<p>Let us first show that the function <span class="math-container">$u$</span> is linear in the sense that for all lotteries <span class="math-container">$L$</span> and <span class="math-container">$L'$</span> and all <span class="math-container">$\alpha \in [0,1]$</span>: <span class="math-container">$u(\alpha L + (1-\alpha) L') = \alpha u(L) + (1-\alpha) u(L')$</span>.</p>
<p>Let <span class="math-container">$L \sim a L_B + (1-a) L_W$</span> and <span class="math-container">$L' \sim b L_B + (1-b) L_W$</span></p>
<p>Then using independence on <span class="math-container">$L \sim a L_B + (1-a) L_W$</span> together with <span class="math-container">$L'$</span> gives
<span class="math-container">$$
\alpha L+ (1-\alpha) L' \sim \alpha \underbrace{[a L_B + (1-a) L_W]}_{=L''} + (1- \alpha) L'. \tag{1}
$$</span>
Next, using independence on <span class="math-container">$L' \sim b L_B + (1-b) L_W$</span> with the lottery <span class="math-container">$L''$</span> gives:
<span class="math-container">$$
\alpha L'' + (1- \alpha) L' \sim \alpha \underbrace{L''}_{=a L_B + (1-a) L_W} + (1-\alpha)[b L_B + (1-b) L_W]. \tag{2}
$$</span>
The lottery on the right hand side equals:
<span class="math-container">$$
(\alpha a + (1-\alpha) b) L_B + (\alpha (1-a) + (1-\alpha)(1- b)) L_B.
$$</span></p>
<p>By transitivity (1) and (2) gives,
<span class="math-container">$$
\alpha L + (1-\alpha) L' \sim (\alpha a + (1-\alpha) b) L_B + (\alpha(1-a) + (1-\alpha)(1-b)) L_B.
$$</span></p>
<p>Replacing <span class="math-container">$a$</span> by <span class="math-container">$u(L)$</span> and <span class="math-container">$b$</span> by <span class="math-container">$u(L')$</span> gives that:
<span class="math-container">$$
u(\alpha L + (1-\alpha) L) = \alpha u(L) + (1-\alpha) u(L').
$$</span></p>
<p>Now to finish the proof, let <span class="math-container">$L = \sum_{i = 1}^n p_i z_i$</span>. Let us show by induction on <span class="math-container">$n$</span> that <span class="math-container">$u(L) = \sum_{i = 1}^n p_i u(z_i)$</span>. If <span class="math-container">$n = 1$</span> then <span class="math-container">$L = z_1$</span> so <span class="math-container">$u(L) = u(z_1)$</span> which gives the base case.</p>
<p>Now for the induction step, assume that <span class="math-container">$L = \sum_{i = 1}^{n+1} p_i z_i$</span>. Define <span class="math-container">$L' = \sum_{i = 1}^n \frac{p_i}{(1 - p_{n+1})} z_i$</span>.</p>
<p>By induction, <span class="math-container">$u(L') = \sum_{i = 1}^n \frac{p_i}{(1 - p_{n+1})} u(z_i)$</span>. Also,
<span class="math-container">$$
L = p_{n+1} z_{n+1} + (1-p_{n+1}) L' 
$$</span>
So using the first part of the proof and the induction hypothesis (linarity of <span class="math-container">$u$</span>):
<span class="math-container">$$
\begin{align*}
u(L) &amp;= u(p_{n+1} z_{n+1} + (1-p_{n+1}) L'),\\
&amp;= p_{n+1} u(z_{n+1}) + (1-p_{n+1}) u(L'),\\
&amp;= p_{n+1} u(z_{n+1}) + \sum_{i = 1}^n p_i u(z_i),\\
&amp;= \sum_{i = 1}^{n+1} p_i u(z_i).
\end{align*}
$$</span></p>

