(My Question)
I solved the following questions. However, if you know the other solutions, please let me know those along with computation processes. Besides, $W_t$ is a S.B.M.
(the details in this model)
See the details in this model at the following cross link site.
https://math.stackexchange.com/questions/3332466/the-ho-lee-model-1986
(Cross-link)
I have posted the same question on https://math.stackexchange.com/questions/3339590/the-ho-lee-model-1986s-bond-option-pricing-solved-by-myself
Lemma for (7)
Let $X$ be a centered Gaussian random variable with variance $v^2$. Namely, $X \sim N(v^2, v^2)$. One has \begin{eqnarray} E^{ \{ \mathbb{Q}, \ X \}} \left[ \left( e^{m+X} - K \right)^+ \right] = e^{m+ v^2/2} N\left( v + \frac{m - \log K}{v} \right) - K N\left( \frac{m - \log K}{v} \right) \end{eqnarray}
(Original Questions)
(1) Derive the S.D.E satisfied by $t \rightarrow P(t, T)$.
(2) Derive the S.D.E satisfied by $t \rightarrow exp( - \int^t_0 r_s ds ) \cdot P(t, T)$.
(3) Express the conditional expectation \begin{eqnarray} E^{\mathbb{P}} \left[ \frac{ d \tilde{ \mathbb{P} } }{d \mathbb{P} } \middle| \mathcal{F}_t \right] \end{eqnarray} in terms of $P(t, T)$, $P(0, T)$ and $exp( -\int^t_0 r_s ds )$, $t \in [0, T]$.
(4) Find a S.D.E satisfied by \begin{eqnarray} t \rightarrow E^{\mathbb{P}} \left[ \frac{ d \tilde{ \mathbb{P} } }{d \mathbb{P} } \middle| \mathcal{F}_t \right] \end{eqnarray} (5) Compute the density $d \tilde{ \mathbb{P} } /d \mathbb{P} $ of the forward measure with respect to $\mathbb{P}$ by solving the S.D.E of (4).
(6) Using the Girsanov Theorem 2.1, compute the dynamics of $r_t$ under the forward measure.
(7) Compute the price \begin{eqnarray} E^{\mathbb{P}} \left[ -\exp \left( \int^t_0 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) - K )^+ \middle| \mathcal{F}_t \right] \end{eqnarray} of a bond call option at time $t \geq 0$.
(1) My answer
Recall what one considered in Exercise 5.1.(1) and the definition of $dr_t=\theta dt+ \sigma dW_t$ at in Exercise 4.1. Moreover, recall the result of Exercise 5.1.(2) that is $r_t = r_0 + \theta t+ \sigma W_t$. \begin{eqnarray} P(t, T) &=&\exp \left( - r_t (T-t) - \frac{1}{2} \theta (T-t)^2 + \frac{ \sigma^2}{6} (T-t)^3 \right) \\ &=&\exp \left( - ( r_0 + \theta t+ \sigma W_t) (T-t) - \frac{1}{2} \theta (T-t)^2 + \frac{ \sigma^2}{6} (T-t)^3 \right) \end{eqnarray}
Use It$\hat{o}$'s formula. \begin{eqnarray} d P(t, T) &=& \left( - \theta (T-t) + ( r_0 + \theta t+ \sigma W_t) \right) P(t, T) dt \nonumber \\ && \qquad + \left( \frac{1}{2} \ 2 \ \theta (T-t) - \frac{ \sigma^2}{6} \ 3 \ (T-t)^2 \right) P(t, T) dt \nonumber \\ && \qquad - \sigma (T-t) P(t, T) dW_t + \frac{1}{2} \sigma^2 (T-t)^2 P(t, T) d [W_t] \\ \frac{ d P(t, T) }{ P(t, T) } &=& \left( - \theta (T-t) + r_t \right) dt %\nonumber \\ %&& \qquad + \left( \ \theta (T-t) - \frac{ \sigma^2}{2} (T-t)^2 \right) dt \nonumber \\ && \qquad \qquad\qquad\qquad\qquad - \sigma (T-t) dW_t + \frac{1}{2} \sigma^2 (T-t)^2 dt \\ &=& r_t dt - \sigma (T-t) dW_t \end{eqnarray}
