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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?

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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.

Addendum

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.

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  • $\begingroup$ 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. $\endgroup$ – A.Oreo Aug 25 '17 at 1:20
  • $\begingroup$ @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 $\frac{X_T}{B_T^d}$. $\endgroup$ – Gordon Aug 25 '17 at 12:57
  • $\begingroup$ 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? $\endgroup$ – A.Oreo Aug 25 '17 at 13:46
  • $\begingroup$ $E$ is not a measure, instead, it is the expectation operator under the domestic risk-neutral measure. Then $B_t^dE\left( \frac{1}{B_T^d}\right) = P^d(t, T)$. $\endgroup$ – Gordon Aug 25 '17 at 14:17
  • $\begingroup$ 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. $\endgroup$ – A.Oreo Aug 26 '17 at 8:25

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