CC BY-SA 4.0
2024-05-30T07:13:17.410Z
Thank you! I hadn't thought to prove the linearity of $u$, but I can see that makes the approach much clearer. Thanks again!
Joseph Basford
CC BY-SA 4.0
2024-05-29T16:20:42.157Z
2024-05-29T16:20:42.157Z
tdm
<p>In my micro class we covered the proof of the existence of a Von Neumann–Morgenstern utility representation of preferences <span class="math-container">$\succeq$</span> over a set of lotteries <span class="math-container">$\Delta(Z)$</span> - where <span class="math-container">$Z$</span> is some finite outcome space - when preferences are rational, continuous, and satisfy the independence axiom. The independence axiom states that for <span class="math-container">$L,L',L''\in \Delta(Z)$</span>, for each <span class="math-container">$\alpha \in (0,1)$</span>, <span class="math-container">$$L\succeq L' \iff \alpha L+(1-\alpha)L'' \succeq \alpha L'+(1-\alpha )L''$$</span>
In proving existence, we showed there existed two lotteries <span class="math-container">$L_B,L_W\in \Delta(Z)$</span> such that <span class="math-container">$L_B \succeq L \succeq L_W$</span> for all <span class="math-container">$L\in \Delta(Z)$</span>. Ignoring the trivial case when <span class="math-container">$L_B\sim L_W$</span>, we defined <span class="math-container">$U(L_B)=1$</span> and <span class="math-container">$U(L_W)=0$</span> as the first step of constructing our vNM representation. We then proved that by continuity of preferences, for each <span class="math-container">$L\in \Delta(Z),\exists !\alpha \in[0,1]$</span> such that <span class="math-container">$L\sim \alpha L_B+(1-\alpha)L_W$</span>. Then, for each outcome <span class="math-container">$z\in Z$</span>, we defined <span class="math-container">$U([z])=\alpha_z$</span>, where <span class="math-container">$\alpha_z$</span> is such that <span class="math-container">$[z]\sim \alpha_z L_B +(1-\alpha)L_W$</span> and <span class="math-container">$[z]$</span> is the degenerate lottery on <span class="math-container">$z$</span>.</p>
<p>The step I don't understand is the following. For any lottery <span class="math-container">$L\in \Delta(Z)$</span>, we can represent it as a compound lottery <span class="math-container">$L\sim\sum_{i=1}^n p_i^L[z_i]$</span> where <span class="math-container">$p_i^L$</span> is the probability of outcome <span class="math-container">$z_i$</span> in the lottery <span class="math-container">$L$</span>. Additionally, <span class="math-container">$[z]\sim u(z)L_B+(1-u(z))L_W$</span>. From here, my professor concludes that by independence <span class="math-container">$$L\sim \left(\sum_{i=1}^n p_i^L u(z_i)\right)L_B+\left(1-\sum_{i=1}^n p_i^L u(z_i)\right)L_W$$</span> I do not understand how this conclusion is reached. This looks like we have just inserted <span class="math-container">$[z]=u(z)L_B+(1-u(z))L_W$</span>  into the compound representation of <span class="math-container">$L$</span>, but I can't see how it is following from independence.</p>
<p>Could someone explain how to get this result just using the definition of independence?</p>
<hr />
<p>I am thinking about approaching the problem like this. Notice that <span class="math-container">$$L\sim\sum_{i=1}^np_i^L[z_i]=p_1^L[z_1]+(1-p_1^L)\sum_{j=2}^n \frac{p_j^L}{1-p_1^L} [z_j]$$</span> and that <span class="math-container">$\tilde{L}=\sum_{j=2}^n \frac{p_j^L}{1-p_1^L} [z_j]$</span> is a lottery itself since <span class="math-container">$0\leq \frac{p_j^L}{1-p_1^L}=\frac{p_j^L}{\sum_{i=2}^np_i^L}\leq 1$</span> and <span class="math-container">$\sum_{j=2}^n\frac{p_j^L}{1-p_1^L}=1$</span>. Since <span class="math-container">$[z_1]\sim u(z_1)L_B+(1-u(z_1))L_W$</span>, by independence, <span class="math-container">$$p_1^L[z_1]+(1-p_1)\tilde{L}\sim p_1^Lu(z_1)L_B+(p_1^L-p_1^Lu(z_1))L_W+\tilde{L}$$</span> therefore <span class="math-container">$$L\sim p_1^Lu(z_1)L_B+(p_1^L-p_1^Lu(z_1))L_W+\sum_{j=2}^n \frac{p_j^L}{1-p_1^L} [z_j]$$</span> I think we can continue iteratively, first taking the <span class="math-container">$[z_2]$</span> element out of the sum, then <span class="math-container">$[z_3]$</span>, etc., but I can't quite figure out how to get the maths to work.</p>

<microeconomics><mathematical-economics><decision-theory><expected-utility>
Independence Axiom and Expected Utility Theorem Proof
Joseph Basford
# Independence Axiom and Expected Utility Theorem Proof

(Asked by: username_0 on 2024-05-28)

In my micro class we covered the proof of the existence of a Von Neumann–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?

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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.

**Tags:** microeconomics, mathematical-economics, decision-theory, expected-utility

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## 1 answers

### Answer by username_1 on 2024-05-29. Score: 1 | Accepted answer

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 $u$):
$$
\begin{align\*}
u(L) &= u(p\_{n+1} z\_{n+1} + (1-p\_{n+1}) L'),\\
&= p\_{n+1} u(z\_{n+1}) + (1-p\_{n+1}) u(L'),\\
&= p\_{n+1} u(z\_{n+1}) + \sum\_{i = 1}^n p\_i u(z\_i),\\
&= \sum\_{i = 1}^{n+1} p\_i u(z\_i).
\end{align\*}
$$

> **Comments:**
> **username_0**: Thank you! I hadn't thought to prove the linearity of $u$, but I can see that makes the approach much clearer. Thanks again!  

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