$\square$
(2) My answer
- Apply It$\hat{o}$'s formula according to the instructions of the question. When applying It$\hat{o}$'s formula, let $f(t, p)=exp( -\int^t_0 r_s ds ) \ p$. \begin{eqnarray} && d \left( \exp \left( -\int^t_0 r_s ds \right) \ P(t, T) \right) \\ && \quad = -r_t \exp \left( -\int^t_0 r_s ds \right) \ P(t, T) dt + \exp\left( -\int^t_0 r_s ds \right) dP(t, T) \nonumber \\ && \qquad + \frac{1}{2} \ 0 \ d \left[ P(t, T) \right] \\ && \quad = -r_t \exp \left( -\int^t_0 r_s ds \right) \ P(t, T) dt \nonumber \\ && \qquad + r_t \exp \left( -\int^t_0 r_s ds \right) \ P(t, T) dt \nonumber \\ && \qquad - \sigma (T-t) \exp \left( -\int^t_0 r_s ds \right) \ P(t, T) dW_t \\ && \quad = - \sigma (T-t) \exp \left( -\int^t_0 r_s ds \right) \ P(t, T) dW_t \end{eqnarray}
$\square$
(3) My answer
Recall (1). $P(t , T)$ has only one time parameter $t$. In other words, $P(t , T)$ is $\mathcal{F}_t$ measurable. Moreover, $P(0 , T)$ is also $\mathcal{F}_t$ measurable. \begin{eqnarray} P(t, T) &=&\exp \left( - ( r_0 + \theta t+ \sigma W_t) (T-t) - \frac{1}{2} \theta (T-t)^2 + \frac{ \sigma^2}{6} (T-t)^3 \right) \end{eqnarray}
Recall the definition of $d \tilde{ \mathbb{P} } / d \mathbb{P} $. \begin{eqnarray} \frac{ d \tilde{ \mathbb{P} } }{d \mathbb{P} } & \equiv& \frac{1}{P(0, T)} \exp \left( -\int^T_0 r_s ds \right) \end{eqnarray}
One calculates this question with the above conditions in mind. \begin{eqnarray} E^{\mathbb{P}} \left[ \frac{ d \tilde{ \mathbb{P} } }{d \mathbb{P} } \middle| \mathcal{F}_t \right] &=& E^{\mathbb{P}} \left[ \frac{1}{P(0, T)} \exp \left( -\int^T_0 r_s ds \right) \middle| \mathcal{F}_t \right] \\ &=& \frac{1}{P(0, T)} E^{\mathbb{P}} \left[ \exp \left( -\int^t_0 r_s ds -\int^T_t r_s ds \right) \middle| \mathcal{F}_t \right] \\ &=& \frac{1}{P(0, T)} \exp \left( -\int^t_0 r_s ds \right) E^{\mathbb{P}} \left[ \exp \left( -\int^T_t r_s ds \right) \middle| \mathcal{F}_t \right] \end{eqnarray}
Here, one computes the expectation value with the policy of eliminating the integral. \begin{eqnarray} E^{\mathbb{P}} \left[ \exp \left( -\int^T_t r_s ds \right) \middle| \mathcal{F}_t \right] &=& E^{\mathbb{P}} \left[ \frac{P(t, T)}{P(t, T)} \exp \left( -\int^T_t r_s ds \right) \middle| \mathcal{F}_t \right] \\ &=& P(t, T) E^{\mathbb{P}} \left[ \frac{1}{P(t, T) \exp \left( \int^T_t r_s ds \right)} \middle| \mathcal{F}_t \right] \\ &=& P(t, T) E^{\mathbb{P}} \left[ \frac{1}{P(T, T) } \middle| \mathcal{F}_t \right] \\ &=& P(t, T) E^{\mathbb{P}} \left[ \ \frac{1}{1} \ \middle| \mathcal{F}_t \right] \\ &=& P(t, T) \end{eqnarray}
