<p>Here was my attempt at a proof:</p>

<p><em>First let us look at the "if" case.</em></p>

<p>Assume $\succcurlyeq$ on $X$ is rational, continuous, additive, monotonic, and non-trivial.</p>

<p>BUT</p>

<p><strong>Case 1:</strong> $x \succcurlyeq y$ and $p \cdot x &lt; p \cdot y \ \forall p$</p>

<p>Suppose $p = \{\frac{1}{s} \cdots \frac{1}{s}\}$</p>

<p>$$p \cdot x &lt; p \cdot y \implies \frac{1}{s} \sum^s_{i=1} x_i &lt; \frac{1}{s} \sum^s_{i=1} y_i \implies \sum^s_{i=1} x_i &lt; \sum^s_{i=1} y_i$$</p>

<p>So $x \succcurlyeq y \iff x \geq y$ by monotonicity</p>

<p>$x \geq y \iff x_i \geq y_i \ \forall i$ by definition</p>

<p>but then $\sum^s_{i=1} x_i \geq \sum^s_{i=1} y_i$ which is a contradiction.</p>

<hr>

<p><strong>Case 2:</strong> $p \cdot x \geq p \cdot y$ and $y \succ x \ \forall p$</p>

<p>Similarly, pick the same $p$.</p>

<p>$$p \cdot x \geq p \cdot y \implies \sum^s_{i=1} x_i \geq \sum^s_{i=1} y_i$$</p>

<p>$y \succ x \iff y &gt; x$ by monotonicity</p>

<p>but then $\sum^s_{i=1} x_i &lt; \sum^s_{i=1} y_i$ which is a contradiction.</p>

<hr>

<p><em>Now, let us look at the "only if" case.</em></p>

<p>Let $u(x) = p \cdot x$ and $\exists \ p$ s.t. $\forall x,y$ then $x \succcurlyeq y \iff p \cdot x \geq p \cdot y$</p>

<p>Now we wish to show our five properties hold and the uniqueness of $p$.</p>

<hr>

<p>The above implies that</p>

<p>$\exists \ p$ s.t. $\forall x,y$, then $x \succcurlyeq y \iff U(x) \geq U(y)$</p>

<p>This we have utility function <strong>representation</strong>, by definition. It is a basic proof from the beginning of your class probably that representation implies <strong>rationality</strong>.</p>

<p>We can also say if $U(x)$ is continuous in $X$ and $U(x)$ represents $\succcurlyeq$ on $X$, this implies the preferences are <strong>continuous</strong>.</p>

<p>$x \succcurlyeq y \iff p \cdot x \geq p \cdot y \iff p \cdot (x + z) \geq p \cdot (y + z) \iff x + z \succcurlyeq y + z \quad \forall x,y,z \in X$</p>

<p>So that shows <strong>additivity</strong>.</p>

<p>Now suppose <strong>monotonicity</strong> does not hold. Suppose $x \succcurlyeq y \implies p \cdot x \geq p \cdot y$</p>

<p>but additionally, $x &lt; y$
$\implies p \cdot x &lt; p \cdot y$ which is a contradiction.</p>

<p>Suppose <strong>non-triviality</strong> does not hold. Suppose $x \sim y \ \forall \ x, y$, where $x \neq y$, so $x$ and $y$ are unique. But we have monotonicity now.</p>

<p>but $x &gt; y \iff x \succ y$
and $y &gt; x \iff y \succ x$</p>

<hr>

<p>For the uniqueness of $p$, building off of the outline you gave pick $x, y \in X$ where $x \sim y$ for both $p$ and $p' = (p + \epsilon, 1-p- \epsilon)$</p>

<p>This can only be true if $\epsilon = 0$ or if $x$ and $y$ are not unique, that is, $x_i = y_i \ \forall i$. We will rule out the trivial case where $x$ and $y$ are not unique and show $\epsilon = 0$ (and thus $p = p'$)</p>

