Financial Mathematics 1 - Sample Final

Instructor: Dr. Le Nhat Tan

1 Problems

1.1 Question 1

Given three assets \(S_1,S_2,S_3\) with annual mean returns \(r_1=4\%,r_2=5\%,r_3=6\%,\) volatilities \(\sigma_1=2\%,\sigma_2=4\%,\sigma_3=8\%\) and correlations \(\rho_{12}=0.1,\rho_{23}=0.2,\rho_{13}=-0.1.\)

  1. (15 pts) If short sell is permitted, find the efficient portfolio \(P_1\) with the mean return \(4.8\%.\)
  2. (15 pts) If short sell is not permitted, find the minimum variance portfolio \(P_2.\)

1.2 Question 2

Consider a \(1-\)year forward contract on a stock of current price $\(100.\) The risk-free rate \(r\) is \(5\%\) p.a.

  1. (10 pts) Assume \(r\) compounds semi-annually. Compute the delivery price of this contract if (i) the stock pays no dividend and (ii) the stock pays a \(4\%\) dividend every \(6\) months (the \(1^{\textrm{st}}\) dividend payment is in \(6\) months).
  2. (10 pts) Assume \(r\) compounds continuously. Compute the delivery price of this contract if (i) the stock pays no dividend and (ii) the stock pays a \(4\%\) dividend p.a. continuously.

1.3 Question 3

Consider a \(1-\)year European put of $\(5\) with strike price $\(50\) on a stock \(S\) of current price $\(52.\) Assume the continuously compounded interest rate \(r\) is \(8\%.\)

  1. (10 pts) Evaluate the value of a \(1-\)year European call option \(C\) on the same stock, with the same strike price.
  2. (6 pts) What is the maximum loss to the buyer of this call?
  3. (10 pts) Assume that a \(1-\)year European call \(C'\) with the strike price $\(49\) is priced at $\(12.\) Show that there exists an arbitrage opportunity. What should you do to earn at least $\(100\) guaranteed profit?

1.4 Question 4

Consider a \(1-\) year European put with strike price $\(100\) on a stock of current price $\(90.\) Assume the stock volatility is \(20\%\) p.a. and the interest rate \(r\) is \(5\%\) p.a. compounded continuously. Consider the \(3-\)step CRR model.

  1. (6 pts) Draw the binomial tree and compute the stock price at each node.
  2. (6 pts) Compute the option price at each node.
  3. (10 pts) Compute the price of the corresponding American put option.
  4. (2 pts) When should the American put option be exercised?

2 Solutions

2.1 Question 1

Note that \[\sigma_1^2=\frac{1}{2500},\sigma_2^2=\frac{1}{625},\sigma_3^2=\frac{4}{625},\] \[\rho_{12}=\frac{1}{12500},\rho_{23}=\frac{2}{3125},\rho_{13}=\frac{-1}{6250}.\] Assume \(P=(\omega_1,\omega_2,1-\omega_1-\omega_2),\) the portfolio return is \[\begin{align*}r_P &= \omega_1r_1+\omega_2r_2+(1-\omega_1-\omega_2)r_3\\ &= \omega_1\left(4\%-6\%\right)+\omega_2\left(5\%-6\%\right)+6\%\\ &= 0.06-0.02\omega_1-0.01\omega_2\end{align*}\] and portfolio variance is \[\begin{align*}\sigma_P^2 &= \omega_1^2\sigma_1^2+\omega_2^2\sigma_2^2+(1-\omega_1-\omega_2)^2\sigma_3^2+2\omega_1\omega_2\rho_{12}+2\omega_2(1-\omega_1-\omega_2)\rho_{23}\\ &+2\omega_1(1-\omega_1-\omega_2)\rho_{13}\\&= \omega_1^2\left(\frac{1}{2500}+\frac{4}{625}+\frac{1}{3125}\right)+\omega_2^2\left(\frac{1}{625}+\frac{4}{625}-\frac{4}{3125}\right)\\&+ \omega_1\omega_2\left(\frac{8}{625}+\frac{1}{6250}-\frac{4}{3125}+\frac{1}{3125}\right)\\&+ \omega_1\left(\frac{-8}{625}-\frac{1}{3125}\right)+\omega_2\left(\frac{-8}{625}+\frac{4}{3125}\right)+\frac{4}{625}\\&= \frac{89\omega_1^2}{12500}+\frac{21\omega_2^2}{3125}+\frac{3\omega_1\omega_2}{250}-\frac{41\omega_1}{3125}-\frac{36\omega_2}{3125}+\frac{4}{625}\end{align*}\]

