(My question)
Please show me how to solve from (2) to (4) with computation processes. These are too difficult to solve.
Thank you for your help in advance.
(Cross-link)
I have posted the same question on https://quant.stackexchange.com/questions/47567/bond-option-hedging
(Original questions)
Exercise 7.4 Bond Option Hedging
Consider a portfolio $(\xi^T_t, \xi^S_t)_{t \in [0, T]}$ made of two bonds with maturities $T$, S, and value \begin{eqnarray} V_t=\xi^T_t P(t, T) + \xi^S_t P(t, S) \end{eqnarray} at time $t$, and assume that it hedges the bond call option payoff $( P(T, S) - \kappa )^+$, so that we have \begin{eqnarray} V_t &=& E \left[ \exp \left( - \int^T_t r_s ds \right) \cdot ( P(T, S) - \kappa )^+ \middle| \mathcal{F}_t \right] \\ &=& P(t, T) E^{ \tilde{\mathbb{P}} } \left[ ( P(T, S) - \kappa )^+ \middle| \mathcal{F}_t \right] \end{eqnarray}
(1) Assume that $( \sigma^T_t)_{t \in [0, T]}$ and $( \sigma^S_t)_{t \in [0, S]}$ are deterministic functions, show that the price of a bond option with strike $\kappa$ can be written as \begin{eqnarray} && E \left[ \exp \left( - \int^T_t r_s ds \right) \cdot ( P(T, S) - \kappa )^+ \middle| \mathcal{F}_t \right] \nonumber \\ && \qquad \qquad = P(t, T) E^{ \tilde{\mathbb{P}} } \left[ ( P(T, S) - \kappa )^+ \middle| \mathcal{F}_t \right] \\ && \qquad \qquad = P(t, T) C(X_t, \kappa, v(t, T) ) \\ && \qquad \qquad = P(t, T) C(X_t, \kappa, \sigma) \end{eqnarray} where $X_t$ is the forward price $X_t \equiv P(t, S)/P(t, T)$, \begin{eqnarray} v^2(t, T) = \int^T_t \left( \sigma^S_u - \sigma^T_u \right)^2 du \end{eqnarray} and $C(X_t, \kappa, \sigma)$ is a function to be determined. Recall that if $X$ is a centered Gaussian random variable with mean $m_t$ and variance $v^2_t$ given $\mathcal{F}_t$, we have \begin{eqnarray} E \left[ \left( e^X - K \right)^+ | \mathcal{F}_t \right] &=& e^{ m_t + v^2_t /2 } N \left( \frac{v_t}{2} + \frac{1}{v_t} \left( m_t + \frac{v^2_t}{2} - \log K \right) \right) \nonumber \\ && \qquad - K N \left( - \frac{v_t}{2} + \frac{1}{v_t} \left( m_t + \frac{v^2_t}{2} - \log K \right) \right) \end{eqnarray} where $N(x)$, $x \in \mathbb{R}$, denotes the Gaussian distribution function, cf. Lemma 2.3.
(2) We assume that the portfolio $(\xi^T_t, \xi^S_t)_{t \in [0, T]}$ is self-financing, i.e. \begin{eqnarray} dV_t=\xi^T_t dP(t, T) + \xi^S_t dP(t, S) \end{eqnarray} Show that the forward portfolio price $\hat{V_t} \equiv V_t/P(t, T)$ satisfies
\begin{eqnarray} d\hat{V_t}=\frac{ \partial C(X_t, \kappa, v(t, T) ) }{ \partial x } d X_t. \end{eqnarray}
(3) Show that we have \begin{eqnarray} dV_t &=& \left( \hat{V_t} - \frac{ P(t, S) }{ P(t, T) } \frac{ \partial C( X_t, \kappa, v(t, T) ) }{ \partial x } \right) dP(t, T) \nonumber \\ && + \frac{ \partial C(X_t, \kappa, v(t, T) ) }{ \partial x } dP(t, S) \end{eqnarray}
(4) Compute the hedging portfolio strategy $(\xi^T_t, \xi^S_t)_{t \in [0, T]}$ of the bond call option on $P(T, S)$.
