# European Call Option on Forward Contract [closed]

I am doing a problem on pricing european call option on forward contract. The forward contract has maturity $$T^{'}$$ and the option has maturity $$T < T^{'}$$ with strike $$K$$. Assuming the underlying asset price process is governed by BS model: $$dS_t = \mu S_tdt + \sigma S_tdB_t.$$ I am confused by the payoff of this option. I checked online and some says it should be $$(\text{difference between forward price at T and K})^{+}$$. However, I was thinking why it is not the difference between the value of forward contract at $$T$$ and $$K$$?

• Review how options on futures/forwards work. A new forward is created iff at the Maturity Date $F>K$, the new forward is marked to market, so its initial value is zero. The call owner will be long the forward and the call writer will end up with the short position in the forward. (This is, for example, how call options on S&P futures work). – noob2 Oct 10 '20 at 9:38
• So long story short a payoff of $(F-K)^+$ is correct. – noob2 Oct 10 '20 at 11:21
• @noob2 Got it, thx! – Van Tom Oct 10 '20 at 12:51
• I beleive your remark is correct @Van Tom, A call option on a forward contract, is the product whose payoff is given by $(\pi_T-K)^+$ where $\pi_T$ is the price of a forward contract. Wheover, it is more convenient to express the price of a call option with respect of the forward price $F(t,T)$ as it gives a more general expression, that is also correct in the case of stochastic rates. – DeepInTheQF Oct 10 '20 at 22:43
• And that is why you might found some texts where the black-scholes closed form formula for a call option is given with respect to $F(t,T)$. Also worth noting that, $(S_T-K)^+=(F(T,T)-K)^+$ which a standard call option on $S$ not to be confused with an option on a forward contract on $S$ – DeepInTheQF Oct 10 '20 at 22:52

## 1 Answer

I think a shortcut to your pricing problem could be to use the following approach: Suppose today is time $$t_0$$. As of today, the Forward value of the stock is:

$$F(t_0,T')=\mathbb{E}[S(T')]=e^{r(T'-t_0)}S_0$$

You can treat the above as an identity and invert the relationship to get the Stock value $$S_0$$ as a function of the Forward value and the discount factor:

$$S_0=F(t_0,T')e^{-r(T'-t_0)}$$

For ease of notation, let's suppose that $$t_0=0$$ so we get: $$S_0=F(t_0,T')e^{-r T'}$$.

The Black-Scholes formula based on the underlying stock is:

$$C=N(d)S_0 - e^{-rT}KN(d - \sigma \sqrt{T})$$

With:

$$d=\frac{ln \left( \frac{S_0}{K} \right)+rT+0.5\sigma^2T}{\sigma \sqrt{T}}$$

Now substitute $$F(t_0,T')e^{-r T'}$$ for $$S_0$$ in the above, to get (I use $$F$$ instead of $$F(t_0,T')$$ for ease of notation) :

$$d=\frac{ln \left( \frac{Fe^{-rT'}}{K} \right)+rT+0.5\sigma^2T}{\sigma \sqrt{T}}=\frac{ln \left( \frac{F}{K} \right)+r(T-T')+0.5\sigma^2T}{\sigma \sqrt{T}}$$

In the Option price, also substitute $$Fe^{-r T'}$$ for $$S_0$$ to get:

$$C=N(d)Fe^{-rT'} - e^{-rT}KN(d - \sigma \sqrt{T})=e^{-rT}\left(N(d)Fe^{-r(T'-T)} - KN(d - \sigma \sqrt{T})\right)$$

And that should give the answer to how to value an option expiring at time $$T$$ on a forward that expires at time $$T'\geq T$$.

• Shouldn't the last term in the numerator in formula for $d$ be $+\frac{\sigma^2}{2}T$? And why simply subsitituting the underlying asset price in the formula by $Fe^{-rT^{'}}$ will give me the price of this option? – Van Tom Oct 10 '20 at 9:56
• @VanTom Thank you, you are correct about the last term in $d$, that was a typo. Substituting the discounted forward price for the underlying price gives the correct answer, because the discounted Forward is identical to the underlying: so expressing the option price in terms of one or the other is equivalent. – Jan Stuller Oct 10 '20 at 10:06