2.1.1 Part 1

The corresponding linear program is \[\begin{matrix}\textrm{minimize} & \sigma_P^2\\ \textrm{subject to} & r_P=0.048\\ & \omega_1,\omega_2\in\mathbb{R}\end{matrix}\] i.e.  \[\begin{matrix}\textrm{minimize} & \frac{89\omega_1^2}{12500}+\frac{21\omega_2^2}{3125}+\frac{3\omega_1\omega_2}{250}-\frac{41\omega_1}{3125}-\frac{36\omega_2}{3125}+\frac{4}{625}\\ \textrm{subject to} & 0.06-0.02\omega_1-0.01\omega_2=0.048\\ & \omega_1,\omega_2\in\mathbb{R}\end{matrix}\] The Lagrangian function is \[\begin{align*}\mathcal{L}(\omega_1,\omega_2,\lambda) &= \frac{89\omega_1^2}{12500}+\frac{21\omega_2^2}{3125}+\frac{3\omega_1\omega_2}{250}-\frac{41\omega_1}{3125}-\frac{36\omega_2}{3125}\\&+\frac{4}{625}+\lambda(0.012-0.02\omega_1-0.01\omega_2)\end{align*}\] with the system \[\left\{\begin{matrix} \frac{89\omega_1}{6250}+\frac{3\omega_2}{250}-\frac{41}{3125}-0.02\lambda=0\\ \frac{42\omega_2}{3125}+\frac{3\omega_1}{250}-\frac{36}{3125}-0.01\lambda=0\\ 0.012-0.02\omega_1-0.01\omega_2=0 \end{matrix}\right.\Rightarrow\left\{\begin{matrix}\omega_1=248/625\\\omega_2=254/625\\\lambda\approx-0.1296\end{matrix}\right..\] The optimal portfolio is \(\left(\frac{248}{625},\frac{254}{625},\frac{123}{625}\right)\) with the variance \(6.783\cdot10^{-4}.\)

2.1.2 Part 2

The corresponding linear program is \[\begin{matrix}\textrm{minimize} & \frac{89\omega_1^2}{12500}+\frac{21\omega_2^2}{3125}+\frac{3\omega_1\omega_2}{250}-\frac{41\omega_1}{3125}-\frac{36\omega_2}{3125}+\frac{4}{625}\\ \textrm{subject to} & \omega_1+\omega_2\leq1\\ & \omega_1,\omega_2\geq0\end{matrix}\] The Lagrangian function is \[\begin{align*}\mathcal{L}(\omega_1,\omega_2,\lambda) &= \frac{89\omega_1^2}{12500}+\frac{21\omega_2^2}{3125}+\frac{3\omega_1\omega_2}{250}-\frac{41\omega_1}{3125}-\frac{36\omega_2}{3125}\\&+\frac{4}{625}- \lambda_1\omega_1-\lambda_2\omega_2+\lambda_3(\omega_1+\omega_2-1)\end{align*}\] with the system \[\left\{\begin{matrix} \frac{89\omega_1}{6250}+\frac{3\omega_2}{250}-\frac{41}{3125}-\lambda_1+\lambda_3=0\\ \frac{42\omega_2}{3125}+\frac{3\omega_1}{250}-\frac{36}{3125}-\lambda_2+\lambda_3=0\\ \lambda_1\omega_1=\lambda_2\omega_2=\lambda_3(\omega_1+\omega_2-1)=0 \end{matrix}\right.\]

