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



I have limited knowledge of financial econ, so excuse me if I miss the ball completely.

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: $$ 1 = (1 - p_2)u + p_2 d = u - p_2(u - d),\\ \to p_2 = \frac{u - 1}{u - d},\\ \to p_1 = \frac{1 - d}{u - d}. $$

  • $\begingroup$ 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) $\endgroup$
    – BCLC
    Sep 17 at 14:54

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