Pricing barrier option with Brownian bridge MC simulation

Monte Carlo simulation for call price

By the law of large number, we can calculate the call option by simulating the stock price.

r <- 0.05
Maturity <- 1
stock_path <- genStock2(mu = r, sigma = 0.2, 
                        S_0 = 100, maturity = Maturity,
                        numPath = 100000, dt = 1)

# filter payoff
stock_path <- stock_path[Maturity +1,]

# PV and take expectation
mean(exp(-r*Maturity)*pmax(stock_path - 100, 0))

## [1] 10.43656

c_0 <- price_Call(S_0 = 100, K = 100, r = 0.05, vol = 0.2, maturity = 1)

Let us figure out the distribtuion of the MC based call option price.

result_mc <- rep(0, 500)
for (i in 1:500){
  stock_path <- genStock2(mu = r, sigma = 0.2, 
                        S_0 = 100, maturity = Maturity,
                        numPath = 10000, dt = 1)

  # filter payoff
  stock_path <- stock_path[Maturity +1,]

  # PV and take expectation
  result_mc[i] <- mean(exp(-r*Maturity)*pmax(stock_path - 100, 0))
}

hist(result_mc, oma=c(2, 3, 3, 2), 
     breaks = seq(min(result_mc), max(result_mc), length.out = 30),
     main = NULL,
     xlab = "Call price",
     xlim = c(10, 11))
mytitle <- "Empirical Dist. of MC based call prices"
mysubtitle <-  paste("Mean:", round(mean(result_mc), 3), ",",
                     "Std:", round(sd(result_mc), 3))
mtext(side=3, line=3, adj=1, cex=1.5, mytitle)
mtext(side=3, line=2, adj=1, cex=1, mysubtitle)
abline(v = c_0, col = "red")
abline(v = mean(result_mc), col = "blue")

In the figure, the blue bar represents the mean of the emprical distribution of the MC based call price while the red bar represents the B-S formular price.

Payoff of the barrier options

Let us denote the following notations:

• Stock price at time \(t\): \(S(t)\)
• Strike price: \(K\)
• Barrier Level: \(L\)
• Maturity: \(T\)

Then, we can write the pay off of the following four barreir call options. Note that \(\unicode{x1D7D9}_{[\cdot]}\) represents the indicator function which returns 1 when the condition in the braket is true.

1. (Knock) Up and in call barrier, assume \(S(0) < L\). \[ (S(T) - K)_{+} \unicode{x1D7D9}_\left[\underset{0 \leq t \leq T}{max}(S(t)) \ge L \right] \]
2. (Knock) Up and out call barrier, assume \(S(0) < L\). \[ (S(T) - K)_{+} \unicode{x1D7D9}_\left[\underset{0 \leq t \leq T}{max}(S(t)) < L \right] \]
3. (Knock) Down and in call barrier, asumme \(S(0) > L\). \[ (S(T) - K)_{+} \unicode{x1D7D9}_\left[\underset{0 \leq t \leq T}{min}(S(t)) \leq L \right] \]
4. (Knock) Down and out call barrier, assume \(S(0) > L\). \[ (S(T) - K)_{+} \unicode{x1D7D9}_\left[\underset{0 \leq t \leq T}{min}(S(t)) > L \right] \]

One of the thing that the readers notice that the vanilla call option pay-off can be decomposed into the in and out barrier options. For example, combining the Up-and-in option and the Up-and-out option, we can make a vanilla option pay-off. This means, the summation of the Up-and-in and out call options should be the same as the price of the vanilla call. This also can be confirm by looking at the fact that the summation of the following indicator functions will be always the same as 1.

\[ \unicode{x1D7D9}_\left[\underset{0 \leq t \leq T}{max}(S(t)) \ge L \right] + \unicode{x1D7D9}_\left[\underset{0 \leq t \leq T}{max}(S(t)) < L \right] = 1 \]

In the other hand, the pay off of the four put barreir options will be as follows:

