<p>First it is important to define our terms.  The concept of "weighted voters" does not necessarily make sense for an arbitrary social choice rule.  So presumably we are interested in social choice rules where, at some point in the aggregation process, we add up a collection of terms, where each term contains information about the preferences or opinion of some voter.  Typically, such rules are "anonymous" in the sense that we can permute the opinions of the voters without changing the answer.  We can easily introduce a "weighted" version of such a rule by computing a <em>weighted</em> sum instead of an unweighted sum.  (Typically, a weighted version of the rule fails to be anonymous.)</p>

<p>Let us say that two social choice rules $F$ and $G$ are <em>different</em> if they sometimes produce different outputs when presented with identical input (e.g. when presented with identical preference profiles).  In that case, it is definitely the case that in general, different voter-weights will yield different social choice rules.  </p>

<p>For example, let's consider "weighted majority voting" with three voters, $A$, $B$ and $C$.  One possibility is that each voter receives equal weight (say, 1/3) ---this is effectively the "unweighted" version, which satisfies anonymity.  Another possibility is that voter $A$ receives weight 5, while voters $B$ and $C$ each receive weight 1.  It is easily verified that this "weighted majority rule" is actually a dictatorship for $A$.</p>

<p>Of course, we might say that all the different weighted majority rules belong to the same "family" of voting rules, and therefore they all have certain properties in common.  For example, there are many theorems which are proved for the family of "all weighted majority rules"  (or "for all weighted scoring rules", or whatever).  But that does not mean that they are the "same" rule.</p>

<p>When you ask, "does a disagreement regarding how to weight the voters entail a disagreement regarding aggregation rule?", the answer is "yes and no".  Clearly, we might agree on the <em>type</em> of voting rule to use (say, we agree to use a weighted majority rule because it satisfies certain desirable properties and we all agree these properties are important), while still disagreeing on the exact <em>weights</em> to use in the rule.</p>

CC BY-SA 4.0
2023-05-25T02:27:34.893Z
Thanks so much for this; I know it’s been years but wanted to ask about part of your answer: 1. It seemed to suggest that all anonymous rules can be given ‘weighted versions’ via computing a weighted sum, is that correct? 2. When you say that "Typically, a weighted version of the rule fails to be anonymous" do you just mean that some are anonymous because the 'weights' in question would be set to one?
Nikelmouse Dylar
CC BY-SA 4.0
2023-05-26T06:19:07.660Z
Hi Nikelmouse.  I will respond to your questions in this comment and the next one.

Q: " It seemed to suggest that all anonymous rules can be given ‘weighted versions’ via computing a weighted sum, is that correct?"

A: No, because not all voting rules involve computing any kind of sum at all.  (For example, some voting rules try to choose the alternative that minimizes some kind of "aggregate distance" to the reported opinions of the voters.  This is not necessarily representable as a sum.)
Marcus Pivato
CC BY-SA 4.0
2023-05-26T06:23:02.690Z
Q: "do you just mean that some are anonymous because the 'weights' in question would be set to one?"

A:  Yes.  If a family of voting rules can be represented in terms of "weighted sums" of the opinions of the voters, then typically the version where all these weights are *equal* is anonymous.
Marcus Pivato
CC BY-SA 4.0
2023-06-01T05:39:53.637Z
Thank you so much for all your help!
Nikelmouse Dylar
CC BY-SA 4.0
2018-07-31T20:20:00.137Z
2018-07-31T20:20:00.137Z
Marcus Pivato
<p>In Christian List's Stanford Encyclopaedia entry <em>Social Choice Theory</em> (2013, <a href="https://plato.stanford.edu/entries/social-choice" rel="nofollow noreferrer">https://plato.stanford.edu/entries/social-choice</a>) he says that:</p>

<blockquote>
  <p>"an aggregation rule is defined for a fixed set of individuals <em>N</em> and a fixed decision problem, so that majority rule in a group of two individuals is a different mathematical object from majority rule in a group of three."</p>
</blockquote>

<p>I was wondering if, in a similar way, social choice aggregation rules are defined for a set of weightings over <em>N</em>? In other words, does a disagreement regarding how to weight the voters entail a disagreement regarding aggregation rule?</p>

<p>E.g. let's say that <em>N</em> = {voter-1, voter-2}. Furthermore, you and I both favour a lexicographic dictatorship. The only difference is that you favour a lexicographic dictatorship which ranks voter-1 over voter-2, and I favour a lexicographic dictatorship which ranks voter-2 over voter-1. Strictly speaking, do we agree on aggregation rule, or do we disagree?</p>

<p>Any help would be greatly appreciated! :)</p>

CC BY-SA 4.0
2018-07-31T00:27:42.143Z
If the aggregation rules potentially produce different results then I would say they were different, even if they are in the same family of rules.  So you disagree. Civil wars and coups have happened because of this disagreement
Henry
<social-choice>
Is a social choice aggregation rule defined for a set of weightings over the set of voters (N)?
Nikelmouse Dylar
# Is a social choice aggregation rule defined for a set of weightings over the set of voters (N)?

