<p>No, not necessarily. Without the independence axiom (or something else to replace it) there is not much you can infer about preferences over (non-degenerate) lotteries from knowing preferences over outcomes only.</p>

<p>For instance, let <span class="math-container">$p^L_n$</span> be the probability of outcomes <span class="math-container">$n \in \{1, 2, 3\}$</span>. Then preferences  over lotteries <span class="math-container">$\succeq^*$</span> represented by the utility function</p>

<p><span class="math-container">$$U(L) = p^L_1 + \beta [p^L_2p^L_3],$$</span></p>

<p>are continuous  and rational, but do not satisfy the independence axiom. For <span class="math-container">$\beta$</span> large enough, it is not even the case that <span class="math-container">$(1,0,0)$</span> is the best lottery, although <span class="math-container">$(1,0,0) \succ^* (0,1,0)$</span> and <span class="math-container">$(1,0,0) \succ^* (0,0,1)$</span>.</p>

<p>To see why, observe that</p>

<p><span class="math-container">$$ U(1,0,0) = 1, $$</span>
<span class="math-container">$$ U(0,1,0) = 0, $$</span>
<span class="math-container">$$ U(0,0,1) = 0, $$</span></p>

<p>However, for <span class="math-container">$\beta &gt; 4$</span>,</p>

<p><span class="math-container">$$ U\left(0,\frac{1}{2},\frac{1}{2}\right) &gt; 1 .$$</span></p>

<p>Violation of the independence axiom can be seen from the fact that, when <span class="math-container">$\beta &gt; 4$</span>,</p>

<p><span class="math-container">$$ [1,0,0] \succ [0,1,0]  ,$$</span></p>

<p>although</p>

<p><span class="math-container">$$ \left[0,\frac{1}{2},\frac{1}{2}\right] \succ \left[ \frac{1}{2}, 0, \frac{1}{2}\right]. $$</span> </p>

CC BY-SA 4.0
2014-12-05T20:33:31.233Z
2019-11-07T00:34:38.633Z
2019-11-07T00:34:38.633Z
Martin Van der Linden
<p>Suppose a set of $N$ outcomes can be ranked in the following order: $1\succ 2\succsim\cdots\succsim N$. Further, suppose a decision maker has preference over lotteries over these outcomes. Assume the preference over lotteries is rational, continuous, but <em>not necessarily consistent with the independence axiom</em>. </p>

<p><strong>Does it follow that the best lottery in this case is the degenerate lottery $(1,0,\dots,0)$?</strong></p>

<p><strong>What if the independence axiom is <em>violated</em>?</strong></p>

CC BY-SA 3.0
2014-12-05T20:24:49.467Z
Shouldn't the title say preferences over lotteries (risk) without independence axiom, since expected utility Von Neumann Morgesten is actually derived from the independence axiom.
user157623
CC BY-SA 3.0
2014-12-05T22:52:29.380Z
@user157623: Title changed. Thanks for the comment.
Herr K.
<decision-theory><expected-utility>
Preference over lotteries without independence axiom
Herr K.
# Preference over lotteries without independence axiom

(Asked by: username_0 on 2014-12-05)

Suppose a set of $N$ outcomes can be ranked in the following order: $1\succ 2\succsim\cdots\succsim N$. Further, suppose a decision maker has preference over lotteries over these outcomes. Assume the preference over lotteries is rational, continuous, but *not necessarily consistent with the independence axiom*.

**Does it follow that the best lottery in this case is the degenerate lottery $(1,0,\dots,0)$?**

**What if the independence axiom is *violated*?**

**Tags:** decision-theory, expected-utility

> **Comments:**
> **username_1**: Shouldn't the title say preferences over lotteries (risk) without independence axiom, since expected utility Von Neumann Morgesten is actually derived from the independence axiom.  
> **username_0**: @username_1: Title changed. Thanks for the comment.  

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## 1 answers

### Answer by username_2 on 2014-12-05. Score: 9 | Accepted answer

No, not necessarily. Without the independence axiom (or something else to replace it) there is not much you can infer about preferences over (non-degenerate) lotteries from knowing preferences over outcomes only.

For instance, let $p^L\_n$ be the probability of outcomes $n \in \{1, 2, 3\}$. Then preferences over lotteries $\succeq^\*$ represented by the utility function

$$U(L) = p^L\_1 + \beta [p^L\_2p^L\_3],$$

are continuous and rational, but do not satisfy the independence axiom. For $\beta$ large enough, it is not even the case that $(1,0,0)$ is the best lottery, although $(1,0,0) \succ^\* (0,1,0)$ and $(1,0,0) \succ^\* (0,0,1)$.

To see why, observe that

$$ U(1,0,0) = 1, $$
$$ U(0,1,0) = 0, $$
$$ U(0,0,1) = 0, $$

However, for $\beta > 4$,

$$ U\left(0,\frac{1}{2},\frac{1}{2}\right) > 1 .$$

Violation of the independence axiom can be seen from the fact that, when $\beta > 4$,

$$ [1,0,0] \succ [0,1,0] ,$$

although

$$ \left[0,\frac{1}{2},\frac{1}{2}\right] \succ \left[ \frac{1}{2}, 0, \frac{1}{2}\right]. $$


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