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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.

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?

Kitsune Cavalry
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    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 Oct 15 '15 at 23:01

1 Answers1

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

$\square$

Kitsune Cavalry
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