We consider the expectation
\begin{align*}
E^{Q_d^{t_f}} \Big(P_d(t_f, T) X_{t_f} \mid \mathcal{F}_t \Big),
\end{align*}
where $Q_d^{t_f}$ is the $t_f$-forward measure, and $P_d(t_f, T)$ is the price at $t_f$ of a domestic zero-coupon bond with maturity $T$. Note that $P_d(t_f, T) X_{t_f}$ is the value at time $t_f$ of the process
\begin{align*}
P_d(t, T) X_t \frac{P_f(t, t_f)}{P_d(t, t_f)},
\end{align*}
where $P_f(t, t_f)$ is the price at $t$ of a foreign zero-coupon bond with maturity $t_f$. If we model the bond prices directly, then we need dynamics for the four processes $X_t$, $P_d(t, t_f)$, $P_d(t, T)$, and $P_f(t, t_f)$. For simplicity, we model the short rates based on the Hull-White interest rate model with corresponding correlated dynamics for $X_t$, $r_t^f$, and $r_t^d$.
Let $X_t$ be the instantaneous exchange rate from one unit foreign currency to units of domestic currency, and
$r^d_t$ and $r^f_t$ be the domestic and the equity's denominated foreign short interest rates at time $t$.
Moreover, let
\begin{align*}
\Xi= (\rho_{i, j})_{i,j=1}^3
\end{align*}
be the correlation matrix of the driving Brownian motions of the 3$-$dimensional random process $\{(X_t, r^f_t, r^d_t),\, t \geq 0\}$, and let
$$
LL^T = \Xi,
$$
where $L=(l_{i,j})_{i,j=1}^3$ is a lower triangular matrix and $l_{i, i} >0$, for $i=1, \ldots, 3$, be the Cholesky decomposition of $\Xi$.
We assume that, under the domestic risk-neutral probability measure $Q_d$,
\begin{align*}
dX_t &= X_t\Big[\big(r^d_t - r^f_t\big) dt + \sigma^X dW_t^1\Big], \\
dr^f_t &= \Big[\big( \theta_t^f - \lambda^f r_t^f \big) - \rho_{2, 1}\sigma^f \sigma^X\, \Big] dt + \sigma^f
\sum_{i=1}^2 l_{2, i} d W_t^i,\\
dr^d_t &= \big( \theta_t^d - \lambda^d r_t^d \big) dt + \sigma^d\sum_{i=1}^3 l_{3, i} d W_t^i,
\end{align*}
where $\{W_t^1,\, t \ge 0\}, \ldots, \{W_t^3,\, t \ge 0\}$ are independent standard Brownian motions, $\theta_{t}^{a},$ are piece-wise constant functions, and $\lambda^{a}$ and $\sigma^{a}$ are positive
constant, for $a=X$, $d$, or $f$. Here, $\lambda^d$ and $\lambda^f$ are the mean-reverting speed of the respective dynamic processes.
Note that, under the foreign risk-neutral probability measure $Q_f$,
$\{\widehat{W}_t^1= W_t^1-\sigma^X t,\, t \ge 0\}$, $\{\widehat{W}_t^2=W_t^2,\, t \ge 0\}$, and $\{\widehat{W}_t^3=W_t^3,\, t \ge 0\}$ are independent standard Brownian motions. Moreover,
\begin{align*}
dX_t &= X_t\Big[\left(r^d_t - r^f_t + (\sigma^X)^2\right) dt + \sigma^X d\widehat{W}_t^1\Big], \\
dr^f_t &= \big( \theta_t^f - \lambda^f r_t^f \big) dt + \sigma_t^f
\sum_{i=1}^2 l_{2, i} d \widehat{W}_t^i,\\
dr^d_t &= \Big[\big( \theta_t^d - \lambda^d r_t^d \big) + \rho_{3, 1}\sigma^d \sigma^X\,\Big] dt + \sigma^d\sum_{i=1}^3 l_{3, i} d \widehat{W}_t^i.
\end{align*}
Let $P_d(t, T)$ and $P_f(t, T)$ denote the prices, at time $t$, of the domestic and foreign zero-coupon bonds with maturity $T$. From this question,
\begin{align*}
P_a(t, T) &= e^{A_a(t, T) - B_a(t, T) r_t^a},
\end{align*}
for $a=d$ or $f$, where
\begin{align*}
\beta_a(t, T) &= \frac{1}{\lambda^a}\left(1-e^{-\lambda^a (T-t)} \right),\\
A_a(t, T) &= -\int_t^T \theta_s^a\beta_a(s, T) ds - \frac{(\sigma^a)^2}{2(\lambda^a)^2}(\beta_a(t, T)-T+t) - \frac{(\sigma^a)^2}{4\lambda^a}\beta_a(t, T)^2.
