# How to regard foreign currency forward as foreign and domestic bonds on VaR

In John Hull's book Options, Futures and Other Derivatives 9th page 507

We want to calculate the VaR of a forward contract of a foreign currency and we should spread forward into two bonds. It's said that

a forward contract to buy a foreign currency. Suppose the contract matures at time T. It can be regarded as the exchange of a foreign zero-coupon bond maturing at time T for a domestic zero-coupon bond maturing at time T.

We know that the payment at $T$ is $$Q_T - K$$ it's easy to understand regarding $K$ as a domestic zero-coupon bond maturing at time $T$ with principle $K.$ But how to regard $Q_T$ as the foreign zero-coupon?

Let $X_t$ be the exchange rate from one unit foreign currency to units domestic currency. Moreover, let $K$ be the forward rate set at the contract inception time, which is usually prior to the VaR calculation date $t_0=0$. Then the value at time $t$, where $t_0 \le T \le T$, of the forward contract is given by \begin{align*} B_t^d\,\mathbb{E}\left(\frac{X_T-K}{B^d_T} \,\big|\, \mathcal{F}_t\right), \end{align*} where $B_t^d$ is the domestic money-market account value, $P^f(t, T)$ is the value at time $t$ of the foreign zero coupon bond with maturity $T$, and $\mathbb{E}$ is the expectation operator under the domestic risk-neutral measure $\mathbb{Q}^d$.

As $P^f(t, T)X_t$ and $P^d(t, T)$ are values at time $t$ of domestic tradable assets. the processes $\left\{\frac{P^f(t, T)X_t}{B_t^d}, \, t_0\le t \le T \right\}$ and $\left\{\frac{P^d(t, T)}{B_t^d}, \, t_0\le t \le T \right\}$ are martingales under the domestic risk-neutral measure. Then, the value at $t$ of the forward contract is given by \begin{align*} B_t^d\,\mathbb{E}\left(\frac{X_T-K}{B^d_T} \,\big|\, \mathcal{F}_t\right) &= B_t^d\,\mathbb{E}\left(\left(\frac{P^f(T, T)X_T}{B^d_T} -K\frac{P^d(T, T)}{B^d_T}\right) \,\big|\, \mathcal{F}_t\right)\\ &=P^f(t, T)X_t - K P^d(t, T). \end{align*} That is, the value of the exchange of certain amount of foreign zero-coupon bond with certain amount of domestic zero-coupon bond.

Here, we provide a mathematical derivation that the process $\left\{\frac{P^f(t, T)X_t}{B_t^d}, \, 0 \le t \le T \right\}$ is a martingale under the domestic risk-neutral measure $\mathbb{Q}^d$.
Let $\mathbb{Q}^f$ be the foreign risk-neutral measure, and $\mathbb{E}^f$ be the corresponding expectation operator. Note that the process $\left\{\frac{P^f(t, T)}{B_t^f}, \, 0 \le t \le T \right\}$ is a martingale under the foreign risk-neutral measure. We denote by $\eta_t$ the Radon–Nikodym derivative \begin{align*} \frac{d\mathbb{Q}^d}{d\mathbb{Q}^f}\big|_t = \frac{B_t^d X_0}{B_t^f X_t}. \end{align*} Then, for $0 \le s \le t \le T$, \begin{align*} \mathbb{E}\left(\frac{P^f(t, T)X_t}{B_t^d}\, \big| \, \mathcal{F}_s\right) &= \mathbb{E}^f\left(\frac{\eta_t}{\eta_s}\frac{P^f(t, T)X_t}{B_t^d}\, \big| \, \mathcal{F}_s\right)\\ &=\mathbb{E}^f\left(\frac{B_t^d}{B_t^f X_t} \frac{B_s^f X_s}{B_s^d}\frac{P^f(t, T)X_t}{B_t^d}\, \big| \, \mathcal{F}_s\right)\\ &=\frac{B_s^f X_s}{B_s^d}\mathbb{E}^f\left(\frac{P^f(t, T)}{B_t^f}\, \big| \, \mathcal{F}_s\right)\\ &=\frac{P^f(s, T)X_s}{B_s^d}. \end{align*} That is, $\left\{\frac{P^f(t, T)X_t}{B_t^d}, \, 0 \le t \le T \right\}$ is a martingale under the domestic risk-neutral measure.
• I think under the domestic risk neutral measure, $\dfrac{B^fX}{B^d}$ is martingale, thus $\dfrac{P^fX}{B^d}$ must not be martingale. – A.Oreo Aug 25 '17 at 1:20
• @A.Oreo: You are right that $\frac{B_t^f X_t}{B_t^d}$ is a martingale under the domestic risk-neutral measure; however, note that the value at maturity $T$ is $\frac{B_T^f X_T}{B_T^d}$, which is not the form $\frac{X_T}{B_T^d}$ that we are going to consider. Instead, the process $\frac{P^f(t, T) X_t}{B_t^d}$ satisfies our requirement, that is, the value at the forward contract maturity $T$ is $\frac{X_T}{B_T^d}$. – Gordon Aug 25 '17 at 12:57
• So, here you no longer use the domestic risk-neutral measure, instead, you use the measure $E$ to make $\dfrac{P^fX}{B^d}$ and $\dfrac{1}{B^d}$ martingale? And under this measure does $B_t^d E[\dfrac{1}{B_T^d}]$ still mean the bond price? – A.Oreo Aug 25 '17 at 13:46
• $E$ is not a measure, instead, it is the expectation operator under the domestic risk-neutral measure. Then $B_t^dE\left( \frac{1}{B_T^d}\right) = P^d(t, T)$. – Gordon Aug 25 '17 at 14:17
• So, the identity $B^d_tE\left(\dfrac{1}{B^d_T}\right) = P^d(t,T)$ only holds under the domestic risk-neutral measure, but $\dfrac{P^fX}{B^d}$ is impossible martingale under the domestic risk-neutral measure. – A.Oreo Aug 26 '17 at 8:25