Substitute the above result into the expectation value. \begin{eqnarray} E^{\mathbb{P}} \left[ \frac{ d \tilde{ \mathbb{P} } }{d \mathbb{P} } \middle| \mathcal{F}_t \right] &=& \frac{1}{P(0, T)} \exp \left( -\int^t_0 r_s ds \right) E^{\mathbb{P}} \left[ \exp \left( -\int^T_t r_s ds \right) \middle| \mathcal{F}_t \right] \\ &=& \frac{ P(t, T) }{P(0, T)} \exp \left( -\int^t_0 r_s ds \right) \end{eqnarray}
$\square$
(4) My answer
- Apply It$\hat{o}$'s formula according to the instructions of the question. When applying It$\hat{o}$'s formula, let the expectation value $\Psi(t)$. \begin{eqnarray} d \Psi(t) &=& d \left( E^{\mathbb{P}} \left[ \frac{ d \tilde{ \mathbb{P} } }{d \mathbb{P} } \middle| \mathcal{F}_t \right] \right) \\ &=& d \left( \frac{ P(t, T) }{P(0, T)} \exp \left( -\int^t_0 r_s ds \right) \right) \\ &=& \frac{ 1 }{P(0, T)} d \left( P(t, T) \exp \left( -\int^t_0 r_s ds \right) \right) \\ &=& \frac{ 1 }{P(0, T)} \left( - \sigma (T-t) \exp \left( -\int^t_0 r_s ds \right) \ P(t, T) dW_t \right) \\ &=& - \sigma (T-t) \left( \frac{ P(t, T) }{P(0, T)} \exp \left( -\int^t_0 r_s ds \right) \right) dW_t \\ &=& - \sigma (T-t) \Psi(t) dW_t \end{eqnarray}
$\square$
(5) My answer
Solve the S.D.E of (4) that is $d \Psi(t) = - \sigma (T-t) \Psi(t) dW_t $ using It$\hat{o}$'s formula. Let $f(t, \Psi)=log \Psi$. \begin{eqnarray} d \log \Psi(t) &=& 0 \ dt + \frac{1}{ \Psi(t) } d \Psi(t) + \frac{1}{2} \left( - \frac{1}{ \Psi(t)^2 } \right) d[ \Psi(t) ] \\ &=& \frac{1}{ \Psi(t) } \left( - \sigma (T-t) \Psi(t) dW_t \right) - \frac{1}{2} \frac{1}{ \Psi(t)^2 } \sigma^2 (T-t)^2 \Psi(t)^2 dt \\ &=& - \frac{1}{2} \sigma^2 (T-t)^2 dt - \sigma (T-t) dW_t \end{eqnarray}
Let $\psi_t = \sigma (T-t) $. \begin{eqnarray} d \log \Psi(t) &=&- \frac{1}{2} \sigma^2 (T-t)^2 dt - \sigma (T-t) dW_t \\ &=& - \frac{1}{2} \psi_t^2 dt -\psi_tdW_t \\ \int^t_0 d \log \Psi(s) &=& - \frac{1}{2} \int^t_0 \psi_s^2 ds - \int^t_0 \psi_s dW_s \\ \Psi(t) &=& \Psi(0) \exp \left( - \frac{1}{2} \int^t_0 \psi_s^2 ds - \int^t_0 \psi_s dW_s \right) \end{eqnarray}
Here, one computes $\Psi(0)$. \begin{eqnarray} \Psi(t) &=& \frac{ P(t, T) }{P(0, T)} \exp \left( -\int^t_0 r_s ds \right) \\ \Psi(0) &=& \frac{ P(0, T) }{P(0, T)} \exp \left( -\int^0_0 r_s ds \right) \\ &=& 1 \ \exp ( 0) \\ &=& 1 \end{eqnarray}
Substitute the above result into the $\Psi(t) $. \begin{eqnarray} \Psi(t) &=& \Psi(0) \exp \left( - \frac{1}{2} \int^t_0 \psi_s^2 ds - \int^t_0 \psi_s dW_s \right) \\ &=& \exp \left( - \frac{1}{2} \int^t_0 \psi_s^2 ds - \int^t_0 \psi_s dW_s \right) \end{eqnarray}
Therefore, one computes $\tilde{ \mathbb{P} }/ \mathbb{P}$ the according to the instructions of the question.
