1 Slide 7

Consider a European down-and-out call at time \(t = 1\). The current stock price: \(S_1 = 8\), the barrier: \(B = 6\), strike price: \(E = 7\), maturity time \(T = 3\). What is the option payoff, given that \(S_{1.5} = 9\), \(S_2 = 7\), \(S_{2.5} = 5.5\), \(S_3 = 7.5\).

The payoff of this call option is \(0\) since \(S_{2.5} = 5.5\) stays below the barrier \(B = 6\).

2 Slide 9

Consider a European down-and-in call at time \(t = 1\). The current stock price: \(S_1 = 8\), the barrier: \(B = 6\), strike price: \(E = 7\), maturity time \(T = 3\). What is the option payoff, given that \(S_{1.5} = 7\), \(S_2 = 6\), \(S_{2.5} = 6.5\), \(S_3 = 7.5\).

This down-and-in call becomes a vanilla call since the stock price \(S_2 = 6\) touches the barrier \(B = 6\) from above. The payoff of this call option is \[\begin{align*} V &= \max(S_3-E,0) \\ &= \max(7.5-7,0) \\ &=0.5 \end{align*}\]

3 Slide 10

Consider a stock with price follows a geometric Brownian motion: \[\,d S_t = 0.05 S_t \,dt + 0.2 S_t \,d W_t,\] where \((W_t)_{t \geq 0}\) is a standard Brownian motion with respect to the risk-neutral measure Q. A European down-and-out call option written on the stock with parameters: exercise price \(K = 10\), time to maturity \(T > t\) (year), the barrier is \(B = 9\), the risk-free interest rate is \(r = 0.05\), \(t=0.5\), \(T=1\), \(S_t =11\).

build_stock_tree = function(S, sigma, delta_t, N) {
  tree = matrix(0, nrow=N+1, ncol=N+1)
  u = exp(sigma*sqrt(delta_t)) ### CRR model
  d = exp(-sigma*sqrt(delta_t))
  for (i in 1:(N+1)) { ## Each i denotes for the period i-1 
    for (j in 1:i) { ## Each j denotes for the node j 
      tree[i,j] = S * u^(j-1) * d^((i-1)-(j-1)) ###
  }
  }
  return(tree)
}

### compute the risk-neutral probability
  q_prob = function(r, delta_t, sigma) { ##delta_t: the time period, sigma: the volatility of the stock
  u = exp(sigma*sqrt(delta_t)) ## up factor
  d = exp(-sigma*sqrt(delta_t)) ## down factor
  R=exp(r*delta_t) ## future factor
  return((R - d)/(u-d))
}
### locial function with output 0 or 1
logical <- function(x,a){
for (i in 1: length(x)){if (x[i]>a) {x[i]=1} else x[i]=0}
return (x)}

### compute the price of European down and out call options
value_binomial_option_down_out = function(tree, sigma, delta_t,
r, X, B) {
## X:the strike price, B: barrier
q = q_prob(r, delta_t, sigma)
option_tree = matrix(0, nrow=nrow(tree), ncol=ncol(tree))
option_tree[nrow(option_tree),] =
logical(tree[nrow(tree),],B)*pmax(tree[nrow(tree),] - X, 0)
for (i in (nrow(tree)-1):1) {
for(j in 1:i) {
option_tree[i, j] = logical(tree[i,j],B)*
((1-q)*option_tree[i+1,j] +
q*option_tree[i+1,j+1])/exp(r*delta_t)}}
return(option_tree)}

### combine every thing
binomial_option_down_out = function( sigma, T,t, r, X, B,S, N) {
q = q_prob(r=r, delta_t=(T-t)/N, sigma=sigma)
tree = build_stock_tree(S=S, sigma=sigma, delta_t=(T-t)/N, N=N)
option = value_binomial_option_down_out(tree, sigma=sigma,
delta_t=(T-t)/N, r=r, X=X, B=B)
return(list(q=q, stock=tree, option=option, price=option[1,1]))
}

binom = binomial_option_down_out(sigma=0.2, 
                         T=1, 
                         t=0.5, 
                         r=0.05,
                         X=10, 
                         B=9, 
                         S=11, 
                         N=4)

