Midterm 2 (Practice 1)

A forward contract is an agreement between two parties \(A\) and \(B\) by which the party \(A\) has to sell the specified security to the party \(B\) at time \(T\) for a price that is determined at the time of signing the contract.

If the price of the underlying security at time \(0\) is \(S(0)\) and the return on the risk-free asset is \(R\), the forward price is given by \(F(0,T)=S(0)\cdot (1+R)\).

The stock price tree is \(\begin{array}{ccc}&&121\newline &110&\newline 100&&99\newline &90&\newline &&81\end{array}\), the call price tree is \(\begin{array}{ccc} &&21\newline &10.5&\newline 5.25&&0\newline &0&\newline &&0 \end{array}\), and the hedge ratio tree is \(\begin{array}{ccc}&0.9545454545\newline 0.525&\newline &0 \end{array}\).

The stock price tree is given below: \[\begin{array}{cccc}&&&109.85\newline &&84.5&\newline &65&&76.05\newline 50&&58.5&\newline &45&&52.65\newline &&40.5&\newline &&&36.45 \end{array}\]

Assume that \(d\geq 1+R\). Then there is a possibility of an arbitrage. One can borrow \(S(0)\) and by one stock. At time \(1\), one owes \(S(0)(1+R)\), but the stock is worth at least \(S(0)d\geq S(0)(1+R)\). With positive probability the stock is worth \(S(0)u > S(0)(1+R)\) which would mean a risk-less profit.

On the other hand, if we assume that \(1+R\geq u\), then the arbitrage can be realized by short-selling one share of stock and investing the money \(S(0)\). At time \(1\), the investment amounts to \(S(0)(1+R)\), while the closure of the short-sale can cost at most \(S(0)u\). Since \(S(0)u\leq S(0)(1+R)\) and \(S(0)d < S(0)(1+R)\), we conclude that there is an arbitrage opportunity.

• (a) Using the put-call parity we obtain \[C_E(120)-P_E(120)=S(0)-X\cdot\frac1{1+R}=100-\frac{120}{1.1} < 0.\] Therefore \(C_E(120) < p_E(120)\) and the put option is more expensive.

• (b) Using the put-call parity we obtain: \[C_E(120)-P_E(120)=100-\frac{120}{1.1^2}=100-\frac{120}{1.21} > 0.\] Therefore \(C_E(120) > P_E(120)\). Since the put option is an increasing function of the strike price we obtain that \(P_E(120) > P_E(110)\). Thus \(C_E(120) > P_E(110)\).

The risk neutral probability is \(q=\frac{1+R-d}{u-d}= \frac12\). The price of the option is given by \[C(0)=P_*\left(\Gamma\right),\] where \(\Gamma\) is the event that for every \(k\), the inequality \(S(k)\leq B_k\) holds. Let \(Z_i\) be the factor by which the price is multiplied at the period \(i\). Then we have \(S(i)=S(0)\cdot Z_1\cdot Z_2\cdots Z_i\), and if we denote \(X_i=\ln { Z_i}\), then we have \(S(i)=S(0)e^{X_1+\cdots+X_i}\) and \(X_i\in\{\ln u,\ln d\}\). Let \(W_i=X_1+\cdots+X_i\). The inequality \(S(k) < B_k\) can be rewritten as \[W_k\leq k\cdot \frac{\ln u+\ln d}2.\] Notice that \(E_*(X_1)=\frac{\ln u+\ln d}2\). Let us denote \(\mu=E_*(X_1)\). Therefore \begin{eqnarray*}C(0)&=&P_*\left(\Gamma\right)=P_*\left(X_1+X_2+\cdots+X_k\leq \frac k2\left(\ln u+\ln d\right)\mbox{ for every }k\right) \newline &=&P_*\left((X_1-\mu)+(X_2-\mu)+\cdots+(X_k-\mu)\leq 0\mbox{ for every }k\right). \end{eqnarray*} Let \(Y_i=\frac{X_i-\mu}{\frac{\ln u-\ln d}2}\). Then \(E_*(Y_i)=0\) and \(Y_i\in\{-1,1\}\). We have \begin{eqnarray*} C(0)&=&P_*\left(Y_1+\cdots+Y_k\leq 0\mbox{ for every }k\right) = P_*\left(Y_1+\cdots+Y_k< 1\mbox{ for every }k\right)=1-P_*\left(Y_1+\cdots+Y_k\geq 1\mbox{ for some }k\right). \end{eqnarray*} We have that \begin{eqnarray*}P_*\left(Y_1+\cdots+Y_k\geq 1\mbox{ for some }k\right)&=&\sum_{j={-98}}^{100}P_*\left(Y_1+\cdots+Y_{100}=j,\;\;Y_1+\cdots+Y_k\geq 1\mbox{ for some }k\right)\newline &=& \sum_{j=1}^{100}P_*\left(Y_1+\cdots+Y_{100}=j,\;\;Y_1+\cdots+Y_k\geq 1\mbox{ for some }k\right)\newline && +\sum_{j={-98}}^0P_*\left(Y_1+\cdots+Y_{100}=j,\;\;Y_1+\cdots+Y_k\geq 1\mbox{ for some }k\right)\newline &=&\sum_{j=1}^{100}P_*\left(Y_1+\cdots+Y_{100}=j\right)+\sum_{j={-98}}^0P_*\left(Y_1+\cdots+Y_{100}=j,\;\;Y_1+\cdots+Y_k\geq 1\mbox{ for some }k\right)\newline \end{eqnarray*} According to reflection principle we have that when \(j\in\{-98,-97,\dots, 0\}\) the following holds: \[P_*\left(Y_1+\cdots+Y_{100}=j,\;\;Y_1+\cdots+Y_k\geq 1\mbox{ for some }k\right)=P_*(Y_1+\cdots+Y_{100}=2-j).\] Let us denote \(M_n=Y_1+\cdots+Y_{n}\). Then we have \begin{eqnarray*}P_*\left(Y_1+\cdots+Y_k\geq 1\mbox{ for some }k\right)&=&\sum_{j=1}^{100}P_*\left(Y_1+\cdots+Y_{100}=j\right)+\sum_{j=-98}^0P_*(Y_1+\cdots+Y_{100}=2-j)\newline &=&P_*\left(M_{100}\geq 1\right)+\sum_{i=0}^{98} P_*\left(M_{100}=i+2\right)\newline &=&P_*\left(M_{100}\geq 1\right)+P_*\left(M_{100}\geq 2\right)=2 P_*\left(M_{100}\geq 2\right),\end{eqnarray*} because \(P_*\left(M_{100}=1\right)=0\) since the walk cannot end at level \(1\).