First, Weber�s utilitly expectation-maximization method:
Because Russ used "candidate j" to represent the candidate for whom we�re considering voting, I used that too in my derivation that I just posted. But here I�ll let i be the candidate for whom we�re considering voting.
By "vote-expectation" I mean the expected benefit of voting for candidate i.
As with the other derivation, there�s the assumption that there are so many voters that your ballot won�t change the probabilities significantly.
There�s also the assumption that there are so many voters that any tie or near-tie will be between only 2 candidates.
And there�s the assumption that Weber�s Pij (which I�ll define soon) = Wi*Wj, the product of the win-probabilities of i & j.
At least as I define it here, Pij is the probability that you can make or break a tie between i and j, in i�s favor, by voting for i and not for j.
Let�s assume that Pji = Pij.
Looking at it with respect to j only, what�s the benefit of voting for i?
If you make or break an ij tie, in i�s favor, that will benefit you by (1/2)(Ui - Uj).
So, with respect to j, the expected benefit of voting for i and not for j is:
(1/2)Pij(Ui - Uj).
The 1/2 is present in all such terms, and so we can drop it.
But what if you�ve voted for j. What then is the benefit, with respect to j, of voting for i?
Well, if you don�t vote for i, then you�re voting for j and not for i, and the formula above applies, negatively. Pij(Uj-Ui), since Pji = Pij. Pij(Uj - Ui) = -Pij(Ui-Uj).
If you also vote for i, then you�re ceasing to vote for j and no for i, so you�re eliminating that
-Pij(Ui - Uj). So the expected benefit, with respect to j, of voting for i, when you�ve voted for j, is Pij(Ui-Uj). The same as if you hadn�t voted for j. So the matter of whether or not you vote for j doesn�t affect the benefit of voting for i, with respect to j.
To repeat, then, with respect to j, the expected benefit of voting for i is Pij(Ui - Uj).
So, to find the overall expected benefit of voting for i, we sum that expression over all j.
Sum, j<>i, Pij(Ui - Uj).
That�s Weber�s formula. I use Merrill�s name for that quantity: Strategic value. That sum is the strategic value of i.
If that sum is positive, then you benefit from voting for i. If it�s negative, then you lose by voting for i.
So, in Approval, vote for i if i�s strategic value is positive.
Incidentally, in Plurality, vote for the candidate with the greatest strategic value.
Now let�s assume that Pij = kWi*Wj, the product of the winning-probabilities of i & j. As I said, this derivation depends on that assumption. That�s the key to this derivation, just as the assumption of a uniform factor of win-probability reduction was key to the derivation in my previous posting.
That�s a reasonable assumption. The more likely a candidate is to win, the better a contender s/he must be. And the better contender s/he is, the more likely s/he is to be in a tie or near-tie.
So let�s replace the Pij by kWi*Wj. But let�s leave out the k, since it�s present in all terms of that type.
So we have:
Sum, j<>i, (Wi*Wj(Ui - Uj) > 0 That�s the condition for voting for i
Multiply it out:
Sum, j<>i(WiWjUi - WiWjUj) > 0
WiUi = Ei, i�s expectation-contribution. WjUj = Ej.
So Sum, j<>i, (WjEi - WiEj) > 0
Writing it as 2 sums:
Sum, j<>i, WjEi - Sum, j<>i, WiEj > 0
In the summation, of course all the non-i candidates take their turn as j. So j changes during the summation, but i doesn�t. For the summation, i just refers to one candidate, because it�s a summation over j. So the things involving only i are constants for the purpose of the summation.
Taking the constants out of the summations:
Ei*Sum, j<>i, Wj - Wi*Sum,j<>i, Ej > 0
Well, for j<>i, the sum of the Wj is 1 - Wi, since the sum of all the win probabilities is 1.
And the sum of the Ej = E - Ei for the same reason. E is the sum of the expectation contributions of all the candidates.
Substituting those in the previous inequality:
Ei(1-Wi) - Wi(E - Ei) > 0
Substituting WiUi for Ei:
WiUi(1 - Ui) - Wi(E - WiUi) > 0
Solve that for Ui. You get:
Ui > E
Mike Ossipoff
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