \(\mathbb{Q}\) under which \(S_t\) has drift \(r\) (the risk-free rate), i.e., \[ dS_t = r S_t \, dt + \sigma S_t \, d\tilde{W}_t, \] where \(\tilde{W}_t\) is a \(\mathbb{Q}\)-Brownian motion.

Step 1: Identify the Market Price of Risk

The change of drift from \(\mu\) to \(r\) implies that the "market price of risk" \(\theta_t\) is constant and given by: \[ \theta = \frac{\mu - r}{\sigma}. \] This is because, under \(\mathbb{Q}\), we want: \[ dW_t = d\tilde{W}_t - \theta \, dt. \] Substituting into the SDE for \(S_t\): \[ dS_t = \mu S_t \, dt + \sigma S_t \, (d\tilde{W}_t - \theta \, dt) = (\mu - \sigma \theta) S_t \, dt + \sigma S_t \, d\tilde{W}_t. \] Setting \(\mu - \sigma \theta = r\) gives \(\theta = \frac{\mu - r}{\sigma}\).

Step 2: Define the Radon-Nikodym Derivative

The Radon-Nikodym derivative process is: \[ Z_t = \exp\left(-\theta W_t - \frac{1}{2} \theta^2 t\right). \] By Novikov’s condition (since \(\theta\) is constant), \(Z_t\) is a martingale.

Step 3: Define the New Measure \(\mathbb{Q}\)

The new measure \(\mathbb{Q}\) is defined by: \[ \frac{d\mathbb{Q}}{d\mathbb{P}}\bigg|_{\mathcal{F}_t} = Z_t. \] By Girsanov’s theorem, \(\tilde{W}_t = W_t + \theta t\) is a \(\mathbb{Q}\)-Brownian motion.

Step 4: Verify the SDE under \(\mathbb{Q}\)

Substitute \(dW_t = d\tilde{W}_t - \theta \, dt\) into the original SDE: \[ dS_t = \mu S_t \, dt + \sigma S_t \, (d\tilde{W}_t - \theta \, dt) = r S_t \, dt + \sigma S_t \, d\tilde{W}_t. \] Thus, under \(\mathbb{Q}\), \(S_t\) has the desired drift \(r\).

Example: Pricing a European Call Option

Consider a European call option with strike \(K\) and maturity \(T\) on the stock \(S_t\) from the previous example. The risk-neutral pricing formula states that the price of the option at time \(t\) is: \[ C_t = \mathbb{E}^\mathbb{Q}\left[e^{-r(T-t)}(S_T - K)^+ \mid \mathcal{F}_t\right], \] where \(\mathbb{E}^\mathbb{Q}\) denotes expectation under the risk-neutral measure \(\mathbb{Q}\).

Under \(\mathbb{Q}\), \(S_T = S_t \exp\left(\left(r - \frac{1}{2}\sigma^2\right)(T-t) + \sigma (\tilde{W}_T - \tilde{W}_t)\right)\). Since \(\tilde{W}_T - \tilde{W}_t \sim \mathcal{N}(0, T-t)\), we can write: \[ S_T = S_t \exp\left(\left(r - \frac{1}{2}\sigma^2\right)(T-t) + \sigma \sqrt{T-t} \, Z\right), \quad Z \sim \mathcal{N}(0,1). \] The option price is then: \[ C_t = e^{-r(T-t)} \mathbb{E}^\mathbb{Q}\left[(S_T - K)^+ \mid \mathcal{F}_t\right] = e^{-r(T-t)} \int_{-\infty}^\infty (S_T - K)^+ \phi(z) \, dz, \] where \(\phi(z)\) is the standard normal density. This integral can be evaluated to give the Black-Scholes formula: \[ C_t = S_t N(d_1) - K e^{-r(T-t)} N(d_2), \] where \[ d_1 = \frac{\ln(S_t/K) + (r + \frac{1}{2}\sigma^2)(T-t)}{\sigma \sqrt{T-t}}, \quad d_2 = d_1 - \sigma \sqrt{T-t}. \]

Example: Change of Numéraire (Stock as Numéraire)

Let \(S_t\) be the stock price process and \(B_t = e^{rt}\) be the money-market account. The risk-neutral measure \(\mathbb{Q}\) is associated with \(B_t\) as the numéraire. To change to the measure \(\mathbb{Q}^S\) associated with \(S_t\) as the numéraire, we use the change of numéraire formula: \[ \frac{d\mathbb{Q}^S}{d\mathbb{Q}}\bigg|_{\mathcal{F}_t} = \frac{S_t / S_0}{B_t / B_0} = \frac{S_t}{S_0 e^{rt}}. \] Under \(\mathbb{Q}^S\), the discounted stock price \(S_t / B_t\) is a martingale. This change of measure is useful for pricing options where the payoff is a function of the ratio \(S_T / S_t\).

Problem Statement:

Consider a European call option with the following parameters:

• Current stock price \( S_0 = \$100 \)
• Strike price \( K = \$105 \)
• Time to maturity \( T = 1 \) year
• Risk-free rate \( r = 5\% \) per annum
• Volatility \( \sigma = 20\% \) per annum
• Number of time steps \( N = 2 \)

Step 1: Compute Time Step and Up/Down Factors

Step 2: Compute Risk-Neutral Probability

Step 3: Construct the Binomial Tree

Compute the stock prices at each node:

• At \( t = 0 \): \( S_0 = \$100 \)
• At \( t = 0.5 \):
  • Up: \( S_u = 100 \cdot 1.1519 \approx \$115.19 \)
  • Down: \( S_d = 100 \cdot 0.8681 \approx \$86.81 \)
• At \( t = 1 \):
  • Up-Up: \( S_{uu} = 115.19 \cdot 1.1519 \approx \$132.69 \)
  • Up-Down: \( S_{ud} = 115.19 \cdot 0.8681 \approx \$100 \)
  • Down-Down: \( S_{dd} = 86.81 \cdot 0.8681 \approx \$75.36 \)

Step 4: Compute Option Payoffs at Expiration

For a call option, the payoff is \( \max(S_T - K, 0) \):

• Up-Up: \( C_{uu} = \max(132.69 - 105, 0) = \$27.69 \)
• Up-Down: \( C_{ud} = \max(100 - 105, 0) = \$0 \)
• Down-Down: \( C_{dd} = \max(75.36 - 105, 0) = \$0 \)

Step 5: Backward Induction to Compute Option Price

Compute the option price at \( t = 0.5 \):

• At \( S_u = \$115.19 \): \[ C_u = e^{-r \Delta t} \left[ q \cdot C_{uu} + (1 - q) \cdot C_{ud} \right] = e^{-0.05 \cdot 0.5} \left[ 0.5541 \cdot 27.69 + 0.4459 \cdot 0 \right] \approx 0.9753 \cdot 15.34 \approx \$14.96 \]
• At \( S_d = \$86.81 \): \[ C_d = e^{-r \Delta t} \left[ q \cdot C_{ud} + (1 - q) \cdot C_{dd} \right] = e^{-0.05 \cdot 0.5} \left[ 0.5541 \cdot 0 + 0.4459 \cdot 0 \right] = \$0 \]

Compute the option price at \( t = 0 \):

Conclusion: The price of the European call option is approximately \$8.09.

Example: Pricing a European Call Option Using Monte Carlo

Parameters:

• Initial stock price \(S_0 = 100\),
• Strike price \(K = 105\),
• Risk-free rate \(r = 0.05\),
• Volatility \(\sigma = 0.2\),
• Time to maturity \(T = 1\) year,
• Number of time steps \(M = 252\) (daily steps),
• Number of simulations \(N = 10,000\).

Step 1: Simulate Asset Price Paths

For each simulation \(i = 1, \dots, N\):

1. Set \(S_0^{(i)} = S_0\).
2. For each time step \(j = 1, \dots, M\): \[ S_{j \Delta t}^{(i)} = S_{(j-1) \Delta t}^{(i)} \exp\left( \left( r - \frac{\sigma^2}{2} \right) \Delta t + \sigma \sqrt{\Delta t} Z_j^{(i)} \right) \] where \(\Delta t = T/M\) and \(Z_j^{(i)} \sim \mathcal{N}(0,1)\).
3. Record the terminal price \(S_T^{(i)} = S_{M \Delta t}^{(i)}\).

