# Bond Option Hedging

## (My question)

Please show me how to solve from (2) to (4) with computation processes. These are too difficult to solve.

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 $$P(T, S)$$.

• 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!

## I solved from (2) to (4) by myself !

• 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, $$\begin{eqnarray} \hat{ V_t } &=& C(X_t, \kappa, v(t, T) ) \\ d \hat{ V_t } &=& d C(X_t, \kappa, v(t, T) ) \\ &=& \partial_x C(X_t, \kappa, v(t, T) ) dX_t \end{eqnarray}$$

$$\square$$

• 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 $$dV_t$$. $$\begin{eqnarray} d V_t &=& \hat{ V_t }dP(t,T) - \partial_x C(X_t, \kappa, v(t, T) ) X_t dP(t, T) \nonumber \\ && \quad + \partial_x C(X_t, \kappa, v(t, T) ) X_t dP(t, T) \nonumber \\ && \quad + \partial_x C(X_t, \kappa, v(t, T) ) P(t, T) dX_t \nonumber \\ && \quad + \partial_x C(X_t, \kappa, v(t, T) ) dP(t, T) dX_t \\ &=& \left( \hat{ V_t } - X_t \partial_x C(X_t, \kappa, v(t, T) ) \right) dP(t, T) \nonumber \\ && \quad + \partial_x C(X_t, \kappa, v(t, T) ) d P(t, S) \\ &=& \left( \hat{ V_t } - \frac{P(t, S)}{P(t, T)} \partial_x C(X_t, \kappa, v(t, T) ) \right) dP(t, T) \nonumber \\ && \quad + \partial_x C(X_t, \kappa, v(t, T) ) d P(t, S) \end{eqnarray}$$

$$\square$$

• (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. $$\begin{eqnarray} \xi^S_t &=& N\left( \frac{v}{2} + \frac{1}{v} \log \frac{X_t}{\kappa } \right) \\ &=& N\left( \frac{v(t, T) }{2} + \frac{1}{v(t, T) } \log \frac{P(t, S)}{\kappa \ P(t, T)} \right) \\ \xi^T_t &=& - \kappa N\left( - \frac{v}{2} + \frac{1}{v} \log \frac{X_t }{\kappa }\right) \\ &=& - \kappa N\left( - \frac{v(t, T)}{2} + \frac{1}{v(t, T)} \log \frac{P(t, S) }{\kappa \ P(t, T) }\right) \end{eqnarray}$$
$$\square$$