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 $W$)?


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.

  • $\begingroup$ You may try to use the Bayes formula: If $d\mathcal{Q}^{d} = L d\mathcal{Q}^{f} $ then ${E_t}^{d}[Z] = \frac{{E_t}^{f}[LZ] }{{E_t}^{f}[Z] }$ $\endgroup$
    – Dark
    Jun 10, 2016 at 12:02
  • $\begingroup$ @Dark: your idea is fine, but needs more work. $\endgroup$
    – Gordon
    Jun 10, 2016 at 19:03
  • 1
    $\begingroup$ That would be another question. $\endgroup$
    – Gordon
    Jul 21, 2016 at 12:57
  • $\begingroup$ @Gordon: Thanks Gordon. Please have a look at my new question about non-deliverable forwards. Link above in the question. $\endgroup$
    – Joe
    Jul 27, 2016 at 21:01

2 Answers 2


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.


I think you should look at it the other way around.

Let $X_t$ denote the FOR/DOM spot exchange rate, i.e. 1 unit of foreign currency = $X_t$ units of domestic currency at time $t$.

The FX forward rate $F^X(t,T)$ is defined as $$ F^X(t,T) = X_t \frac{B_f(t,T)}{B_d(t,T)} $$ by basence of arbitrage opportunity.

To understand this, consider the following thought experiment. Let $C_0$ represent a cash amount denominated in the foreign currency. At $t=0$, you - an investor with initial capital $C_0$ - basically face a choice:

  • Invest in the foreign economy: buy $C_0/B^f(0,T)$ units of foreign zero-coupon bonds at $t=0$. This costs you $C_0$. At $t=T$, you shall receive (assuming no default): $C_0/B^f(0,T)$.
  • Invest in the domestic economy: exchange your $C_0$ units of the foreign currency for $C_0 X_0$ units of domestic currency and buy $C_0 X_0/B^d(0,T)$ units of domestic zero-coupon bonds at $t=0$. This costs you $C_0 X_0$. At $t=T$, you shall receive (assuming no default): $C_0 X_0 / B^d(0,T)$.

Now to preclude arbitrage opportunities, it is clear that the two strategies should be equivalent as seen of today. In other words, the fair forward exchange rate should be such that, what you receive when you invest $C_0$ in the foreign economy, converted at the fair forward rate, should match what you receive when you invest $C_0$ in the domestic economy. With the previous notations this writes: $$ \left( C_0/B^f(0,T) \right) \times F^X(0,T) = C_0 X_0 / B^d(0,T)$$ hence $$ F^X(0,T) = X_0 \frac{B^f(0,T)}{B^d(0,T)} $$

[REM 1]: you need to consider bonds vs investing in the respective money market accounts when interest rates are stochastic, because the evolution of the money market accounts are not deterministic anymore: when you invest, you don't know what you'll get in the end. Buying bonds and assuming zero-default saves the day.

[REM 2]: As explained in Gordon's answer if you want to write the forward exchange rate as a martingale under a certain measure $\mathbb{X}$ $$ F^X(0,T) = X_0 \frac{B^f(0,T)}{B^d(0,T)} = \mathbb{E}^{\mathbb{X}}_0 \left[ X_T \underbrace{\frac{B^f(T,T)}{B^d(T,T)}}_{=1} \right] = \mathbb{E}_0^{\mathbb{X}} [ F(T,T) ]$$ you see that this measure is associated to the numéraire $B^d(t,T)$ since the last expression is equivalent to writing that: $$ F(t,T) = \frac{X_t B^f(t,T)}{B^d(t,T)} \text{ is a } \mathbb{X} \text{ martingale} $$ and $\mathbb{X}$ is called the $T$-forward domestic measure, usually denoted $\mathbb{Q}^d_T$.

[REM3]: If you're wondering whether your proposed SDE is consistent with the above results, here's how you can convince yourself. Applying Itô's lemma to your SDE (expressed under the domestic risk-neutral measure $\mathbb{Q}^d$) and integrating from $t=0$ to $t=T$ gets you $$ X_T = X_0 \exp\left(\int_0^T (r^d_t-r^f_t) dt - \frac{1}{2}\sigma^2 T + \sigma W_T^{\mathbb{Q}^d} \right) = X_0 \frac{B^d(T)}{B^f(T)}\mathcal{E}[\sigma W_T^{\mathbb{Q}^d}]$$ where I've used the notation $B(t)$ for the $t$-value of a money market account in the domestic/foreign economies where $1$ unit of currency has been invested at $t=0$.

The quanto change of measure then classically writes (see related SE questions) $$ \left. \frac{d\mathbb{Q}^f}{d\mathbb{Q}^d} \right\vert_{\mathcal{F}_T} = \frac{X_T B^f(T)}{X_0 B^d(T)} = \mathcal{E}[\sigma W_T^{\mathbb{Q}^d}] $$ given the expression of $X_T$ found above (note how this is a Doléans-Dade exponential).

Now we would like to compute the fair forward price as an expectation under the domestic $T$-forward measure $\mathbb{Q}^d_T$ whose Radon-Nikodym derivative wrt $\mathbb{Q}^d$ reads: $$ \left. \frac{d\mathbb{Q}^d_T}{d\mathbb{Q}^d} \right\vert_{\mathcal{F}_T} = \frac{1/B^d(T)}{B^d(0,T)} $$ (notice how this equals one for deterministic rates). We then have: \begin{align} F^X(0,T) &= \mathbb{E}^{\mathbb{Q}^d_T}_0[ X_T ] \\ &= \mathbb{E}^{\mathbb{Q}^d}_0 \left[ X_T \left. \frac{d\mathbb{Q}^d_T}{d\mathbb{Q}^d} \right\vert_{\mathcal{F}_T} \right] = \mathbb{E}^{\mathbb{Q}^d}_0 \left[ X_T \frac{1/B^d(T)}{B^d(0,T)} \right] \tag{1} \\ &= \frac{1}{B^d(0,T)} X_0 \mathbb{E}^{\mathbb{Q}^d}_0 \left[ \frac{1}{B^f(T)} \mathcal{E}[\sigma W_T^{\mathbb{Q}^d}] \right] \tag{2} \\ &= \frac{1}{B^d(0,T)} X_0 \mathbb{E}^{\mathbb{Q}^f}_0 \left[ \frac{1}{B^f(T)} \underbrace{\mathcal{E}[\sigma W_T^{\mathbb{Q}^d}] \left. \left(\frac{d\mathbb{Q}^f}{d\mathbb{Q}^d}\right)^{-1} \right\vert_{\mathcal{F}_T}}_{=1} \right] \tag{3} \\ &= X_0 \frac{B^f(0,T)}{B^d(0,T)} \tag{4} \end{align} where we have used:

  1. Change of measure from the $T$-forward domestic measure $\mathbb{Q}^d_T$ to the dometic measure $\mathbb{Q}^d$ + expression of the corresponding Radon-Nikodym derivative
  2. Definition of $X_t$ solution of the original SDE provided under the domestic measure $\mathbb{Q}^d$
  3. Change of measure from domestic measure $\mathbb{Q}^d$ to foreign measure $\mathbb{Q}^f$ + expression of the corresponding Randon-Nikodym derivative
  4. Definition of the price of a zero-coupon bond under the foreign measure $\mathbb{Q}^f$

Conclusion you can safely use that SDE as it implies no blatant arbitrage issues.

  • $\begingroup$ Can I ask you why you did not discount the FX rate to obtain the fair price of forward contract? I mean equation (1). $\endgroup$
    – user53249
    Jun 4, 2021 at 11:27

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