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

<p>$$\alpha L$$</p>

<p>when $L$ is a "lottery". How is a lottery denoted mathematically? Authors do not agree on that: for example, the way <strong><a href="http://rads.stackoverflow.com/amzn/click/0273731912" rel="nofollow">Jahle and Reny</a></strong> define a lottery (a "gamble" in their terminology), a lottery can be written as a vector whose elements are bivariate vectors themselves:</p>

<p>$$L=\{(p_1,w_1),...,(p_n, w_n)\}$$</p>

<p>where $p_i$ are probabilities, and $w_i$ are quantitative outcomes. </p>

<p>But <strong><a href="http://rads.stackoverflow.com/amzn/click/0195073401" rel="nofollow">MasColell, Whinston and Green</a></strong>, define a lottery as a vector <em>containing only the probabilities</em>:</p>

<p>$$L=\{p_1,...,p_n\}$$</p>

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

<p>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 <em>multiplication</em> of only the <em>probabilities</em> linked with $L$ by <em>another probability</em>, $\alpha$. This is implementing the reduction of compound lotteries to simple ones. <strong>JR</strong> describe it as "the decision maker cares only about <em>effective</em> probabilities of each outcome". <strong>MWG</strong> call it the "consequentialist premise", that allows them to work only with simple lotteries.</p>

<p>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).  </p>

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

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

CC BY-SA 3.0
2015-10-13T17:21:29.057Z
This is a splendid answer!
Kenneth Chen
CC BY-SA 3.0
2015-10-14T21:04:29.130Z
@KennethChen Thanks. My main concern was to make clear that we fool around with probabilities, not with outcomes.
Alecos Papadopoulos
CC BY-SA 3.0
2015-10-13T15:23:22.857Z
2015-10-14T21:03:36.813Z
2015-10-14T21:03:36.813Z
Alecos Papadopoulos
<p>When studying preference over lotteries we learned the independence axiom which goes like this:</p>
<blockquote>
<p>The preference relation <span class="math-container">$\succsim$</span> on the space of simple lotteries <span class="math-container">$\mathscr{L}$</span> satisfies the independence axiom if for all <span class="math-container">$L$</span>，<span class="math-container">$L^\prime$</span>，<span class="math-container">$L^{\prime \prime} \in \mathscr{L}$</span> and <span class="math-container">$\alpha \in (0,1)$</span> we have</p>
<p><span class="math-container">$$L \succsim L^\prime \iff \alpha L + (1 - \alpha) L^{\prime \prime} \succsim \alpha L^{\prime} + (1 - \alpha) L^{\prime \prime} $$</span></p>
</blockquote>
<p>What if <span class="math-container">$\alpha \ge 1$</span>?  I guess that will be a quick extension for the case of <span class="math-container">$\alpha \in (0,1)$</span> but I don't know how to show it.</p>

CC BY-SA 3.0
2015-10-13T02:39:01.313Z
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.
Kitsune Cavalry
CC BY-SA 3.0
2015-10-13T03:40:02.087Z
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.
Kenneth Chen
CC BY-SA 3.0
2015-10-13T03:45:03.060Z
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.
Kitsune Cavalry
CC BY-SA 3.0
2015-10-13T03:47:54.707Z
I'm thinking of definite new lottery when $\alpha \ge 1$, for example let $\alpha L$ be a new lottery. Then by doing some trick the relation will be again compatible with a new $\alpha^\prime \in (0,1)$. But I'm not successful in this attempt.
Kenneth Chen
CC BY-SA 3.0
2015-10-13T05:17:47.017Z
@KitsuneCavalry Seems to me that selling some $L''$ and buying more than one unit of $L$ is a possible economic interpretation for $\alpha > 1$.
Giskard
CC BY-SA 3.0
2015-10-13T06:54:20.720Z
$\alpha\ge 1$ 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.
Herr K.
CC BY-SA 3.0
2015-10-13T13:37:10.070Z
Thanks, that's what I thought. I would reach out to my lecturer since it's confusing.
Kenneth Chen
CC BY-SA 3.0
2015-10-13T13:52:36.733Z
@denesp selling a stake in a lottery into debt is odd, but I suppose feasible
Kitsune Cavalry
CC BY-SA 3.0
2015-10-13T14:07:43.637Z
@denesp In that case, the economics makes sense, but can we use the aforementioned definition of independence to prove that the case of $\alpha \ge 1$ will still hold?
Kenneth Chen
CC BY-SA 3.0
2015-10-13T14:41:06.177Z
@KennethChen Sorry but I have no clue. Otherwise I would have answered. I guess it depends on your lottery space: If for all $\alpha \in \mathbb{R}, \ L \in \mathscr{L}$ you have $-L \in \mathscr{L}$ and $\alpha L \in \mathscr{L}$ then I think this 'extended independence' would follow from normal independence. Also see the comment of Herr K.
Giskard
<microeconomics><decision-theory><risk><uncertainty>
Independence axiom of lottery when $\alpha \ge 1$
Kenneth Chen
# Independence axiom of lottery when $\alpha \ge 1$

(Asked by: username_0 on 2015-10-13)

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.

**Tags:** microeconomics, decision-theory, risk, uncertainty
> **Comments:**
> **username_1**: 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.  
> **username_0**: 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.  
> **username_1**: 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.  
> **username_0**: I'm thinking of definite new lottery when $\alpha \ge 1$, for example let $\alpha L$ be a new lottery. Then by doing some trick the relation will be again compatible with a new $\alpha^\prime \in (0,1)$. But I'm not successful in this attempt.  
> **username_2**: @KitsuneCavalry Seems to me that selling some $L''$ and buying more than one unit of $L$ is a possible economic interpretation for $\alpha > 1$.  
> **username_3**: $\alpha\ge 1$ 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.  
> **username_0**: Thanks, that's what I thought. I would reach out to my lecturer since it's confusing.  
> **username_1**: @denesp selling a stake in a lottery into debt is odd, but I suppose feasible  
> **username_0**: @denesp In that case, the economics makes sense, but can we use the aforementioned definition of independence to prove that the case of $\alpha \ge 1$ will still hold?  
> **username_2**: @KennethChen Sorry but I have no clue. Otherwise I would have answered. I guess it depends on your lottery space: If for all $\alpha \in \mathbb{R}, \ L \in \mathscr{L}$ you have $-L \in \mathscr{L}$ and $\alpha L \in \mathscr{L}$ then I think this 'extended independence' would follow from normal independence. Also see the comment of Herr K.  

---

## 1 answers

### Answer by username_4 on 2015-10-13. Score: 8 | Accepted answer

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](http://rads.stackoverflow.com/amzn/click/0273731912)** 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](http://rads.stackoverflow.com/amzn/click/0195073401)**, define a lottery as a vector *containing only the probabilities*:

$$L=\{p\_1,...,p\_n\}$$

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?

> **Comments:**
> **username_0**: This is a splendid answer!  
> **username_4**: @KennethChen Thanks. My main concern was to make clear that we fool around with probabilities, not with outcomes.  

---

