Proving the De Finetti Theorem 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 $X$.

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 $px &lt; py \ \forall \ p$. 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.
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 $p \cdot x &lt; p \cdot y \ \forall p$

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

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

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 $p$.

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

$y \succ x \iff y &gt; x$ by monotonicity

but then $\sum^s_{i=1} x_i &lt; \sum^s_{i=1} y_i$ 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, $x &lt; y$
$\implies p \cdot x &lt; p \cdot y$ 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 $x &gt; y \iff x \succ y$
and $y &gt; x \iff y \succ x$



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 $\delta &gt; 0$.

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 $\epsilon = 0$.

$\square$