My question relates to my understanding of Arrow's Impossibility Theorem and ranked choice. It seems to me that the requirements on social choice functions are too strict. A social choice function cannot result in ties. This sounds unreasonable to me.

Let two voters with preferences $a > b > c > d$ and $a > c > b > d$. It seems that any "reasonable" choice function would choose $a > b = c > d$. Any other output would be making an arbitrary distinction between $b$ and $c$. So my question is: if we relax the output of our social choice functions to allow for ties, is there a social choice function that satisfies Arrow's criteria? Is there proof that no such function exists?

As an addendum, I tried a "trivial" such function but it doesn't seem to work. Let $f$ a social choice function which if every voter prefers $c_1 > c_2$ then in the final output $c_1 > c_2$; otherwise $c_1 = c_2$.

This seems to fail by the transitivity of ties. So, for two preferences $a > b > c$ and $c > a > b$ then $a > b$ but $c = a$ and $c = b$ which is a contradiction. If we were to choose, $a > b = c$ we would be violating the independence of irrelevant alternatives. As, if we were to delete $b$ the result "should" be $a = c$. Is there a way to work around this?

  • $\begingroup$ I'm pretty confident Arrow's result says even if we allow ties in the social choice function, it is impossible. $\endgroup$ – Walrasian Auctioneer Dec 31 '20 at 18:57
  • $\begingroup$ Notation varies in social choice theory and the term "social choice function" is often used for a function from preference profiles to alternatives. What Arrow looked at were functions from preference profiles to preference orderings over alternatives. These preference orderings can have nontrivial indifferences. $\endgroup$ – Michael Greinecker Jan 1 at 23:18

Allowing indifference will not solve the problem of the effect of the independence of irrelevant alternatives (combined with the unanimity and transitivity requirements).

You have said $b >_1 c$ and $c >_2 b$ should lead to $b =_s c$

Now consider $a >_1 b$ and $a >_2 b$, which should lead to $a >_s b$ by unanimity

  • so $a >_1 b >_1 c$ and $c >_2 a >_2 b$ should lead to $a >_s b =_s c$ by transitivity
  • and thus $a >_1 c$ and $c >_2 a $ should lead to $a >_s c$ by the independence of irrelevant alternatives

And also consider $c >_1 a$ and $c >_2 a$, which should lead to $c >_s a$

  • so $b >_1 c >_1 a$ and $c >_2 a >_2 b$ should lead to $b =_s c >_s a$
  • and thus $b >_1 a$ and $a >_2 b $ should lead to $b >_s a$

Combining these means $b >_1 a >_1 c$ and $c >_2 a >_2 b$ should lead to $b >_s a >_s c$

  • but that implies $b >_1 c$ and $c >_2 b$ should lead to $b >_s c$

  • inconsistent with your $b >_1 c$ and $c >_2 b$ leading to $b =_s c$


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