From: zwiebel@leland.Stanford.EDU (Jeffrey Zwiebel)
Subject: Re: Samuelson (was Re: Yeah... ANOTHER variance vs expectancy posting.)
Organization: DSG, Stanford University, CA 94305, USA
Date: Thu, 23 Jul 92 03:17:36 GMT

In article <1992Jul22.233732.19558@fid.morgan.com> sethb@fid.morgan.com (Seth Breidbart) writes:
>>> In article <1992Jul15.235907.17186@leland.Stanford.EDU> zwiebel@leland.Stanford.EDU (Jeffrey Zwiebel) writes:


>
>>>>What is true, is that maximizing the expected log (Kelly) will, with
>>>>probability approaching 1 as the number of gambles grows large, yield
>>>>a higher return than any other betting strategy.
>
>As I showed in an earlier post, this is only true under certain
>circumstances.

You are correct and your example is a very nice one.  There are strategies
(such as the one you suggest) that say "play different from Kelly for a 
finite number of periods, and play Kelly all other periods"  (this isn't
quite what your rule says, more precisely one should word this as "with
probability 1 play Kelly all but a finite number of periods") that may beat 
Kelly with a positive probability at the number of periods grows large
(even with probability approaching 1 as in your example!).  This will more
or less ONLY be true for such rules that differ from Kelly only for a finite
number of periods.  (I say more or less because to be precise, one must 
also throw out rules that differ from Kelly each period, but by a very small 
amount that converges to 0 over time).  This is a limitation to the Kelly 
beats all other strategies with probability approaching 1 result, but a 
relatively minor one - it beats all strategies that do not approach the 
Kelly strategy itself.  And as you point out, this includes all strategies
for which the bet depends solely on your bankroll and the odds.  And once
again, for the benefit of anyone who hasn't read previous posts, this by
no means implies Kelly is optimal.


>
>>>>The reconciliation of these two apparently contradictory facts is actually
>>>>quite simple.  While it is true that if one plays long enough, with a very
>>>>high probability one will do better with Kelly, in the (very unlikely)
>>>>states where one does worse, one does much worse, enough so to outweigh
>>>>doing better most of the time.  
>>>
>>> With probability exactly *1.0*, one will do better with Kelly, for any
>>> final state utility function that holds that the more money the better,
>>> assuming one has unlimited time (and a few other less constraining
>>> assumptions.)
>>
>>Jeez Michael, that _exactly_ is only reached after an infinite number of
>>bets.
>
>That's the key point.  The probability of having more money by using
>Kelly approaches 1.  Since the probability <1 after finite time, if
>the excess utility in that fraction is large enough (e.g. linear
>utility), Kelly is not optimal.

The point is, to say correct things about what happens after many periods 
when looking at the limit, one has to take the correct limit.  The answer to
the question, "which of two strategies yields more utility in the long
run" can be answered by looking at how utilities under the two strategies
converge as the number of periods goes to infinity.  This will not give the
same answer as taking the limits of the distributions and evaluating the 
utilities under the limiting distributions.  (This, as I mentioned before, is 
because in the current setting the necessary conditions to bring the limit
into the expectation operator do not hold).  When Michael refers to outcome 
under an infinite number of periods it is ambiguous, it could mean either
of these two limits.  However, only one, the former, gives any guide to 
which strategy is superior after an arbitrarily large (but finite) number
of periods.  And using the former clearly shows that Kelly will not
be near one's optimal strategy unless, one utility is near log. 


>Kelly will also optimize in one other case: if you
>put a sufficiently high utility on bragging rights.
>

Sorry to nitpick here, because the great majority of the points you made
in your set of posts, Seth, were right on the money.  But bragging rights
won't justify playing Kelly unless (almost tautologically) you want to brag
"I made more money than you" (and you don't care by how much) or "I played
Kelly with little justification and still made money" etc ... And this is 
then equivalent to specifying preferences that imply you have log utility.
If you instead want to brag "I made money this trip", or "I made $500 this
trip" or almost any other brag you can imagine (including "I made $x more
than you" where the larger x the more the satisfaction) then your utility
function will not be logarithmic, and you should not bet Kelly.   

>
>
>I just realized that there is another situation in which Kelly might
>be optimal.  Under all admissable strategies, I think Kelly has the
>highest probability of ending up ahead after a large number of bets.
>(An admissible strategy is one where the percentage of bankroll to bet
>is a function of the odds and payoffs.  Otherwise, a strategy like
>"Play Kelly until you're ahead, then play 0" will obviously dominate
>Kelly.)
>

See above.  Kelly will with probability approaching 1 beat all these 
(and almost all other strategies as well); but, and hopefully this is 
clear by now, it sure isn't optimal.  


				Jeff







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