Final (Practice 2)

The risk-neutral probabilities are \( p^*=\frac{1+r-d}{u-d}=\frac{4}{5} \) and \( q^*=\frac{u-(1+r)}{u-d}=\frac{1}{5} \). The payoffs of the call option at time 1 are \( C_{u}=Su-K=120 \) and \( C_d=0 \) depending on whether the price of the stock goes up or down.

Therefore the price of the option is: \[ C_0=\frac1{1+r}\left(C_{u}\cdot p^*+C_{d}\cdot q^*\right)=80.\]

The expected return of the security is \( \mu=\frac{-2\cdot 6+8\cdot 4}{100}=20\% \). The risk of the security is \( \sigma=\sqrt{\mathbb E[K^2]-\mu^2}=\sqrt{\frac{2400}{10000}}= 20\sqrt 6\% \).

We will use the formula for the weights of the market portfolio: \[ w_{M}=\frac{(m-Ru)C^{-1}}{(m-Ru)C^{-1}u^T}.\] Using the theorem for the inverse of block matrices whose off-diagonal blocks are \( 0 \) we obtain: \[ C^{-1}=\left[\begin{array}{ccc}2 & -3&0 \newline -3&5&0 \newline 0&0&1\end{array}\right].\] We now have \[ (m-Ru)C^{-1}u^T=\frac1{10}\left[\begin{array}{ccc} 0&2&3 \end{array}\right]\cdot \left[\begin{array}{ccc}2 & -3&0\newline -3&5&0\newline 0&0&1\end{array}\right] \cdot \left[\begin{array}{c} 1\newline 1\newline 1\end{array}\right]=\frac{7}{10}.\] We therefore get \[ w_{M}=\frac{1}{7}\left[\begin{array}{ccc} -6&10&3\end{array}\right].\]

The inequality \( S(9)< 17S(4) \) is equivalent to \( S_0e^{7B_{9}}< S_0\cdot17e^{7B_{4}} \) which after simplification becomes \[ B_{9}-B_{4} < \frac{\ln 17}{7}.\] The random variable \( B_{9}-B_{4} \) has normal distribution with parameters \( 0 \) and \( 5 \). Therefore there exists \( Z\sim N(0,1) \) such that \( B_{9}-B_{4}=\sqrt{5}Z \). We now have \[ \mathbb P\left(S\left( 9\right)< 17 S\left(4\right)\right)= \mathbb P\left(\sqrt{5} Z< \frac{\ln 17}{7}\right)=\mathbb P\left(Z< \frac{\ln 17}{7\sqrt{5}}\right)= \Phi\left(\frac{\ln 17}{7\sqrt{5}} \right).\]

Notice that this derived security can be realized as a portfolio that consists of purchasing one European call with strike \(K_1\) and writing one European call with strike \(K_2\). Therefore, the price of this derived security is \(C_E(K_1)-C_E(K_2)\). According to the Black-Scholes formula, this price is equal to \begin{eqnarray*}C(0)&=&S_0\Phi\left(d_{K_1,+}\right)-K_1e^{-rT}\Phi\left(d_{K_1,-}\right)-S_0\Phi\left(d_{K_2,+}\right) +K_2e^{-rT}\Phi\left(d_{K_2,-}\right),\end{eqnarray*} \begin{eqnarray*} &&\mbox{where }\quad d_{K_i,\pm}=\frac{\ln\frac{S_0}{K_i}+\left(r\pm\frac{\sigma^2}2\right)T}{\sigma\sqrt T}\mbox{ for }i\in\{1,2\}. \end{eqnarray*}

Notice that the function \( F \) can be expressed in the following form: \[ F(w)=wCw^T+12ww^T=w\left(C+12I\right)w^T.\] Therefore, \( F(w) \) represents the variance of the portfolio of securities whose covariance matrix is \( C+12I \). We know that the minimal variance portfolio has the weights: \[ w=\frac{u\left(C+12I\right)^{-1}}{u\left(C+12I\right)^{-1}u^T}, \mbox{ where } u=\left[\begin{array}{cccc} 1&1&\cdots &1\end{array}\right].\]

The issuer of the bond with maturity \(3\) receives \(B(0,3)\) at time \(0\). Then the issuer is required to return \(1\) dollar at time \(3\). The issuer can invest the received amount of \(B(0,3)\) in bonds with maturity \(2\) and bonds with maturity \(1\).