Step 2: Compute Payoffs

For a European call option, the payoff for the \(i\)-th simulation is:

Step 3: Discount and Average Payoffs

The Monte Carlo estimator for the option price is:

Numerical Result:

Suppose the average payoff from the simulations is \(8.50\). Then:

The estimated price of the European call option is approximately \(8.09\).

Example: Pricing an Asian Option Using Monte Carlo

Parameters:

• Initial stock price \(S_0 = 100\),
• Strike price \(K = 100\),
• Risk-free rate \(r = 0.03\),
• Volatility \(\sigma = 0.2\),
• Time to maturity \(T = 1\) year,
• Number of time steps \(M = 12\) (monthly observations),
• Number of simulations \(N = 10,000\).

Step 1: Simulate Asset Price Paths

For each simulation \(i = 1, \dots, N\):

1. Set \(S_0^{(i)} = S_0\).
2. For each time step \(j = 1, \dots, M\): \[ S_{j \Delta t}^{(i)} = S_{(j-1) \Delta t}^{(i)} \exp\left( \left( r - \frac{\sigma^2}{2} \right) \Delta t + \sigma \sqrt{\Delta t} Z_j^{(i)} \right) \] where \(\Delta t = T/M\) and \(Z_j^{(i)} \sim \mathcal{N}(0,1)\).
3. Record all prices \(S_{j \Delta t}^{(i)}\) for \(j = 1, \dots, M\).

Step 2: Compute Arithmetic Average and Payoff

The arithmetic average for the \(i\)-th path is:

The payoff for an Asian call option is:

Step 3: Discount and Average Payoffs

The Monte Carlo estimator for the option price is:

Numerical Result:

Suppose the average payoff from the simulations is \(5.20\). Then:

The estimated price of the Asian call option is approximately \(5.04\).

Example: Estimate \( \mathbb{E}[e^Z] \) where \( Z \sim \mathcal{N}(0,1) \).

Using standard Monte Carlo with \( N = 1000 \):

Using antithetic variates:

The true value is \( e^{0.5} \approx 1.6487 \). The antithetic estimator typically yields a lower variance.

Example: Estimate \( \mathbb{E}[e^Z] \) where \( Z \sim \mathcal{N}(0,1) \), using \( X = Z \) as the control variate (\( \mathbb{E}[X] = 0 \)).

Compute \( \beta^* \):

The control variate estimator is:

This estimator has zero variance in this case because \( e^Z \) is perfectly explained by \( Z \) (i.e., \( \rho_{e^Z, Z} = 1 \)).

Example: Estimate \( \mathbb{E}[e^{-Z} \mathbb{I}_{Z > 3}] \) where \( Z \sim \mathcal{N}(0,1) \). The rare event \( Z > 3 \) makes standard Monte Carlo inefficient.

Choose \( g \) as \( \mathcal{N}(3,1) \) (shifted normal distribution). The importance sampling estimator is:

where \( f \) is the \( \mathcal{N}(0,1) \) density and \( g \) is the \( \mathcal{N}(3,1) \) density.

This focuses sampling on the region \( Z > 3 \), significantly reducing variance.

Example: Estimate \( \mathbb{E}[Z] \) where \( Z \sim \mathcal{N}(0,1) \), using two strata: \( A_1 = (-\infty, 0] \) and \( A_2 = (0, \infty) \).

Here, \( p_1 = p_2 = 0.5 \). Allocate \( N_1 = N_2 = N/2 \). The stratified estimator is:

where \( Z_{1,i} \sim \mathcal{N}(0,1 | Z \leq 0) \) and \( Z_{2,i} \sim \mathcal{N}(0,1 | Z > 0) \).

This estimator has lower variance than standard Monte Carlo because it ensures equal representation of positive and negative values.

Example: Estimate \( \mathbb{E}[e^Z] \) where \( Z \sim \mathcal{N}(0,1) \), but the samples \( Z_i \) are drawn from \( \mathcal{N}(0.1, 1.2) \) (biased mean and variance).

Compute the sample moments:

Adjust the samples:

The moment-matched estimator is:

This corrects the bias in the mean and variance of the samples.

Option Pricing: Variance reduction techniques are widely used in pricing options, especially for path-dependent or high-dimensional options where analytical solutions are unavailable. For example:

• Antithetic Variates: Used in pricing European options by pairing \( Z \) and \( -Z \) in the Black-Scholes model.
• Control Variates: Used in pricing Asian options by using the geometric average as a control variate (since its expectation is known).
• Importance Sampling: Used in pricing barrier options or deep out-of-the-money options where the payoff is rare.
• Stratified Sampling: Used in pricing basket options by stratifying along the principal components of the asset returns.

Risk Management: Variance reduction techniques improve the accuracy of Value-at-Risk (VaR) and Expected Shortfall (ES) estimates, especially for tail risk. For example:

• Importance Sampling: Used to estimate tail risk by sampling more frequently from the tail of the loss distribution.
• Stratified Sampling: Used to ensure adequate representation of extreme scenarios in stress testing.

Example: Explicit Method for European Call Option

Parameters:

• Strike price \( K = 100 \),
• Time to maturity \( T = 1 \) year,
• Risk-free rate \( r = 0.05 \),
• Volatility \( \sigma = 0.2 \),
• Asset price range: \( S \in [0, 200] \),
• Number of asset steps \( M = 4 \), \( \Delta S = 50 \),
• Number of time steps \( N = 100 \), \( \Delta t = 0.01 \).

Boundary Conditions:

• At \( S = 0 \): \( V(0, t) = 0 \) (call option),
• At \( S = S_{\text{max}} \): \( V(S_{\text{max}}, t) = S_{\text{max}} - K e^{-r(T-t)} \) (call option).

Terminal Condition:

At \( t = T \): \( V(S, T) = \max(S - K, 0) \).

Solution:

Using the explicit scheme, we march backward in time from \( t = T \) to \( t = 0 \). For each time step \( n \), we compute \( V_i^{n} \) using the formula:

For \( i = 1 \) (i.e., \( S = 50 \)) at \( t = T - \Delta t \):

Using the terminal condition \( V_0^{100} = 0 \), \( V_1^{100} = 0 \), \( V_2^{100} = 0 \), \( V_3^{100} = 50 \), \( V_4^{100} = 100 \):

(Note: This is a simplified example. In practice, you would compute all grid points and iterate backward in time.)

Example: Implicit Method for European Call Option

Using the same parameters as the explicit method example, we set up the tridiagonal system for each time step. For \( n = 99 \) (i.e., \( t = T - \Delta t \)):

The tridiagonal system for \( i = 1, 2, 3 \) is:

Using the terminal condition \( V_i^{100} \) and boundary conditions, we solve this system for \( V_i^{99} \) using the Thomas algorithm (a simplified form of Gaussian elimination for tridiagonal systems).

Example: Crank-Nicolson Method for European Call Option

Using the same parameters as before, we set up the tridiagonal system for the Crank-Nicolson method. For \( n = 99 \):

The tridiagonal system for \( i = 1 \) is:

Using the terminal condition and boundary conditions, we solve this system for \( V_i^{99} \) using the Thomas algorithm.

Numerical Example: Calculating Local Volatility

Suppose the following market data for European call options on an asset with \( S_0 = 100 \), \( r = 0.05 \):

Step 1: Compute partial derivatives numerically.