\begin{eqnarray} \frac{\tilde{ \mathbb{P} }}{ \mathbb{P} }&=& \frac{1}{P(0, T)} \exp \left( - \int^T_0 r_s ds \right) \\ &=& E^{\mathbb{P}} \left[ \frac{1}{P(0, T)} \exp \left( - \int^T_0 r_s ds \right) \middle| \mathcal{F}_T \right] \\ &=& E^{\mathbb{P}} \left[ \frac{P(T, T)}{P(0, T)} \exp \left( - \int^T_0 r_s ds \right) \middle| \mathcal{F}_T \right] \\ &=& E^{\mathbb{P}} \left[ \frac{ d \tilde{ \mathbb{P} } }{d \mathbb{P} } \middle| \mathcal{F}_T \right] \\ &=& \Psi(T) \\ &=& \exp \left( - \frac{1}{2} \int^T_0 \psi_s^2 ds - \int^T_0 \psi_s dW_s \right) \end{eqnarray}
$\square$
(6) My answer
One reaches the following equation by Girsanov-Maruyama Theorem. \begin{eqnarray} \tilde{ W_t } &=& W_t + \int^t_0 \psi_s ds \\ d\tilde{ W_t } &=& dW_t + \psi_t dt \\ dW_t &=& d\tilde{ W_t } - \psi_t dt \end{eqnarray}
Recall the definition of $dr_t=\theta dt+ \sigma dW_t$ at in Exercise 4.1. Moreover, substitute the above result into the $dr_t$. \begin{eqnarray} dr_t&=&\theta dt+ \sigma dW_t \\ &=&\theta dt+ \sigma (d\tilde{ W_t } - \psi_t dt) \\ &=& ( \theta - \sigma \psi_t ) dt + \sigma d\tilde{ W_t } \\ &=& \left( \theta - \sigma \ \sigma (T-t) \right) dt + \sigma d\tilde{ W_t } \\ &=& \left( \theta - \sigma^2 (T-t) \right) dt + \sigma d\tilde{ W_t } \end{eqnarray}
$\square$
(7) My answer (under consultation)
One shows the equation. \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 = 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) - K )^+ \middle| \mathcal{F}_t \right] \\ && \quad = P(t, T) E^{\tilde{\mathbb{P}}} \left[ \frac{1}{P(T, T)} \cdot ( P(T,S) - K )^+ \middle| \mathcal{F}_t \right] \\ && \quad = P(t, T) E^{\tilde{\mathbb{P}}} \left[ ( P(T,S) - K )^+ \middle| \mathcal{F}_t \right] \end{eqnarray}
Here, one computes $P(T,S)$. Recall (1). \begin{eqnarray} P(t, T) &=&\exp \left( - ( r_0 + \theta t+ \sigma W_t ) (T-t) - \frac{1}{2} \theta (T-t)^2 + \frac{ \sigma^2}{6} (T-t)^3 \right) \\ P(T, S) &=&\exp \left( - ( r_0 + \theta T+ \sigma W_T) (S-T) \right) \nonumber \\ && \qquad \cdot \exp \left( - \frac{1}{2} \theta (S-T)^2 + \frac{ \sigma^2}{6} (S-T)^3 \right) \\ &=&\exp \left( - ( r_0 + \theta T+ \sigma W_T) (S-T) \right) \nonumber \\ && \qquad \cdot \exp \left( - \frac{1}{2} \theta (S-T)^2 + \frac{ \sigma^2}{6} (S-T)^3 \right) \end{eqnarray}
