Re-evaluate old beliefs

post by PhilGoetz · 2011-10-05T01:18:10.054Z · LW · GW · Legacy · 1 comments

Contents

  Model 1:  Static (or, constant re-evaluation of all beliefs)
  Model 2: No re-evaluation of old beliefs
None
1 comment

I've noticed that, although people can become more rational, they don't win noticeably more.  We usually re-calibrate our self-confidence, become more stubborn, and make bigger errors.

Is it possible that the benefit from increasing your prediction accuracy is no greater than the loss incurred from taking riskier bets due to greater self-confidence?

Model 1:  Static (or, constant re-evaluation of all beliefs)

For any yes/no question q, person i has probability p_i of being right.  Suppose person i always knows the true value of p_i.

Suppose, without loss of generality, that we consider only questions where person i decides the answer is "yes".  (The situation is symmetric for cases where they answer "no", so we will get the same results considering only the "yes" questions.)

Suppose that, for every question q, there is some current odds offered by the world, described by society's accepted value y for P(q).  Society sets odds to have an expected profit near zero.  So the bet will cost the player C if q is false, and will give a payout D if q is true, such that y(D) = (1-y)C.  Set D = 1-y, C = y.

Person i then takes the bet for each question q for which i thinks q is true, and Dp_i > C(1-p_i), or (1-y)p_i > y(1-p_i), 1/y - y/y > 1/p - p/p, y < p.  As they are right with probability p_i, they win p_i of these bets.

Suppose y is distributed uniformly from zero to one.  This is the most-suspicious assumption.

Person i's profit F(p_i) is their payouts minus their losses.  So F(p_i) is the integral, from y = 0 to p_i, of [(1-y)p_i - y(1-p_i) = p_i - y].  This integral equals p_i² / 2.

Good!  Your winnings should be proportional to the square of your accuracy!

But that doesn't seem to match what I observe.

Model 2: No re-evaluation of old beliefs

Now suppose that a person improves their accuracy p_i over time, and assumes that p_i is the accuracy for all their beliefs, but doesn't constantly re-evaluate all old beliefs.  In this model, time t will be equal to p_i (representing linear increase in accuracy, which may or may not be realistic, but is simple), ranging from p = p₀ to p_i, over time t=p₀ to t=p_i.

At time t, person i will take all bets with y < t for all their beliefs, although some of these beliefs were formed earlier, when person i had less accuracy.  Their profit F(i, t) is now the integral, from x = p₀ to t (x representing the time a belief was formed and its probability of being correct), of the integral, from y = 0 to t, of [(1-y)x - y(1-x) = x - y].  The inner integral evaluates to xt - tt/2, the outer integral evaluates to p_ip₀(p_i-p₀)/2, and we will divide the whole mess by (p_i-p₀) to normalize for the number of chances person i had to place bets.

Now, the expected profit is p_ip₀/2.  This is only linear in person i's current accuracy.  Perhaps more interesting, if we set p₀ = 0, expected profit is always zero.  The new bets taken based on new, more accurate beliefs are exactly balanced out by the losses from bets taken at increasingly worse odds on old, inaccurate beliefs.

p₀ ~ 0 is not as unreasonable as it sounds.  This is because we are not presented in life with a random sample of all possible questions.  We are presented with questions that have been filtered by other peoples' uncertainty about these questions.  Questions on which an answer is commonly accepted get that answer worked into the fabric of society (so that the decision is usually made for you), and then get ignored.  I expect for that reason that the average person has p_i ~ .5, and some people have p_i << .5.

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