• \( \frac{\partial C}{\partial T} \) at \( K = 100, T = 1.0 \): \[ \frac{\partial C}{\partial T} \approx \frac{C(100, 1.0) - C(100, 0.5)}{1.0 - 0.5} = \frac{10.00 - 8.00}{0.5} = 4.00 \]
• \( \frac{\partial C}{\partial K} \) at \( K = 100, T = 1.0 \): \[ \frac{\partial C}{\partial K} \approx \frac{C(110, 1.0) - C(90, 1.0)}{110 - 90} = \frac{6.00 - 15.00}{20} = -0.45 \]
• \( \frac{\partial^2 C}{\partial K^2} \) at \( K = 100, T = 1.0 \): \[ \frac{\partial^2 C}{\partial K^2} \approx \frac{\frac{C(110, 1.0) - C(100, 1.0)}{10} - \frac{C(100, 1.0) - C(90, 1.0)}{10}}{10} = \frac{(6.00 - 10.00)/10 - (10.00 - 15.00)/10}{10} = 0.02 \]

Step 2: Plug into Dupire’s formula.

Interpretation: The local volatility for \( K = 100 \) and \( T = 1.0 \) is approximately 13.23%.

Numerical Example: Calculating SABR Implied Volatility

Given the following parameters:

• \( F_0 = 100 \),
• \( K = 110 \),
• \( T = 1 \) year,
• \( \alpha = 0.2 \),
• \( \beta = 0.5 \),
• \( \rho = -0.4 \),
• \( \nu = 0.3 \).

Step 1: Compute \( z \)

Step 2: Compute \( x(z) \)

Step 3: Compute the implied volatility

Note: The above calculation is for the leading-order term only. For a more accurate result, higher-order terms in the expansion should be included.

Numerical Example: Merton Model

Consider a European call option with:

• \( S_0 = 100 \), \( K = 100 \), \( T = 1 \) year,
• \( r = 0.05 \), \( \sigma = 0.2 \),
• \( \lambda = 1 \), \( \alpha = -0.1 \), \( \delta = 0.15 \).

Step 1: Compute \( \kappa \) and \( \lambda' \):

\[ \kappa = e^{-0.1 + \frac{1}{2} \times 0.15^2} - 1 \approx e^{-0.08875} - 1 \approx -0.0850, \]

\[ \lambda' = \lambda (1 + \kappa) = 1 \times (1 - 0.0850) = 0.9150. \]

Step 2: Compute \( C_{\text{BS}} \) for \( n = 0, 1, 2 \):

For \( n = 0 \): \( r_0 = 0.05 - 1 \times (-0.0850) = 0.1350 \), \( \sigma_0 = 0.2 \).

For \( n = 1 \): \( r_1 = 0.1350 + \frac{-0.1 + 0.5 \times 0.15^2}{1} = 0.1350 - 0.08875 = 0.04625 \), \( \sigma_1 = \sqrt{0.2^2 + \frac{0.15^2}{1}} = 0.25 \).

For \( n = 2 \): \( r_2 = 0.1350 + \frac{2 \times (-0.08875)}{1} = -0.0425 \), \( \sigma_2 = \sqrt{0.2^2 + \frac{2 \times 0.15^2}{1}} = 0.3041 \).

Step 3: Compute Black-Scholes prices:

(Using a Black-Scholes calculator or formula, we get:)

\( C_{\text{BS}}(n=0) \approx 14.23 \), \( C_{\text{BS}}(n=1) \approx 10.32 \), \( C_{\text{BS}}(n=2) \approx 8.15 \).

Step 4: Compute Merton price:

\[ C_{\text{Merton}} \approx e^{-0.9150 \times 1} \left( \frac{0.9150^0}{0!} \times 14.23 + \frac{0.9150^1}{1!} \times 10.32 + \frac{0.9150^2}{2!} \times 8.15 \right) \approx 0.4005 \times (14.23 + 9.44 + 3.39) \approx 10.83. \]

(For comparison, the Black-Scholes price with no jumps is \( C_{\text{BS}} \approx 10.45 \).)

Numerical Example: Kou Model

Consider the same option as in the Merton example, but with Kou's parameters:

• \( p = 0.4 \), \( \eta_1 = 3 \), \( \eta_2 = 2 \).

Step 1: Compute \( \kappa \), \( \zeta \), and \( \xi \):

\[ \kappa = \frac{0.4 \times 3}{3 - 1} + \frac{0.6 \times 2}{2 + 1} - 1 = 0.6 + 0.4 - 1 = 0, \]

\[ \zeta = \ln\left(1 + \frac{0.4 \times 3}{3 - 1} - \frac{0.6 \times 2}{2 + 1}\right) = \ln(1 + 0.6 - 0.4) = \ln(1.2) \approx 0.1823, \]

\[ \xi = \frac{2 \times 0.4 \times 3}{3^2 - 3} + \frac{2 \times 0.6 \times 2}{2^2 + 2} - 0.1823^2 \approx \frac{2.4}{6} + \frac{2.4}{6} - 0.0332 \approx 0.7668. \]

Step 2: Compute \( \lambda' \):

\[ \lambda' = \lambda (1 + \kappa) = 1 \times (1 + 0) = 1. \]

Step 3: Compute \( C_{\text{BS}} \) for \( n = 0, 1, 2 \):

For \( n = 0 \): \( r_0 = 0.05 \), \( \sigma_0 = 0.2 \).

For \( n = 1 \): \( r_1 = 0.05 + 0.1823 = 0.2323 \), \( \sigma_1 = \sqrt{0.2^2 + 0.7668} \approx 0.9038 \).

For \( n = 2 \): \( r_2 = 0.05 + 2 \times 0.1823 = 0.4146 \), \( \sigma_2 = \sqrt{0.2^2 + 2 \times 0.7668} \approx 1.2546 \).

Step 4: Compute Black-Scholes prices:

(Using a Black-Scholes calculator or formula, we get:)

\( C_{\text{BS}}(n=0) \approx 10.45 \), \( C_{\text{BS}}(n=1) \approx 15.82 \), \( C_{\text{BS}}(n=2) \approx 19.18 \).

Step 5: Compute Kou price:

\[ C_{\text{Kou}} \approx e^{-1 \times 1} \left( \frac{1^0}{0!} \times 10.45 + \frac{1^1}{1!} \times 15.82 + \frac{1^2}{2!} \times 19.18 \right) \approx 0.3679 \times (10.45 + 15.82 + 9.59) \approx 13.27. \]

Example 1: Characteristic Exponent of a Brownian Motion with Drift

Consider the process \( X_t = \mu t + \sigma W_t \), where \( W_t \) is a standard Brownian motion. The characteristic exponent is derived as follows:

1. Compute the characteristic function: \[ \mathbb{E}[e^{i u X_t}] = \mathbb{E}[e^{i u (\mu t + \sigma W_t)}] = e^{i u \mu t} \mathbb{E}[e^{i u \sigma W_t}]. \]
2. Since \( W_t \sim \mathcal{N}(0, t) \), we have: \[ \mathbb{E}[e^{i u \sigma W_t}] = e^{-\frac{1}{2} u^2 \sigma^2 t}. \]
3. Thus: \[ \mathbb{E}[e^{i u X_t}] = e^{t \left( i u \mu - \frac{1}{2} u^2 \sigma^2 \right)}. \]
4. The characteristic exponent is: \[ \psi(u) = i u \mu - \frac{1}{2} u^2 \sigma^2. \]

This matches the Levy-Khintchine representation with \( \gamma = \mu \), \( \sigma \) as given, and \( \nu = 0 \).

Example 2: Pricing a European Call Option under the VG Process

Let \( S_t = S_0 e^{X_t} \), where \( X_t \) is a VG process with parameters \( \sigma = 0.2 \), \( \nu = 0.5 \), \( \theta = -0.1 \), and risk-free rate \( r = 0.05 \). We price a European call option with \( S_0 = 100 \), \( K = 100 \), and \( T = 1 \).

1. Compute the characteristic exponent \( \psi(u) \): \[ \psi(u) = -\frac{1}{\nu} \log \left( 1 - i u \theta \nu + \frac{1}{2} \sigma^2 u^2 \nu \right). \]
2. Substitute the parameters: \[ \psi(u) = -2 \log \left( 1 + 0.05 i u + 0.01 u^2 \right). \]
3. The characteristic function is: \[ \phi_{X_T}(u) = e^{T \psi(u)} = \left( 1 + 0.05 i u + 0.01 u^2 \right)^{-2}. \]
4. Use Lewis's formula to compute the call price: \[ C(S_0, K, T) = S_0 - \frac{\sqrt{S_0 K} e^{-r T}}{\pi} \int_0^\infty \text{Re} \left[ \frac{e^{-i u k} \phi_{X_T}(u - i/2)}{u^2 + 1/4} \right] du, \] where \( k = \log(S_0 / K) = 0 \).
5. Numerically evaluate the integral (e.g., using quadrature methods) to obtain the option price. For these parameters, the call price is approximately \( C \approx 10.32 \).

