yes it very much does imply that; that's how math works
Conversation
I think you think I'm objecting to the first step (EV[wealth]-maximizing -> EV[utility]-maximizing when utility linear in wealth) when actually I'm objecting to the latter step (EV[utility]-maximizing -> utility-maximizing)
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It is a tautology that I prefer having greater utility to having less.
It is NOT a tautology that I prefer a strategy with higher EV(utility) to one with lower
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ok so what do you mean by "utility" here?
I guess, if we want, we can sidestep this and blacklist the world 'utility'.
I'm going to define 'qwer' to be "the thing that I'm trying to maximize the EV of".
if qwer is linear in wealth then the paper doesn't apply.
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My point is that you're not trying to maximize EV of qwer at all.
You are trying to maximize qwer.
And whatever qwer is, maximizing expected growth in qwer means that you will ALMOST SURELY maximize qwer itself in the long run
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dude I literally tried to define a made up word because you kept bending words to mean nonstandard things and you responded by disagreeing with the definition of a word I just made up?
also:
almost surely is not the only thing that matters!
outliers matter.
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I'm not disagreeing with the definition of the word qwer! I am disagreeing that you are trying to maximize the EV of anything.
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Like, fine, to be pedantic:
a) if only one sequence will end up happening, there are no interaction effects between them
b) So you have a set of possible outcomes and probabilities o_i and p_i
c) because of (a), U = sum_i[p_i*f(o_i)] for some function f
d) (cont'd)
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f here is whatever wacky nonlinear function you want.
d) define qwer == f(o_i)
e) you are trying to maximize the EV of qwer
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The key thing here is that, e.g. in St Petersberg, exactly one sequence of coin flips will happen. You'll get some ultimate value v from that sequence. Now just let f be the transformation that takes monotonically ordered v_i to a linear function in probability tradeoff space.
If this isn't enough:
Take 4 possible outcomes, a/b/c/d, ordered from worst to best.
Option 1: 50% B, 50% C
Option 2: X% A, (1-X)% D
For what X are you indifferent?
Because exactly 1 must happen, there is no interaction effect, and so this will determine your linear EV to max
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