  1. If \(\omega_1=0,\) the system becomes \[\left\{\begin{matrix} \frac{3\omega_2}{250}-\frac{41}{3125}-\lambda_1+\lambda_3=0\\ \frac{42\omega_2}{3125}-\frac{36}{3125}-\lambda_2+\lambda_3=0\\ \lambda_2\omega_2=\lambda_3(\omega_2-1)=0 \end{matrix}\right.\] If \(\omega_2=0,\) then \(\lambda_3=0\) and so \[0=-\frac{36}{3125}-\lambda_2\leq-\frac{36}{3125}<0,\] a contradiction, hence \(\omega_2\neq0\Rightarrow\lambda_2=0.\) The system now is \[\left\{\begin{matrix} \frac{3\omega_2}{250}-\frac{41}{3125}-\lambda_1+\lambda_3=0\\ \frac{42\omega_2}{3125}-\frac{36}{3125}+\lambda_3=0\\ \lambda_3(\omega_2-1)=0 \end{matrix}\right.\] If \(\omega_2=1\) then \[0=\frac{42}{3125}-\frac{36}{3125}+\lambda_3\geq\frac{6}{3125}>0,\] a contradiction, hence \(\omega_2\neq1\Rightarrow\lambda_3=0\) and \[\omega_2=\frac{6}{7},\lambda_1=\frac{62}{21875}.\]
  2. If \(\omega_1\neq0,\) then \(\lambda_1=0.\) The system becomes \[\left\{\begin{matrix} \frac{89\omega_1}{6250}+\frac{3\omega_2}{250}-\frac{41}{3125}+\lambda_3=0\\ \frac{42\omega_2}{3125}+\frac{3\omega_1}{250}-\frac{36}{3125}-\lambda_2+\lambda_3=0\\ \lambda_2\omega_2=\lambda_3(\omega_1+\omega_2-1)=0 \end{matrix}\right.\]

If \(\omega_2=0,\) the system becomes \[\left\{\begin{matrix} \frac{89\omega_1}{6250}-\frac{41}{3125}+\lambda_3=0\\ \frac{3\omega_1}{250}-\frac{36}{3125}-\lambda_2+\lambda_3=0\\ \lambda_3(\omega_1-1)=0 \end{matrix}\right.\] If \(\omega_1=1,\) then \[0=\frac{89}{6250}-\frac{41}{3125}+\lambda_3\geq\frac{7}{6250}>0,\] a contradiction. Hence \(\omega_1\neq1\Rightarrow\lambda_3=0\) and \[\omega_1=\frac{82}{89},\lambda_2\approx-4.638\cdot10^{-4}<0,\] a contradiction. Hence \(\omega_2\neq0\Rightarrow\lambda_2=0\) and the system becomes \[\left\{\begin{matrix} \frac{89\omega_1}{6250}+\frac{3\omega_2}{250}-\frac{41}{3125}+\lambda_3=0\\ \frac{42\omega_2}{3125}+\frac{3\omega_1}{250}-\frac{36}{3125}+\lambda_3=0\\ \lambda_3(\omega_1+\omega_2-1)=0 \end{matrix}\right.\] If \(\lambda_3\neq0,\) the system becomes \[\left\{\begin{matrix} \frac{89\omega_1}{6250}+\frac{3\omega_2}{250}-\frac{41}{3125}+\lambda_3=0\\ \frac{42\omega_2}{3125}+\frac{3\omega_1}{250}-\frac{36}{3125}+\lambda_3=0\\ \omega_1+\omega_2-1=0 \end{matrix}\right.\Rightarrow\left\{\begin{matrix}\omega_1=19/23\\\omega_2=4/23\\\lambda_3=-21/28570<0\end{matrix}\right.,\] a contradiction. Hence \(\lambda_3=0\) and the system becomes \[\left\{\begin{matrix} \frac{89\omega_1}{6250}+\frac{3\omega_2}{250}-\frac{41}{3125}=0\\ \frac{42\omega_2}{3125}+\frac{3\omega_1}{250}-\frac{36}{3125}=0\\ \end{matrix}\right.\Rightarrow\left\{\begin{matrix}\omega_1=496/617\\\omega_2=86/617\end{matrix}\right..\]