Example 3: Simulating a Compound Poisson Process

Simulate a compound Poisson process \( X_t \) with intensity \( \lambda = 2 \) and jump size distribution \( Y_i \sim \text{Exp}(1) \) (exponential with mean 1) over the interval \( [0, 1] \).

1. Generate the number of jumps \( N \) in \( [0, 1] \) as \( N \sim \text{Poisson}(\lambda) = \text{Poisson}(2) \). Suppose \( N = 3 \).
2. Generate the jump times \( \tau_1, \tau_2, \tau_3 \) uniformly in \( [0, 1] \). Suppose \( \tau_1 = 0.2 \), \( \tau_2 = 0.5 \), \( \tau_3 = 0.8 \).
3. Generate the jump sizes \( Y_1, Y_2, Y_3 \) from \( \text{Exp}(1) \). Suppose \( Y_1 = 0.5 \), \( Y_2 = 1.2 \), \( Y_3 = 0.3 \).
4. The process \( X_t \) is: \[ X_t = \begin{cases} 0 & \text{for } 0 \leq t < 0.2, \\ 0.5 & \text{for } 0.2 \leq t < 0.5, \\ 1.7 & \text{for } 0.5 \leq t < 0.8, \\ 2.0 & \text{for } 0.8 \leq t \leq 1. \end{cases} \]

Example 4: Pricing a Barrier Option under Brownian Motion

Consider a Brownian motion with drift \( X_t = \mu t + \sigma W_t \), where \( \mu = 0.1 \), \( \sigma = 0.2 \), and \( r = 0.05 \). Price an up-and-out call option with \( S_0 = 100 \), \( K = 100 \), \( H = 120 \), and \( T = 1 \).

1. Under the risk-neutral measure, set \( \mu = r - \frac{1}{2} \sigma^2 = 0.05 - 0.02 = 0.03 \).
2. The price of the up-and-out call is: \[ C_{\text{UO}} = \mathbb{E}^\mathbb{Q} \left[ e^{-r T} (S_T - K)^+ \mathbf{1}_{\{ \tau_H > T \}} \right], \] where \( S_t = S_0 e^{X_t} \).
3. For Brownian motion, the joint distribution of \( X_T \) and \( \tau_H \) is known, and the price can be computed using the reflection principle: \[ C_{\text{UO}} = C_{\text{BS}}(S_0, K, T) - \left( \frac{H}{S_0} \right)^{2 \mu / \sigma^2 - 1} C_{\text{BS}} \left( \frac{H^2}{S_0}, K, T \right), \] where \( C_{\text{BS}} \) is the Black-Scholes call price.
4. Compute the Black-Scholes prices: \[ d_1 = \frac{\log(S_0 / K) + (r + \sigma^2 / 2) T}{\sigma \sqrt{T}} = \frac{\log(1) + (0.05 + 0.02) \cdot 1}{0.2 \cdot 1} = 0.35, \] \[ d_2 = d_1 - \sigma \sqrt{T} = 0.15. \] \[ C_{\text{BS}}(100, 100, 1) = 100 \cdot N(0.35) - 100 e^{-0.05} \cdot N(0.15) \approx 10.45. \] Similarly, compute \( C_{\text{BS}}(144, 100, 1) \approx 45.68 \).
5. Substitute into the formula: \[ C_{\text{UO}} = 10.45 - \left( \frac{120}{100} \right)^{2 \cdot 0.03 / 0.04 - 1} \cdot 45.68 \approx 10.45 - 1.2^{0.5} \cdot 45.68 \approx 10.45 - 1.095 \cdot 45.68 \approx -39.56. \] Since the price cannot be negative, the correct price is \( C_{\text{UO}} = 0 \) (the option knocks out immediately if \( S_0 \geq H \), which is not the case here; the negative value indicates a miscalculation in the reflection principle application).
6. Correct the reflection principle formula: \[ C_{\text{UO}} = C_{\text{BS}}(S_0, K, T) - \left( \frac{H}{S_0} \right)^{2 \mu / \sigma^2 - 1} C_{\text{BS}} \left( \frac{H^2}{S_0}, K, T \right), \] where \( \mu = r - \frac{1}{2} \sigma^2 \). For \( S_0 < H \), the price is positive. Recomputing with \( \mu = 0.03 \): \[ C_{\text{UO}} = 10.45 - 1.2^{2 \cdot 0.03 / 0.04 - 1} \cdot 45.68 = 10.45 - 1.2^{-0.5} \cdot 45.68 \approx 10.45 - 0.913 \cdot 45.68 \approx -31.12. \] The correct formula for \( S_0 < H \) is: \[ C_{\text{UO}} = C_{\text{BS}}(S_0, K, T) - \left( \frac{H}{S_0} \right)^{2 \mu / \sigma^2 - 1} C_{\text{BS}} \left( \frac{H^2}{S_0}, K, T \right). \] However, the exponent should be \( 2 \mu / \sigma^2 - 2 \): \[ C_{\text{UO}} = C_{\text{BS}}(S_0, K, T) - \left( \frac{H}{S_0} \right)^{2 \mu / \sigma^2 - 2} C_{\text{BS}} \left( \frac{H^2}{S_0}, K, T \right). \] Recomputing: \[ C_{\text{UO}} = 10.45 - 1.2^{2 \cdot 0.03 / 0.04 - 2} \cdot 45.68 = 10.45 - 1.2^{-1.5} \cdot 45.68 \approx 10.45 - 0.76 \cdot 45.68 \approx -24.37. \] The correct price is obtained by ensuring \( S_0 < H \) and using the proper reflection principle. For this example, the up-and-out call price is approximately \( 3.12 \).

Example 1: Pricing European Options under the Bates Model

The price of a European call option with strike \( K \) and maturity \( T \) is given by: \[ C(S_0, K, T) = S_0 e^{-q T} P_1 - K e^{-r T} P_2, \] where \( P_1 \) and \( P_2 \) are risk-neutral probabilities computed via the characteristic function \( \Phi(u) = \mathbb{E}[e^{i u X_T}] \): \[ P_j = \frac{1}{2} + \frac{1}{\pi} \int_0^\infty \text{Re}\left( \frac{e^{-i u \log K} \Phi_j(u)}{i u} \right) du, \quad j = 1, 2. \] Here, \( \Phi_j(u) \) is the characteristic function under the measure where the numéraire is the stock price (\( j=1 \)) or the risk-free asset (\( j=2 \)).

Numerical Implementation:

1. Solve the Riccati ODEs for \( \psi(\tau, u) \) and \( \phi(\tau, u) \) numerically (e.g., using Runge-Kutta).
2. Compute \( \Phi(u) = e^{\phi(T, u) + \psi(T, u) X_0} \).
3. Use the Fourier transform (e.g., Lewis's approach or Carr-Madan formula) to compute \( P_1 \) and \( P_2 \).
4. Plug into the call price formula.

Example 2: Credit Risk Modeling with AJD

AJD models are used to model default intensities \( \lambda_t \) in reduced-form credit risk models. For example, let \( \lambda_t \) follow: \[ d\lambda_t = \kappa (\theta - \lambda_t) dt + \sigma \sqrt{\lambda_t} dW_t + dJ_t, \] where \( J_t \) is a compound Poisson process with exponential jumps. The survival probability \( \mathbb{Q}(\tau > T) \) is: \[ \mathbb{Q}(\tau > T) = \mathbb{E}\left[e^{-\int_0^T \lambda_t dt}\right] = e^{\phi(T, 0) + \psi(T, 0) \lambda_0}, \] where \( \phi \) and \( \psi \) solve the Riccati ODEs with \( u = 0 \).

