FX forward with stochastic interest rates pricing

I would like to extend the following question about FX Forward rates in stochastic interest rate setup: "Expectation" of a FX Forward

We consider a FX process $X_t = X_0 \exp( \int_0^t(r^d_s-r^f_s)ds -\frac{\sigma^2}{2}t+ \sigma W_t)$ where $r^d$ and $r^f$ are stochastic processes not independent of the Brownian motion $W$. As we know the FX Forward rate is $F^X(t,T) = E_t^d\left[X_T \right]$ under the domestic risk-neutral measure.

The question is how to show that $F^X(t,T) = X_t \frac{B_f(t,T)}{B_d(t,T)}$ where $B_d(t,T)$ and $B_f(t,T)$ are respective the domestic and foreing zero-coupon bond prices of maturity $T$ at time $t$.

Since $X_T = X_t \exp\left( \int_t^T(r^d_s-r^f_s)ds+ \sigma (W_T-W_t)\right)$

$$\begin{align} F^X(t,T) &= X_t E_t^d\left[\exp\left( \int_t^T(r^d_s-r^f_s)ds-\frac{\sigma^2}{2}(T-t) + \sigma (W_T-W_t)\right)\right] \\& =X_t E_t^d\left[\exp\left( \int_t^T(r^d_s-r^f_s)ds \right) \frac{\mathcal E_T(\sigma W )}{\mathcal E_t(\sigma W )}\right] \\&= X_t E_t^d\left[\exp\left( \int_t^T(r^d_s-r^f_s)ds \right) \frac{d\mathcal Q^f}{d\mathcal Q^d} \frac{1}{E_t^d \left[\frac{d\mathcal Q^f}{d\mathcal Q^d}\right]}\right] \\&= X_t E_t^f\left[\exp\left( \int_t^T(r^d_s-r^f_s)ds \right) \right] \end{align}$$

Now how to conclude given that $r^d$ and $r^f$ are not necessarilly independent of each other since they both depend on the Brownian motion $W$ (by the way let's assume we working in the natural filtration of $W$)?

Edit

I would like to extend my question to the pricing of non-deliverable FX forwards. I posted a new question for that here : FX Forward pricing with correlation between FX and Zero-Cupon.

Answer

The formula $F^X(t,T) = E_t^d\left(X_T \right)$, under the domestic risk-neutral measure, is problematic. Note that, at time $t$, the forward exchange rate $F^X(t,T)$, for maturity $T$, is the exchange rate such that the payoff $X_T-F^X(t,T)$ has a zero value at $t$. That is,

$$\begin{align*} B_t^d E_d\left(\frac{X_T-F^X(t,T)}{B_T^d} \mid \mathcal{F}_t\right)=0, \end{align*}$$ where $E_d$ is the expectation under the domestic risk-neutral measure $Q_d$. Here, $B_t^d$ and $B_t^f$ denote respectively the domestic and foreign money market account values. Then, $$\begin{align*} F^X(t,T) &= \frac{1}{B_d(t, T)}B_t^d E_d\left(\frac{X_T}{B_T^d} \mid \mathcal{F}_t\right) \tag{1}\\ &\neq E_d(X_T \mid \mathcal{F}_t), \end{align*}$$ under the stochastic interest rate assumption.

Let $Q_d^T$ be the domestic $T$-forward measure, and $E_d^T$ be the corresponding expectation operator. Then, for $0 \le t \le T$,

$$\begin{align*} \frac{dQ_d}{dQ_d^T}\big|_t = \frac{B_t^d B_d(0, T)}{B_d(t, T)}. \end{align*}$$ From $(1)$, $$\begin{align*} F^X(t,T) &= \frac{1}{B_d(t, T)}B_t^d E_d\left(\frac{X_T}{B_T^d} \mid \mathcal{F}_t\right)\\ &=\frac{1}{B_d(t, T)}B_t^d E_d^T\left(\frac{X_T}{B_T^d} \frac{\frac{dQ_d}{dQ_d^T}\big|_T}{\frac{dQ_d}{dQ_d^T}\big|_t}\mid \mathcal{F}_t\right)\\ &=\frac{1}{B_d(t, T)}B_t^d E_d^T\left(\frac{X_T}{B_T^d} \frac{B_T^dB_d(t, T)}{B_t^d}\mid \mathcal{F}_t\right)\\ &= E_d^T\left(X_T\mid \mathcal{F}_t\right).\tag{2} \end{align*}$$ That is, it is the expectation of the spot exchange rate at maturity $T$, under the $T$-forward measure rather than the risk-neutral measure.

Back to Formula $(1)$. Let $Q_f$ be the foreign risk-neutral measure and $E_f$ be the corresponding expectation operator. Then, for $t\ge 0$,

$$\begin{align*} \frac{dQ_d}{dQ_f}\big|_t = \frac{B_t^d X_0}{B_t^f X_t}. \end{align*}$$ Moreover, $$\begin{align*} F^X(t,T) &= \frac{1}{B_d(t, T)}B_t^d E_d\left(\frac{X_T}{B_T^d} \mid \mathcal{F}_t\right)\\ &=\frac{1}{B_d(t, T)}B_t^d E_f\left(\frac{X_T}{B_T^d} \frac{\frac{dQ_d}{dQ_f}\big|_T}{\frac{dQ_d}{dQ_f}\big|_t}\mid \mathcal{F}_t \right)\\ &=\frac{1}{B_d(t, T)}B_t^d E_f\left(\frac{X_T}{B_T^d} \frac{B_T^d}{B_T^f X_T} \frac{B_t^fX_t}{B_t^d}\mid \mathcal{F}_t\right)\\ &=\frac{X_t}{B_d(t, T)}E_f\left(\frac{B_t^f}{B_T^f}\mid \mathcal{F}_t \right)\\ &=X_t\frac{B_f(t, T)}{B_d(t, T)}. \end{align*}$$

Additional information.

Combining with Formula $(2)$,

$$\begin{align*} E_d^T(X_T \mid \mathcal{F}_t) &= E_d^T\left(X_T\frac{B^f(T, T)}{B_d(T, T)} \mid \mathcal{F}_t\right)\\ &=X_t\frac{B_f(t, T)}{B_d(t, T)}. \end{align*}$$ That is, the forward exchange rate process $\left\{X_t\frac{B_f(t, T)}{B_d(t, T)}, 0 \le t \le T \right\}$ is a martingale under the domestic $T$-forward measure.