In summary, we obtain two solutions \[(\omega_1,\omega_2,1-\omega_1-\omega_2)=\left(0,\frac{6}{7},\frac{1}{7}\right),\left(\frac{496}{617},\frac{86}{617},\frac{35}{617}\right)\] with the variances \(0.00146,3.236\cdot10^{-4}.\) The optimal portfolio is \(\left(\frac{496}{617},\frac{86}{617},\frac{35}{617}\right).\)

2.2 Question 2

2.2.1 Part 1 (i)

We set up a portfolio \(V\) as follows:

  1. Today \((t=0):\)
  • Borrow \(\$100\) to buy the share.
  • Enter the contract to sell the share (at maturity) for \(\$K.\)
  • Additional fees: \(\$0.\)
  • Contract value: \(V_0=\$100-\$100+\$0=\$0.\)
  1. At maturity \((t=2):\)
  • Repay the loan with \(\$100\cdot(1+5\%/2)^2=\$105.0625.\)
  • Hand over one share and receive the delivery price \(\$K.\)
  • Additional fees: \(\$0.\)
  • Contract value: \(V_2=-\$105.0625+\$K+\$0=\$K-\$105.0625.\)
  1. Under the assumption of no arbitrage, \(V_2=\$0\Rightarrow K=\$105.0625.\)
  2. The delivery price is, therefore, \(\$105.0625.\)

2.2.2 Part 1 (ii)

We set up a portfolio \(V\) as follows:

  1. Today \((t=0):\)
  • Borrow \(\$3.9024\) for \(6\) months and \(\$96.0976\) for \(1\) year to buy the share.
  • Enter the contract to sell the share (at maturity) for \(\$K.\)
  • Additional fees: \(\$0.\)
  • Contract value: \(V_0=\$100-\$100+\$0=\$0.\)
  1. At \(t=1:\) use the dividend \(\$4\) to pay the first loan.
  2. At maturity \((t=2):\)
  • Repay the second loan with \(\$96.0976\cdot(1+5\%/2)^2=\$100.9625.\)
  • Receive the dividend \(\$4.\)
  • Hand over one share and receive the delivery price \(\$K.\)
  • Additional fees: \(\$0.\)
  • Contract value: \(V_2=-\$100.9625+\$4+\$K+\$0=\$K-\$96.9625.\)
  1. Under the assumption of no arbitrage, \(V_2=\$0\Rightarrow K=\$96.9625.\)
  2. The delivery price is, therefore, \(\$96.9625.\)

2.2.3 Part 2 (i)

We set up a portfolio \(V\) as follows:

  1. Today \((t=0):\)
  • Borrow \(\$100\) to buy the share.
  • Enter the contract to sell the share (at maturity) for \(\$K.\)
  • Additional fees: \(\$0.\)
  • Contract value: \(V_0=\$100-\$100+\$0=\$0.\)
  1. At maturity \((t=2):\)
  • Repay the loan with \(\$100\cdot e^{5\%}\approx\$105.1271.\)
  • Hand over one share and receive the delivery price \(\$K.\)
  • Additional fees: \(\$0.\)
  • Contract value: \(V_2=-\$105.1271+\$K+\$0=\$K-\$105.1271.\)
  1. Under the assumption of no arbitrage, \(V_2=\$0\Rightarrow K=\$105.1271.\)
  2. The delivery price is, therefore, \(\$105.1271.\)