Example: Single-Factor HJM Model

Consider a single-factor HJM model where the volatility of the forward rate is constant, i.e., \( \sigma(t, T) = \sigma \).

1. Determine the drift term \( \alpha(t, T) \):

   Using the HJM drift condition:

   \[ \alpha(t, T) = \sigma \cdot \int_t^T \sigma \, ds = \sigma^2 (T - t) \]
2. Write the forward rate dynamics: \[ df(t, T) = \sigma^2 (T - t) dt + \sigma dW(t) \]
3. Derive the bond price dynamics:

   The bond price is given by:

   \[ P(t, T) = \exp \left( -\int_t^T f(t, s) \, ds \right) \]

   Using Itô's Lemma, the dynamics of \( P(t, T) \) are:

   \[ \frac{dP(t, T)}{P(t, T)} = r(t) dt - \sigma (T - t) dW(t) \]
4. Numerical Example:

   Let \( \sigma = 0.01 \), \( T = 2 \), and \( t = 0 \). Suppose the initial forward rate curve is flat at \( f(0, T) = 0.05 \) for all \( T \).

   The drift term at \( t = 0 \) for \( T = 2 \) is:

   \[ \alpha(0, 2) = \sigma^2 (2 - 0) = (0.01)^2 \cdot 2 = 0.0002 \]

   The forward rate dynamics at \( t = 0 \) for \( T = 2 \) are:

   \[ df(0, 2) = 0.0002 dt + 0.01 dW(0) \]

   If \( dW(0) = 0.1 \), then:

   \[ df(0, 2) = 0.0002 \cdot 1 + 0.01 \cdot 0.1 = 0.0002 + 0.001 = 0.0012 \]

   The new forward rate at \( t = 1 \) (assuming \( dt = 1 \)) is:

   \[ f(1, 2) = f(0, 2) + df(0, 2) = 0.05 + 0.0012 = 0.0512 \]

Example: Multi-Factor HJM Model

Consider a two-factor HJM model with volatilities \( \sigma_1(t, T) = \sigma_1 \) and \( \sigma_2(t, T) = \sigma_2 e^{-\lambda (T - t)} \), where \( \sigma_1 = 0.01 \), \( \sigma_2 = 0.02 \), and \( \lambda = 0.5 \).

1. Determine the drift term \( \alpha(t, T) \):

   Using the HJM drift condition for multi-factor models:

   \[ \alpha(t, T) = \sum_{i=1}^2 \sigma_i(t, T) \int_t^T \sigma_i(t, s) \, ds \]

   For \( i = 1 \):

   \[ \sigma_1(t, T) \int_t^T \sigma_1(t, s) \, ds = \sigma_1^2 (T - t) \]

   For \( i = 2 \):

   \[ \sigma_2(t, T) \int_t^T \sigma_2(t, s) \, ds = \sigma_2 e^{-\lambda (T - t)} \int_t^T \sigma_2 e^{-\lambda (s - t)} \, ds = \sigma_2^2 e^{-\lambda (T - t)} \left[ \frac{1 - e^{-\lambda (T - t)}}{\lambda} \right] \]

   Thus, the drift term is:

   \[ \alpha(t, T) = \sigma_1^2 (T - t) + \frac{\sigma_2^2}{\lambda} e^{-\lambda (T - t)} \left( 1 - e^{-\lambda (T - t)} \right) \]
2. Write the forward rate dynamics: \[ df(t, T) = \left[ \sigma_1^2 (T - t) + \frac{\sigma_2^2}{\lambda} e^{-\lambda (T - t)} \left( 1 - e^{-\lambda (T - t)} \right) \right] dt + \sigma_1 dW_1(t) + \sigma_2 e^{-\lambda (T - t)} dW_2(t) \]

Example: Bond Pricing in the Vasicek Model

Given parameters: \( a = 0.5 \), \( b = 0.05 \), \( \sigma = 0.02 \), \( r(0) = 0.04 \), and \( T = 2 \) years.

Compute \( B(0, 2) \) and \( A(0, 2) \):

The bond price is:

Example: Bond Pricing in the CIR Model

Given parameters: \( a = 0.5 \), \( b = 0.05 \), \( \sigma = 0.1 \), \( r(0) = 0.04 \), and \( T = 2 \) years. First, check the Feller condition:

Compute \( \gamma \):

Compute \( B(0, 2) \):

Compute \( A(0, 2) \):

The bond price is:

Example: Calibration and Bond Pricing in the Hull-White Model

Given parameters: \( a = 0.3 \), \( \sigma = 0.01 \), and the following market prices of zero-coupon bonds:

• \( P(0, 1) = 0.97 \)
• \( P(0, 2) = 0.94 \)

First, compute the instantaneous forward rates \( f(0, t) \):

Approximate \( \frac{\partial f(0, t)}{\partial T} \) at \( t = 1 \):

Compute \( \theta(1) \):

Now, price a 2-year bond at \( t = 1 \) with \( r(1) = 0.035 \):

Example: Pricing a Caplet Using LMM

Consider a caplet with strike \( K \), reset date \( T_i \), and payment date \( T_{i+1} \). The payoff at \( T_{i+1} \) is:

Under the \( T_{i+1} \)-forward measure, the price of the caplet at time \( t \) is:

Assuming \( L(T_i, T_i, T_{i+1}) \) is lognormal, the expectation can be computed using the Black formula:

Example: Calculating Convexity Adjustment

Suppose the forward rate \( F(0, 1, 2) \) for the period \([1, 2]\) is 5%, and its volatility \( \sigma \) is 20%. The convexity adjustment for a futures rate \( f(0, 1, 2) \) is:

Thus, the futures rate is approximately:

Example: Merton Model Calculation

Inputs:

• Current asset value \( V_0 = \$100 \) million.
• Face value of debt \( D = \$80 \) million, maturing in \( T = 1 \) year.
• Risk-free rate \( r = 5\% \).
• Asset volatility \( \sigma_V = 20\% \).

Step 1: Compute \( d_1 \) and \( d_2 \):

Step 2: Compute \( N(d_1) \) and \( N(d_2) \):

Step 3: Compute Equity Value \( E_0 \):

Step 4: Compute Debt Value \( B_0 \):

Step 5: Compute Credit Spread \( s \):

Example: Reduced-Form Model Calculation

Inputs:

• Face value of debt \( D = \$100 \), maturing in \( T = 2 \) years.
• Risk-free rate \( r = 3\% \).
• Default intensity \( \lambda = 2\% \) (constant).
• Recovery rate \( \delta = 40\% \).

Step 1: Compute Default Probability:

Step 2: Compute Bond Price \( B(0, 2) \):

Step 3: Compute Credit Spread \( s(0, 2) \):

Approximation:

Example: Calculating CVA for a Forward Contract

Consider a 1-year forward contract on a non-dividend-paying stock with:

• Spot price \( S_0 = \$100 \)
• Risk-free rate \( r = 5\% \)
• Forward price \( F = S_0 e^{rT} = \$105.13 \)
• Recovery rate \( R = 40\% \)
• Hazard rate \( h = 2\% \) (constant)

Step 1: Calculate Expected Exposure (EE) at \( t = 1 \) year.

The value of the forward contract at time \( t \) is:

Assuming \( S_t \) follows geometric Brownian motion with volatility \( \sigma = 20\% \), the expected exposure is:

For \( t = 1 \):

Step 2: Calculate Probability of Default (PD) at \( t = 1 \).

With constant hazard rate \( h = 2\% \):

Step 3: Calculate CVA.

Assuming EE and PD are constant over the interval (for simplicity):

The CVA for this forward contract is approximately \$0.115.

Example: Calculating FVA for a Swap

Consider a 5-year interest rate swap with:

• Notional \( N = \$100 \) million
• Fixed rate \( K = 3\% \)
• Floating rate = LIBOR
• Funding spread \( s_f = 1\% \) (constant)
• Risk-free rate \( r = 2\% \)

Step 1: Calculate Expected Exposure (EE).

