<p>So I figured I'd post the proof for the third condition now that I know it.</p>

<blockquote>
  <p>We want to show that if $U$ and $V$ are linear representations of $\succeq$ on  $\mathscr{L}$, then $\exists \ \beta &gt; 0, \gamma \in \mathbb{R}$ such that $V = \beta U + \gamma$</p>
</blockquote>

<p><strong>Case 1:</strong> All the lotteries are all equally preferred.</p>

<p>If $\forall \ L_1, L_2 \in \mathscr{L}$ we also have $L_1 \sim L_2$, then $U$ is constant.</p>

<p>$V = \beta U$ holds $\forall \beta &gt; 0, \gamma  &gt; 0$</p>

<p><strong>Case 2:</strong> All the lotteries are not equally preferred.</p>

<p>$\forall L \in \mathscr{L}$ where $L \neq L_b, L_w$
$$L_b \succeq L \succeq L_w$$
$$L_b \succ L_w$$</p>

<p>For a lottery $L \in \mathscr{L}$,</p>

<p>$$\lambda_L \equiv \frac{U(L) - U(L_w)}{U(L_b) - U(L_w)}$$</p>

<p>for $\lambda \in [0,1]$</p>

<p>$$1 - \lambda_L \equiv \frac{U(L_b) - U(L)}{U(L_b) - U(L_w)}$$</p>

<p>So now we can write $U(L), \ \forall L \in \mathscr{L}$ as</p>

<p>$$U(L) = \lambda_L U(L_b) + (1-\lambda_L)U(L_w)$$
$$\implies L \sim \lambda_L L_b + (1-\lambda_L)L_w$$</p>

<p>For $V$ to also represent $\succeq$,</p>

<p>$$V(L) = \lambda_L V(L_b) + (1-\lambda_L)V(L_w)$$</p>

<p>$$V(L) = \bigg[\frac{U(L) - U(L_w)}{U(L_b) - U(L_w)}\bigg] V(L_b) + \bigg[\frac{U(L_b) - U(L)}{U(L_b) - U(L_w)}\bigg] V(L_w)$$</p>

<p>$$V(L) = \frac{U(L)V(L_b) - U(L_w)V(L_b)}{U(L_b) - U(L_w)} + \frac{U(L_b)V(L_w) - U(L)V(L_w)}{U(L_b) - U(L_w)}$$</p>

<p>$$V(L) = \frac{V(L_b) - V(L_w)}{U(L_b) - U(L_w)} \cdot U(L) + \frac{U(L_b)V(L_w) - U(L)V(L_b)}{U(L_b) - U(L_w)}$$</p>

<p>Define $\frac{V(L_b) - V(L_w)}{U(L_b) - U(L_w)} \equiv \beta$ and $\frac{U(L_b)V(L_w) - U(L)V(L_b)}{U(L_b) - U(L_w)} \equiv \gamma$</p>

<p>Thus $V(L) = \beta U(L) + \gamma$, $\forall L$</p>

<p>$\square$</p>

CC BY-SA 3.0
2015-10-23T00:00:47.327Z
2015-10-23T00:00:47.327Z
Kitsune Cavalry
<p>We went over a lemma in class leading up to a larger theorem. The Lemma states:</p>

<blockquote>
  <p>Let $\succeq$ be a rational preference relation on $\mathscr{L}$ and let $\succeq$ admit utility representation under Von-Neumann-Morgenstern expectations. Then</p>
  
  <ol>
  <li>$U(\sum_{k=1}^{K} \alpha_k L_k) = \sum_{k=1}^{K} \alpha_k \ U(L_k)$</li>
  <li>$\succeq$ satisfies independence</li>
  <li>Every linear representation $V$ of $\succeq$ is a positive affine transformation of $U$. So $V = \beta U + \gamma$ where $\beta &gt; 0$</li>
  </ol>
</blockquote>

