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

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Comment author: DanielLC 10 December 2014 07:45:38PM 3 points [-]

Yes, but what maximizes the expected value of one does not necessarily maximize the expected value of the other. The average of zero and four is two. The average of log(zero) and log(four) is negative infinity. Ergo, if you have one dollar and you have the option of betting it so you get four dollars if you win and nothing if you lose, betting increases the expected value of your bankroll, but reduces the expected value of the log of your bankroll.

Comment author: Lumifer comment score below threshold [+] (11 children)

Comment author: DanielLC 10 December 2014 09:26:46PM 3 points [-]

Average of 1 and 100 is 50.5. Average of log(1) and log(100) is log(10). If you're offered a choice of a 50:50 chance of 1 vs 10 and a sure amount between 10 and 50.5, then you're better off with the bet for maximizing expected value, but you're better off with the sure thing for maximizing the expected log value.

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.

Comment author: Lumifer comment score below threshold [+] (3 children)

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

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