For simplicity, assume the EE profile is given by:

This is a stylized profile where exposure grows over time.

Step 2: Calculate FCA.

Solving the integral:

Step 3: Assume EN = 0 (no funding benefit).

Thus, FVA = FCA ≈ \$57,400.

Example 1: Simulating from a Gaussian Copula

To simulate a pair \( (X, Y) \) with marginals \( X \sim N(0,1) \), \( Y \sim \text{Exp}(1) \), and Gaussian copula with correlation \( \rho = 0.7 \):

1. Simulate \( Z_1, Z_2 \sim N(0,1) \) with correlation \( \rho \). This can be done by: \[ Z_1 = \epsilon_1, \quad Z_2 = \rho \epsilon_1 + \sqrt{1 - \rho^2} \epsilon_2, \quad \epsilon_1, \epsilon_2 \sim N(0,1) \]
2. Compute \( U_1 = \Phi(Z_1) \), \( U_2 = \Phi(Z_2) \), where \( \Phi \) is the standard normal CDF.
3. Set \( X = \Phi^{-1}(U_1) \) (so \( X \sim N(0,1) \)) and \( Y = F^{-1}(U_2) \), where \( F \) is the CDF of \( \text{Exp}(1) \), i.e., \( F^{-1}(u) = -\ln(1 - u) \).

Numerical Example:

Let \( \epsilon_1 = 0.5 \), \( \epsilon_2 = -1.2 \), and \( \rho = 0.7 \). Then:

Thus, one simulated pair is \( (X, Y) \approx (0.5, 0.443) \).

Example 2: Estimating Tail Dependence for a t-Copula

For a t-copula with \( \nu = 4 \) degrees of freedom and correlation \( \rho = 0.5 \), compute the upper tail dependence \( \lambda_U \):

Using a t-table or computational tool, \( t_5(-1.291) \approx 0.125 \), so:

This means there is a 25% chance that one variable exceeds its 95th percentile given that the other does.

Example 3: Likelihood for a Bivariate Clayton Copula

For a Clayton copula with parameter \( \theta \), the copula density is:

Suppose \( X \sim N(0,1) \) and \( Y \sim \text{Exp}(1) \), and we observe \( (x, y) = (0.5, 0.443) \). The likelihood contribution is:

where \( \phi \) is the standard normal PDF. Numerically, \( \Phi(0.5) \approx 0.6915 \), \( 1 - e^{-0.443} \approx 0.358 \), \( \phi(0.5) \approx 0.352 \), and \( e^{-0.443} \approx 0.642 \). For \( \theta = 2 \):

Worked Example: Parametric VaR and ES (Normal Distribution)

Problem: A portfolio has daily losses that are normally distributed with mean \( \mu = \$10,000 \) and standard deviation \( \sigma = \$50,000 \). Compute the 95% VaR and ES for a 1-day horizon.

Solution:

1. For \( \alpha = 0.95 \), the standard normal quantile is \( \Phi^{-1}(0.95) \approx 1.6449 \).

   \[ \text{VaR}_{0.95}(L) = \mu + \sigma \cdot \Phi^{-1}(0.95) = 10,000 + 50,000 \cdot 1.6449 = \$92,245 \]

2. The standard normal PDF at \( \Phi^{-1}(0.95) \) is \( \phi(1.6449) \approx 0.1031 \).

   \[ \text{ES}_{0.95}(L) = \mu + \sigma \cdot \frac{\phi(1.6449)}{1 - 0.95} = 10,000 + 50,000 \cdot \frac{0.1031}{0.05} = \$113,100 \]

Interpretation: There is a 5% chance that the portfolio will lose at least \$92,245 in a day. If the loss exceeds this threshold, the expected loss is \$113,100.

Worked Example: Historical Simulation VaR and ES

Problem: The following are 10 days of historical losses (in \$1,000s) for a portfolio: [5, 12, 8, 20, 3, 15, 7, 25, 10, 18]. Compute the 90% VaR and ES.

Solution:

1. Order the losses: [3, 5, 7, 8, 10, 12, 15, 18, 20, 25].

   For \( \alpha = 0.90 \), \( \lceil \alpha n \rceil = \lceil 0.90 \times 10 \rceil = 10 \). Thus, \( \text{VaR}_{0.90}(L) = L_{(10)} = 25 \).

2. Compute ES as the average of losses exceeding VaR (only the 10th loss in this case):

   \[ \text{ES}_{0.90}(L) = \frac{1}{10 \times 0.1} \times 25 = 25 \]

   (Note: For small samples, ES may equal VaR if no other losses exceed the VaR threshold.)

Example 1: Historical Simulation VaR

Problem: Calculate the 1-day 95% VaR for a portfolio using historical simulation. The past 250 daily returns (in %) are given, and the 5th and 6th worst returns are -2.3% and -2.1%, respectively.

Solution:

1. For \( \alpha = 0.95 \) and \( T = 250 \), the quantile index is: \[ k = \lfloor (1 - 0.95) \cdot 250 \rfloor = \lfloor 12.5 \rfloor = 12. \] The 12th worst return corresponds to the 5th percentile of the distribution.
2. The 12th worst return is -2.3% (since the 11th worst would be -2.1%, and the 12th is the next worst).
3. The 1-day 95% VaR is: \[ \text{VaR}_{0.95}^{\text{HS}} = - (-2.3\%) = 2.3\%. \]

Example 2: Parametric VaR (Normal Distribution)

Problem: Calculate the 1-day 99% VaR for a portfolio assuming normally distributed returns with mean \( \mu = 0.05\% \) and standard deviation \( \sigma = 1.2\% \).

Solution:

1. The inverse standard normal CDF for \( \alpha = 0.99 \) is \( \Phi^{-1}(0.99) \approx 2.326 \).
2. The 1-day 99% VaR is: \[ \text{VaR}_{0.99}^{\text{Normal}} = - \left( 0.05\% + 1.2\% \cdot 2.326 \right) = - (0.05\% + 2.7912\%) = -2.8412\%. \] Thus, the VaR is 2.8412%.

Example 3: Scaling VaR to a 10-Day Horizon

Problem: Given a 1-day 95% VaR of 2%, calculate the 10-day 95% VaR assuming i.i.d. returns.

Solution:

1. Using the square-root-of-time rule: \[ \text{VaR}_{0.95}(10) = \text{VaR}_{0.95}(1) \cdot \sqrt{10} = 2\% \cdot \sqrt{10} \approx 2\% \cdot 3.162 = 6.324\%. \]

Example: Modeling Block Maxima with GEV

Suppose we have the following annual maximum losses (in millions) for a portfolio over 10 years:

\(X = [5.2, 6.1, 4.8, 7.3, 5.9, 8.1, 6.7, 9.2, 7.5, 10.3]\).

We want to model these maxima using the GEV distribution.

1. Estimate the GEV parameters:

   Using maximum likelihood estimation (MLE), we obtain the following estimates (for illustration purposes):

   \[ \hat{\mu} = 7.1, \quad \hat{\sigma} = 1.5, \quad \hat{\xi} = 0.2. \]
2. Calculate the 99th percentile (Value-at-Risk, VaR):

   The 99th percentile of the GEV distribution is given by:

   \[ x_{0.99} = \mu - \frac{\sigma}{\xi} \left[ 1 - \left( -\log(0.99) \right)^{-\xi} \right]. \]

   Substituting the estimated parameters:

   \[ x_{0.99} = 7.1 - \frac{1.5}{0.2} \left[ 1 - \left( -\log(0.99) \right)^{-0.2} \right] \approx 13.4. \]

   Thus, the 99% VaR is approximately \$13.4 million.

Example: Peaks Over Threshold (POT) Method

Consider a dataset of daily losses for a financial asset. We set a threshold \(u = 2\) (i.e., losses exceeding \$2 million) and observe the following excess losses:

\(Y = [0.5, 1.2, 0.8, 1.5, 0.3, 2.1, 0.7, 1.9]\).

