Bond Option Hedging (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!
Thank you for your help in advance.
 A: I solved (2) by myself!
(2) My answer

*

*Use the result of (1) with keeping in mind that the following R.H.S is $\mathcal{F}_t $ measurable.
\begin{eqnarray}
V_t &=& E^{\mathbb{P}} \left[ \exp \left(- \int^T_t r_s ds \right) \cdot ( P(T,S) - K )^+  \middle| \mathcal{F}_t \right] \\
&=& P(t, T) E^{ \tilde{\mathbb{P}} } \left[  ( P(T, S) - \kappa )^+ \middle| \mathcal{F}_t \right]    \\
&=&  P(t, T) C(X_t, \kappa, v(t, T) ) 
\end{eqnarray}


*Therefore,
\begin{eqnarray}
\hat{ V_t }&=& \frac{ V_t }{P(t, T)} \\
&=&  E^{ \tilde{\mathbb{P}} } \left[  ( P(T, S) - \kappa )^+ \middle| \mathcal{F}_t \right]    \\
&=& C(X_t, \kappa, v(t, T) ) 
\end{eqnarray}


*First, use It$\hat{o}$'s formula on the L.H.S.
\begin{eqnarray}
d \hat{ V_t }&=& 0 \ dt + d \hat{ V_t } + \frac{1}{2} \ 0 \ d [ \hat{ V_t }] \\
&=& d \hat{ V_t } 
\end{eqnarray}


*Second, recall the result of (1), $C(X_t, \kappa, v(t, T) ) $.
\begin{eqnarray}
C(X_t, \kappa, v(t, T) ) &=& \frac{ P(t,S) }{ P(t, T) }  N\left(   \frac{v}{2}  +  \frac{1}{v} \log \frac{P(t,S) }{\kappa \ P(t,T)} \right)  \nonumber \\
&& \qquad -  \kappa N\left( - \frac{v}{2}  +  \frac{1}{v} \log \frac{P(t,S) }{\kappa\ P(t,T)}\right)  \\
&=& X_t  N\left(   \frac{v}{2}  +  \frac{1}{v} \log \frac{X_t}{\kappa } \right)  \nonumber \\
&& \qquad - \kappa N\left( - \frac{v}{2}  +  \frac{1}{v} \log \frac{X_t }{\kappa }\right)  
\end{eqnarray}


*Third, use It$\hat{o}$'s formula on the R.H.S.
\begin{eqnarray}
d C(X_s, \kappa, v(s, T) )  &=& 0 \ ds +  \partial_x C(X_s, \kappa, v(s, T) )  dX_s + \frac{1}{2} \ 0 \ d [ X_s ] \\
\int^t_0 d C(X_s, \kappa, v(s, T) )  &=& \int^t_0 \partial_x C(X_s, \kappa, v(s, T) )  dX_s \\
C(X_t, \kappa, v(t, T) ) &=& C(X_0, \kappa, v(0, T) )  \nonumber \\
&& \qquad+ \int^t_0 \partial_x C(X_s, \kappa, v(s, T) )  dX_s \\
&=&  E^{ \tilde{\mathbb{P}} } \left[  ( P(T, S) - \kappa )^+ \middle| \mathcal{F}_t \right]   \nonumber \\
&& \qquad + \int^t_0 \partial_x C(X_s, \kappa, v(s, T) )  dX_s 
\end{eqnarray}


*Substitute the above result into $\hat{ V_t }$.
\begin{eqnarray}
\hat{ V_t }&=&C(X_t, \kappa, v(t, T) ) \\
&=&  E^{ \tilde{\mathbb{P}} } \left[  ( P(T, S) - \kappa )^+ \middle| \mathcal{F}_t \right]   + \int^t_0 \partial_x C(X_s, \kappa, v(s, T) )  dX_s 
\end{eqnarray}


*Moreover,
\begin{eqnarray}
E^{ \tilde{\mathbb{P}} } \left[  ( P(T, S) - \kappa )^+ \middle| \mathcal{F}_t \right]  &=& E^{ \tilde{\mathbb{P}} } \left[  ( P(T, S) - \kappa )^+ \middle| \mathcal{F}_t \right]   \nonumber \\
&& \qquad + \int^t_0 \partial_x C(X_s, \kappa, v(s, T) )  dX_s \\
\int^t_0 \partial_x C(X_s, \kappa, v(s, T) )  dX_s &=&0
\end{eqnarray}