<p>So in proving this, here is some work so far:</p>

<p>1:</p>

<p>Let $L_k = (\Pi^k_1, \Pi^k_2, ..., \Pi_s^k)$
$$U(\sum_{k=1}^{K} \alpha_k L_k) = \sum^s_{i=1}\sum^K_{k=1}\Pi^k_i \alpha_k u_i$$ by Von-Neumann-Morgenstern, where $U(L) = \sum^s_{i=1}\Pi_i u_i $</p>

<p>$$\sum^s_{i=1}\sum^K_{k=1}\Pi^k_i \alpha_k u_i =  \sum^K_{k=1} \alpha_k \sum^s_{i=1}\Pi^k_i u_i = \sum_{k=1}^{K} \alpha_k \ U(L_k)$$</p>

<p>2:</p>

<p>Consider $L_1, L_2, L_3 \in \mathscr{L}$ and say $L_1 \succeq L_2$.</p>

<p>$L_1 \succeq L_2 \iff U(L_1) \geq U(L_2)$</p>

<p>Take $\alpha \in (0,1)$ and define</p>

<p>$$L_4 = \alpha L_1 + (1-\alpha)L_3$$
$$L_5 = \alpha L_2 + (1-\alpha)L_3$$</p>

<p>Say $L_5 \succ L_4$</p>

<p>$\implies U(L_5) &gt; U(L_4)$
and from 1.)</p>

<p>$$U(L_5) = \alpha U(L_2) + (1-\alpha) U(L_3)$$ 
$$U(L_4) = \alpha U(L_1) + (1-\alpha) U(L_3)$$</p>

<p>$$\implies \alpha U(L_2) + (1-\alpha) U(L_3) &gt; \alpha U(L_1) + (1-\alpha) U(L_3) \implies U(L_2) &gt; U(L_1)$$</p>

<p>which is a contradiction, so independence must hold.</p>

<hr>

<p>I assume 3. is some sort of "monotonicity" condition. How would I approach the proof for this condition? Any hints would be appreciated. I also am wondering what this Lemma is leading up to, so I can study for that in advance. Does anyone have any idea?</p>

CC BY-SA 3.0
2015-10-15T23:01:36.317Z
This lemma might be leading up to the "expected utility theorem" (MWG pg. 176). The expected utility theorem says that if rational $\succsim$ satisfies continuity and independence, then it has an expected utility representation.
möbius
<microeconomics><utility><preferences>
Utility Representation of Lotteries
Kitsune Cavalry
# Utility Representation of Lotteries

(Asked by: username_0 on 2015-10-15)

We went over a lemma in class leading up to a larger theorem. The Lemma states:

> Let $\succeq$ be a rational preference relation on $\mathscr{L}$ and let $\succeq$ admit utility representation under Von-Neumann-Morgenstern expectations. Then
>
> 1. $U(\sum\_{k=1}^{K} \alpha\_k L\_k) = \sum\_{k=1}^{K} \alpha\_k \ U(L\_k)$
> 2. $\succeq$ satisfies independence
> 3. Every linear representation $V$ of $\succeq$ is a positive affine transformation of $U$. So $V = \beta U + \gamma$ where $\beta > 0$

So in proving this, here is some work so far:

1:

Let $L\_k = (\Pi^k\_1, \Pi^k\_2, ..., \Pi\_s^k)$
$$U(\sum\_{k=1}^{K} \alpha\_k L\_k) = \sum^s\_{i=1}\sum^K\_{k=1}\Pi^k\_i \alpha\_k u\_i$$ by Von-Neumann-Morgenstern, where $U(L) = \sum^s\_{i=1}\Pi\_i u\_i $

$$\sum^s\_{i=1}\sum^K\_{k=1}\Pi^k\_i \alpha\_k u\_i = \sum^K\_{k=1} \alpha\_k \sum^s\_{i=1}\Pi^k\_i u\_i = \sum\_{k=1}^{K} \alpha\_k \ U(L\_k)$$

2:

Consider $L\_1, L\_2, L\_3 \in \mathscr{L}$ and say $L\_1 \succeq L\_2$.

