How can the forward risk neutral measure be used to derive Black's model?

Question

In the Hull textbook's derivation of Black's model (Section 27.6), they apply equation (27.20), which is $f_0 = P(0,T)E_T(f_T)$, where $P(0,T)$ is the value of a zero coupon bond at time $0$ expiring at $T$, and $E_T$ is the expectation with respect to the forward risk neutral measure of the zero coupon bond.

They set $f_T=\max(S_T-K,0)$ to the call payoff, and go from there.

However, $f_0 = P(0,T),E_T(f_T)$ was derived in Section 27.3 assuming that $f_t$ satisfies $df = \mu f dt + \sigma f dz$.

My question:

1. Why is it valid to set $f_T=\max(S_T-K,0)$? That is, how do we show that $f_t$ satisfies $df = \mu f dt + \sigma f dz$?

2. Are all European derivatives $f_T$, not necessarily a call, also of this form?

Answer 1

The equation $f_T = \max \{ S_T - K, 0 \}$ is not an assumption, this is true by definition of what a call option is. It's an option which, at the time of maturity $T$, gives the value $\max\{ S_T - K, 0\}$ to the holder.

And yes, options $f_t$ follow the diffusion $dF_t = \mu dt + \sigma dW_t$ because the underlying stock (or forward) also follows an Ito process, and since the option is a function of that underlying, you can apply Ito's formula to figure out that the option also follows an Ito diffusion.

From Wikipedia:

Here, your underlying spot or forward is represented by $X_t$ and your option is a function $F(X, t)$, which, as the Lemma says, also follows an Ito diffusion.