I am self-studying for an actuarial exam on models for financial economics. I am having difficulty thinking about the put-call parity for currency options, specifically how use the notation. Here is the problem:

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My book uses the notation $C(x_0, K, T)$ to mean a call option on currency with a spot exchange rate $x_0$ to purchase it at exchange rate $K$ at time $T$, and $P(x_0, K, T)$ the corresponding put option.

Please tell me if my interpretation is correct:

I interpreted (iv) to mean $P(1 £, 1.5\frac{\$}{£}, 0.5) = \$0.03.$

I interpreted that the problem is asking us to find $C(1\$, \frac{1}{1.5} \frac{£}{\$}, 0.5)$.

The problem is that the first argument of those options do not appear to be rates.

This may not help, but by duality, $P(1 £, 1.5\frac{\$}{£}, 0.5) = C(1.5\frac{\$}{£}, 1 £, 0.5)$.

I don't see how to take what we're given, and convert to what the problem is asking us to find.


Let $\{X_t \mid t \ge 0\}$ be the foreign exchange rate rate from $£$ to $\$$. Moreover, let $C(X_0, K, T)$ and $P(X_0, K, T)$ be the prices of the respective call and put options with strike $K$ and maturity $T$. Then \begin{align*} \frac{1}{X_0}P(X_0,\, K,\, T) = K C\left(\frac{1}{X_0},\, \frac{1}{K},\, T \right). \end{align*} Based on the given condition, \begin{align*} P\left(1.5 \$/£, \, 1.5 \$/£,\, 0.5 \right) = 0.03. \end{align*} Then, \begin{align*} C\left(\frac{1}{1.5} £/\$, \, \frac{1}{1.5} £/\$,\, 0.5 \right) &= \frac{1}{1.5\times 1.5}\times P\left(1.5 \$/£, \, 1.5 \$/£,\, 0.5 \right)\\ &\approx 0.01333. \end{align*}

$$ $$ The above duality formula can be derived as follows. Note that, for the put option payoff at maturity $T$, \begin{align*} (K-X_T)^+ = KX_T\left(\frac{1}{X_T} - \frac{1}{K} \right)^+. \end{align*} Let $P_d$ and $P_f$ denote, respectively, the USD and GBP risk-neutral measures. Moreover, let $E_d$ and $E_f$ denote the expectation operators corresponding to $P_d$ and $P_f$. Note that, \begin{align*} \frac{dP_d}{dP_f}\big|_T=\frac{X_0 e^{r_d T}}{X_T e^{r_f T}}, \end{align*} where $r_d$ and $r_f$ are the respective USD and GBP interest rates. Then, \begin{align*} P(X_0, K, T) &= E_d\left(\frac{1}{e^{r_d T}} (K-X_T)^+\right)\\ &=E_d\left(\frac{KX_T}{e^{r_d T}} \left(\frac{1}{X_T} - \frac{1}{K} \right)^+\right)\\ &= E_f\left(\frac{dP_d}{dP_f}\big|_T\frac{KX_T}{e^{r_d T}} \left(\frac{1}{X_T} - \frac{1}{K} \right)^+\right)\\ &= X_0 K E_f\left(\frac{1}{e^{r_f T}} \left(\frac{1}{X_T} - \frac{1}{K} \right)^+\right)\\ &= X_0 K C\left(\frac{1}{X_0},\, \frac{1}{K},\, T \right). \end{align*} That is, \begin{align*} \frac{1}{X_0}P(X_0,\, K,\, T) = K C\left(\frac{1}{X_0},\, \frac{1}{K},\, T \right). \end{align*}


It costs 0.03 dollars for the option to (sell 1 pound/buy 1.5 dollars. Now divide everything by 1.5:
It costs 0.02 dollars for the option to (sell 2/3 pound / buy 1 dollar). Now convert to pounds at spot rate:
It costs 0.0133 pounds for the option to (sell 2/3 pound / buy 1 dollar). Done


A call lets to purchase one unit of underlying for some strike price x. So a call on GBP in USD lets us buy 1 unit of GBP for price x. However, since this is FX, lets clarify this to be USD x and USD 1 gets us GBP 1/x.

A put lets you sell one unit of underlying for some strike price y (= 1/x). So a put on USD in GBP lets us sell 1 unit of USD for price 1/x. Again, to be specifc, this is GBP 1/x. so, for x units of USD, we get 1 GBP

Both cases above are the same transaction (outflow USD for inflow GBP with options). Based on the base currency, we can view it as a call (USD base) or a put (GBP base). This is what @Alex C mentioned in his answer.


In foreign exchange a contract can equally be seen as a put or a call, depending on the point of view: a call on dollars or a put on sterling. This is not Put-call-parity, which is not needed for this problem, it is just two names for the same thing. All you need to do is to invert the strike and convert the price to the other currency: 0.03 usd is 0.02 gbp.

  • $\begingroup$ I guess I'm not seeing why inverting the strike price and converting the currency will convert between a call of one denomination and a put of the other denomination. $\endgroup$ – user2521987 Nov 12 '15 at 1:21

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