Here, recall the following equation by Girsanov-Maruyama Theorem. \begin{eqnarray} dW_t &=& d\tilde{ W_t } - \psi_t dt \\ \int^T_0 dW_t &=& \int^T_0 d\tilde{ W_t } - \int^T_0 \sigma (T-t) dt \\ W_T &=& \tilde{ W_T } - \sigma \left[ Tt - \frac{1}{2} t^2 \right]^T_0 \\ &=& \tilde{ W_T } - \sigma \left( T^2 - \frac{1}{2} T^2 \right) \\ &=& \tilde{ W_T } - \sigma \frac{1}{2} T^2 \end{eqnarray}
Substitute the above result into the $P(T, S) $. \begin{eqnarray} P(T, S) &=&\exp \left( - ( r_0 + \theta T+ \sigma W_T ) (S-T) \right) \nonumber \\ && \qquad \cdot \exp \left( - \frac{1}{2} \theta (S-T)^2 + \frac{ \sigma^2}{6} (S-T)^3 \right) \\ &=&\exp \left( - \left( r_0 + \theta T+ \sigma \left( \tilde{ W_T } - \sigma \frac{1}{2} T^2 \right) \right) (S-T) \right) \nonumber \\ && \qquad \cdot \exp \left( - \frac{1}{2} \theta (S-T)^2 + \frac{ \sigma^2}{6} (S-T)^3 \right) \end{eqnarray}
Here, one computes the exponent part. Let the exponent part $\mathscr{P}(t, W)$. \begin{eqnarray} && \mathscr{P}(t, W) \nonumber \\ && \quad = - \left( r_0 + \theta T+ \sigma \left( \tilde{ W_T } - \sigma \frac{1}{2} T^2 \right) \right) (S-T) \nonumber \\ && \qquad - \frac{1}{2} \theta (S-T)^2 + \frac{ \sigma^2}{6} (S-T)^3 \\ && \quad = -r_0 (S-T) - \theta T(S-T) - \sigma \tilde{ W_T } (S-T) \nonumber \\ && \qquad + \frac{1}{2}\sigma^2 T^2 (S-T) - \frac{1}{2} \theta (S-T)^2 + \frac{ \sigma^2}{6} (S-T)^3 \\ && \quad = -r_0 (S-T) - \theta (S-T) \left( T+ \frac{S-T}{2} \right) + \frac{ \sigma^2}{6} (S-T)^3 \nonumber \\ && \qquad + \frac{1}{2}\sigma^2 T^2 (S-T) - \sigma \tilde{ W_T } (S-T) \\ && \quad = -r_0 (S-T) - \theta (S-T) \left( T+ \frac{S-T}{2} \right) + \frac{ \sigma^2}{6} (S-T)^3 \nonumber \\ && \qquad + \frac{1}{2}\sigma^2 T^2 (S-T) - \sigma \tilde{ W_t } (S-T) - \sigma \tilde{ W_T } (S-T) + \sigma \tilde{ W_t } (S-T) \\ && \quad = -r_0 (S-T) - \frac{1}{2} \theta (S-T) (S+T) + \frac{ \sigma^2}{6} (S-T)^3 \nonumber \\ && \qquad - \sigma \left( - \sigma \frac{1}{2}T^2 + \tilde{ W_t } \right) (S-T) - \sigma \left( \tilde{ W_T } - \tilde{ W_t } \right) (S-T) \\ && \quad = -r_0 (S-T) - \frac{1}{2} \theta (S^2-T^2) + \frac{ \sigma^2}{6} (S-T)^3 \nonumber \\ && \qquad - \sigma \left( - \frac{1}{2 }\sigma T^2 + W_t + \frac{1}{2}\sigma t^2 \right) (S-T) \nonumber \\ && \qquad - \sigma \left( W_T + \frac{1}{2 } \sigma T^2 - W_t - \frac{1}{2}\sigma t^2 \right) (S-T) \end{eqnarray}
