Am I stupid? I can't fit these two concepts into any type of harmony.

[BillC]

Well, OK you have a point--maybe it is more precise to say

the rate of growth of the expected bankroll is zero at twice Kelly

where the bankroll B(t) is geometric Brownian motion, which is at the foundation of Kelly betting theory, and is a standard topic in graduate courses in stochastic processes. See our paper cited in my July magazine article.

[PairTheBoard]

[ QUOTE ]
Well, OK you have a point--maybe it is more precise to say

the rate of growth of the expected bankroll is zero at twice Kelly

where the bankroll B(t) is geometric Brownian motion, which is at the foundation of Kelly betting theory, and is a standard topic in graduate courses in stochastic processes. See our paper cited in my July magazine article.

[/ QUOTE ]

If you have a July magazine article please provide the link here, and if you want to cite a paper cite it here.


It's not the "the rate of growth of the expected bankroll" either.

This is always assuming you are playing with an edge. Take the case where proportional betting is More than twice Kelly. Yes, in the long run, your Bankroll will converge in Probabilty to a negatively sloping curve that asymtopes to zero. But your Expected Bankroll will always be Greater than your Initial Bankroll. Your Expected Bankroll will Always be your Inititial Bankroll plus your Edge times your Action.

Again, playing with an edge. Consider the simple case of suicide Martingale Betting where you bet your bankroll until you lose. Clearly, your Bankroll converges in probabilty to zero. ie. Your Wong Win Rate is negative. But even as your Bankroll is converging in Probabilty to Zero, your Expected Bankroll climbs throughout.

Or are you now using the term "Expected Bankroll" to mean something other than the Expected Value of your Bankroll?

PairTheBoard

[PairTheBoard]

[ QUOTE ]
BillC --
the rate of growth of the expected bankroll is zero at twice Kelly

where the bankroll B(t) is geometric Brownian motion, which is at the foundation of Kelly betting theory, and is a standard topic in graduate courses in stochastic processes. See our paper cited in my July magazine article.


[/ QUOTE ]

I understand that under proportional betting the Bankroll can be aproximated by geometric brownian motion.

Again, you are wrong in saying, "the rate of growth of the expected bankroll is zero at twice Kelly".

However, I should correct one of my statements as well. It's Not the case that the Bankroll converges in probabilty to the curve defined by the Wong Win Rate.

Wong's Win Rate

Normalizing the initial bankroll to be 1 unit the convergence in probabilty is for this expression:

ln(Bn)/n

where Bn is the bankroll after n plays.

This converges in probabilty to what Wong calls the Bankroll's Exponential Growth Rate, G. Wong goes on to define the Win Rate, r, to be

r = exp(G) -1

The idea is that the nth root of your Bankroll is behaving like exp(G). For example, if your Bankroll was increasing by a factor of .05 on each play, after n plays your Bankroll would be (1.05)^n, just like n years of 5% compound interest. With continuous compound interest of 5% you get the exponential growth of exp(.05t). One year of 5% continuous compound interest gives you exp(.05)-1 for your Annual Percentage Rate. Inutitively, r is acting like you think of your edge acting. But it's not your edge.

For double Kelly betting r is zero. For over double Kelly betting r is negative. For Kelly Betting, r is Half your Edge. And here Wong says, "when you bet the optimal proportion of your bankroll, your expected win divided by your bet size is half of your expected arithmetic win rate."

At this point I am throwing up my hands. When he refers to "bet size" he is referring to the "proportion" not the actual action. Your expected win divided by your actual action remains equal to your edge. And to make matters worse he calls r/f your "rate of return on action under proportional betting", where f is your Kelly proportion.

This mish mash of terminology including terms like "win rate" and "expected win" as he uses it above has got me shaking my head.



PairTheBoard

[PairTheBoard]

[ QUOTE ]
Well, OK you have a point--maybe it is more precise to say

the rate of growth of the expected bankroll is zero at twice Kelly

where the bankroll B(t) is geometric Brownian motion, which is at the foundation of Kelly betting theory, and is a standard topic in graduate courses in stochastic processes. See our paper cited in my July magazine article.

[/ QUOTE ]

Ok Bill. After looking at your article Risk Formulas I'll settle for the term "Expected Growth Rate" as in the Expected Growth Rate for the Bankroll. However, I don't see what the "Expected" can refer to except the Expected Value of a specially defined "Growth Rate" random variable.

PairTheBoard

[BillC]

The expected value of the bankroll is a decreasing function of time. It has a negative derivative, i.e., a negative growth rate.

Do not confuse the expected bankroll's rate of change with positive expectation on a particular bet, or series of bets.

[PairTheBoard]

[ QUOTE ]
The expected value of the bankroll is a decreasing function of time. It has a negative derivative, i.e., a negative growth rate.

Do not confuse the expected bankroll's rate of change with positive expectation on a particular bet, or series of bets.

[/ QUOTE ]

Just to be clear here. Playing with a fixed positive edge and proportional betting which exceeds twice the Kelly optimum proportion. You say,

" The expected value of the bankroll is a decreasing function of time. "

Calling David Sklansky. Do you still agree with Bill here David? Anybody else?

PairTheBoard

[PairTheBoard]

[ QUOTE ]
BillC --
<font color="white"> ,
</font> The expected value of the bankroll is a decreasing function of time. It has a negative derivative, i.e., a negative growth rate.



[/ QUOTE ]

FALSE

Let me give a simple example that proves you are wrong here Bill.

Consider a weighted coinflip where you have a 55% chance of winning. Regular 1-1 payoff. Let your initial bankroll be 1 unit and make one coinflip bet per unit of time. Your proportional betting will be 100% of your current bankroll.

What is the expected value of your bankroll at time t=n?

E[Bn] = (.55)^n * 2^n = 1.1^n

The "expected value of your bankroll as a function of time" not only is NOT decreasing, it is an Increasing Function of Time. In fact, the expected value of your bankroll as a function of time Grows Exponentially to Infinity.

This is true even though the "Growth Rate" for your Bankroll is Negative and your Bankroll at time t=n converges in probabilty to zero as n--&gt;infinity.

PairTheBoard

[PairTheBoard]

btw, If you insist on the proportional betting being strictly less than 100% of the Bankroll, then in the above example make it 99%. The expected value of your bankroll at time t=n flips is then greater than the RHS below:

E[Bn] &gt; .55^n * 1.99^n = 1.0945^n

Now you never go broke, the "Growth Rate" of your bankroll is negative, your bankroll still converges in probabilty asymptotically to zero, but the Expected Value of your Bankroll grows exponentially toward infinity.

PairTheBoard

[David Sklansky]

Please let your uncle use the computer again.