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

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

?

The no-arbitrage condition implies that the value of an asset should be a weighted sum of the payoffs of the asset over the various states of the world. Here you have two states, the good one and the bad one. If $$F_0$$ is the price of the asset, $$F_0 u$$ the payoff in the good state and $$F_0d$$ the payoff in the bad state, then we get: $$F_0 = p_1 F_0 u + p_2 F_0 d.$$ Here $$p_1$$ and $$p_2$$ are the Arrow-security prices.
Dividing by $$F_0$$ you get: $$1 = p_1 u + p_2 d \tag{1}$$ As another asset, you can also invest your money in a riskless bond with zero interest. If $$B$$ is the price of the bond, then this gives $$B$$ in state 1 and $$B$$ in state 2. As such: $$B = p_1 B + p_2 B$$ Dividing by $$B$$ gives: $$1 = p_1 + p_2 \tag{2}$$ Combining $$(1)$$ and $$(2)$$ gives: $$1 = (1 - p_2)u + p_2 d = u - p_2(u - d),\\ \to p_2 = \frac{u - 1}{u - d},\\ \to p_1 = \frac{1 - d}{u - d}.$$
• what? i'm asking if we indeed get $p_1$ as you have computed instead of that $p_1$ is given and then we compute something else (the expectation)