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
- $U(\sum_{k=1}^{K} \alpha_k L_k) = \sum_{k=1}^{K} \alpha_k \ U(L_k)$
- $\succeq$ satisfies independence
- 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?