<p>$$px_1 + (1-p)x_2 = py_1 + (1-p)y_2$$
and
$$(p+\epsilon)x_1 + (1-p-\epsilon)x_2 = (p+\epsilon)y_1 + (1-p-\epsilon)y_2$$
The second equation implies
$$px_1 + \epsilon x_1 + (1-p)x_2 - \epsilon x_2 = py_1 + \epsilon y_1 + (1-p)y_2 - \epsilon y_2$$</p>

<p>Subtract the first equation from this equation.</p>

<p>$$\epsilon x_1 - \epsilon x_2 = \epsilon y_1 - \epsilon y_2$$
$$\epsilon (x_1 - x_2) = \epsilon (y_1 - y_2)$$</p>

<p>Consider if $x_1 + \delta = y_1$ and $x_2 + \delta = y_2$ for any $\delta &gt; 0$.</p>

<p>Then the differences $x_1 - x_2$ and $y_1 - y_2$ are equal, but bundle x will have higher monetary outcomes in whichever state occurs, so $x \succ y$. Contradiction.</p>

<p>Similarly, if $x_1 = y_1 + \delta$ and $x_2 = y_2 + \delta$, then $y \succ x$, which is again a contradiction with $x \sim y$</p>

<p>Thus $\epsilon = 0$.</p>

<p>$\square$</p>

CC BY-SA 3.0
2015-11-16T19:15:22.600Z
2015-11-16T19:15:22.600Z
Kitsune Cavalry
<p>Let us have a finite state space, $\Omega = {\omega_1,\cdots,\omega_s}$, where $2 \leq s &lt; \infty$. Define a bet as a function $x:\Omega \rightarrow X$, where $X \subseteq \mathbb{R}^s$ is the set of monetary outcomes. Assume that the agent only cares about expected payoffs. Let $\succcurlyeq$ be a rational, continuous preference relation on $X$.</p>

<p><strong>Additivity:</strong>
$$\forall x, y, z \in X, \text{then} \ x \succcurlyeq y \iff x+z \succcurlyeq y+z$$</p>

<p><strong>Monotonicity</strong>
$$\forall x, y \in X, \text{then} \ x \geq y \iff x \succcurlyeq y$$</p>

<p><strong>Non-Triviality</strong>
$$\exists x, y \in X, \text{s.t.} \ x \succcurlyeq y$$</p>

<hr>

<p><strong><em>De Finetti Theorem</em></strong></p>

<p>$\succcurlyeq$ on $X$ is rational, continuous, additive, monotonic, and nontrivial if and only if</p>

<p>$$\exists p \in \mathbb{R}^s \setminus {0} = \{p \in \mathbb{R}^s \mid \sum^s_{i=1} p_i = 1, \ p_i \in [0,1] \ \forall \ i \}$$</p>

<p>s.t. $\forall x,y \in X$, we have $x \succcurlyeq y \iff p \cdot x \geq p \cdot y$</p>

<p>Moreover, p is unique.</p>

<hr>

<p>So my professor has asked us to prove the De Finetti Theorem, in which she told us that:</p>

<p>First, prove that $\succcurlyeq$ is rational, continuous, additive, monotone, and nontrivial.  </p>

<p>$$( \exists \ p \quad \text{s.t.} \ \forall x,y \ \text{and} \ x \succ y \Leftrightarrow px \geq py )$$</p>

<p>Use the uniqueness of p with a proof of contradiction using:</p>

<p>Assume $\succcurlyeq$, then suppose $x \succcurlyeq y$ but $px &lt; py \ \forall \ p$. Then suppose $px \geq py, \ y$  but $y \succ x \ \forall \ p$.</p>

<p>Second, let $U(\lambda) = px$, then show the property holds.</p>

<p>Then with the uniqueness of $p$, let $s = 2 \rightarrow ( p , 1- p)$</p>

<p>Consider $x \sim y$, and then suppose $p$ is not unique $\rightarrow$ $( p + \epsilon , 1 - p - \epsilon )$.</p>

<p>$\epsilon \in \mathbb{R}$</p>

<p>So the question is: how do I construct $\epsilon$ for it to be consistent with the model? Any help would be appreciated in applying the outline of this proof.</p>

