I agree that you will stop voting on a post when the marginal cost = marginal benefit. This means if I take my votes from above:
Post A: 5 votes
Post B: 8 votes
Post C: 1 vote
This means I valued a vote on post A at price 5, post B at price 8, and post C at price 1, where the prices are all relative to each other.
So my noob question is: Don’t I know this relative pricing having to say what the marginal cost function is? If each marginal vote had costed 2n, wouldn’t I still have stopped at the same relative prices? Like, if the above votes were done with the marginal vote costing n, here’s what it would look like if the marginal vote cost 2n.
Post A: 2 votes
Post B: 4 votes
Post C: 0 votes
Which are all basically the same ratios (with a little bit less resolution), right?
If you scale it by a constant k that will happen (as the constant will just stick around in the derivative, and so you’ll buy votes until marginal cost = marginal benefit / k).
If you were to use like f(V)=V33 then each marginal vote would cost f′(V)=V2 , and so you’d buy a number of votes V such that V=√B (where B is your marginal benefit).
Some of the QV papers have uniqueness proofs that quadratic voting is the only voting scheme that satisfies some of their desiderata for optimality. I haven’t read it and don’t know exactly what it shows.
Yeah, I guess I’m hearing that all versions of it still cause the voter to work out prices, and that the information is findable by transforming their votes, but that quadratic doesn’t require doing any transformation and makes things simple.
I think that it’s not just about having an easier time reverse engineering people’s values from their votes. It might be deeper. Different rules might cause different equilibria/different proposals to win, etc. However I’m not sure and should probably just read the paper to find out the details.
I agree that you will stop voting on a post when the marginal cost = marginal benefit. This means if I take my votes from above:
Post A: 5 votes
Post B: 8 votes
Post C: 1 vote
This means I valued a vote on post A at price 5, post B at price 8, and post C at price 1, where the prices are all relative to each other.
So my noob question is: Don’t I know this relative pricing having to say what the marginal cost function is? If each marginal vote had costed 2n, wouldn’t I still have stopped at the same relative prices? Like, if the above votes were done with the marginal vote costing n, here’s what it would look like if the marginal vote cost 2n.
Post A: 2 votes
Post B: 4 votes
Post C: 0 votes
Which are all basically the same ratios (with a little bit less resolution), right?
If you scale it by a constant k that will happen (as the constant will just stick around in the derivative, and so you’ll buy votes until marginal cost = marginal benefit / k).
If you were to use like f(V)=V33 then each marginal vote would cost f′(V)=V2 , and so you’d buy a number of votes V such that V=√B (where B is your marginal benefit).
Some of the QV papers have uniqueness proofs that quadratic voting is the only voting scheme that satisfies some of their desiderata for optimality. I haven’t read it and don’t know exactly what it shows.
Yeah, I guess I’m hearing that all versions of it still cause the voter to work out prices, and that the information is findable by transforming their votes, but that quadratic doesn’t require doing any transformation and makes things simple.
I think that it’s not just about having an easier time reverse engineering people’s values from their votes. It might be deeper. Different rules might cause different equilibria/different proposals to win, etc. However I’m not sure and should probably just read the paper to find out the details.