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

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?

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.
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.
• This seems a bit circular. Ito's lemma requires that $f_t$ is twice differentiable. How do we know that $f_t$ is twice differentiable without knowing the form of $f_t$ (we're not assuming yet that $f_t$ follows the Black-Scholes formula). Is there a theorem that says assuming some regularity of $f_T$, then Ito's lemma is applicable to $f_t$? – leacorv Jul 2 '19 at 1:19