CC BY-SA 3.0
2015-11-11T07:34:28.237Z
Perhaps my ignorance is to blame but I was unable to find a "De Finetti Theorem" that was not about probability, while this question is not. Without the theorem this question is a bit murky. For example I suspect an "if" is missing from the second sentence. From the information available it seems that any $\epsilon$ is consistent, but consistent may not be the word you are looking for here.
Giskard
CC BY-SA 3.0
2015-11-11T19:56:10.623Z
I know the theorem this person is talking about. I'll make some edits.
Kitsune Cavalry
<preferences><expected-utility>
Proving the De Finetti Theorem
Spector White
# Proving the De Finetti Theorem

(Asked by: username_0 on 2015-11-11)

Let us have a finite state space, $\Omega = {\omega\_1,\cdots,\omega\_s}$, where $2 \leq s < \infty$. Define a bet as a function $x:\Omega \rightarrow X$, where $X \subseteq \mathbb{R}^s$ is the set of monetary outcomes. Assume that the agent only cares about expected payoffs. Let $\succcurlyeq$ be a rational, continuous preference relation on $X$.

**Additivity:**
$$\forall x, y, z \in X, \text{then} \ x \succcurlyeq y \iff x+z \succcurlyeq y+z$$

**Monotonicity**
$$\forall x, y \in X, \text{then} \ x \geq y \iff x \succcurlyeq y$$

**Non-Triviality**
$$\exists x, y \in X, \text{s.t.} \ x \succcurlyeq y$$

---

***De Finetti Theorem***

$\succcurlyeq$ on $X$ is rational, continuous, additive, monotonic, and nontrivial if and only if

$$\exists p \in \mathbb{R}^s \setminus {0} = \{p \in \mathbb{R}^s \mid \sum^s\_{i=1} p\_i = 1, \ p\_i \in [0,1] \ \forall \ i \}$$

s.t. $\forall x,y \in X$, we have $x \succcurlyeq y \iff p \cdot x \geq p \cdot y$

Moreover, p is unique.

---

So my professor has asked us to prove the De Finetti Theorem, in which she told us that:

First, prove that $\succcurlyeq$ is rational, continuous, additive, monotone, and nontrivial.

$$( \exists \ p \quad \text{s.t.} \ \forall x,y \ \text{and} \ x \succ y \Leftrightarrow px \geq py )$$

Use the uniqueness of p with a proof of contradiction using:

Assume $\succcurlyeq$, then suppose $x \succcurlyeq y$ but $px < py \ \forall \ p$. Then suppose $px \geq py, \ y$ but $y \succ x \ \forall \ p$.

Second, let $U(\lambda) = px$, then show the property holds.

Then with the uniqueness of $p$, let $s = 2 \rightarrow ( p , 1- p)$

Consider $x \sim y$, and then suppose $p$ is not unique $\rightarrow$ $( p + \epsilon , 1 - p - \epsilon )$.

$\epsilon \in \mathbb{R}$

So the question is: how do I construct $\epsilon$ for it to be consistent with the model? Any help would be appreciated in applying the outline of this proof.

**Tags:** preferences, expected-utility

> **Comments:**
> **username_1**: Perhaps my ignorance is to blame but I was unable to find a "De Finetti Theorem" that was not about probability, while this question is not. Without the theorem this question is a bit murky. For example I suspect an "if" is missing from the second sentence. From the information available it seems that any $\epsilon$ is consistent, but consistent may not be the word you are looking for here.  
> **username_2**: I know the theorem this person is talking about. I'll make some edits.  

---

## 1 answers

### Answer by username_2 on 2015-11-16. Score: 3 | Accepted answer

Here was my attempt at a proof:

*First let us look at the "if" case.*

Assume $\succcurlyeq$ on $X$ is rational, continuous, additive, monotonic, and non-trivial.

