$E[F_T] = F_0$ implies $p = \frac{1-d}{u-d}$? or is implied by?

Question

From Ch 12 in Hull's OFOD, we compute the risk-neutral probabilities for a futures contract:

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Later in Ch 17, futures options are valued, and we have the same result:

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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: $p = \frac{1-d}{u-d} \to E[F_T] = F_0$

?

Answer 1

As I commented, I think this is simply a way to prove that both statements are equivalents, that is when the implication goes in both directions. There are no such things as a definition, it's all about the assumption that you make.

Actually, a more general point could be the following: $u$ and $d$ are define such that after one period the asset gets $u$ with probability $\hat{p}$ and $d$ with probability $(1-\hat{p})$. Then the following proposition holds :

\begin{equation} \forall \hat{p}\hspace{0.5cm}\exists p\in\mathbf{R}\hspace{0.2cm};\hspace{0.2cm}puF_0 + (1-p)dF_0 = F_0 \end{equation}

This is called change of measure in mathematical term. Then risk neutral measure is then obtained by setting $\hat{p}=\frac{1}{2}$.

You should view that as a tool rather than something that is "true" because actually this is a change of measure like any other, there are no risk-neutral probability in the "real" world, it's artificial.

Answer 2

Let's first make your problem more rigorous. Suppose, $F_t$ is the future price of underlying security $S_t$, maturing at time $T$. Now you need to prove that, $$\mathbb{E}(F_\tau)=F_t, \quad \forall \tau \in [t, T]$$

By using no arbitrage principle (creating replicating portfolio), it can be easily prove that: $$F_t = S_t e^{r(T-t)}=\mathbb{E}_\mathbb{Q}[S_T|S_t]$$ where, $\mathbb{Q}$ represent risk neutral measure. We can write, future price at time $\tau \in [t,T]$ as, $$F_\tau=S_\tau e^{r(T-\tau)}$$

The above future price $F_\tau$ represents actual future price at time $\tau$. We want expression for $\mathbb{E}(F_\tau|\mathscr{F}_t)$. Just take expectation on both side in the last equation assuming we are still at time $t$, so both $F_\tau$ and $S_\tau$ is random variable. We have; \begin{align} \mathbb{E}(F_\tau)&=e^{r(T-\tau)}\mathbb{E}_\mathbb{Q}(S_\tau)\\ &=e^{r(T-\tau)}S_te^{r(\tau -t)}\\ &=S_te^{r(T-t)}\\ &=F_t \end{align}

NB: $\mathbb{E}(F_\tau)$ is conditional on the filtration upto time $t$. It must be written as $\mathbb{E}(F_\tau|\mathscr{F}_t)$, instead of $\mathbb{E}(F_\tau)$.