# Independence Axiom and Expected Utility Theorem Proof

In my micro class we covered the proof of the existence of a Von Neumannâ€“Morgenstern utility representation of preferences $$\succeq$$ over a set of lotteries $$\Delta(Z)$$ - where $$Z$$ is some finite outcome space - when preferences are rational, continuous, and satisfy the independence axiom. The independence axiom states that for $$L,L',L''\in \Delta(Z)$$, for each $$\alpha \in (0,1)$$, $$L\succeq L' \iff \alpha L+(1-\alpha)L'' \succeq \alpha L'+(1-\alpha )L''$$ In proving existence, we showed there existed two lotteries $$L_B,L_W\in \Delta(Z)$$ such that $$L_B \succeq L \succeq L_W$$ for all $$L\in \Delta(Z)$$. Ignoring the trivial case when $$L_B\sim L_W$$, we defined $$U(L_B)=1$$ and $$U(L_W)=0$$ as the first step of constructing our vNM representation. We then proved that by continuity of preferences, for each $$L\in \Delta(Z),\exists !\alpha \in[0,1]$$ such that $$L\sim \alpha L_B+(1-\alpha)L_W$$. Then, for each outcome $$z\in Z$$, we defined $$U([z])=\alpha_z$$, where $$\alpha_z$$ is such that $$[z]\sim \alpha_z L_B +(1-\alpha)L_W$$ and $$[z]$$ is the degenerate lottery on $$z$$.

The step I don't understand is the following. For any lottery $$L\in \Delta(Z)$$, we can represent it as a compound lottery $$L\sim\sum_{i=1}^n p_i^L[z_i]$$ where $$p_i^L$$ is the probability of outcome $$z_i$$ in the lottery $$L$$. Additionally, $$[z]\sim u(z)L_B+(1-u(z))L_W$$. From here, my professor concludes that by independence $$L\sim \left(\sum_{i=1}^n p_i^L u(z_i)\right)L_B+\left(1-\sum_{i=1}^n p_i^L u(z_i)\right)L_W$$ I do not understand how this conclusion is reached. This looks like we have just inserted $$[z]=u(z)L_B+(1-u(z))L_W$$ into the compound representation of $$L$$, but I can't see how it is following from independence.

Could someone explain how to get this result just using the definition of independence?

I am thinking about approaching the problem like this. Notice that $$L\sim\sum_{i=1}^np_i^L[z_i]=p_1^L[z_1]+(1-p_1^L)\sum_{j=2}^n \frac{p_j^L}{1-p_1^L} [z_j]$$ and that $$\tilde{L}=\sum_{j=2}^n \frac{p_j^L}{1-p_1^L} [z_j]$$ is a lottery itself since $$0\leq \frac{p_j^L}{1-p_1^L}=\frac{p_j^L}{\sum_{i=2}^np_i^L}\leq 1$$ and $$\sum_{j=2}^n\frac{p_j^L}{1-p_1^L}=1$$. Since $$[z_1]\sim u(z_1)L_B+(1-u(z_1))L_W$$, by independence, $$p_1^L[z_1]+(1-p_1)\tilde{L}\sim p_1^Lu(z_1)L_B+(p_1^L-p_1^Lu(z_1))L_W+\tilde{L}$$ therefore $$L\sim p_1^Lu(z_1)L_B+(p_1^L-p_1^Lu(z_1))L_W+\sum_{j=2}^n \frac{p_j^L}{1-p_1^L} [z_j]$$ I think we can continue iteratively, first taking the $$[z_2]$$ element out of the sum, then $$[z_3]$$, etc., but I can't quite figure out how to get the maths to work.

Let us first show that the function $$u$$ is linear in the sense that for all lotteries $$L$$ and $$L'$$ and all $$\alpha \in [0,1]$$: $$u(\alpha L + (1-\alpha) L') = \alpha u(L) + (1-\alpha) u(L')$$.

Let $$L \sim a L_B + (1-a) L_W$$ and $$L' \sim b L_B + (1-b) L_W$$

Then using independence on $$L \sim a L_B + (1-a) L_W$$ together with $$L'$$ gives $$\alpha L+ (1-\alpha) L' \sim \alpha \underbrace{[a L_B + (1-a) L_W]}_{=L''} + (1- \alpha) L'. \tag{1}$$ Next, using independence on $$L' \sim b L_B + (1-b) L_W$$ with the lottery $$L''$$ gives: $$\alpha L'' + (1- \alpha) L' \sim \alpha \underbrace{L''}_{=a L_B + (1-a) L_W} + (1-\alpha)[b L_B + (1-b) L_W]. \tag{2}$$ The lottery on the right hand side equals: $$(\alpha a + (1-\alpha) b) L_B + (\alpha (1-a) + (1-\alpha)(1- b)) L_B.$$

By transitivity (1) and (2) gives, $$\alpha L + (1-\alpha) L' \sim (\alpha a + (1-\alpha) b) L_B + (\alpha(1-a) + (1-\alpha)(1-b)) L_B.$$

Replacing $$a$$ by $$u(L)$$ and $$b$$ by $$u(L')$$ gives that: $$u(\alpha L + (1-\alpha) L) = \alpha u(L) + (1-\alpha) u(L').$$

Now to finish the proof, let $$L = \sum_{i = 1}^n p_i z_i$$. Let us show by induction on $$n$$ that $$u(L) = \sum_{i = 1}^n p_i u(z_i)$$. If $$n = 1$$ then $$L = z_1$$ so $$u(L) = u(z_1)$$ which gives the base case.

Now for the induction step, assume that $$L = \sum_{i = 1}^{n+1} p_i z_i$$. Define $$L' = \sum_{i = 1}^n \frac{p_i}{(1 - p_{n+1})} z_i$$.

By induction, $$u(L') = \sum_{i = 1}^n \frac{p_i}{(1 - p_{n+1})} u(z_i)$$. Also, $$L = p_{n+1} z_{n+1} + (1-p_{n+1}) L'$$ So using the first part of the proof and the induction hypothesis (linarity of $$u$$): \begin{align*} u(L) &= u(p_{n+1} z_{n+1} + (1-p_{n+1}) L'),\\ &= p_{n+1} u(z_{n+1}) + (1-p_{n+1}) u(L'),\\ &= p_{n+1} u(z_{n+1}) + \sum_{i = 1}^n p_i u(z_i),\\ &= \sum_{i = 1}^{n+1} p_i u(z_i). \end{align*}

• Thank you! I hadn't thought to prove the linearity of $u$, but I can see that makes the approach much clearer. Thanks again! Commented May 30 at 7:13