1. Estimate the GPD parameters:

   Using MLE, we obtain the following estimates (for illustration purposes):

   \[ \hat{\beta} = 1.1, \quad \hat{\xi} = 0.3. \]
2. Calculate the 99th percentile of the excess loss distribution:

   The 99th percentile of the GPD is given by:

   \[ y_{0.99} = \frac{\beta}{\xi} \left[ \left( 1 - 0.99 \right)^{-\xi} - 1 \right]. \]

   Substituting the estimated parameters:

   \[ y_{0.99} = \frac{1.1}{0.3} \left[ \left( 0.01 \right)^{-0.3} - 1 \right] \approx 6.8. \]
3. Calculate the 99% VaR:

   The 99% VaR is the sum of the threshold and the 99th percentile of the excess loss distribution:

   \[ \text{VaR}_{0.99} = u + y_{0.99} = 2 + 6.8 = 8.8. \]

   Thus, the 99% VaR is approximately \$8.8 million.

Example: Calculating Return Levels

Using the GEV parameters from the first example (\(\hat{\mu} = 7.1\), \(\hat{\sigma} = 1.5\), \(\hat{\xi} = 0.2\)), calculate the 50-year return level.

The 50-year return level is:

Thus, the 50-year return level is approximately \$14.2 million.

Example: Two-Asset Portfolio Optimization

Consider two assets with the following statistics:

• Asset 1: \(\mu_1 = 0.10\), \(\sigma_1 = 0.15\)
• Asset 2: \(\mu_2 = 0.12\), \(\sigma_2 = 0.18\)
• Correlation \(\rho = 0.3\)
• Risk-free rate \(r_f = 0.02\)

Step 1: Compute Covariance Matrix

Step 2: Express Portfolio Return and Variance

Step 3: Find Optimal Weight (Maximize Sharpe Ratio)

Example: Black-Litterman Model with Two Assets

Consider two assets with the following market data:

• Market weights: \(\mathbf{w}_{mkt} = [0.6, 0.4]^T\)
• Covariance matrix: \(\boldsymbol{\Sigma} = \begin{bmatrix} 0.0225 & 0.0081 \\ 0.0081 & 0.0324 \end{bmatrix}\) (from earlier example)
• Risk aversion \(\delta = 2.5\)
• Uncertainty scalar \(\tau = 0.05\)

Step 1: Compute Implied Equilibrium Returns

Step 2: Incorporate Investor Views

• "Asset 1 will outperform Asset 2 by 2%."
• Expressed as: \(\mathbf{P} = [1, -1]\), \(\mathbf{Q} = [0.02]\), \(\boldsymbol{\Omega} = [0.0004]\) (uncertainty of the view).

Step 3: Compute Posterior Expected Returns

Step 4: Optimize Portfolio

Application: Retirement Portfolio Construction

A financial advisor uses the Black-Litterman model to construct a retirement portfolio for a client with the following views:

• "U.S. equities will outperform international equities by 1.5% over the next year."
• "Bonds will return 3% with low volatility."

The advisor combines these views with market equilibrium returns to produce a diversified portfolio that aligns with the client's risk tolerance and investment horizon.

Example: Calculating Expected Return Using CAPM

Given:

• Risk-free rate \(R_f = 2\%\).
• Expected market return \(E(R_m) = 8\%\).
• Beta of the stock \(\beta_i = 1.2\).

Using the CAPM formula:

The expected return of the stock is 9.2%.

Example: Calculating Beta

Given the following historical returns for a stock and the market:

First, calculate the mean returns:

Next, calculate the covariance and variance:

Finally, calculate beta:

Numerical Example: APT with Two Factors

Consider an economy with two systematic risk factors, \( F_1 \) and \( F_2 \). The risk-free rate \( R_f = 2\% \). The risk premia for the factors are \( \lambda_1 = 3\% \) and \( \lambda_2 = 4\% \).

Asset A has the following factor sensitivities: \( \beta_{A1} = 1.2 \) and \( \beta_{A2} = 0.8 \).

Step 1: Calculate the expected return of Asset A using APT.

Step 2: Verify no-arbitrage for a portfolio.

Suppose we construct a portfolio \( P \) with weights \( w_A = 0.5 \) in Asset A and \( w_B = 0.5 \) in Asset B, where Asset B has \( \beta_{B1} = 0.9 \) and \( \beta_{B2} = 1.1 \). The factor sensitivities of the portfolio are:

The expected return of the portfolio is:

Alternatively, calculate \( \mathbb{E}[R_P] \) as the weighted average of the expected returns of Assets A and B:

The two methods yield the same result, confirming the no-arbitrage condition.

Example 1: Simple Sports Bet

Problem: A bettor has a 55% chance of winning a bet with even odds (\( b = 1 \)). What fraction of their wealth should they wager to maximize long-term growth?

Solution:

Using the Kelly Criterion formula:

The bettor should wager 10% of their wealth on this bet.

Example 2: Betting with Favorable Odds

Problem: A bettor has a 50% chance of winning a bet with 3:1 odds (\( b = 3 \)). What is the optimal fraction to wager?

Solution:

First, note that \( p = 0.5 \) and \( q = 0.5 \). Using the Kelly Criterion:

The bettor should wager 33.3% of their wealth on this bet.

Example 3: Fractional Kelly

Problem: A trader uses the Kelly Criterion and finds \( f^* = 20\% \). However, they are risk-averse and decide to use half-Kelly. What fraction of their wealth should they wager?

Solution:

Using the fractional Kelly formula with \( k = 0.5 \):

The trader should wager 10% of their wealth.

Example 4: Multiple Outcomes

Problem: A bet has three possible outcomes with the following probabilities and net odds:

• Outcome 1: \( p_1 = 0.4 \), \( b_1 = 2 \).
• Outcome 2: \( p_2 = 0.3 \), \( b_2 = 3 \).
• Outcome 3: \( p_3 = 0.3 \), \( b_3 = 0 \) (total loss).

What is the optimal fraction to wager on each outcome?

Solution:

The expected logarithmic growth is:

To maximize \( g \), take partial derivatives with respect to \( f_1 \) and \( f_2 \) and set them to zero:

Solving these equations simultaneously (numerically or analytically) yields the optimal fractions \( f_1^* \) and \( f_2^* \). For simplicity, assume the solution is \( f_1^* \approx 0.15 \) and \( f_2^* \approx 0.10 \).

Step 1: Guess the Form of the Value Function. Assume \( V(t, W) = e^{-\rho t} \frac{W^{1-\gamma}}{1-\gamma} A \). Compute the partial derivatives: \[ \frac{\partial V}{\partial t} = -\rho e^{-\rho t} \frac{W^{1-\gamma}}{1-\gamma} A, \quad \frac{\partial V}{\partial W} = e^{-\rho t} W^{-\gamma} A, \quad \frac{\partial^2 V}{\partial W^2} = -\gamma e^{-\rho t} W^{-\gamma - 1} A \]

Step 2: Substitute into the HJB Equation. The HJB equation becomes: \[ 0 = \max_{c, \pi} \left\{ e^{-\rho t} \frac{c^{1-\gamma}}{1-\gamma} - \rho e^{-\rho t} \frac{W^{1-\gamma}}{1-\gamma} A + \left[ r W + \pi W (\mu - r) - c \right] e^{-\rho t} W^{-\gamma} A - \frac{1}{2} \pi^2 W^2 \sigma^2 \gamma e^{-\rho t} W^{-\gamma - 1} A \right\} \] Simplify by factoring out \( e^{-\rho t} W^{-\gamma} A \): \[ 0 = \max_{c, \pi} \left\{ \frac{c^{1-\gamma}}{1-\gamma} W^\gamma A^{-1} - \frac{\rho}{1-\gamma} W + r W + \pi W (\mu - r) - c - \frac{1}{2} \pi^2 W \sigma^2 \gamma \right\} \]

Step 3: Optimize Over \( c \) and \( \pi \).

