From Ch 12 in Hull's OFOD, we compute the risk-neutral probabilities for a futures contract:
Later in Ch 17, futures options are valued, and we have the same result:
In relation to Chapter 16 and 17, my Derivatives Pricing prof gave us this exercise:
Show that, in the Risk-Neutral World, $E[F_T] = F_0$
I guess, $F_T$ is the random variable s.t.
$$F_T = 1_{A}F_0u + 1_{A^C}F_0d$$
where $A$ is the event corresponding to case 1.
The solution:
$$E[F_T] = pF_0u + (1-p)F_0d$$
$$= \frac{1-d}{u-d}F_0u + \frac{u-1}{u-d}F_0d = F_0$$
That seems strange. To me it seems that the reason why we know that $p = \frac{1-d}{u-d}$ is because $E[F_T] = F_0$ based on 'If $F_0$ is the initial futures price, the expected futures price at the end of one time step of length $\Delta t$ should also be $F_0$' from Ch 12.
Iirc, my prof said that the reason why we have 'If $F_0$ is the initial futures price, the expected futures price at the end of one time step of length $\Delta t$ should also be $F_0$' is because of said exercise which comes from $p = \frac{1-d}{u-d}$.
So how do we get $p = \frac{1-d}{u-d}$ without $E[F_T] = F_0$?
In both texts from Ch 12 and 17, it seems that $E[F_T] = F_0$ is an assumption. Am I wrong? Is $E[F_T] = F_0$ not an assumption in Ch 17? So $E[F_T] = F_0$ comes from Ch 17? That seems very inconsistent of Hull:
Ch 12 proposition: $E[F_T] = F_0 \to p = \frac{1-d}{u-d}$
Ch 17 proposition: $p = \frac{1-d}{u-d} \to E[F_T] = F_0$
?