Does anyone know of a decent proof for the spot-forward relationship of a currency? I've been looking on Google for hours and I'm not getting anywhere. My lecture notes are useless in that they don't even tell us what the spot-forward relationship is. I'm presuming it is in a Black-Scholes setting with constant interest rates but who knows because my lecturer doesn't feel the need to include information like that.

This is the question: Write and prove in details the spot-forward relationship satisfied at date $t$ by the price of a currency $X(t)$ and the forward price $f(t,T)$.

That's all I have to go on. Any help will be gratefully received.


Let's go for a detailed and rigorous proof.

Let us define our local currency $Y$ as the numéraire, i.e. the asset in terms of whose price the relative prices of all other tradeables are expressed. $X$ is therefore the foreign currency, whose price in terms of $Y$ is $X(t)$ at any time $t$.

Let $r_X(t,T)$ be the risk-free interest rate in currency $X$ and $r_Y(t,T)$ the risk-free interest rate in currency $Y$, for maturity $T-t$ and at time $t$. Both are continuously coumpounded. Let two portfolios of value $1$ (in terms of currency $Y$) at time $t$: $P_1(t)=P_2(t)=1$.

The first portfolio $P_1$ consists in buying currency $X$ at spot price $X(t)$ and investing this amount in the risk-free rate of currency $X$. The final value at time $T$ in currency $Y$ is therefore: $P_1(T)=X(t)*e^{(T-t)r_X(t,T)}/X(T)$.

The second portfolio $P_2$ consists in buying currency $X$ in the future at time $T$, and at a forward price $f(t,T)$ determined at $t$. In the meanwhile, the $1$ is invested in the risk-free rate in currency $Y$. The final value at time $T$ in currency $Y$ is therefore: $P_2(T)=e^{(T-t)r_Y(t,T)}*f(t,T)/X(T)$.

Note that the two portfolios are riskless and have the same initial value. Hence, by no-arbitrage we must have $P_1(s)=P_2(s), \forall s\geq t$, and in particular:

$$P_1(T)=P_2(T)$$ $$X(t)*e^{(T-t)r_X(t,T)}/X(T)=e^{(T-t)r_Y(t,T)}*f(t,T)/X(T)$$ $$f(t,T)=X(t)*e^{(T-t)(r_X(t,T)-r_Y(t,T))}$$

Hope this helps!

  • $\begingroup$ Perfect - that all makes sense. Exactly what I was looking for. Thanks. It's been surprisingly difficult to find such an elegant explanation. $\endgroup$ – Joe Bloggs Feb 25 '18 at 20:26
  • $\begingroup$ You're welcome. $\endgroup$ – Soumirai Feb 25 '18 at 20:28
  • $\begingroup$ Sorry - i tried to edit the comment to tag you but it wouldn't let me. Wanted to make sure you saw it. Thanks again. Much appreciated. $\endgroup$ – Joe Bloggs Feb 25 '18 at 20:29

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