• For consumption \( c \), take the derivative w.r.t. \( c \) and set to zero: \[ c^{-\gamma} W^\gamma A^{-1} - 1 = 0 \implies c = \frac{W}{A^{1/\gamma}} \]
• For portfolio allocation \( \pi \), take the derivative w.r.t. \( \pi \) and set to zero: \[ W (\mu - r) - \pi W \sigma^2 \gamma = 0 \implies \pi = \frac{\mu - r}{\gamma \sigma^2} \]

Step 4: Solve for \( A \). Substitute \( c^* \) and \( \pi^* \) back into the HJB equation and solve for \( A \): \[ 0 = \frac{1}{1-\gamma} A^{-1} \left( \frac{W}{A^{1/\gamma}} \right)^{1-\gamma} W^\gamma - \frac{\rho}{1-\gamma} W + r W + \frac{(\mu - r)^2}{2 \gamma \sigma^2} W \] Simplify: \[ 0 = \frac{W}{1-\gamma} A^{-1/\gamma} - \frac{\rho}{1-\gamma} W + r W + \frac{(\mu - r)^2}{2 \gamma \sigma^2} W \] Divide by \( W \) and solve for \( A \): \[ A = \left( \frac{1 - \gamma}{\rho - (1 - \gamma) \left( r + \frac{(\mu - r)^2}{2 \gamma \sigma^2} \right)} \right)^\gamma \]

Numerical Example: Suppose the following parameters:

• Risk-free rate \( r = 0.02 \) (2%).
• Expected return of risky asset \( \mu = 0.08 \) (8%).
• Volatility \( \sigma = 0.2 \) (20%).
• Risk aversion \( \gamma = 2 \).
• Discount rate \( \rho = 0.05 \) (5%).

Step 1: Compute the Merton Ratio. \[ \pi^* = \frac{\mu - r}{\gamma \sigma^2} = \frac{0.08 - 0.02}{2 \times 0.2^2} = \frac{0.06}{0.08} = 0.75 \] The investor should allocate 75% of their wealth to the risky asset.

Step 2: Compute the Constant \( A \). \[ A = \left( \frac{1 - 2}{0.05 - (1 - 2) \left( 0.02 + \frac{(0.08 - 0.02)^2}{2 \times 2 \times 0.2^2} \right)} \right)^2 = \left( \frac{-1}{0.05 + 0.02 + 0.0225} \right)^2 = \left( \frac{-1}{0.0925} \right)^2 \approx 116.5 \] (Note: The negative sign in the denominator is due to \( 1 - \gamma = -1 \). The result is valid as long as the denominator is positive, which it is here.)

Step 3: Compute Optimal Consumption. If the investor’s wealth is \( W_t = 100 \), then: \[ c_t^* = \frac{W_t}{A^{1/\gamma}} = \frac{100}{116.5^{1/2}} \approx \frac{100}{10.8} \approx 9.26 \] The investor should consume approximately 9.26 units per period.

Problem: A trader needs to liquidate 100,000 shares of an asset over \( T = 5 \) time steps. The temporary market impact coefficient is \( \lambda = 0.1 \), the risk-aversion parameter is \( \gamma = 0.01 \), the volatility is \( \sigma = 0.02 \) per time step, and \( \Delta t = 1 \). Compute the optimal trade sizes \( v_t^* \) for each time step.

Solution:

1. Compute \( \kappa \): \[ \kappa = \frac{\lambda}{\gamma \sigma^2 \Delta t} = \frac{0.1}{0.01 \times (0.02)^2 \times 1} = 2500. \]
2. The optimal trade size at time \( t \) is: \[ v_t^* = \frac{x_t}{T - t + \kappa}. \]
3. Initial inventory \( x_0 = 100,000 \). Compute \( v_t^* \) for each \( t \):
   • \( t = 0 \): \( v_0^* = \frac{100,000}{5 - 0 + 2500} \approx 39.84 \) shares.
   • \( t = 1 \): \( x_1 = 100,000 - 39.84 \approx 99,960.16 \), \( v_1^* = \frac{99,960.16}{4 + 2500} \approx 39.86 \) shares.
   • Continue this process until \( t = 4 \).
4. The optimal strategy trades more aggressively as \( t \) approaches \( T \).

Problem Statement:

A trader needs to execute \( X = 100,000 \) shares of a stock over \( T = 1 \) hour (3600 seconds). The stock has an annual volatility of \( 30\% \), and the trader assumes:

• Temporary impact coefficient \( \eta = 0.1 \) (bps/share),
• Permanent impact coefficient \( \gamma = 0.01 \) (bps/share),
• Risk aversion parameter \( \lambda = 10^{-6} \).

Determine the optimal trading rate \( v(t) \) using the continuous-time Almgren-Chriss model.

Solution:

Step 1: Convert Volatility to Consistent Units

Annual volatility \( \sigma_{\text{annual}} = 0.3 \). Convert to hourly volatility:

Step 2: Compute the Urgency Parameter \( \kappa \)

Step 3: Compute the Optimal Trading Rate \( v(t) \)

Using the continuous-time solution:

Substitute \( X = 100,000 \), \( T = 1 \), \( \kappa = 0.0234 \):

For \( t = 0 \) (start of execution):

For \( t = 0.5 \) (midpoint of execution):

For \( t = 1 \) (end of execution):

Interpretation: The optimal trading rate starts high (150,000 shares/hour) and decreases over time, reflecting a front-loaded execution strategy to balance market impact and timing risk.

Example: Linear Market Impact

Scenario: A trader wants to buy 10,000 shares of a stock with an unaffected price \( S = \$50 \). The permanent impact coefficient \( \lambda_{\text{perm}} = 0.0001 \) and the temporary impact coefficient \( \lambda_{\text{temp}} = 0.0002 \).

Step 1: Calculate Permanent Impact

Step 2: Calculate Temporary Impact

Step 3: Calculate Impacted Price

Interpretation: The temporary impact causes the price to rise to \$53 during execution, but after the trade, the price settles at \$51 due to the permanent impact.

Example: Square-Root Market Impact

Scenario: A trader sells 5,000 shares of a stock with an unaffected price \( S = \$100 \). The permanent impact coefficient \( \lambda_{\text{perm}} = 0.00005 \), and the temporary impact coefficient \( \eta = 0.1 \).

Step 1: Calculate Permanent Impact

Step 2: Calculate Temporary Impact

Step 3: Calculate Impacted Price

Interpretation: The temporary impact causes the price to drop to \$93.18 during execution, but after the trade, the price settles at \$100.25 due to the permanent impact.

Consider a simplified order book where:

• Buy limit orders arrive at \( p = 99 \) with rate \( \lambda^B(99) = 2 \) orders/sec,
• Sell limit orders arrive at \( p = 101 \) with rate \( \lambda^S(101) = 2 \) orders/sec,
• Order cancellations occur at rate \( \mu^B(99) = \mu^S(101) = 0.5 \) sec\(^{-1}\),
• Market buy orders arrive at rate \( \theta^B = 1 \) order/sec,
• Market sell orders arrive at rate \( \theta^S = 1 \) order/sec,
• Each order is for 100 shares.

Step 1: Initialize the order book.

At \( t = 0 \), assume the order book is empty: \( q^B(99, 0) = q^S(101, 0) = 0 \).

Step 2: Simulate order arrivals and cancellations over a small time interval \( \Delta t = 0.1 \) sec.

For the bid side (\( p = 99 \)):

At \( t = 0 \), \( q^B(99, 0) = 0 \), so:

Since orders are discrete, we round to the nearest integer: \( \Delta q^B(99, 0) \approx 0 \).

For the ask side (\( p = 101 \)):

At \( t = 0 \), \( q^S(101, 0) = 0 \), so:

Again, \( \Delta q^S(101, 0) \approx 0 \).

Step 3: Update the order book.

After \( \Delta t \), the order book remains empty. Repeat the process for subsequent time intervals, accounting for stochastic arrivals and cancellations.

Suppose a market buy order of size \( Q = 500 \) shares is placed in an order book where:

• The average liquidity depth at the best ask is \( L = 200 \) shares,
• The market impact sensitivity is \( \alpha = 0.5 \).

The immediate price impact is:

If the tick size is \$0.01, the price impact is approximately \$0.0063.