\end{align*}
Then, under the domestic risk-neutral measure,
\begin{align*}
\frac{dP_d(t, T)}{P_d(t, T)} &= r_t^d dt - \beta_d(t, T)\sigma^d \sum_{i=1}^3 l_{3, i} d W_t^i,\\
\frac{dP_f(t, T)}{P_f(t, T)} &= r_t^f dt - \beta_f(t, T)\sigma^f \sum_{i=1}^2 l_{2, i} d \widehat{W}_t^i\\
&=\left(r_t^f + \rho_{2,1}\beta_f(t, T)\sigma^f \sigma^X \right)dt - \beta_f(t, T)\sigma^f \sum_{i=1}^2 l_{2, i} d W_t^i.
\end{align*}
Let $Q_d^{t_f}$ be the domestic $t_f$-forward measure. Moreover, let $B_d(t)=e^{\int_0^t r_s^d ds}$ be the domestic money market account value at time $t$. Then,
\begin{align*}
\frac{dQ_d^{t_f}}{dQ_d}\big|_t &= \frac{P_d(t, t_f)}{P_d(0, t_f) B_d(t)}\\
&=\exp\left(-\frac{1}{2}\int_0^t (\sigma^d \beta_d(s, t_f))^2ds - \int_0^t\sigma^d \beta_d(s, t_f) \sum_{i=1}^3 l_{3, i} d W_s^i\right).
\end{align*}
Then, $\{\widetilde{W}_t^1,\, 0 \le t \le t_f\}$, $\{\widetilde{W}_t^2,\, 0 \le t \le t_f\}$, and $\{\widetilde{W}_t^3,\, 0 \le t \le t_f\}$, where, for $i=1, \ldots, 3,$
\begin{align*}
\widetilde{W}_t^i = W_t^i + \int_0^t l_{3, i} \sigma^d\beta_d(u, t_f) du,
\end{align*}
are independent standard Brownian motions.
Let $F_X(t, t_f)=X_t\frac{P_f(t, t_f)}{P_d(t, t_f)}$ be the forward exchange rate at time $t$ for maturity $t_f$. Then,
\begin{align*}
\frac{dF_X(t, t_f)}{F_X(t, t_f)} = \sigma^X d\widetilde{W}_t^1 + \beta_d(t, t_f)\sigma^d \sum_{i=1}^3 l_{3, i} d \widetilde{W}_t^i- \beta_f(t, t_f)\sigma^f \sum_{i=1}^2 l_{2, i} d \widetilde{W}_t^i.
\end{align*}
Moreover, let
\begin{align*}
M(t, t_f) &= -\frac{1}{2}\int_t^{t_f} \bigg((\sigma^X)^2 + (\beta_d(u, t_f)\sigma^d)^2 + (\beta_f(u, t_f)\sigma^f)^2\\
&\quad + 2 \rho_{1, 3}\sigma^X \sigma^d\beta_d(u, t_f)- 2 \rho_{1, 2}\sigma^X \sigma^f\beta_f(u, t_f) - 2 \rho_{2, 3}\sigma^d \sigma^f \beta_d(u, t_f)\beta_f(u, t_f)\bigg) du.
\end{align*}
Then,
\begin{align*}
X_{t_f} &= F_X(t_f, t_f)\\
&=F_X(t, t_f) \exp\bigg(M(t, t_f) \\
&\quad + \int_t^{t_f} \Big( \sigma^X d\widetilde{W}_u^1 + \beta_d(u, t_f)\sigma^d \sum_{i=1}^3 l_{3, i} d \widetilde{W}_u^i- \beta_f(u, t_f)\sigma^f \sum_{i=1}^2 l_{2, i} d \widetilde{W}_u^i\Big)\bigg).
\end{align*}
Furthermore, let
\begin{align*}
N(t, t_f) &= e^{-\lambda^d (t_f-t)} r_t + \int_t^{t_f} \theta_u^d e^{-\lambda^d(t_f-u)} du -\int_t^{t_f}(\sigma^d)^2 e^{-\lambda^d(t_f-u)}\beta_d(u, t_f) du.
\end{align*}
Then,
\begin{align*}
r_{t_f}^d &= e^{-\lambda^d (t_f-t)} r_t + \int_t^{t_f} \theta_u^d e^{-\lambda^d(t_f-u)} du + \int_t^{t_f} \sigma^d e^{-\lambda^d(t_f-u)} \sum_{i=1}^3 l_{3, i} d W_u^i\\
&= N(t, t_f)+ \int_t^{t_f} \sigma^d e^{-\lambda^d(t_f-u)} \sum_{i=1}^3 l_{3, i} d \widetilde{W}_u^i.
