33 05 August 2007 08:49PM

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Comment author: 19 June 2011 07:13:59PM *  1 point [-]

Eliezer wrote: "...you do work out that, if some particular outcome occurs, then your utility function is logarithmic in time spent preparing the excuse." That kind of dropped out of the sky, didn't it?

The only way that I can make sense of the line you quote is to assume that the pundit already identifies "the probability that bond prices go up" with "the fraction of the 100 minutes that I ought to spend on a story explaining why bond prices went up".

For simplicity, suppose that there are only two possible outcomes, UP and DOWN. Let p be the probability of UP, where 0 < p < 1. Let U(x) be the utility of having spent 100x minutes on an explanation for an outcome, given that that outcome occurs. (So, we are assuming that the utility of spending 100x minutes on a story depends only on whether you get to use that story, not on whether the story explains UP or DOWN. In other words, it is equally easy to concoct equally good stories of either kind.) Assume that the utility function U is differentiable.

The pundit is trying to maximize the expected utility

EU(x) = U(x) p + U(1−x) (1−p).

But it is given that the pundit ought to spend 100p minutes on UP. That is, the expected utility attains its maximum when x = p. Equivalently, the utility function U must satisfy

EU′(p) = U′(p) pU′(1−p) (1−p) = 0.

That is,

U′(p) / U′(1−p) = (1/p) / (1/(1−p)).

This equation should hold regardless of the value of p. In other words, the conditions are equivalent to saying that U is a solution to the DE

U′(x) / U′(1−x) = (1/x) / (1/(1−x)).

It's natural enough to notice that this DE holds if U′(x) = 1/x. That is, setting U(x) = ln(x) yields the desired behavior.

More generally, the DE says that U′(x) = (1/x) g(x) for some function g satisfying g(x) = g(1−x). But if you are only interested in finding some model of the pundit's behavior that predicts what the pundit does for all values of p, you can set g(x) = 1.