[EM] Square Vote: an alternative to the Clarke-tax

Peter Barath peb at freemail.hu
Thu Dec 4 14:46:15 PST 2008

No mistake: I'm a big fan of the Clarke-tax. And I don't think
the fact that in the Clarke-tax real payment is rare, is a really
big problem. But here is a proposal to solve this problem:

I think the attribute of the Clarke tax that you pay only
if you become a pivotal voter, can be substituted with every
voter on the winning side having to pay, but only a small amount.

The following are for only the situation with two options. Why
work out more if nobody is interested? Besides, I don't yet have
any idea how to do for more than two.

We ask the voter to declare on the ballot, which is her option,
and how much more money that option is worth for her, compared
to the other one. In other words she have to price the difference,
exactly as in the Clarke-tax.

The option with the bigger sum of declared money wins and the
winners pay for the treasury.

But how much? If they have to pay the whole amount, voting will be
uneconomic for them. What if we divide the amount with a constant?
That wouldn't be good, this is why:

We can make an estimation amout the probability distribution.
Let D be the  (money for option A) - (money for option B)  difference.
Let d be the probability of D being between $0 and $1. I call it
the probability density, although I guess it has already a name
but I don't know it.

In a short area the density can be considered to be constant.
So the probability of D being somewhere between $0 and $a is d*a

If we make the winning voter to pay her declared money divided
by a fixed number, her decision will be uncertain, because if
for example she doubles the declared, the probability of making
her option win is also doubles, and the money she pays in case
of winning also doubles. No countable optimum.

So let's make the declared -> payed function non-linear.
It seems good if the function starts from the origo horizontally
and its steepness grows. With very low declarations, the payment
is almost 0, so it seems rational to declare more than 0. But
in high numbers, a small increase of the declaration leads
to very big increase of payment. An optimum must exist between.

The simplest such function is the square ( x*x or x^2 ) function.

Let the method be:

- The voting authority makes an estimation for the  d  density
and makes it public.
- Every voter declares which option she wants and the value for that.
- The option with the bigger sum wins.
- If a voter is on the winning side, and she declared value  a,
she must pay to the treasury   d*a^2

This will usually be a very small payment, because  d  is a
very small number - the probability of a $0-$1 margin.

What is the rational decision for the declared  a  money?

Let  P  be the probability that her choice wins even without her.
Let  V  be the value of winning for her.

There are three cases:

- Her side loses. Probability:  1-P-d*a  She does not win the
V  value and doesn't pay anything.
- Her side wins and would have won even without her. Probability:  P
She didn't win the  V  value (because her vote was unnecessary for
that). And she have to pay  d*a^2  so her gain is  -d*a^2
- She becomes a pivotal voter. Probability:  d*a   Now she really
gained the  V  value, and also have to pay so her gain is  V-d*a^2

So her expected gain, the sum of gains multiplied with their
respective probabilities is   d*a*(V - d*a^2) - P*d*a^2

If we consider it a function of  a  and search for the maximum
gain, we derivate (or how it called in English) it and search
the  a  at which the derivated function has 0 value. It will
be an equasion of a second degree, and the solution:

a = ( SQRT( P^2 + 3*d*V ) - P ) / (3*d)

There is a possible approximation that if  X>>y  then

SQRT(X^2+y)-X  approximately equal to   y/(2*x)

so here the optimal  a  approximately equal to  ((3*d*V)/(2*P))/(3*d)
that is  V/(2*P)

In a relatively close election  P  can be considered around 1/2
so the optimal declared  a  is closely equal to the real  V  value
of the choice.

I think we can not leave out the approximation that  P  is about 1/2.
It would be somehow impolite if the election authority predicted
which choice were more likely to win.

But we can adjust the approximation in the optimum calculation
if we use a more precisely chosen payment function, an infinite sum.

instead of the  d*a^2  the winner payment would be:

(1/(4*d)) * ((2*d*a)^2 - (2*d*a)^3 + (2*d*a)^4 - (2*d*a)^5 + ....)

Peter Barath

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