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?

  • 2
    $\begingroup$ 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. $\endgroup$
    – user157623
    Dec 5, 2014 at 20:24
  • $\begingroup$ @user157623: Title changed. Thanks for the comment. $\endgroup$
    – Herr K.
    Dec 5, 2014 at 22:52

1 Answer 1


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 $\beta > 4$,

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


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


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