FX Futures pricing formula

Question

I'm reading Paul Wilmott's Introduces Quantitative Finance and stuck a bit with formula $F = S(t)e^{(r-r_f)(T-t)}$ for FX futures pricing. I don't get how to incorporate $r_f$ into the formula, could you please help? There is a description in the book for simple futures:

We know all of $F, S(t), t$ and $T$, but is there any relationship between them? You might think not, since the forward contract entitles us to receive an amount $S(T) − F$ at expiry and this is unknown. However, by entering into a special portfolio of trades now we can eliminate all randomness in the future. This is done as follows.

Enter into the forward contract. This costs us nothing up front but exposes us to the uncertainty in the value of the asset at maturity. Simultaneously sell the asset. It is called going short when you sell something you don’t own. This is possible in many markets, but with some timing restrictions. We now have an amount $S(t)$ in cash due to the sale of the asset, a forward contract, and a short asset position. But our net position is zero. Put the cash in the bank, to receive interest.

When we get to maturity we hand over the amount F and receive the asset, this cancels our short asset position regardless of the value of $S(T)$. At maturity we are left with a guaranteed $−F$ in cash as well as the bank account. The word guaranteed is important because it emphasizes that it is independent of the value of the asset. The bank account contains the initial investment of an amount $S(t)$ with added interest, this has a value at maturity of $$S(t)e^{r(T-t)}$$ Our net position at maturity is therefore $$S(t)e^{r(T-t)} - F$$

Since we began with a portfolio worth zero and we end up with a predictable amount, that predictable amount should also be zero. We can conclude that $$F = S(t)e^{r(T-t)}$$

Answer 1

Have you seen formulae for stocks where a dividend yield $q$ pops up? The same idea applies here and is more general a part of the notion of cost of carry. In general, $$F_t=S_te^{b(T-t)},$$ where $b$ is the cost of carry, i.e. $$b=\text{Cost from holding the asset} - \text{benefits from holding the asset}.$$

For a stock, you gain the dividend yield $q$ but you have to ``pay'' the risk-free rate $r$ since you cannot invest in a risk-free bank account (so called opportunity cost). Thus, $b=r-q$.

For an investment in a foreign currency, $r=r$ and $q=r_f$. This means that if you live in the UK and you invest in the US and if \$/£ increases, you make extra money! If however £/\$ rises, you lose money. Thus, your opportunity cost is still the investment in your domestic market with yield $r$. However, you do obtain $r_f$ in the foreign market, just as you obtain dividends from a stock. Thus, $b=r-r_f$.

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Edit

To boil down Paul Wilmott’s example, consider the following example. Suppose you are from the US but happen to have £1. What can you do with that money?

• Case 1) Exchange it immediately into USD and keep USD until the end.
• Case 2) Keep the GBP and exchange it later using a forward.

What happens in Case 1)?

• At time $t$: You have $\$1\cdot S_t$, where $S_t$ is the current exchange rate
• At time $T$: You have $\$1\cdot S_t\cdot e^{r\cdot (T-t)}$ since you received risk-free interest rate in your domestic US currency

What happens in Case 2)?

• At time $t$: You simply have £1 and you enter a forward contract that costs nothing upright
• At time $T$: You have \$1$\cdot e^{r_f\cdot (T-t)} \cdot F_t$ where $F_t$ is the forward exchange rate and £1$\cdot e^{r_f(T-t)}$ is accumulated capital

Since both future cash flows were generated with the same initial wealth, they need to be equal, \begin{align*} e^{r_f\cdot(T-t)}\cdot F_t = S_t \cdot e^{r\cdot (T-t)} \implies F_t = S_t\cdot e^{(r-r_f) \cdot (T-t)}. \end{align*}

The Wilmott example is then a combination of both cases such that your initial wealth is zero. This works always by borrowing or lending some money.