Put-Call relationship for Option on Forward

The forward price of a forward contract maturing at time T on an asset with price St at time t is,

$$ F=S_te^{(r-q)(T-t)} $$

where $r$ is the risk free rate and $q$ is the continuous dividend rate for $S_t$.

The Black Scholes equation for an option contingent on F is, $$ \frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2F^2\frac{\partial ^2V}{\partial F^2} -rV = 0 $$

How do i show that the prices of European call, C, and put options, P, on the forward F, with the same strike K and expiry date $T_1$, where $T_1 < T$ (ie, the options expire before the forward matures), are related by

$$ C(F,t)=\frac{F}{K}P(\frac{K^2}{F},t) $$

Thanks!

Answer

Let $\{F(t, T), 0 \leq t \leq T\}$ be the forward process that satisfies an SDE of the form \begin{align*} dF(t, T) = \sigma F(t, T) dW_t, \end{align*} where $\sigma$ is the constant volatility, $\{W_t, t>0\}$ is a standard Brownian motion. The payoff at time $T_1$, where $0 < T_1 \leq T$, of a vanilla European forward option is of the form \begin{align*} \max(\psi (F(T_1, T)-K), \, 0), \end{align*} where $\psi = 1$, for a call option, and $-1$, for a put option. Note that, for any $0\leq t \leq T_1$, \begin{align*} F(T_1, T) = F(t, T) \exp\Big(-\frac{\sigma^2}{2} (T_1 -t) + \sigma \sqrt{T_1 -t} \xi \Big), \end{align*} where $\xi$ is a standard normal random variable. Then the value at time $t$ of the option payoff above is given by \begin{align*} d(t, T_1)\psi\Big[F(t, T) \Phi\big(\psi d_1(F)\big) -K \Phi\big(\psi d_2(F)\big) \Big], \end{align*} where $d(t, T_1)$ is the discount factor, \begin{align*} d_1 (F) = \frac{\ln \frac{F(t, T)}{K} + \frac{\sigma^2}{2} (T_1 -t)}{\sqrt{T_1-t}\,\sigma}, \end{align*} and \begin{align*} d_2 (F) = \frac{\ln \frac{F(t, T)}{K} - \frac{\sigma^2}{2} (T_1 -t)}{\sqrt{T_1-t}\,\sigma}. \end{align*} That is, \begin{align*} C(F, t) = d(t, T_1)\Big[F(t, T) \Phi\big(d_1(F)\big) -K \Phi\big(d_2(F)\big) \Big], \end{align*} and \begin{align*} P(F, t) = d(t, T_1)\Big[K \Phi\big(-d_2(F)\big) -F(t, T) \Phi\big(-d_1(F)\big)\Big], \end{align*} Note that, by replacing $F$ in $d_1$ with $K^2/F(t, T)$, \begin{align*} d_1 \Big(\frac{K^2}{F}\Big) &= \frac{\ln \frac{K^2/F(t, T)}{K} +\frac{\sigma^2}{2} (T_1 -t)}{\sqrt{T_1-t}\,\sigma}\\ &= \frac{-\ln \frac{F(t, T)}{K} + \frac{\sigma^2}{2} (T_1 -t)}{\sqrt{T_1-t}\,\sigma}\\ &= -d_2(F). \end{align*} Similarly, \begin{align*} d_2 \Big(\frac{K^2}{F}\Big) &= \frac{\ln \frac{K^2/F(t, T)}{K} -\frac{\sigma^2}{2} (T_1 -t)}{\sqrt{T_1-t}\,\sigma}\\ &= \frac{-\ln \frac{F(t, T)}{K} - \frac{\sigma^2}{2} (T_1 -t)}{\sqrt{T_1-t}\,\sigma}\\ &= -d_1(F). \end{align*} Then \begin{align*} \frac{F}{K}P\bigg(\frac{K^2}{F}, t \bigg) &= d(t, T_1)\frac{F}{K}\Bigg[K \Phi\bigg(-d_2\bigg(\frac{K^2}{F}\bigg)\bigg) -\frac{K^2}{F} \Phi\bigg(-d_1\bigg(\frac{K^2}{F}\bigg)\bigg)\Bigg]\\ &= d(t, T_1)\bigg[F \Phi\Bigg(-d_2\bigg(\frac{K^2}{F}\bigg)\bigg) -K \Phi\bigg(-d_1\bigg(\frac{K^2}{F}\bigg)\bigg)\Bigg]\\ &= d(t, T_1)\Big[F(t, T) \Phi\big(d_1(F)\big) -K \Phi\big(d_2(F)\big) \Big]\\ &= C(F, t). \end{align*}