$L\_1 \succeq L\_2 \iff U(L\_1) \geq U(L\_2)$

Take $\alpha \in (0,1)$ and define

$$L\_4 = \alpha L\_1 + (1-\alpha)L\_3$$
$$L\_5 = \alpha L\_2 + (1-\alpha)L\_3$$

Say $L\_5 \succ L\_4$

$\implies U(L\_5) > U(L\_4)$
and from 1.)

$$U(L\_5) = \alpha U(L\_2) + (1-\alpha) U(L\_3)$$
$$U(L\_4) = \alpha U(L\_1) + (1-\alpha) U(L\_3)$$

$$\implies \alpha U(L\_2) + (1-\alpha) U(L\_3) > \alpha U(L\_1) + (1-\alpha) U(L\_3) \implies U(L\_2) > U(L\_1)$$

which is a contradiction, so independence must hold.

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I assume 3. is some sort of "monotonicity" condition. How would I approach the proof for this condition? Any hints would be appreciated. I also am wondering what this Lemma is leading up to, so I can study for that in advance. Does anyone have any idea?

**Tags:** microeconomics, utility, preferences
> **Comments:**
> **username_1**: This lemma might be leading up to the "expected utility theorem" (MWG pg. 176). The expected utility theorem says that if rational $\succsim$ satisfies continuity and independence, then it has an expected utility representation.  

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## 1 answers

### Answer by username_0 on 2015-10-23. Score: 0 | Accepted answer

So I figured I'd post the proof for the third condition now that I know it.

> We want to show that if $U$ and $V$ are linear representations of $\succeq$ on $\mathscr{L}$, then $\exists \ \beta > 0, \gamma \in \mathbb{R}$ such that $V = \beta U + \gamma$

**Case 1:** All the lotteries are all equally preferred.

If $\forall \ L\_1, L\_2 \in \mathscr{L}$ we also have $L\_1 \sim L\_2$, then $U$ is constant.

$V = \beta U$ holds $\forall \beta > 0, \gamma > 0$

**Case 2:** All the lotteries are not equally preferred.

$\forall L \in \mathscr{L}$ where $L \neq L\_b, L\_w$
$$L\_b \succeq L \succeq L\_w$$
$$L\_b \succ L\_w$$

For a lottery $L \in \mathscr{L}$,

$$\lambda\_L \equiv \frac{U(L) - U(L\_w)}{U(L\_b) - U(L\_w)}$$

for $\lambda \in [0,1]$

$$1 - \lambda\_L \equiv \frac{U(L\_b) - U(L)}{U(L\_b) - U(L\_w)}$$

So now we can write $U(L), \ \forall L \in \mathscr{L}$ as

$$U(L) = \lambda\_L U(L\_b) + (1-\lambda\_L)U(L\_w)$$
$$\implies L \sim \lambda\_L L\_b + (1-\lambda\_L)L\_w$$

For $V$ to also represent $\succeq$,

$$V(L) = \lambda\_L V(L\_b) + (1-\lambda\_L)V(L\_w)$$

$$V(L) = \bigg[\frac{U(L) - U(L\_w)}{U(L\_b) - U(L\_w)}\bigg] V(L\_b) + \bigg[\frac{U(L\_b) - U(L)}{U(L\_b) - U(L\_w)}\bigg] V(L\_w)$$

$$V(L) = \frac{U(L)V(L\_b) - U(L\_w)V(L\_b)}{U(L\_b) - U(L\_w)} + \frac{U(L\_b)V(L\_w) - U(L)V(L\_w)}{U(L\_b) - U(L\_w)}$$

$$V(L) = \frac{V(L\_b) - V(L\_w)}{U(L\_b) - U(L\_w)} \cdot U(L) + \frac{U(L\_b)V(L\_w) - U(L)V(L\_b)}{U(L\_b) - U(L\_w)}$$

Define $\frac{V(L\_b) - V(L\_w)}{U(L\_b) - U(L\_w)} \equiv \beta$ and $\frac{U(L\_b)V(L\_w) - U(L)V(L\_b)}{U(L\_b) - U(L\_w)} \equiv \gamma$

Thus $V(L) = \beta U(L) + \gamma$, $\forall L$

$\square$


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