BUT

**Case 1:** $x \succcurlyeq y$ and $p \cdot x < p \cdot y \ \forall p$

Suppose $p = \{\frac{1}{s} \cdots \frac{1}{s}\}$

$$p \cdot x < p \cdot y \implies \frac{1}{s} \sum^s\_{i=1} x\_i < \frac{1}{s} \sum^s\_{i=1} y\_i \implies \sum^s\_{i=1} x\_i < \sum^s\_{i=1} y\_i$$

So $x \succcurlyeq y \iff x \geq y$ by monotonicity

$x \geq y \iff x\_i \geq y\_i \ \forall i$ by definition

but then $\sum^s\_{i=1} x\_i \geq \sum^s\_{i=1} y\_i$ which is a contradiction.

---

**Case 2:** $p \cdot x \geq p \cdot y$ and $y \succ x \ \forall p$

Similarly, pick the same $p$.

$$p \cdot x \geq p \cdot y \implies \sum^s\_{i=1} x\_i \geq \sum^s\_{i=1} y\_i$$

$y \succ x \iff y > x$ by monotonicity

but then $\sum^s\_{i=1} x\_i < \sum^s\_{i=1} y\_i$ which is a contradiction.

---

*Now, let us look at the "only if" case.*

Let $u(x) = p \cdot x$ and $\exists \ p$ s.t. $\forall x,y$ then $x \succcurlyeq y \iff p \cdot x \geq p \cdot y$

Now we wish to show our five properties hold and the uniqueness of $p$.

---

The above implies that

$\exists \ p$ s.t. $\forall x,y$, then $x \succcurlyeq y \iff U(x) \geq U(y)$

This we have utility function **representation**, by definition. It is a basic proof from the beginning of your class probably that representation implies **rationality**.

We can also say if $U(x)$ is continuous in $X$ and $U(x)$ represents $\succcurlyeq$ on $X$, this implies the preferences are **continuous**.

$x \succcurlyeq y \iff p \cdot x \geq p \cdot y \iff p \cdot (x + z) \geq p \cdot (y + z) \iff x + z \succcurlyeq y + z \quad \forall x,y,z \in X$

So that shows **additivity**.

Now suppose **monotonicity** does not hold. Suppose $x \succcurlyeq y \implies p \cdot x \geq p \cdot y$

but additionally, $x < y$
$\implies p \cdot x < p \cdot y$ which is a contradiction.

Suppose **non-triviality** does not hold. Suppose $x \sim y \ \forall \ x, y$, where $x \neq y$, so $x$ and $y$ are unique. But we have monotonicity now.

but $x > y \iff x \succ y$
and $y > x \iff y \succ x$

---

For the uniqueness of $p$, building off of the outline you gave pick $x, y \in X$ where $x \sim y$ for both $p$ and $p' = (p + \epsilon, 1-p- \epsilon)$

This can only be true if $\epsilon = 0$ or if $x$ and $y$ are not unique, that is, $x\_i = y\_i \ \forall i$. We will rule out the trivial case where $x$ and $y$ are not unique and show $\epsilon = 0$ (and thus $p = p'$)

$$px\_1 + (1-p)x\_2 = py\_1 + (1-p)y\_2$$
and
$$(p+\epsilon)x\_1 + (1-p-\epsilon)x\_2 = (p+\epsilon)y\_1 + (1-p-\epsilon)y\_2$$
The second equation implies
$$px\_1 + \epsilon x\_1 + (1-p)x\_2 - \epsilon x\_2 = py\_1 + \epsilon y\_1 + (1-p)y\_2 - \epsilon y\_2$$

Subtract the first equation from this equation.

$$\epsilon x\_1 - \epsilon x\_2 = \epsilon y\_1 - \epsilon y\_2$$
$$\epsilon (x\_1 - x\_2) = \epsilon (y\_1 - y\_2)$$

Consider if $x\_1 + \delta = y\_1$ and $x\_2 + \delta = y\_2$ for any $\delta > 0$.

Then the differences $x\_1 - x\_2$ and $y\_1 - y\_2$ are equal, but bundle x will have higher monetary outcomes in whichever state occurs, so $x \succ y$. Contradiction.

Similarly, if $x\_1 = y\_1 + \delta$ and $x\_2 = y\_2 + \delta$, then $y \succ x$, which is again a contradiction with $x \sim y$

Thus $\epsilon = 0$.

$\square$


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