Here, one computes the last 2 terms with letting those $\mathscr{Q}(t, W)$. \begin{eqnarray} \mathscr{Q}(t, W) &=& - \sigma \left( - \frac{1}{2}\sigma T^2 + W_t + \frac{1}{2 }\sigma t^2 \right) (S-T) \nonumber \\ && \qquad - \sigma \left( W_T + \frac{1}{2}\sigma T^2 - W_t - \frac{1}{2 }\sigma t^2 \right) (S-T) \\ %&=& - \sigma \left( - \frac{1}{2 }\sigma T^2 + \frac{1}{2 }\sigma t^2 + W_t \right) (S-T) \nonumber \\ %&& \qquad - \sigma \left( (W_T - W_t) + \frac{1}{2 }\sigma(T-t)^2 +\frac{1}{2 }\sigma2Tt \right) (S-T) \\ &=& - \sigma \left( - \frac{1}{2 }\sigma T^2 + \frac{1}{2 }\sigma t^2 + W_t \right) (S-T) \nonumber \\ && \qquad - \sigma \left( (W_T - W_t) + \frac{1}{2 }\sigma(T-t)^2 + \sigma Tt \right) (S-T) \\ &=& - \sigma \left( - \frac{1}{2 }\sigma T^2 + \frac{1}{2 }\sigma t^2 + \sigma Tt + W_t \right) (S-T) \nonumber \\ && \qquad - \sigma \left( (W_T - W_t) + \frac{1}{2 }\sigma(T-t)^2 \right) (S-T) \\ &=& \sigma \left( \frac{1}{2 }\sigma T^2 - \frac{1}{2 }\sigma t^2 - \sigma Tt - W_t \right) (S-T) \nonumber \\ && \qquad - \sigma \left( (W_T - W_t) + \frac{1}{2 }\sigma(T-t)^2 \right) (S-T) \end{eqnarray}
Substitute the above result into the $\mathscr{P}(t, W)$. \begin{eqnarray} \mathscr{P}(t, W) &=& -r_0 (S-T) - \frac{1}{2} \theta (S^2-T^2) + \frac{ \sigma^2}{6} (S-T)^3 \nonumber \\ && \qquad - \sigma \left( - \frac{1}{2 }\sigma T^2 + W_t + \frac{1}{2}\sigma t^2 \right) (S-T) \nonumber \\ && \qquad - \sigma \left( W_T + \frac{1}{2 } \sigma T^2 - W_t - \frac{1}{2}\sigma t^2 \right) (S-T) \\ &=& -r_0 (S-T) - \frac{1}{2} \theta (S^2-T^2) + \frac{ \sigma^2}{6} (S-T)^3 \nonumber \\ && \qquad +\sigma \left( \frac{1}{2 }\sigma T^2 + \frac{1}{2 }\sigma t^2 - \sigma Tt - W_t \right) (S-T) \nonumber \\ && \qquad - \sigma \left( (W_T - W_t) + \frac{1}{2 }\sigma(T-t)^2 \right) (S-T) \\ &=& m +X \end{eqnarray}
Here, let $m$ and $X$ as below, respectively. \begin{eqnarray} m&=& -r_0 (S-T) - \frac{1}{2} \theta (S^2-T^2) + \frac{ \sigma^2}{6} (S-T)^3 \nonumber \\ && \qquad +\sigma \left( \frac{1}{2 }\sigma T^2 + \frac{1}{2 }\sigma t^2 - \sigma Tt - W_t \right) (S-T) \\ X&=& - \sigma \left( (W_T - W_t) + \frac{1}{2 }\sigma(T-t)^2 \right) (S-T) \end{eqnarray}
