When studying preference over lotteries we learned the independence axiom which goes like this:

The preference relation $\succsim$ on the space of simple lotteries $\mathscr{L}$ satisfies the independence axiom if for all $L$$L^\prime$$L^{\prime \prime} \in \mathscr{L}$ and $\alpha \in (0,1)$ we have

$$L \succsim L^\prime \iff \alpha L + (1 - \alpha) L^{\prime \prime} \succsim \alpha L^{\prime} + (1 - \alpha) L^{\prime \prime} $$

What if $\alpha \ge 1$? I guess that will be a quick extension for the case of $\alpha \in (0,1)$ but I don't know how to show it.

  • 2
    $\begingroup$ It doesn't really seem like the case where $\alpha \geq 1$ has an economic interpretation. The point of $\alpha \in (0, 1)$ is that $\alpha$ is a weight between the different lotteries. $\endgroup$
    – Kitsune Cavalry
    Oct 13, 2015 at 2:39
  • $\begingroup$ Yes, from my understanding $\alpha$ come into place to make $\alpha L +(1 - \alpha) L^{\prime \prime}$ a compound lottery. I'm just wondering if it's possible to allow $\alpha \ge 1$ mathematically, and if so how can we use the definition of independence to prove that. $\endgroup$ Oct 13, 2015 at 3:40
  • $\begingroup$ Well the definition you are given requires $\alpha \in (0, 1)$, so as for proving that, in this particular case, $\alpha \geq 1$ could still satisfy independence, that is a futile cause I think. As for whether you could create some weird alternative definition for independence, even that would seem suspect, but I'm not sure. $\endgroup$
    – Kitsune Cavalry
    Oct 13, 2015 at 3:45
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    $\begingroup$ @KitsuneCavalry Seems to me that selling some $L''$ and buying more than one unit of $L$ is a possible economic interpretation for $\alpha > 1$. $\endgroup$
    – Giskard
    Oct 13, 2015 at 5:17
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    $\begingroup$ $\alpha\ge 1$ raises question to the consequentialism of the expected utility framework, i.e. a compound lottery is treated the same as its reduced simple lottery. If $\alpha\ge1$, then the object $\alpha L+(1-\alpha)L''$ need not be a lottery any more, and thus it is outside the admissible set of objects over which the preference $\succsim$ is defined. $\endgroup$
    – Herr K.
    Oct 13, 2015 at 6:54

1 Answer 1


To understand why $\alpha$ must be constrained in $(0,1)$, one has to contemplate the meaning of the expression

$$\alpha L$$

when $L$ is a "lottery". How is a lottery denoted mathematically? Authors do not agree on that: for example, the way Jahle and Reny define a lottery (a "gamble" in their terminology), a lottery can be written as a vector whose elements are bivariate vectors themselves:

$$L=\{(p_1,w_1),...,(p_n, w_n)\}$$

where $p_i$ are probabilities, and $w_i$ are quantitative outcomes.

But MasColell, Whinston and Green, define a lottery as a vector containing only the probabilities:


and so the "Lottery Space" is a vector space, including vectors each containing only probabilities.

But in both cases, the authors make clear that an expression like $\alpha L$, when the time comes to translate it into a mathematical operation, denotes a multiplication of only the probabilities linked with $L$ by another probability, $\alpha$. This is implementing the reduction of compound lotteries to simple ones. JR describe it as "the decision maker cares only about effective probabilities of each outcome". MWG call it the "consequentialist premise", that allows them to work only with simple lotteries.

So it is not valid to consider $\alpha$ outside $(0,1)$, because it is defined as a probability (the open bounds are used in order to avoid triviality in the statement of the Independence axiom).

Also, the above imply that $\alpha$ does not directly interact with the quantitative outcomes of the lottery/gamble...

...which points to a (perhaps interesting, perhaps not) research direction: What can we say (if anything), if we start tweaking the outcomes linked with a lottery? Will "attitude towards risk" get in the way and prevent us from drawing any general conclusions? Or not?

  • $\begingroup$ This is a splendid answer! $\endgroup$ Oct 13, 2015 at 17:21
  • $\begingroup$ @KennethChen Thanks. My main concern was to make clear that we fool around with probabilities, not with outcomes. $\endgroup$ Oct 14, 2015 at 21:04

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