[EM] Saari reply

[Blake Cretney]

On Sun, 2002-06-23 at 16:17, Blake Cretney wrote:
> On Thu, 2002-06-20 at 13:45, Tarr, Adam wrote:
> > Dave wrote:
> > 
> > > I missed any believable proof of this equivalence.
> > >  
> 

Here's a more general proof than I gave yesterday.  BTW, I don't claim
that this is original.  It's just easier to do the algebra than search
for the paper.

p[1]..p[n] are the probabilities of each candidate being in the lead
before I vote.  u[1]..u[n] are the utilities I assign to each candidate.

The expected utility of the outcome without my vote is therefore given
by avr_u.

avr_u= p[1]*u[1] + p[2]*u[2] ... p[n]*u[n]

Now, let's say I am thinking about how to rate candidate i on the [0,1]
scale.  I was thinking about giving candidate i a rating of r, some
value other than 1, but I am considering giving it a rating of 1
instead.  I want to know how this will affect the result.

To find the change in expected utility caused by this change in my vote,
I have to sum the change in expected utility when each individual
candidate (j) is in front.  I will assume that the chance of this effect
is independent of which other candidate is front-runner.  This gives
 
(Probability of candidate being in front)*(probability I change the
result)*(utility of this change)

The probability of candidate j being in front is p[j].  I will assume
that the chance my rating change will change the result is proportional
to the size of my change in ratings.  So, it will be p_unk*(1-r), since
r was my old rating and 1 the new.

If there is to be any change in result caused by this change of vote it
will be to cause candidate i to win instead of j.  The benefit of this
is u[i]-u[j].  Of course, this may be a negative benefit.

Since I don't have a sigma, I'll use Sum(j,f(j)) to mean
f(1)+f(2)...+f(n).

change_u= Sum(j,p[j]*p_unk*(1-r)(u[i]-u[j]))

change_u= p_unk*Sum(j,p[j]u[i]-p[j]u[j]-r*p[j]u[i]+r*p[j]u[j])

of course, Sum(j,p[j])=1, since someone always wins.

change_u= p_unk * [ u[i]-Sum(j,p[j]u[j])-r*u[i]+r*Sum(j,p[j]u[j]) ]

But Sum(j,p[j]u[j])=avr_u

change_u= p_unk * [u[i]-avr_u-r*u[i]+r*avr_u]

change_u= (p_unk)(u[i]-avr_u)(1-r)

Now, all we really care about is whether change_u is positive.  Clearly
1-r and p_unk are positive, so the only question is whether u[i]-avr_u
is positive.  In other words, the strategy is good, as long as
u[i]>avr_u.  If u[i]=avr_u, it is neutral, otherwise it is bad.

So, what should we do if u[i]<avr_u, perhaps we could try setting the
rating to 0.  This creates the following equations.

change_u= Sum(j,p[j]*p_unk*(r)(u[j]-u[i]))

change_u= p_unk*r*Sum(j,p[j]u[j]-p[j]u[i])

By the same chicanery as I used before,

change_u= p_unk*r*(avr_u-u[i])

Well, since p_unk and r are greater than 0, this turns out to be a good
strategy if avr_u<u[i].  Neutral if avr_u=u[i]

So, my conclusion is that the best strategy in usual situations is to
find avr_u, then rate all candidates above avr_u at maximum, and all
those below at minimum.  That goes for Approval as well, BTW.

As for what usual situations are, a very small electorate might
invalidate the assumption of a proportional effect for a change in
ratings (I haven't thought too much about it).  It is also possible that
for some reason the probability of front-runners and close-seconds isn't
independent, so if X is winning, Y is likely to be in second, but if W
is winning, then Z is likely to be in second.  That would also
invalidate my argument.  Also, algebraic errors may invalidate my
argument ;-)

---
Blake Cretney


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