1. (Knock) Up and in put barrier, assume \(S(0) < L\). \[ (K - S(T))_{+} \unicode{x1D7D9}_\left[\underset{0 \leq t \leq T}{max}(S(t)) \ge L \right] \]
2. (Knock) Up and out put barrier, assume \(S(0) < L\). \[ (K - S(T))_{+} \unicode{x1D7D9}_\left[\underset{0 \leq t \leq T}{max}(S(t)) < L \right] \]
3. (Knock) Down and in put barrier, asumme \(S(0) > L\). \[ (K - S(T))_{+} \unicode{x1D7D9}_\left[\underset{0 \leq t \leq T}{min}(S(t)) \leq L \right] \]
4. (Knock) Down and out put barrier, assume \(S(0) > L\). \[ (K - S(T))_{+} \unicode{x1D7D9}_\left[\underset{0 \leq t \leq T}{min}(S(t)) > L \right] \]

Theoretical price of up-and-in put barrier

The theoretical price of up or down barrier is little complicated since the pay-off is the function of maximum or minimum of the stock prices during time from \(0\) to \(T\). Among these, I choose price formula of the up-and-out put barrier in this section as a reference value of the MC simulation value.

\[ P_0^{UI} = e^{-rT}\mathbb{E}^{Q}\left[ (K - S(T))_{+} \unicode{x1D7D9}_\left[\underset{0 \leq t \leq T}{max}(S(t)) \ge L \right] \right] \]

In the actuarial scinece field, this expectation can be cleverly calculated using exponential tilting called the Esscher transform.

Note that when the Brownian motion \(B(t)\) with drift \(\mu\) and \(\sigma\) and \(M(T)\) represents the maximum of the Brownian motions during time from \(0\) to \(T\), it satisfies the following equation by the reflection principle.

\[ \begin{align} P_{UO}(x,m;\mu,\sigma, T) &= P\left(B(T) \leq x, M(T) \leq m \right) \\ &= \Phi\left(\frac{x-\mu T}{\sigma \sqrt{T}}\right) - e^{\frac{2\mu}{\sigma^2}m}\Phi\left(\frac{x-2m-\mu T}{\sigma \sqrt{T}}\right) \end{align} \]

Note that \(\Phi\) indicates the cdf of the standard normal distribution. Then, the price of the up-and-out put barrier option with parameters.

• Interest rate: \(r\)
• Strike price: \(K\)
• Barrier level: \(L\)
• Stock volatility: \(\sigma\)
• Stock drift: \(\mu\)
• Maturity: \(T\)
• Stock value at time \(0\): \(S(0)\)

can be written as the following compact form using the above probability function \(P_{UO}\):

\[ \begin{align} P_0^{UO} = K e^{-rT}P_{UO}&\left(a, b; r-\frac{1}{2}\sigma^2, \sigma, T\right) - \\ &S(0)P_{UO}\left(a, b;r+\frac{1}{2}\sigma^2, \sigma, T\right) \end{align} \] where \(a\) and \(b\) represent the \(a = ln\left(\frac{K}{S(0)}\right), b = ln\left(\frac{L}{S(0)}\right)\).

Let us define p_UO function.

p_UO <- function(x, m, mu, sigma, maturity){
  d_a <- (x - mu * maturity) / (sigma * sqrt(maturity))
  d_b <- (x - 2*m - mu * maturity) / (sigma * sqrt(maturity))
  pnorm(d_a) - exp(2*mu*m / sigma^2) * pnorm(d_b)
}

The price of the up-and-out put option with the parameters

• Interest rate: \(r = 0.05\)
• Strike price: \(K = 110\)
• Barrier level: \(L = 200\)
• Stock volatility: \(\sigma = 0.2\)
• Stock drift: \(\mu = 0.126\)
• Maturity: \(T = 1\)
• Stock value at time \(0\): \(S(0)=100\)

will be the same as follows:

S_0 <- 100
r <- 0.05
K <- 110
L <- 110
mu <- 0.126
sigma <- 0.2
maturity <- 1
delta_t <- 1/365

put_UpOut <- function(S_0, K, L, r, sigma, maturity){
  a <- log(K / S_0)
  b <- log(L / S_0)
  N_da <- p_UO(a, b, r - 0.5 * sigma^2, sigma, maturity)
  N_db <- p_UO(a, b, r + 0.5 * sigma^2, sigma, maturity)
  K*exp(-r*maturity)*N_da - S_0*N_db
}

put_uo <- put_UpOut(S_0 = 100, K = 110, L = 110, 
                    r = 0.05, sigma = 0.2, maturity = 1)
put_uo

