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(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://math.stackexchange.com/questions/3347148/bond-option-hedging/3349320#3349320


(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.

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I solved from (2) to (4) 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}

$\square$


(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}

$\square$


(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}

$\square$

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