Therefore, $v^2$ is computed as below. \begin{eqnarray} && E^{ \{ \mathbb{P}, X \}} \left[ X \middle| \mathcal{F}_t \right] \nonumber \\ && \qquad = E^{ \{ \mathbb{P}, X\}} \left[ - \sigma \left( (W_T - W_t) + \frac{1}{2 }\sigma(T-t)^2 \right) (S-T)\middle| \mathcal{F}_t \right] \\ && \qquad = - \frac{1}{2 }\sigma^2 (T-t)^2 (S-T) \\ && \qquad = \frac{1}{2 }\sigma^2 (T-t)^2 (T-S) \\ && E^{ \{ \mathbb{P}, X \}} \left[ X^2 \middle| \mathcal{F}_t \right] \nonumber \\ && \qquad = E^{ \{ \mathbb{P}, X\}} \left[ \left( - \sigma \left( (W_T - W_t) + \frac{1}{2 }\sigma(T-t)^2 \right) (S-T) \right)^2 \middle| \mathcal{F}_t \right] \\ && \qquad = E^{ \{ \mathbb{P}, X\}} \left[ \left( \sigma \left( (W_T - W_t) + \frac{1}{2 }\sigma(T-t)^2 \right) (T-S) \right)^2 \middle| \mathcal{F}_t \right] \\ && \qquad = E^{ \{ \mathbb{P}, X\}} \left[ \sigma^2 (W_T - W_t) ^2 (T-S)^2 \middle| \mathcal{F}_t \right] \nonumber \\ && \qquad \qquad + E^{ \{ \mathbb{P}, X\}} \left[ \sigma^2 (W_T - W_t) \frac{1}{2 } \ 2 \ \sigma(T-t)^2 (T-S) \middle| \mathcal{F}_t \right] \nonumber \\ && \qquad \qquad + E^{ \{ \mathbb{P}, X\}} \left[ \sigma^2 \frac{1}{4 }\sigma^2 (T-t)^4 (T-S) ^2 \middle| \mathcal{F}_t \right] \\ && \qquad = \sigma^2 (T - t) (T-S)^2 + \frac{1}{4 }\sigma^4 (T-t)^4 (T-S) ^2 \\ && Var^{ \{ \mathbb{P}, X \}} \left( X \middle| \mathcal{F}_t \right) \nonumber \\ && \qquad \equiv E^{ \{ \mathbb{P}, X \}} \left[ X^2 \middle| \mathcal{F}_t \right] -\left( E^{ \{ \mathbb{P}, X \}} \left[ X \middle| \mathcal{F}_t \right] \right)^2 \\ && \qquad = \sigma^2 (T - t) (T-S)^2 + \frac{1}{4 }\sigma^4 (T-t)^4 (T-S) ^2 \nonumber \\ && \qquad \qquad - \left( - \frac{1}{2 }\sigma^2 (T-t)^2 (T-S) \right)^2 \\ && \qquad = \sigma^2 (T - t) (T-S)^2 = v^2 \end{eqnarray}
Hence, one reaches below equations. \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^{\tilde{\mathbb{P}}} \left[ ( P(T,S) - K )^+ \middle| \mathcal{F}_t \right] \\ && \quad = P(t, T) E^{ \{ \mathbb{P}, X \} } \left[ (e^{m+X} - K )^+ \middle| \mathcal{F}_t \right] \\ && \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) \end{eqnarray}
where \begin{eqnarray} \phi(y) &=& \frac{1}{\sqrt{2 \pi}} e^{-y^2/2} \\ N(z)&=& \int^z_{ - \infty} \phi(y) dy \end{eqnarray}