2.2.4 Part 2 (ii)

We set up a portfolio \(V\) as follows:

  1. Today \((t=0):\)
  • Borrow \(\$100\cdot e^{-4\%}\approx\$96.0789\) to buy \(e^{-4\%}\) share.
  • Enter the contract to sell \(1\) share (at maturity) for \(\$K.\)
  • Additional fees: \(\$0.\)
  • Contract value: \(V_0=\$96.0789-\$96.0789+\$0=\$0.\)
  1. At maturity \((t=2):\)
  • Repay the loan with \(\$96.0789\cdot e^{5\%}\approx\$101.005.\)
  • Hand over one share and receive the delivery price \(\$K.\)
  • Additional fees: \(\$0.\)
  • Contract value: \(V_2=-\$101.005+\$K+\$0=\$K-\$101.005.\)
  1. Under the assumption of no arbitrage, \(V_2=\$0\Rightarrow K=\$101.005.\)
  2. The delivery price is, therefore, \(\$101.005.\)

2.3 Question 3

  1. The put\(-\)call parity implies \[C_t+50\cdot e^{-8\%\cdot1}=5+52\Rightarrow C_t\approx10.844.\]
  2. Consider two cases:
  1. If the stock price at maturity \(S_1>50,\) then the buyer exercises the option and gain \((S_1-50)-10.844>-10.844.\)
  2. If \(S_t\leq50,\) then the buyer does not exercise the option and lose $\(10.844.\) In both cases, the maximum loss for the call buyer is $\(10.844.\)
  1. We sell 1 option \(C',\) buy 1 option \(C\) and earns $\(1.156.\) Consider two cases:
  1. If \(C'\) is exercised, then we exercise \(C\) to buy the stock \(S\) with price $\(50,\) hand the stock over and receive $\(49.\) The remaining profit is $\(0.156.\)
  2. If \(C'\) is not exercised, then we do nothing. The profit stays at $\(1.156.\) In both cases, we are ensured a profit of at least $\(0.156.\) Notice that \(100/0.156\approx641.0256,\) a $\(100\) profit can be guaranteed by selling \(642\) options \(C'\) and buying \(642\) options \(C.\)

2.4 Question 4

time = 1
delta_t = time / 3
sigma = 0.2
u = exp(sigma * sqrt(delta_t))
d = exp(-sigma * sqrt(delta_t))
price = 90
st = stock_tree(price, u, d, 4)
st
##          [,1]      [,2]     [,3]     [,4]
## [1,] 90.00000   0.00000   0.0000   0.0000
## [2,] 80.18525 101.01608   0.0000   0.0000
## [3,] 71.44083  90.00000 113.3805   0.0000
## [4,] 63.65001  80.18525 101.0161 127.2584
k = 100
r = 0.05
p = (exp(r * delta_t) - d) / (u - d)
European_put(st, p, r, delta_t, k)
##           [,1]      [,2] [,3] [,4]
## [1,]  9.683252  0.000000    0    0
## [2,] 16.826956  3.989029    0    0
## [3,] 26.906315  8.890534    0    0
## [4,] 36.349988 19.814747    0    0
a = American_put(st, p, r, delta_t, k)
a
##          [,1]      [,2] [,3] [,4]
## [1,] 11.29004  0.000000    0    0
## [2,] 19.81475  4.486827    0    0
## [3,] 28.55917 10.000000    0    0
## [4,] 36.34999 19.814747    0    0
bool = (k - st == a)
for (i in 1:nrow(bool)) {
  for (j in i:nrow(bool)) {
    bool[i, j] = FALSE
  }
}
bool
##       [,1]  [,2]  [,3]  [,4]
## [1,] FALSE FALSE FALSE FALSE
## [2,]  TRUE FALSE FALSE FALSE
## [3,]  TRUE  TRUE FALSE FALSE
## [4,]  TRUE  TRUE FALSE FALSE