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

?

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    $\begingroup$ I agree with you. I guess this is just a way to prove the implication in both directions : if $p=(1-d)/(u-d)$ then $E[F_t]=F_0$ and if $E[F_t]=F_0$ then $p=(1-d)/(u-d)$. These are two different propositions. $\endgroup$
    – Louis. B
    Commented Dec 11, 2015 at 0:05
  • $\begingroup$ @Louis.B thanks. So which one is true by definition and implies the other? $\endgroup$
    – BCLC
    Commented Dec 11, 2015 at 5:45
  • $\begingroup$ There is no definition, it's just something that you demonstrate like a theorem in math. Let me put that in an answer. $\endgroup$
    – Louis. B
    Commented Dec 11, 2015 at 7:50
  • $\begingroup$ By proving is both ways you show that $E[F_t] = F_o \iff p = \frac{1-d}{u-d}$, I think that's his point. $\endgroup$
    – SRKX
    Commented Dec 11, 2015 at 9:22
  • $\begingroup$ @SRKX his ? :P $\endgroup$
    – BCLC
    Commented Dec 11, 2015 at 9:38

2 Answers 2

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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.

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  • $\begingroup$ Thanks Louis. B. So Hull is being confusing when he seems to claim $E[F_T] = F_0$ is true by definition? Or am I understanding him wrong? $\endgroup$
    – BCLC
    Commented Dec 11, 2015 at 9:45
  • $\begingroup$ Is it really that in the Chapter 17 text above $E[F_T] = F_0$ is not an assumption? In that case I guess my prof is right $\endgroup$
    – BCLC
    Commented Dec 11, 2015 at 9:47
  • $\begingroup$ @BCLC I don't remember everything in the chapter 17 and I don't know what your prof said before that, but these are really two equivalent propositions. This just means that when you are building your model, whenever you get one of these you can equivalently replace by the other one, and nothing else. My personal view is that, often, you want the expectation to be $F_0$ so you pick that value for $p$ which then yields the risk-neutral world. $\endgroup$
    – Louis. B
    Commented Dec 11, 2015 at 19:15
  • $\begingroup$ Louis. B, ch 17 text is the one quoted above. Is there a $E[F_T] = F_0$ being assumed there? $\endgroup$
    – BCLC
    Commented Dec 11, 2015 at 23:42
  • $\begingroup$ ? en.wikipedia.org/wiki/Futures_contract#Pricing_via_expectation $\endgroup$
    – BCLC
    Commented Dec 11, 2015 at 23:44
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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 $\mathbb{E}(F_\tau)$.

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