Pricing a Call Option on a Forward That Expires Later

# European Call Option on Forward Contract


# European Call Option on Forward Contract












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 $K$?

## Answer by Jan Stuller (score 1)

https://quant.stackexchange.com/a/58593

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 $T'\geq T$.