von Neumann-Morgenstern utility criterion
Here's a possible interesting criterion for rated methods:
Suppose the ratings are on a continuum (discrete versions automatically
fail). Then the winner should not change if any voter's rating vector is
subjected to some arbitrary nonnegative affine scaling, as long as the
resulting values do not go outside the scale.
The idea is that lottery utilities (which I understand are called von
Neumann-Morgenstern utilities) are only defined up to two constants per
voter, which we may call constants of incommensurability or affine
scaling constants. And if the method is serious about being utilitarian,
it shouldn't make any assumptions about what these scaling constants are.
Obviously, ranked methods pass this criterion since affine scalings are
monotone. But some rated methods might also do so. Range clearly doesn't.
I would suspect that methods that pass this criterion will fail IIA. But
in a sense, that's more honest than technically passing IIA but still
having the election outcome depend on losers (like Range does when the
voters do the normalization themselves).
-km
El sáb., 29 de ene. de 2022 2:54 p. m., Kristofer Munsterhjelm <
km_elmet@t-online.de> escribió:
Here's a possible interesting criterion for rated methods:
Suppose the ratings are on a continuum (discrete versions automatically
fail). Then the winner should not change if any voter's rating vector is
subjected to some arbitrary nonnegative affine scaling, as long as the
resulting values do not go outside the scale.
The idea is that lottery utilities (which I understand are called von
Neumann-Morgenstern utilities) are only defined up to two constants per
voter, which we may call constants of incommensurability or affine
scaling constants. And if the method is serious about being utilitarian,
it shouldn't make any assumptions about what these scaling constants are.
I think invariance under affine transformation is too strong a condition.
Invariance under scaling by a positive constant should be enough.
Also, as in symmetrical MJ there should be an understood zero, a transition
between goods and bads.
However, negative ratings may not work in a political context.
Obviously, ranked methods pass this criterion since affine scalings are
monotone. But some rated methods might also do so. Range clearly doesn't.
I would suspect that methods that pass this criterion will fail IIA. But
in a sense, that's more honest than technically passing IIA but still
having the election outcome depend on losers (like Range does when the
voters do the normalization themselves).
-km
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On 30.01.2022 06:49, Forest Simmons wrote:
El sáb., 29 de ene. de 2022 2:54 p. m., Kristofer Munsterhjelm
<km_elmet@t-online.de mailto:km_elmet@t-online.de> escribió:
Here's a possible interesting criterion for rated methods:
Suppose the ratings are on a continuum (discrete versions automatically
fail). Then the winner should not change if any voter's rating vector is
subjected to some arbitrary nonnegative affine scaling, as long as the
resulting values do not go outside the scale.
The idea is that lottery utilities (which I understand are called von
Neumann-Morgenstern utilities) are only defined up to two constants per
voter, which we may call constants of incommensurability or affine
scaling constants. And if the method is serious about being utilitarian,
it shouldn't make any assumptions about what these scaling constants
are.
I think invariance under affine transformation is too strong a
condition. Invariance under scaling by a positive constant should be enough.
Also, as in symmetrical MJ there should be an understood zero, a
transition between goods and bads.
It might be possible to define away one of the two constants in a
particular setting by setting a privileged zero, although I don't know
if the resulting system have all the properties of VNM.
Suppose you have a one-shot event: either X happens or X doesn't happen,
and you can choose whether it does. If you prefer it happening to not
happening, then your utility is positive, otherwise it's negative.
So in common-sense logic: if you'd rather not be around for it to happen
(as is the case with e.g. most pain), then your utility is negative,
otherwise it's positive (or zero if you're indifferent).
The problem, as someone (it might have been you) pointed out, is that in
matters like politics, the counterfactual is not possible: someone has
to be elected as governor (or president or MP or what have you). Even if
there was a NOTA option, not electing anyone doesn't make everything
else equal: ungoverned, the country would still go on changing.
(By the way, could you set gmail to text mode for EM? Then I wouldn't
have to add more > marks for nested quotes.)
I would suspect that methods that pass this criterion will fail IIA. But
in a sense, that's more honest than technically passing IIA but still
having the election outcome depend on losers (like Range does when the
voters do the normalization themselves).
One note about this: now that I've thought about it a bit more, I think
it's clear that it fails IIA. Linear scaling invariance implies that
two-candidate elections are determined by majority rule. If we have a
three-candidate election that's a Condorcet cycle, then we may engineer
the lottery information so that there's a definite winner according to
the method in question. Then eliminating one of the other two candidates
(which one depends on who the winner is) will lead to IIA failure.
-km