[EM] 0-info approval voting, repeated polling, and adjusting priors

[Jobst Heitzig]

Hello,

some days ago I wrote:
> A point which troubles me is this: The justification of placing the
> approval cutoff at the expected or median utility according to the
> current priors is based upon the two assumptions that (1) given a top
> tie between two candidates, the events "x is among the two tied
> candidates" and "y is among the two tied candidates" are
> *independent* and that (2) the priors give the respective
> probabilities of these events. But neither do I believe that this
> independence can be assumed, nor do I believe that the probability of
> winning the poll is the same as the probability of being one of the
> two tied candidates given a top tie...

In the meantime I have thought about this more deeply.


0-info approval strategy (and analogously Weinstein's strategy) is based
upon the following fundamental thought: When I ask whether I should
approve of a candidate x, I must look at those possible situations in
which my approving or not approving of x makes a difference. In approval
voting, this is the case when the other voters produce a tie or near tie
between x and one or more other candidates.

  Assuming a large enough electorate, the probability of three- or more
tied candidates is so small that I may ignore this possibility and only
look at situations in which there is a top two-way tie between x and
some other candidate y. In each such situation, whether I approve of y
or not, the gain of switching from not approving of x to approving of x
is proportional to the difference in utility  u(x)-u(y)  according to my
personal utility function.

  In order to decide whether to approve of x or not, I can follow the
rationale of maximizing either expected utility or median utility. I
both cases, I need an estimate  p(x,y)  of the probability that the
two-way tie between x and y occurs. For expected utility, I approve of
 x iff the expected gain is positive, that is, iff the weighted sum

    sum { p(x,y)*(u(x)-u(y)) | all y }

is positive. For median utility, I approve of x iff the gain is more
probably positive than negative, that is, iff the weighted sum

    sum { p(x,y)*sign(u(x)-u(y)) | all y }

is positive. Both sums can be simplified when we assume that  p(x,y)  is
a product of two independent factors p(x), p(y):

    p(x,y) = p(x)*p(y) ???

This would be the case when the probability of any candidate y belonging
to a two-way tie was independent of what the other candidate x was. This
would imply that p(x) is proportional to the probability that x belongs
to a two-way tie. (But it would not necessarily imply that p(x) is
proportional to the probability that x wins without a tie!)


The above assumption of independence of the events "x in tie" and "y in
tie" is not only comfortable but also implies that the resulting ballot
is sincere in the sense that when the strategy tells me to approve of x,
then it also tells me to approve of all candidates I prefer to x.

  Unfortunately, this need not be the case when the events "x in tie"
and "y in tie" are not assumed to be independent. For example, assume
that my sincere preferences are  A>B>C>D  and that I estimate the
probabilities  p(A,B)  and  p(C,D)  as large compared to all other
p(x,y).  Then both 0-info and Weinstein's strategy tell me to approve of
A and C but not of B or D -- although I prefer B to C. This is because
given "C in tie", it seems much more probable that "D in tie" than
either "A in tie" or "B in tie", whereas given "B in tie", it seems much
more probable that "A in tie" than eiter "C in tie" or "D in tie". This
situation could easily arise when A and B are almost clones, and also C
and D are almost clones.


In the above example, I explicitly assume a correlation of the outcomes
of A and B and of the outcomes of C and D by assuming that they are
pairs of almost clones. The following example shows that a similar
suggestion for insincere approval of A and C can result even when the
outcomes of A,B,C, and D are modeled as explicitly *independent*:

  Let us assume that a number of polls have been taken which give
approval values  a(x,t)  for  x=A,B,C,D  and  t=1,...,n  which show some
variance but no time trend. Then it is a natural idea to model the
outcome approval a(x) of the candidates as independently distributed
random variables. In order to keep computation simple, let us first
assume normal distributions:

    a(x) ~ N ( m(x), s(x) ),

where m(x) and s(x) are the mean and standard deviation of the poll
outcomes a(x,t). In other words, we estimate the probability density of
the event "x gets approval score a" as

    f(x,a) = exp { - ((a-m(x))/s(x))^2 / 2 } / sqrt(2pi)s(x).

Consequently, the probability of the event "x gets approval score at
most a" is

    F(x,a) = integral { f(x,b) | b=-infinity..a }.

Then it is easy to see that the probability density of the event "x and
y tied" is

    integral { f(x,a) * f(y,a) * product { F(z,a) | all z except x,y }
               | a=-infinity..infinity }

Now assume that the polls show the following means and standard
deviations, listed along with my utilities for the candidates:

    x  m(x)  s(x)  u(x)
    A   65    5    100
    B   60    5     95
    C   20    5     90
    D   50    5      0

Then the above formula gives the following tie probabilities  p(x,y):

           y
           A      B      C      D
    x A    -      .043   8e-14  .002
      B    .043   -      6e-14  .001
      C    8e-14  6e-14  -      2e-14
      D    .002   .001   2e-14  -

So, although the outcome distributions are independent, the
probabilities of occurring in a tie are not.

  0-info strategy here tells me not to approve of B since given a B-tie,
it is much more probable that the tie is with A than with either C or D,
implying that the expected gain of approving of B is about

    ( 43(95-100) + 0(95-90) + 1(95-0) ) / 44 = -30/11 < 0

However, it still tells me to approve of C since given a C-tie, my
expected gain of approving of C is

    ( 8(90-100) + 6(90-95) + 2(90-0) ) / 16 = 35/8 > 0.

Isn't this strange?


Jobst