(Asked by: username_0 on 2018-07-30)

In Christian List's Stanford Encyclopaedia entry *Social Choice Theory* (2013, <https://plato.stanford.edu/entries/social-choice>) he says that:

> "an aggregation rule is defined for a fixed set of individuals *N* and a fixed decision problem, so that majority rule in a group of two individuals is a different mathematical object from majority rule in a group of three."

I was wondering if, in a similar way, social choice aggregation rules are defined for a set of weightings over *N*? In other words, does a disagreement regarding how to weight the voters entail a disagreement regarding aggregation rule?

E.g. let's say that *N* = {voter-1, voter-2}. Furthermore, you and I both favour a lexicographic dictatorship. The only difference is that you favour a lexicographic dictatorship which ranks voter-1 over voter-2, and I favour a lexicographic dictatorship which ranks voter-2 over voter-1. Strictly speaking, do we agree on aggregation rule, or do we disagree?

Any help would be greatly appreciated! :)

**Tags:** social-choice

> **Comments:**
> **username_1**: If the aggregation rules potentially produce different results then I would say they were different, even if they are in the same family of rules.  So you disagree. Civil wars and coups have happened because of this disagreement  

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## 1 answers

### Answer by username_2 on 2018-07-31. Score: 2 | Accepted answer

First it is important to define our terms. The concept of "weighted voters" does not necessarily make sense for an arbitrary social choice rule. So presumably we are interested in social choice rules where, at some point in the aggregation process, we add up a collection of terms, where each term contains information about the preferences or opinion of some voter. Typically, such rules are "anonymous" in the sense that we can permute the opinions of the voters without changing the answer. We can easily introduce a "weighted" version of such a rule by computing a *weighted* sum instead of an unweighted sum. (Typically, a weighted version of the rule fails to be anonymous.)

Let us say that two social choice rules $F$ and $G$ are *different* if they sometimes produce different outputs when presented with identical input (e.g. when presented with identical preference profiles). In that case, it is definitely the case that in general, different voter-weights will yield different social choice rules.

For example, let's consider "weighted majority voting" with three voters, $A$, $B$ and $C$. One possibility is that each voter receives equal weight (say, 1/3) ---this is effectively the "unweighted" version, which satisfies anonymity. Another possibility is that voter $A$ receives weight 5, while voters $B$ and $C$ each receive weight 1. It is easily verified that this "weighted majority rule" is actually a dictatorship for $A$.

Of course, we might say that all the different weighted majority rules belong to the same "family" of voting rules, and therefore they all have certain properties in common. For example, there are many theorems which are proved for the family of "all weighted majority rules" (or "for all weighted scoring rules", or whatever). But that does not mean that they are the "same" rule.

When you ask, "does a disagreement regarding how to weight the voters entail a disagreement regarding aggregation rule?", the answer is "yes and no". Clearly, we might agree on the *type* of voting rule to use (say, we agree to use a weighted majority rule because it satisfies certain desirable properties and we all agree these properties are important), while still disagreeing on the exact *weights* to use in the rule.

> **Comments:**
> **username_0**: Thanks so much for this; I know it’s been years but wanted to ask about part of your answer: 1. It seemed to suggest that all anonymous rules can be given ‘weighted versions’ via computing a weighted sum, is that correct? 2. When you say that "Typically, a weighted version of the rule fails to be anonymous" do you just mean that some are anonymous because the 'weights' in question would be set to one?  
> **username_2**: Hi Nikelmouse.  I will respond to your questions in this comment and the next one.  Q: " It seemed to suggest that all anonymous rules can be given ‘weighted versions’ via computing a weighted sum, is that correct?"  A: No, because not all voting rules involve computing any kind of sum at all.  (For example, some voting rules try to choose the alternative that minimizes some kind of "aggregate distance" to the reported opinions of the voters.  This is not necessarily representable as a sum.)  
> **username_2**: Q: "do you just mean that some are anonymous because the 'weights' in question would be set to one?"  A:  Yes.  If a family of voting rules can be represented in terms of "weighted sums" of the opinions of the voters, then typically the version where all these weights are *equal* is anonymous.  
> **username_0**: Thank you so much for all your help!  

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