3.1 Question 1

Compute the option price using the 4-step binomial tree

\[\begin{align*} u&=e^{\sigma \sqrt{\frac{T-t}{N}}} \\ &=e^{0.2\sqrt{\frac{2-0.5}{4}}} \\ &=1.07 \\ d&=0.93 \\ p &= \frac{e^{r \cdot \frac{T-t}{N}}-d}{u-d} \\ &= \frac{e^{0.05 \cdot \frac{1-0.5}{4}}-0.93}{1.07-0.93} \\ &=0.53 \\ q &=0.47 \\ \Delta &=\frac{T-t}{N} \\ &= \frac{1-0.5}{4} \\ &= 0.125 \end{align*}\]

\[\begin{align*} S_{04} &= S_0 \cdot u^4 \\ &= 11 \cdot 1.07^4 \\ &= 14.42 \\ S_{14} &= S_0 \cdot u^3d \\ &= 11 \cdot 1.07^2 \\ &= 12.59 \\ S_{24} &= S_0 \cdot u^2d^2 \\ &= 11 \cdot 1.07^0 \\ &= 11.00 \\ S_{34} &= S_0 \cdot ud^3 \\ &= 11 \cdot 1.07^{-2} \\ &= 9.61 \\ S_{44} &= S_0 \cdot d^4 \\ &= 11 \cdot 1.07^{-4} \\ &= 8.39 \end{align*}\]

\[\begin{align*} V_{04}&= \begin{cases} \max(S_{04}-E,0) & S_{04} \geq 9 \\ 0 & S_{04} < 9 \end{cases} \\ &=\max(14.42-10,0) \\ &= 4.42 \\ V_{14}&= \begin{cases} \max(S_{14}-E,0) & S_{14} \geq 9 \\ 0 & S_{14} < 9 \end{cases} \\ &=\max(12.59-10,0) \\ &= 2.59 \\ V_{24}&= \begin{cases} \max(S_{24}-E,0) & S_{24} \geq 9 \\ 0 & S_{24} < 9 \end{cases} \\ &=\max(11.00-10,0) \\ &= 1 \\ V_{34}&= \begin{cases} \max(S_{34}-E,0) & S_{34} \geq 9 \\ 0 & S_{34} < 9 \end{cases} \\ &=\max(9.61-10,0) \\ &= 0 \\ V_{44}&= \begin{cases} \max(S_0 \cdot d^4-E,0) & S_{44} \geq 9 \\ 0 & S_{44} < 9 \end{cases} \\ &= 0 \\ \end{align*}\]

\[\begin{align*} S_{03} &= S_0 \cdot u^3 \\ &= 11 \cdot 1.07^3 \\ &= 13.48 \\ S_{13} &= S_0 \cdot u^2d \\ &= 11 \cdot 1.07 \\ &= 11.77 \\ S_{23} &= S_0 \cdot ud^2 \\ &= 11 \cdot 1.07^{-1} \\ &= 10.28 \\ S_{33} &= S_0 \cdot d^3 \\ &= 11 \cdot 1.07^{-3} \\ &= 8.98 \end{align*}\]

\[\begin{align*} V_{03}&= \begin{cases} e^{-r \cdot \Delta}(V_{04}\cdot p + V_{14}\cdot q) & S_{03} \geq 9 \\ 0 & S_{03} < 9 \end{cases} \\ &=e^{-0.05 \cdot 0.125}(4.42\cdot 0.53 + 2.59\cdot 0.47) \\ &= 3.54 \\ V_{13}&= \begin{cases} e^{-r \cdot \Delta}(V_{14}\cdot p + V_{24}\cdot q) & S_{13} \geq 9 \\ 0 & S_{13} < 9 \end{cases} \\ &=e^{-0.05 \cdot 0.125}(2.59\cdot 0.53 + 1\cdot 0.47) \\ &= 1.83 \\ V_{23}&= \begin{cases} e^{-r \cdot \Delta}(V_{24}\cdot p + V_{34}\cdot q) & S_{23} \geq 9 \\ 0 & S_{23} < 9 \end{cases} \\ &=e^{-0.05 \cdot 0.125}(1\cdot 0.53 + 0\cdot 0.47) \\ &= 0.53 \\ V_{33}&= \begin{cases} e^{-r \cdot \Delta}(V_{34}\cdot p + V_{44}\cdot q) & S_{33} \geq 9 \\ 0 & S_{33} < 9 \end{cases} \\ &= 0 \\ \end{align*}\]

\[\begin{align*} S_{02} &= S_0 \cdot u^2 \\ &= 11 \cdot 1.07^2 \\ &= 12.59 \\ S_{12} &= S_0 \cdot ud \\ &= 11 \cdot 1.07^0 \\ &= 11.00 \\ S_{22} &= S_0 \cdot d^2 \\ &= 11 \cdot 1.07^{-2} \\ &= 9.60 \end{align*}\]