Next, one computes $E^{ \{ \mathbb{P}, X \} } \left[ e^{m+X} \middle| \mathcal{F}_t \right] $. \begin{eqnarray} E^{ \{ \mathbb{P}, X \} } \left[ e^{m+X} \middle| \mathcal{F}_t \right] &=& e^m E^{ \{ \mathbb{P}, X \} } \left[ e^X \middle| \mathcal{F}_t \right] \\ &=& e^m \int^{}_{\mathbb{R}} \frac{1}{ \sqrt{2 \pi v^2} } e^x \ e^{-x^2/(2v^2)} dx \\ &=& e^m \int^{}_{\mathbb{R}} \frac{1}{ \sqrt{2 \pi v^2} } e^{ -(x^2-2v^2x)/(2v^2)} dx\\ &=& e^{m+v^2/2} \int^{}_{\mathbb{R}} \frac{1}{ \sqrt{2 \pi v^2} } e^{ -(x^2 - v^2)/(2v^2)} dx \\ &=& e^{m+v^2/2} \end{eqnarray}
Therefore, one reaches the following equations. \begin{eqnarray} E^{\tilde{\mathbb{P}}} \left[ P(T,S) \middle| \mathcal{F}_t \right] &=& E^{\tilde{\mathbb{P}}} \left[ \frac{P(T,S)}{1} \middle| \mathcal{F}_t \right] \\ &=& E^{\tilde{\mathbb{P}}} \left[ \frac{P(T,S)}{P(T,T)} \middle| \mathcal{F}_t \right] \\ &=& E^{\mathbb{P}} \left[ \frac{\exp \left( -\int^T_0 r_s ds \right)}{P(0,T)} P(T,S) \middle| \mathcal{F}_t \right] \\ &=& \frac{\exp \left( -\int^t_0 r_s ds \right)}{P(0,T)} E^{\mathbb{P}} \left[ \exp \left( -\int^T_t r_s ds \right) P(T,S) \middle| \mathcal{F}_t \right] \\ &=& \frac{1}{P(t,T)} E^{\mathbb{P}} \left[ P(t,S) \middle| \mathcal{F}_t \right] \\ &=& \frac{P(t,S) }{P(t,T)} E^{\mathbb{P}} \left[ \ 1 \ \middle| \mathcal{F}_t \right] \\ &=& \frac{P(t,S) }{P(t,T)} \\ E^{\tilde{\mathbb{P}}} \left[ P(T,S) \middle| \mathcal{F}_t \right] &=& E^{ \{ \mathbb{P}, X \} } \left[ e^{m+X} \middle| \mathcal{F}_t \right] \\ &=& e^{m+v^2/2} \end{eqnarray}
Then, one reaches the following equations. \begin{eqnarray} e^{m+v^2/2} &=& \frac{P(t,S) }{P(t,T)} \\ m+ \frac{v^2}{2} &=& \log \frac{P(t,S) }{P(t,T)} \\ \end{eqnarray} \item Moreover, one reaches the following equations. \begin{eqnarray} \frac{m}{v} + \frac{v}{2} &=& \frac{1}{v} \log \frac{P(t,S) }{P(t,T)} \\ \frac{m}{v} &=& - \frac{v}{2} + \frac{1}{v} \log \frac{P(t,S) }{P(t,T)} \\ \frac{m}{v} -\frac{\log K}{v} &=& - \frac{v}{2} + \frac{1}{v} \log \frac{P(t,S) }{P(t,T)} -\frac{\log K}{v}\\ \frac{m - \log K}{v} &=& - \frac{v}{2} + \frac{1}{v} \log \frac{P(t,S) }{K \ P(t,T)} \end{eqnarray}
Hence, substitute the above result into the expectation value. One reaches this answer. \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) \end{eqnarray}
where $ v = \sigma (T - S) \sqrt{T-t} $.
$\square$