*Therefore,
\begin{eqnarray}
\hat{ V_t }&=&C(X_t, \kappa, v(t, T) ) \\
&=&  E^{ \tilde{\mathbb{P}} } \left[  ( P(T, S) - \kappa )^+ \middle| \mathcal{F}_t \right]   
\end{eqnarray}


*On the other hand, hence,
\begin{eqnarray}
\hat{ V_t } &=& C(X_t, \kappa, v(t, T) ) \\
d \hat{ V_t } &=& d C(X_t, \kappa, v(t, T) ) \\
&=& \partial_x C(X_t, \kappa, v(t, T) )  dX_t 
\end{eqnarray}
$\square$
A: I also solved (3) by myself !
(3) My answer

*

*Use It$\hat{o}$'s formula.
\begin{eqnarray}
d V_t &=& d (P(t, T) \cdot \hat{ V_t } ) \\
&=& \partial_t (P(t, T) \cdot \hat{ V_t } ) dt  \nonumber \\
&& \quad  +  \hat{ V_t } \partial_x P(t, T) |_{x=P(t,T)} dP(t,T)  + P(t, T) \partial_y \hat{ V_t } |_{y=\hat{ V_t } } d \hat{ V_t } \nonumber \\
&& \quad  +  \frac{1}{2}  \hat{ V_t } \partial_{xx}  P(t, T) |_{x=P(t,T)} d[ P(t,T) ] +   \frac{1}{2} P(t,T) \partial_{yy}  \hat{ V_t } |_{y=\hat{ V_t }} d[ \hat{ V_t } ]  \nonumber \\
&& \quad  +  \frac{1}{2} \partial_{xy} (P(t, T) \cdot \hat{ V_t } ) |_{x=P(t,T), y=\hat{ V_t }} d[ P(t,T) ,  \hat{ V_t } ]  \nonumber \\
&& \quad+  \frac{1}{2} \partial_{yx} (P(t, T) \cdot \hat{ V_t } ) |_{ y=\hat{ V_t }, x=P(t,T)} d[ \hat{ V_t }, P(t,T)  ] \\ 
&=&  \hat{ V_t }dP(t,T)  + P(t, T) d \hat{ V_t } + d[ \hat{ V_t }, P(t,T)  ]
\end{eqnarray}


*Use the result of (2).
\begin{eqnarray}
d V_t &=& \hat{ V_t }dP(t,T)  + P(t, T) d \hat{ V_t } + d[ \hat{ V_t }, P(t,T)  ] \\
&=& \hat{ V_t }dP(t,T)  + P(t, T) \partial_x C(X_t, \kappa, v(t, T) )  dX_t  \nonumber \\
&& \quad   + \partial_x C(X_t, \kappa, v(t, T) ) dP(t, T) dX_t \\
&=& \hat{ V_t }dP(t,T)  -  \partial_x C(X_t, \kappa, v(t, T) )  X_t dP(t, T) \nonumber \\
&& \quad + \partial_x C(X_t, \kappa, v(t, T) )  X_t dP(t, T) \nonumber \\
&& \quad + \partial_x C(X_t, \kappa, v(t, T) )  P(t, T)  dX_t  \nonumber \\
&& \quad + \partial_x C(X_t, \kappa, v(t, T) ) dP(t, T) dX_t 
\end{eqnarray}


*Here, one computes the dynamics of $P(t, S)$ using the definition of $X_t=P(t, S)/P(t, T)$ by It$\hat{o}$'s formula.
\begin{eqnarray}
d P(t, S) &=& d(X_t P(t, T)) \\
&=& \partial_t  (X_t P(t, T)) dt  \nonumber \\
&& \quad  + P(t, T) \partial_x X_t |_{x=X_t} dX_t + X_t \partial_y P(t, T) |_{y=P(t, T)} dP(t, T)  \nonumber \\
&& \quad  + \frac{1}{2} P(t, T) \partial_{xx}  X_t |_{x=X_t} d [X_t ]  \nonumber \\
&& \quad  + \frac{1}{2} X_t  \partial_{yy}  P(t, T) |_{y=P(t, T)} d [P(t, T) ]  \nonumber \\
&& \quad  + \frac{1}{2} \partial_{xy} (X_t P(t, T))  |_{x=X_t, y=P(t,T)} d[ X_t, P(t,T) ] \nonumber \\
&& \quad  + \frac{1}{2} \partial_{yx} (X_t P(t, T))  |_{y=P(t,T), x=X_t} d[P(t,T),  X_t ]  \\
&=&  P(t, T) dX_t + X_t dP(t, T)  + d[ X_t, P(t,T) ]  \\
&=&  P(t, T) dX_t + X_t dP(t, T)  + d X_t  dP(t,T)   \\ 
&=&  P(t, T) dX_t + X_t dP(t, T)  + dP(t,T) d X_t
\end{eqnarray}