\[\begin{align*} V_{02}&= \begin{cases} e^{-r \cdot \Delta}(V_{03}\cdot p + V_{13}\cdot q) & S_{02} \geq 9 \\ 0 & S_{02} < 9 \end{cases} \\ &=e^{-0.05 \cdot 0.125}(3.54\cdot 0.53 + 1.83\cdot 0.47) \\ &= 2.72 \\ V_{12}&= \begin{cases} e^{-r \cdot \Delta}(V_{13}\cdot p + V_{23}\cdot q) & S_{12} \geq 9 \\ 0 & S_{12} < 9 \end{cases} \\ &=e^{-0.05 \cdot 0.125}(1.83\cdot 0.53 + 0.53\cdot 0.47) \\ &= 1.21 \\ V_{22}&= \begin{cases} e^{-r \cdot \Delta}(V_{23}\cdot p + V_{33}\cdot q) & S_{22} \geq 9 \\ 0 & S_{22} < 9 \end{cases} \\ &=e^{-0.05 \cdot 0.125}(0.53\cdot 0.53 + 0\cdot 0.47) \\ &= 0.28 \end{align*}\]

\[\begin{align*} S_{01} &= S_0 \cdot u \\ &= 11 \cdot 1.07 \\ &= 11.77 \\ S_{11} &= S_0 \cdot d \\ &= 11 \cdot 1.07^{-1} \\ &= 10.28 \end{align*}\]

\[\begin{align*} V_{01}&= \begin{cases} e^{-r \cdot \Delta}(V_{02}\cdot p + V_{12}\cdot q) & S_{01} \geq 9 \\ 0 & S_{01} < 9 \end{cases} \\ &=e^{-0.05 \cdot 0.125}(2.72\cdot 0.53 + 1.21\cdot 0.47) \\ &= 2.00 \\ V_{11}&= \begin{cases} e^{-r \cdot \Delta}(V_{12}\cdot p + V_{22}\cdot q) & S_{11} \geq 9 \\ 0 & S_{11} < 9 \end{cases} \\ &=e^{-0.05 \cdot 0.125}(1.21\cdot 0.53 + 0.28\cdot 0.47) \\ &= 0.77 \end{align*}\]

\[\begin{align*} V_{0}&= \begin{cases} e^{-r \cdot \Delta}(V_{01}\cdot p + V_{11}\cdot q) & S_{0} \geq 9 \\ 0 & S_{0} < 9 \end{cases} \\ &=e^{-0.05 \cdot 0.125}(2.00\cdot 0.53 + 0.77\cdot 0.47) \\ &= 1.41 \end{align*}\]

binom$stock
##           [,1]      [,2]     [,3]     [,4]     [,5]
## [1,] 11.000000  0.000000  0.00000  0.00000  0.00000
## [2,] 10.249046 11.805977  0.00000  0.00000  0.00000
## [3,]  9.549358 11.000000 12.67101  0.00000  0.00000
## [4,]  8.897437 10.249046 11.80598 13.59942  0.00000
## [5,]  8.290021  9.549358 11.00000 12.67101 14.59586
binom$option
##           [,1]      [,2]     [,3]     [,4]     [,5]
## [1,] 1.4288538 0.0000000 0.000000 0.000000 0.000000
## [2,] 0.7693910 2.0386445 0.000000 0.000000 0.000000
## [3,] 0.2738891 1.2239491 2.795231 0.000000 0.000000
## [4,] 0.0000000 0.5233442 1.868282 3.661727 0.000000
## [5,] 0.0000000 0.0000000 1.000000 2.671009 4.595861
binom$price
## [1] 1.428854

Answer: The fair price of the option is \(1.41\)

3.2 Question 2

Compute the option price using the Monte Carlo simulation

MC_down_out <- function(St , K,B,t , T , r , sigma ,M, N ){
  dt=(T-t)/N
  payoff=matrix(0,nrow=M,ncol=1)
  S =matrix(0,nrow=M, ncol=N+1)
  for (i in 1:M){
  S[i,1] = St
  for (j in 1:N){
    S[i,j+1]=S[i,j]*exp((r-sigma^2/2)*dt+sigma*sqrt(dt)*rnorm(1,mean=0,sd=1))
    if (S[i,j+1]<=B ){
    break
    }
  }
  payoff[i]=max(S[i,N+1]-K,0)}
  price <- exp(-r * (T - t))*mean(payoff)
  return(price)
}

MC_down_out(St=11, 
            K=10, 
            B=9, 
            t=0.5, 
            T=1, 
            r=0.05,
            sigma=0.2, 
            M=10000, 
            N=1000 )
## [1] 1.396451