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Lumifer comments on An investment analogy for Pascal's Mugging - Less Wrong Discussion

5 [deleted] 09 December 2014 07:50AM

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Comment author: Lumifer 10 December 2014 09:34:18PM 0 points [-]

You are confused between maximizing the log of the expected value of your bets and maximizing the expected value of the log of your bets. These are, of course, not the same.

In Kelly Rule bets you do not get paid the log of the outcomes.

Comment author: DanielLC 10 December 2014 09:51:45PM *  2 points [-]

The Kelly rule maximizes the expected value of the log of your bets. You get paid the outcome, but you presumably value the log of what you get paid.

The proof section on the Wikipedia article gives the derivation using Python and SymPy:

>>> from sympy import *
>>> x,b,p = symbols('x b p')
>>> y = p*log(1+b*x) + (1-p)*log(1-x)
>>> solve(diff(y,x), x)
[-(1 - p - b*p)/b]

It is maximizing the function y = p*log(1+b*x) + (1-p)*log(1-x). Maximizing the function y = p*(1+b*x) + (1-p)*(1-x) with the restriction that p is between 0 and 1 will give p as 0 or 1, since it's a linear function.

Comment author: Lumifer 10 December 2014 10:20:37PM *  -1 points [-]

The Kelly rule maximizes the expected value of the log of your bets.

The Kelly rule maximizes the log of your bankroll as the number of trials goes to infinity. Note that Wikipedia says:

one would set the derivative y'(x) of the expected value of the logarithmic bankroll y(x) to 0 and solve for x:

You're maximizing "the expected value of the logarithmic bankroll y(x)".

Comment author: DanielLC 10 December 2014 10:34:28PM 1 point [-]

If you take any of the bets, your bankroll is a probability distribution. Probability distributions have no standard ordering, and cannot be maximized.

Comment author: Lumifer 11 December 2014 01:05:13AM -1 points [-]

Yes, that's why you're maximizing the expected value and not an entire probability distribution.

Comment author: DanielLC 11 December 2014 06:31:47AM 2 points [-]

The Kelly rule maximizes the log of your bankroll as the number of trials goes to infinity.

I seem to have misread the second thing you said, which is more helpful.

You're maximizing "the expected value of the logarithmic bankroll y(x)".

Yes. The Kelly criterion maximizes the expected value of the logarithmic bankroll. Not the expected value of the bankroll.

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