At time 1 the bonds with maturity \(1\) bought at time \(0\) become worth USD 1 each. The bonds with maturity \(2\) are either worth \(B^u(1,2)\) in the \(u\) scenario or \(B^d(1,2)\) in the \(d\) scenario. The issuer of the bond can now protect its position if he/she is able to purchase one bond of maturity \(3\). Then at time \(3\), the issuer will receive \(1\) dollar from this bond and thus will be able to give that dollar back to the purchaser of the bond.

This means that we can regard the bond with maturity \(3\) as a derived security in the one-period binomial model with the following parameters:

• The underlying security has initial price \(S_0=B(0,2)\);

• The price of the underlying security is multiplied either by factor \(u=\frac{B^u(1,2)}{B(0,2)}\) or by factor \(d=\frac{B^d(1,2)}{B(0,2)}\);

• The payoff on the derived security is \(B^u(1,3)\) if the underlying security reached the price \(S_0\cdot u\), or \(B^d(1,3)\) if the underlying security reached the price \(S_0\cdot d\);

• The risk-free interest rate is \(r=\frac1{B(0,1)}-1\).

We first calculate the risk-neutral probability \[p^*=\frac{1+r-d}{u-d}=\frac{\frac1{B(0,1)}-\frac{B^d(1,2)}{B(0,2)}}{ \frac{B^u(1,2)}{B(0,2)}-\frac{B^d(1,2)}{B(0,2)} } =\frac{\frac{B(0,2)}{B(0,1)}-B^d(1,2)}{B^u(1,2)-B^d(1,2)} \] and then we may use the standard formula for the price of the derived security: \begin{eqnarray*}B(0,3)&=&\frac1{1+r}\mathbb E_*\left[\mbox{Payoff}\right]=B(0,1)\left[B^u(1,3)\cdot p^*+B^d(1,3)\cdot \left(1-p^*\right)\right] \newline &=&\frac{\left(B^u(1,3)-B^d(1,3)\right)B(0,2)+B(0,1)\left(B^d(1,3)B^u(1,2)-B^u(1,3)B^d(1,2)\right) }{B^u(1,2)-B^d(1,2)} . \end{eqnarray*}

The payoff is equal to \(1\) if and only if \begin{eqnarray*} && S(T)^2+4K^2\leq 5KS(T)\\ \Longleftrightarrow && S(T)^2-5KS(T)+4K^2\leq 0\\ \Longleftrightarrow && (S(T)-4K)\cdot (S(T)-K)\leq 0\\ \Longleftrightarrow && K\leq S(T)\leq 4K. \end{eqnarray*} The price \(P_0\) of derived security at time \(0\) is \begin{eqnarray*} P_0&=&e^{-rT}\mathbb E_*\left[ 1_{S(T)\in [K, 4K]}\right]= e^{-rT}\mathbb P_*\left( S(T)\in [K,4K]\right)\\ &=&e^{-rT}\mathbb P_*\left(Se^{\left(r-\frac{\sigma^2}2\right)T+\sigma \sqrt TZ_*}\in [K,4K]\right)\\ &=&e^{-rT}\mathbb P_*\left(e^{\left(r-\frac{\sigma^2}2\right)T+\sigma \sqrt TZ_*}\in \left[\frac{K}{S},\frac{4K}{S}\right]\right)\\ &=&e^{-rT}\mathbb P_*\left( \left(r-\frac{\sigma^2}2\right)T+\sigma \sqrt TZ_* \in \left[\ln\frac{K}{S}, \ln\frac{4K}{S}\right]\right)\\ &=&e^{-rT}\mathbb P_*\left( \sigma \sqrt TZ_* \in \left[\ln\frac{K}{S}-\left(r-\frac{\sigma^2}2\right)T,\right.\right. \\ && \quad\quad\quad\left.\left.\ln\frac{4K}{S}-\left(r-\frac{\sigma^2}2\right)T\right]\right)\\ &=&e^{-rT}\mathbb P_*\left( Z_* \in \left[\frac{\ln\frac{K}{S}-\left(r-\frac{\sigma^2}2\right)T}{\sigma \sqrt T},\right.\right. \\ && \quad\quad\quad\left.\left.\frac{\ln\frac{4K}{S}-\left(r-\frac{\sigma^2}2\right)T}{\sigma \sqrt T}\right]\right)\\ &=&e^{-rT}\left(\Phi\left(\frac{\ln\frac{4K}{S}-\left(r-\frac{\sigma^2}2\right)T}{\sigma \sqrt T}\right) \right.\\ &&\quad\quad\quad\left. - \Phi\left(\frac{\ln\frac{K}{S}-\left(r-\frac{\sigma^2}2\right)T}{\sigma \sqrt T}\right)\right). \end{eqnarray*}