(1) My answer
This dynamics of $dP(t, T)$ uses $\sigma^T_t$ as its volatility instead of $\zeta^T_t$ on the text page 89. Namely, the dynamics of $dP(t, T)$ is a same type of Exercise 7.3. Therefore, recall the result of Exercise 7.3.(1). On the other words, recall the results of Exercise 4.3.(5). Besides, $d B^T_t = d B_t - \sigma^T_t dt $. Or, recall Exercise 7.1.(4) and Exercise 7.1.(7). Exercise 7.1 uses $\zeta_t$ instead of $\zeta^T_t$ as the volatility on its dynamics of $dP(t, T)$. \begin{eqnarray} \frac{ dP(t, T)}{P(t, T)} &=& r_t dt + \sigma^T_t dB_t \\ \frac{P(T, S)}{P(T, T)}&=&\frac{P(t, S)}{P(t, T)} \exp \left( \int^T_t \left( \sigma^S_u - \sigma^T_u \right) d B^T_u \right) \nonumber \\ && \qquad \qquad \cdot \exp \left( - \frac{1}{2} \int^T_t \left( \sigma^S_u -\sigma^T_u \right)^2 du \right) \\ P(T, S)&=&\frac{P(t, S)}{P(t, T)} \exp \left( \int^T_t \left( \sigma^S_u - \sigma^T_u \right) d B^T_u \right) \nonumber \\ && \qquad \qquad \cdot \exp \left( - \frac{1}{2} \int^T_t \left( \sigma^S_u -\sigma^T_u \right)^2 du \right) \end{eqnarray}
Let $m(t, T)$ and $v^2(t, T)$ as below. \begin{eqnarray} m(t, T) &=& \log \frac{P(t, S)}{P(t, T)} - \frac{1}{2} \int^T_t \left( \sigma^S_u -\sigma^T_u \right)^2 du \\ v^2(t, T) &=& \left( \int^T_t \left( \sigma^S_u - \sigma^T_u \right) d B^T_u \right)^2 \\ &=& \int^T_t \left( \sigma^S_u - \sigma^T_u \right)^2 du \\ m(t, T) + \frac{ v^2(t, T) }{2} &=& \log \frac{P(t, S)}{P(t, T)} \end{eqnarray}
Substitute the above result into the expectation value. \begin{eqnarray} && E \left[ \exp \left( - \int^T_t r_s ds \right) \cdot ( P(T, S) - \kappa )^+ \middle| \mathcal{F}_t \right] \nonumber \\ && \qquad \qquad = E^{ \tilde{\mathbb{P}} } \left[ \frac{ P(t, T) }{ P(t, T) } \exp \left( - \int^T_t r_s ds \right) \cdot ( P(T, S) - \kappa )^+ \middle| \mathcal{F}_t \right] \\ && \qquad \qquad = P(t, T) E^{ \tilde{\mathbb{P}} } \left[ \frac{ 1 }{ P(T, T) } ( P(T, S) - \kappa )^+ \middle| \mathcal{F}_t \right] \\ && \qquad \qquad = P(t, T) E^{ \tilde{\mathbb{P}} } \left[ ( P(T, S) - \kappa )^+ \middle| \mathcal{F}_t \right] \end{eqnarray}
Recall the result of Exercise 7.1.(7). Here, let $m(t, T) =m$, $v(t, T)=v$, and $\kappa=K$. \begin{eqnarray} && E^{\mathbb{P}} \left[ \exp \left(- \int^T_t r_s ds \right) \cdot ( P(T,S) - K )^+ \middle| \mathcal{F}_t \right] \nonumber \\ && \quad = P(t, T) e^{m+ v^2/2} N\left( v + \frac{m - \log K}{v} \right) -P(t, T) K N\left( \frac{m - \log K}{v} \right) \\ && \quad = P(t, T) \frac{P(t,S) }{P(t,T)} N\left( v - \frac{v}{2} + \frac{1}{v} \log \frac{P(t,S) }{K \ P(t,T)} \right) \nonumber \\ && \qquad -P(t, T) K N\left( - \frac{v}{2} + \frac{1}{v} \log \frac{P(t,S) }{K \ P(t,T)}\right) \\ && \quad = P(t,S) N\left( \frac{v}{2} + \frac{1}{v} \log \frac{P(t,S) }{K \ P(t,T)} \right) \nonumber \\ && \qquad -P(t, T) K N\left( - \frac{v}{2} + \frac{1}{v} \log \frac{P(t,S) }{K \ P(t,T)}\right) \\ && \quad = P(t, T) \frac{ P(t,S) }{ P(t, T) } N\left( \frac{v}{2} + \frac{1}{v} \log \frac{P(t,S) }{K \ P(t,T)} \right) \nonumber \\ && \qquad -P(t, T) K N\left( - \frac{v}{2} + \frac{1}{v} \log \frac{P(t,S) }{K \ P(t,T)}\right) \\ && \quad = P(t, T) C\left( \frac{ P(t,S) }{ P(t, T) } , K, v \right) \\ && \quad = P(t, T) C(X_t, \kappa, v(t, T) ) \\ && \quad = P(t, T) C(X_t, \kappa, \sigma) \end{eqnarray}
$\square$
(2) ??? This is too difficult to solve!
