<p>I have limited knowledge of financial econ, so excuse me if I miss the ball completely.</p>
<p>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 <span class="math-container">$F_0$</span> is the price of the asset, <span class="math-container">$F_0 u$</span> the payoff in the good state and <span class="math-container">$F_0d$</span> the payoff in the bad state, then we get:
<span class="math-container">$$
F_0 = p_1 F_0 u + p_2 F_0 d.
$$</span>
Here <span class="math-container">$p_1$</span> and <span class="math-container">$p_2$</span> are the Arrow-security prices.</p>
<p>Dividing by <span class="math-container">$F_0$</span> you get:
<span class="math-container">$$
1 = p_1 u + p_2 d \tag{1}
$$</span>
As another asset, you can also invest your money in a riskless bond with zero interest. If <span class="math-container">$B$</span> is the price of the bond, then this gives <span class="math-container">$B$</span> in state 1 and <span class="math-container">$B$</span> in state 2. As such:
<span class="math-container">$$
B = p_1 B + p_2 B
$$</span>
Dividing by <span class="math-container">$B$</span> gives:
<span class="math-container">$$
1 = p_1 + p_2 \tag{2}
$$</span>
Combining <span class="math-container">$(1)$</span> and <span class="math-container">$(2)$</span> gives:
<span class="math-container">$$
1 = (1 - p_2)u + p_2 d = u - p_2(u - d),\\
\to p_2 = \frac{u - 1}{u - d},\\
\to p_1 = \frac{1 - d}{u - d}.
$$</span></p>

CC BY-SA 4.0
2021-09-17T14:54:20.850Z
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)
BCLC
CC BY-SA 4.0
2021-09-17T12:46:39.090Z
2021-09-17T12:46:39.090Z
tdm
<p>From Ch 12 in <a href="http://workofark.com/pdf/options.PDF" rel="nofollow noreferrer">Hull's OFOD</a>, we compute the risk-neutral probabilities for a futures contract:</p>
<hr />
<p><a href="https://i.sstatic.net/cr6e0.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/cr6e0.png" alt="enter image description here" /></a></p>
<hr />
<p>Later in Ch 17, futures options are valued, and we have the same result:</p>
<hr />
<p><a href="https://i.sstatic.net/aWPPm.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/aWPPm.png" alt="enter image description here" /></a></p>
<hr />
<p>In relation to Chapter 16 and 17, my Derivatives Pricing prof gave us this <strong>exercise</strong>:</p>
<blockquote>
<p>Show that, in the Risk-Neutral World, <span class="math-container">$E[F_T] = F_0$</span></p>
</blockquote>
<p>I guess, <span class="math-container">$F_T$</span> is the random variable s.t.</p>
<p><span class="math-container">$$F_T = 1_{A}F_0u + 1_{A^C}F_0d$$</span></p>
<p>where <span class="math-container">$A$</span> is the event corresponding to case 1.</p>
<p>The <strong>solution</strong>:</p>
<blockquote>
<p><span class="math-container">$$E[F_T] = pF_0u + (1-p)F_0d$$</span></p>
<p><span class="math-container">$$= \frac{1-d}{u-d}F_0u + \frac{u-1}{u-d}F_0d = F_0$$</span></p>
</blockquote>
<p><strong>That seems strange</strong>. To me it seems that the reason why we know that <span class="math-container">$p = \frac{1-d}{u-d}$</span> is because <span class="math-container">$E[F_T] = F_0$</span> based on 'If <span class="math-container">$F_0$</span> is the initial futures price, the expected futures price at the end of one time step of length <span class="math-container">$\Delta t$</span> should also be <span class="math-container">$F_0$</span>' from Ch 12.</p>
<p>Iirc, my prof said that the reason why we have 'If <span class="math-container">$F_0$</span> is the initial futures price, the expected futures price at the end of one time step of length <span class="math-container">$\Delta t$</span> should also be <span class="math-container">$F_0$</span>' is because of said exercise which comes from <span class="math-container">$p = \frac{1-d}{u-d}$</span>.</p>
<p><strong>So how do we get</strong> <span class="math-container">$p = \frac{1-d}{u-d}$</span> without <span class="math-container">$E[F_T] = F_0$</span>?</p>
<p>In both texts from Ch 12 and 17, it seems that <span class="math-container">$E[F_T] = F_0$</span> is an assumption. Am I wrong? Is <span class="math-container">$E[F_T] = F_0$</span> not an assumption in Ch 17? So <span class="math-container">$E[F_T] = F_0$</span> comes from Ch 17? That seems very inconsistent of Hull:</p>
<p>Ch 12 proposition: <span class="math-container">$E[F_T] = F_0 \to p = \frac{1-d}{u-d}$</span></p>
<p>Ch 17 proposition: <span class="math-container">$p = \frac{1-d}{u-d} \to E[F_T] = F_0$</span></p>
<p>?</p>

<finance><financial-economics><asset-pricing><risk><arbitrage>
$E[F_T] = F_0$ implies $p = \frac{1-d}{u-d}$? or is implied by?
BCLC
# $E[F_T] = F_0$ implies $p = \frac{1-d}{u-d}$? or is implied by?

(Asked by: username_0 on 2021-09-14)

From Ch 12 in [Hull's OFOD](http://workofark.com/pdf/options.PDF), we compute the risk-neutral probabilities for a futures contract:

---

[![enter image description here](https://i.sstatic.net/cr6e0.png)](https://i.sstatic.net/cr6e0.png)

---

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

---

[![enter image description here](https://i.sstatic.net/aWPPm.png)](https://i.sstatic.net/aWPPm.png)

---

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: $p = \frac{1-d}{u-d} \to E[F\_T] = F\_0$

?

**Tags:** finance, financial-economics, asset-pricing, risk, arbitrage


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## 1 answers

### Answer by username_1 on 2021-09-17. Score: 1

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:
$$
1 = (1 - p\_2)u + p\_2 d = u - p\_2(u - d),\\
\to p\_2 = \frac{u - 1}{u - d},\\
\to p\_1 = \frac{1 - d}{u - d}.
$$

> **Comments:**
> **username_0**: 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)  

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