# FX Forward pricing with correlation between FX and Zero-Cupon

I would like to extend my question about about FX Forward rates in stochastic interest rate setup: FX forward with stochastic interest rates pricing

We consider a FX process $X_t = X_0 \exp( \int_0^t(r^d_s-r^f_s)ds -\frac{\sigma^2}{2}t+ \sigma W_t^2)$ where $r^d$ and $r^f$ are stochastic processes not independent of the Brownian motion $W$. The domestic risk-neutral measure is denoted by $\mathbb Q^d$.

The domestic and foreign bank accounts are $\beta^d$ and $\beta^f$ respectively. The domestic and foreign zero-coupon bond prices of maturity $T$ at time $t$ are respective $B_d(t,T)$ and $B_f(t,T)$.

The domestic bond follows the SDE

$$\frac{dB_d(t,T)}{B_d(t,T)} = r^d_t \ dt + \sigma(t,T) \ dW^1_t$$

with deterministic initial conditions $B_d(0,T)$. The drift and volatility functions in the SDEs are all deterministic functions and $W^1$ and $W^2$ are standard Brownian motions such that $\langle W^1, W^2\rangle_t =\rho \ dt$.

Now consider domestic $\tau$-forward measure $\mathbb Q^{d,\tau}$ $$\left. \frac{d\mathbb Q^{d,\tau}}{d\mathbb Q^d}\right|_{\mathcal F_t} = \frac{B_d(t,\tau)}{\beta_t^dB_d(0,\tau)} = \mathcal E_t \left( \int_0 ^. \sigma(s,\tau) \ dW^1_s\right)$$

and the $\mathbb Q^{\tau}$-Brownian motions $$W^{1,\tau}_. := W^1_. + \int_0 ^. \sigma(s,\tau) \ ds$$

$$W^{2,\tau}_. := W^2_. + \rho \int_0 ^. \sigma(s,\tau) \ ds$$

Question

I would like to calculate the non-deliverable FX forward rate. Since the fixing date $t_f$ is such that $t_f< T$, where $T$ is the settlement date, it implies to pass by the calculation following expectation: $$\mathbb E^{\mathbb Q^d} _t \left[ \exp(-\int_t^T r^d_s ~ds)\ X_{t_f}\right]$$

I can get to this point

$$\mathbb E^{\mathbb Q^d} _t \left[ \exp(-\int_t^T r^d_s ~ds)\ X_{t_f}\right]= B_d(t,t_f)\ \mathbb E^{\mathbb Q^{d,t_f}}_t\left[ B_b(t_f,T)X_{t_f}\right].$$

From that point I am struggling to get through all the calculations. Is there a smart way to compute the last expectation?

• Your problem formulation seems pretty weird to me. Jul 28, 2016 at 8:15
• @Quantuple: Can you develope please? What particularly sounds weird for you?
– Joe
Jul 28, 2016 at 8:20
• The Radon-Nikodym derivative characterising the domestic to domestic $T$-forward measure for instance. Jul 28, 2016 at 8:41
• There is a typo, it should be $\tau$ instead of $T$. Is that what you mean ?
– Joe
Jul 28, 2016 at 8:46
• For the SDE formulation you can assume a HJM setup. So you can have such a representation.
– Joe
Jul 28, 2016 at 9:10

## 1 Answer

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*}

• @Joe, any comments or feedback? Aug 5, 2016 at 16:40
• I hadn't seen this question. I think your solution is true. indeed, it is so nice (+1)
– user16651
Aug 28, 2016 at 20:45
• @Joe Your behavior is not fairly.
– user16651
Aug 28, 2016 at 20:52
• @BehrouzMaleki and Joe, generally, a poster should at least question and challenge the answer, so that we can improve and fine tune the answer together. Aug 29, 2016 at 12:01
• I do not assume that, but you can treat it as an approximation. Apr 18, 2019 at 16:27