## [1] 7.139859

Plain MC simulation the price of up-and-out put barrier

By generating the stock path during the period from zero to maturity, the price of the up-and-out put barrier can be obtained as follows:

stock_path <- genStock2(S_0 = 100, mu = r,
                        sigma = 0.2, maturity = 1,
                        numPath = 100000, dt = delta_t)

# checking out the hitting barrier
check_barrier <- as.integer(apply(stock_path, 2, max) < L)
activated_paths <- stock_path[,which(check_barrier == 1)]

# filter payoff
activated_paths <- activated_paths[maturity * delta_t^(-1) + 1,]
activated_payoff <- pmax(K - activated_paths, 0)

# attached knock out payoff
activated_payoff <- c(activated_payoff, rep(0, sum(!check_barrier)))

# PV and take expectation
plain_mc <- mean(exp(-r*maturity) * activated_payoff)
plain_mc

## [1] 7.467147

# theoretical value
put_uo

## [1] 7.139859

However, this method is little slow and efficient, but also inaccurate. This is because, since we are gerating the stock path discritely, we can not determine that wheter the stock hit the barrier or not eventhough the two consecutive stock prices are close to the barrier. For example, let us assume we have observed the following for \(\epsilon > 0\):

\[ L - \epsilon < S(t_i) < L \text{ and } L - \epsilon < S(t_{i+1}) < L \]

Even if the epsilon is really small, we cannot say the stock hit the barrier level \(L\) until we generate the stock path during the interval. Thus, the accuracy of the plain MC method for the up-and-out put barrier depends on the sample size but also the step size.

MC simulation with the Brownian bridge

This method does not use the brownian bridge directly, however, it is quite clever since it allow us to make the accuracy of the MC only to depend on the sample size \(n\). From the Brownian bridge property, we know that the following probability holds;

Note that the probability that the brownian bridge crossing the barrier does not depends on the drift term. Now, we know that

\[ \frac{dS(t)}{S(t)} = (r - \frac{1}{2}\sigma^2)dt + \sigma dB(t) \] which means, under \(Q\) measure

\[ S(t) = S(0)e^{X(t)}, \quad X(t) \sim \mathcal{N}((r - \frac{1}{2}\sigma^2)t, \sigma^2 t) \]

thus, the crossing probability for the pinned geometric brownian motion for the barrier level \(L\) will be

\[ \begin{align} &P\left(\underset{0\leq t \leq T}{max} (S(t)) > L \Bigg\vert S(0) = s_0, S(T) = s_T \right) \\ =&P\left( \underset{0\leq t \leq T}{max} (X(t)) > log(L / s_0) \Bigg\vert X(0) = 0, X(T) = log(s_T / s_0) \right) \\ \end{align} \]

Let us define the function as follows:

crossp <- function(a, b, m, sigma, maturity){
  pmin(exp(-2 * (m - a)*(m - b) / (sigma^2 * maturity)), 1)
}

crosspStock <- function(s_0, s_T, L, sigma, maturity){
  m <- log(L / s_0)
  a <- 0
  b <- log(s_T / s_0)
  cross_prob <- crossp(a, b, m, sigma, maturity)
  # prob. of hitting the barrier
  cross_prob
}

We can confirm the brownian bridge MC generates more accurate pricing value for the same stock path as follows:

# calculate final stock price & calculate crossing prob.
payoff <- stock_path[maturity * delta_t^(-1) + 1,]
cross_prob <- crosspStock(100, payoff, L, 0.2, 1)

# calculate vanilla put option payoff
payoff <- pmax(K - payoff, 0)

# turn off payoff when it hits the barrier
u <- runif(length(cross_prob))
check_barrier <- rep(1, length(u))
check_barrier[u < cross_prob] <- 0
payoff <- payoff * check_barrier

# calculate price
bridge_mc <- mean(exp(-r*maturity) * payoff)
bridge_mc

## [1] 7.172284

# plain MC
plain_mc

## [1] 7.467147

# theoretical value
put_uo

## [1] 7.139859

Note that stock_path variable was generated in the Plain MC section for up-and-out put barrier.