2
$\begingroup$

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


enter image description here


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


enter image description here


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

?

$\endgroup$

1 Answer 1

1
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ 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
    Commented Sep 17, 2021 at 14:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.