*Substitute the above result into $dV_t $.
\begin{eqnarray}
d V_t &=& \hat{ V_t }dP(t,T)  -  \partial_x C(X_t, \kappa, v(t, T) )  X_t dP(t, T) \nonumber \\
&& \quad + \partial_x C(X_t, \kappa, v(t, T) )  X_t dP(t, T) \nonumber \\
&& \quad + \partial_x C(X_t, \kappa, v(t, T) )  P(t, T)  dX_t  \nonumber \\
&& \quad + \partial_x C(X_t, \kappa, v(t, T) ) dP(t, T) dX_t \\
&=& \left( \hat{ V_t } - X_t \partial_x C(X_t, \kappa, v(t, T) ) \right) dP(t, T) \nonumber \\
&& \quad + \partial_x C(X_t, \kappa, v(t, T) ) d P(t, S) \\
&=& \left( \hat{ V_t } - \frac{P(t, S)}{P(t, T)}  \partial_x C(X_t, \kappa, v(t, T) ) \right) dP(t, T) \nonumber \\
&& \quad + \partial_x C(X_t, \kappa, v(t, T) ) d P(t, S) 
\end{eqnarray}
$\square$
A: I also solved (4) by myself !
(4) My answer

*

*(2) assumes 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}
　


*One also has another expression of $dV_t$ by (3).
\begin{eqnarray}
d V_t &=&\left( \hat{ V_t } - \frac{P(t, S)}{P(t, T)}  \partial_x C(X_t, \kappa, v(t, T) ) \right) dP(t, T) \nonumber \\
&& \quad + \partial_x C(X_t, \kappa, v(t, T) ) d P(t, S) 
\end{eqnarray}


*item Comparing above two equations, one reaches the following equations.
\begin{eqnarray}
\xi^S_t  &=&  \partial_x C(X_t, \kappa, v(t, T) ) \\
\xi^T_t &=&  \hat{ V_t } - \frac{P(t, S)}{P(t, T)}  \partial_x C(X_t, \kappa, v(t, T) ) \\
&=&  \hat{ V_t } - X_t  \partial_x C(X_t, \kappa, v(t, T) ) \\
&=&  \hat{ V_t } - X_t \xi^S_t 
\end{eqnarray}


*Recall the result of (1), $C(X_t, \kappa, v(t, T) ) $ and the result of (2), $\hat{ V_t }=C(X_t, \kappa, v(t, T) ) $.
\begin{eqnarray}
C(X_t, \kappa, v(t, T) ) &=& \frac{ P(t,S) }{ P(t, T) }  N\left(   \frac{v}{2}  +  \frac{1}{v} \log \frac{P(t,S) }{\kappa \ P(t,T)} \right)  \nonumber \\
&& \qquad -  \kappa N\left( - \frac{v}{2}  +  \frac{1}{v} \log \frac{P(t,S) }{\kappa\ P(t,T)}\right)  \\
&=& X_t  N\left(   \frac{v}{2}  +  \frac{1}{v} \log \frac{X_t}{\kappa } \right)  \nonumber \\
&& \qquad - \kappa N\left( - \frac{v}{2}  +  \frac{1}{v} \log \frac{X_t }{\kappa }\right)  \\
\hat{ V_t } - X_t N\left(   \frac{v}{2}  +  \frac{1}{v} \log \frac{X_t}{\kappa } \right) &=& - \kappa N\left( - \frac{v}{2}  +  \frac{1}{v} \log \frac{X_t }{\kappa }\right)  
\end{eqnarray}


*Comparing the above equation to $\xi^T_t $, one reaches the following equations.
\begin{eqnarray}
\xi^S_t  &=&  N\left(   \frac{v}{2}  +  \frac{1}{v} \log \frac{X_t}{\kappa } \right) \\
&=&  N\left(   \frac{v(t, T) }{2}  +  \frac{1}{v(t, T) } \log \frac{P(t, S)}{\kappa \ P(t, T)} \right) \\
\xi^T_t &=&  - \kappa N\left( - \frac{v}{2}  +  \frac{1}{v} \log \frac{X_t }{\kappa }\right)  \\
 &=&  - \kappa N\left( - \frac{v(t, T)}{2}  +  \frac{1}{v(t, T)} \log \frac{P(t, S) }{\kappa \ P(t, T) }\right) 
\end{eqnarray}
$\square$