\end{align*}
That is,
\begin{align*}
P_d(t_f, T) X_{t_f} &= \exp\bigg(A_d(t_f, T) - B_d(t_f, T) r_{t_f}^d \bigg)X_{t_f}\\
&= F_X(t, t_f) \exp\bigg(A_d(t_f, T) - B_d(t_f, T)N(t, t_f) + M(t, t_f)\\
&\quad - B_d(t_f, T) \int_t^{t_f} \sigma^d e^{-\lambda^d(t_f-u)} \sum_{i=1}^3 l_{3, i} d \widetilde{W}_u^i \\
&\quad + \int_t^{t_f} \Big( \sigma^X d\widetilde{W}_u^1 + \beta_d(u, t_f)\sigma^d \sum_{i=1}^3 l_{3, i} d \widetilde{W}_u^i- \beta_f(u, t_f)\sigma^f \sum_{i=1}^2 l_{2, i} d \widetilde{W}_u^i\Big) \bigg).
\end{align*}
Let
\begin{align*}
V(t, t_f, T) &= \int_t^{t_f} \bigg((\sigma^X)^2 + \Big(\beta_d(u, t_f)\sigma^d - B_d(t_f, T)\sigma^d e^{-\lambda^d(t_f-u)} \Big)^2 + (\beta_f(u, t_f)\sigma^f)^2\\
&\quad + 2 \rho_{1, 3}\sigma^X \Big(\beta_d(u, t_f)\sigma^d - B_d(t_f, T)\sigma^d e^{-\lambda^d(t_f-u)} \Big)- 2 \rho_{1, 2}\sigma^X \sigma^f\beta_f(u, t_f) \\
&\quad - 2 \rho_{2, 3}\sigma^f \beta_f(u, t_f)\Big(\beta_d(u, t_f)\sigma^d - B_d(t_f, T)\sigma^d e^{-\lambda^d(t_f-u)} \Big)\bigg) du\\
&= -2M(t, t_f) + B_d(t_f, T)\int_t^{t_f}\bigg[ B_d(t_f, T)\Big( \sigma^d e^{-\lambda^d(t_f-u)} \Big)^2 \\
&\qquad\qquad\qquad\qquad\qquad\qquad - 2 (\sigma^d)^2\beta_d(u, t_f)e^{-\lambda^d(t_f-u)} -2 \rho_{1, 3}\sigma^X \sigma^d e^{-\lambda^d(t_f-u)}\\
&\qquad\qquad\qquad\qquad\qquad\qquad +2 \rho_{2, 3}\sigma^d \sigma^f \beta_f(u, t_f) e^{-\lambda^d(t_f-u)} \bigg]du.
\end{align*}
Moreover, let
\begin{align*}
V_0(t, t_f, T) &=B_d(t_f, T)\int_t^{t_f}\bigg[ B_d(t_f, T)\Big( \sigma^d e^{-\lambda^d(t_f-u)} \Big)^2 \\
&\qquad\qquad\qquad\qquad - 2 (\sigma^d)^2\beta_d(u, t_f)e^{-\lambda^d(t_f-u)} -2 \rho_{1, 3}\sigma^X \sigma^d e^{-\lambda^d(t_f-u)} \\
&\qquad\qquad\qquad\qquad +2 \rho_{2, 3}\sigma^d \sigma^f \beta_f(u, t_f) e^{-\lambda^d(t_f-u)} \bigg]du.
\end{align*}
Then
\begin{align*}
&\ E^{Q_d^{t_f}} \Big(P_d(t_f, T) X_{t_f} \mid \mathcal{F}_t \Big) \\
=&\ F_X(t, t_f) \exp\bigg(A_d(t_f, T) - B_d(t_f, T)N(t, t_f) + M(t, t_f) +\frac{1}{2} V(t, t_f, T) \bigg)\\
=&\ F_X(t, t_f) \exp\bigg(A_d(t_f, T) - B_d(t_f, T)N(t, t_f) + \frac{1}{2} V_0(t, t_f, T) \bigg).
\end{align*}
But anyhow, do you agree with my RHS so that: $$ \mathbb{E}^{\mathbb Q^d}t \left[ \exp(-\int_t^T r^d_s ds) X{t_f} \right] = \mathbb{E}^{\mathbb Q^{d,t_f}}t\left[ \frac{B_t^d}{B_T^d} X{t_f} \left( \frac{B^d_{t_f} B_d(t,t_f)}{B_t^d} \right) \right] $$ and thus $$ \mathbb{E}^{\mathbb{ Q}^d}t \left[ \exp(-\int_t^T r^d_s ds) X{t_f}\right] = B_d(t,t_f) \mathbb{E}^{\mathbb{Q}^{d,t_f}}t \left[ X{t_f} \exp\left(-\int_{t_f}^T r_s ds\right) \right] $$
– Quantuple Jul 28 '16 at 14:27