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A model for FX, presented in Interest Rate Models, Brigo and Mercurio (2006), has the following dynamics: \begin{align} dr_t^d&= \lambda_d(\theta_d(t)-r_t^d)dt+\eta_d dW_t^d\\ dr_t^f&= [\lambda_d(\theta_d(t)-r_t^d)-\eta_f\rho_{S,f}\sigma]dt+\eta_f dW_t^f\\ \frac{dS_t}{S_t}& =(r_t^d-r_t^f)dt +\sigma dW_t^S \end{align} where $S_t$ is the spot exchange rate, the brownian motions are defined under the domestic measure and $\rho_{S,f}$ is the instantaneous correlation of $W_t^S$ and $W_t^f$.

The covered interest parity is given by the no arbitrage argument, hence the forward exchange rate is $F_t(T)=S_t\frac{P_t^f(t,T)}{P_t^d(t,T)}$ with $P_t(t,T)$ the prices of the ZCBs.

Is the uncovered interest parity also respected by the model?

$$E_t[S_T]=?F_t(T)$$

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  • $\begingroup$ What do you mean the covered or uncovered interest parity? $\endgroup$ – Gordon Apr 14 '17 at 14:53
  • $\begingroup$ covered interest parity: $F_t(T)=S_t\frac{P_t^f(t,T)}{P_t^d(t,T)}$ uncovered interest parity: $E_t[S_T]=F_t(T)$ $\endgroup$ – NSZ Apr 14 '17 at 14:57
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Let $K$ be the forward exchange rate determined at time $t$ for maturity $T$. Then the payoff at time $T$ is given by $S_T-K$, which has zero value at time $t$. Let $Q$ and $Q^f$ be the respective domestic and foreign risk-neutral measures, and $E^Q$ and $E^{Q^f}$ be the corresponding expectation operators. Moreover, let $B^d_T = e^{\int_0^t r^d_sds}$ and $B^f_T = e^{\int_0^t r^f_sds}$ be the respective domestic and foreign money market account values at time $t$. Then, \begin{align*} B^d_tE^Q\left(\frac{S_T-K}{B^d_T} \mid \mathcal{F}_t\right) = 0. \end{align*} That is, \begin{align*} K = \frac{B^d_tE^Q\left(\frac{S_T}{B^d_T} \mid \mathcal{F}_t\right)}{E^Q\left(\frac{B^d_t}{B^d_T} \mid \mathcal{F}_t\right)}= \frac{B^d_tE^Q\left(\frac{S_T}{B^d_T} \mid \mathcal{F}_t\right)}{P^d(t, T)}. \tag{1} \end{align*} Let $Q^T$ be the domestic $T$-forward measure and $E^{Q^T}$ be the corresponding expectation operator. Then \begin{align*} \frac{dQ}{dQ^T}\big|_{[t, T]} = \frac{B^d_T}{B^d_tP^d(t, T)}. \end{align*} From $(1)$, \begin{align*} K &= \frac{B^d_t E^Q\left(\frac{S_T}{B^d_T} \mid \mathcal{F}_t\right)}{P^d(t, T)}\\ &=E^{Q^T}(S_T \mid \mathcal{F}_t). \end{align*} That is, \begin{align*} E^{Q^T}(S_T \mid \mathcal{F}_t) = F_t(T). \end{align*} On the other hand, we note that \begin{align*} \frac{dQ}{dQ^f}\big|_{[t, T]} = \frac{B^d_TB^f_t S_t}{B^f_T B^d_tS_T}. \end{align*} Then \begin{align*} E^Q\left(\frac{S_T}{B^d_T} \mid \mathcal{F}_t\right) &= E^{Q^f}\left(\frac{B^d_TB^f_t S_t}{B^f_T B^d_tS_T}\frac{S_T}{B^d_T} \mid \mathcal{F}_t\right)\\ &=\frac{S_t}{B^d_t}E^{Q^f}\left(\frac{B^f_t}{B^f_T} \mid \mathcal{F}_t\right)\\ &=\frac{S_t}{B^d_t}P^f(t, T). \end{align*} Moreover, from $(1)$, \begin{align*} K &= \frac{B^d_t E^Q\left(\frac{S_T}{B^d_T} \mid \mathcal{F}_t\right)}{P^d(t, T)}\\ &=S_t \frac{P^f(t, T)}{P^d(t, T)}. \end{align*} That is, \begin{align*} E^{Q^T}(S_T \mid \mathcal{F}_t) = F_t(T) = S_t \frac{P^f(t, T)}{P^d(t, T)}.\tag{2} \end{align*}

However, we note that, with stochastic interest rates, generally, \begin{align*} E^{Q}(S_T \mid \mathcal{F}_t) \ne F_t(T). \end{align*}

$$$$ As OP has already pointed out, Fromula $(2)$ can also be shown by no arbitrage argument. Specifically, at time $t$, while entering a forward contract with forward exchange rate $F_t(T)$, we borrow one unit domestic currency (which can be used to buy $\frac{1}{P^d(t, T)}$ units domestic zero-coupon bond with maturity $T$), convert into $\frac{1}{S_t}$ units foreign currency, and buy $\frac{1}{S_t P^f(t, T)}$ units foreign zero-coupon bond with maturity $T$. The net value of this trading strategy is zero.

At maturity $T$, in domestic currency, the above trading strategy has value $$F_t(T)\frac{1}{S_tP^f(t, T)}-\frac{1}{P^d(t, T)},$$ which should also have zero value. That is, \begin{align*} F_t(T) &=S_t \frac{P^f(t, T)}{P^d(t, T)}. \end{align*}

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