$Reconciling Kelly Criterion derivatives$

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$Reconciling Kelly Criterion derivatives$

There are various derivatives of the Kelly Criterion, depending on the context (e.g. Betting, Investing, other)

I'm trying to reconcile the different approaches to get a Kelly Criterion for a case where:

There's a 55% chance of winning, 45% chance of losing.
A win will result in 35% increase (e.g. 100−>100−>135).
A loss will result in a 30% loss (100−>100−>70).

Equation 1. Generally: all variations seem to derived from:
$ f^* = \frac{(bp−q)}{b} $

p: probability of a win
q: probability of a loss
b: a win of "b" units for every unit waged

Equation 2. Per this:

$ f^* = \frac{(p \times \text{win} - q \times \text{loss})}{win} $

Equation 3. From Wikipedia, per Thorp (page 7, lines 25-28)
$ f^* = \frac{bp − aq}{ab} = \frac{p}{a} - \frac{q}{b} $

where a unit wager wins "b" and loses "a"

The issue is that these equations seem to not properly reconcile, and leave me uncertain what the correct formula is. I'd like to understand why, ultimately get to a formula to confidently use.

Notice for the initial example I posed:

Table 1: output of equations for above posed case

Eq #	p	q	a	b	$f^*$
1	55%	45%	n/a	$\frac{.35}{.30}$ per this (I'm not certain if correctly captures partial losses)	$ \frac{\frac{0.35}{0.30} \times 0.55-0.45}{\frac{.35}{.30}} = 0.164 $
2	55%	45%	0.30	0.35	$ \frac{(0.55 \times \text{0.35} - 0.45 \times \text{0.30})}{0.35} = 0.164$
3	55%	45%	0.30 (uncertain if these are correct values given the case, see table 3)	0.35 (uncertain if these are correct values given the case, see table 3)	$ \frac{0.55}{0.30} - \frac{0.45}{0.35} = 0.547 $

If we assume a full loss however, everything matches up.

Table 2: adjustment to posed case, to show a full loss reconciles equations

Eq #	p	q	a	b	$f^*$
1	55%	45%	n/a	$\frac{.35}{1}$	-0.7357143
2	55%	45%	1	0.35	-0.7357143
3	55%	45%	1	0.35	-0.7357143

Further adding to confusion, Wikipedia had a revision (see under "Investment formula", which I've for now corrected on the latest revision) indicating that the values of "a" and "b" for the formula of $ f^* = \frac{bp − aq}{ab} = \frac{p}{a} - \frac{q}{b} $ be determined by adding a 1 to the percentage price chanages. Thus this previous revision indicates:

Table 3: wikipedia revision suggests a, and b should be derived differently

Eq #	p	q	a	b	$f^*$
1	55%	45%	n/a	$\frac{.35}{.30}$	$ \frac{\frac{0.35}{0.30} \times 0.55-0.45}{\frac{.35}{.30}} = 0.164 $
2	55%	45%	0.30	0.35	$ \frac{(0.55 \times \text{0.35} - 0.45 \times \text{0.30})}{0.35} = 0.164$
3	55%	45%	1.30	1.35	$ \frac{0.55}{1.30} - \frac{0.45}{1.35} = 0.089 $

Math Finance

Dforootan

24

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The ones on Wikipedia are correct. Note that (1) is just (3) in the case that $a = 1$, meaning that a loss causes you to lose the entire wager. The first equation only applies in the case that there is a total loss. In your case, you must use (3). The second equation is completely incorrect and you should ignore it.

Let us make use of the Kelly criterion for your scenario. The fraction gained in the positive outcome is $b = 35\% = 0.35$, and the fraction lost in the negative outcome is $a = 30\% = 0.30$. The probability of a positive outcome is $p = 55\% = 0.55$ and the probability of a negative outcomes is $q = 1 - p = 0.45$. Therefore, the fraction of your assets to apply is
$$f^* = \frac{p}{a} - \frac{q}{b} = 0.548.$$

From Wikipedia (if you need help understanding specifics of the paper or proof I suggest making an additional question, since you originally did not seem to be asking about the proof of this result):

The heuristic proof for the general case proceeds as follows. In a single trial, if one invests the fraction f of their capital, if the strategy succeeds, the capital at the end of the trial increases by the factor $1 − f + f ( 1 + b ) = 1 + f b$ , and, likewise, if the strategy fails, the capital is decreased by the factor $1 − f a$ . Thus at the end of N trials (with $p N$ successes and $q N$ failures), the starting capital of $\$1$ yields

$$C_N = ( 1 + f b )^{p N} ( 1 − f a )^{q N}.$$
Maximizing $\log ⁡ ( C_N ) / N$ , and consequently $C_N$ , with respect to f leads to the desired result

$$f^∗ = p / a − q / b.$$

Edward O. Thorp provided a more detailed discussion of this formula for the general case.

Blue

166

Dforootan

0

Thanks. Can you show the translation of Eq1 into Eq3?
- Dforootan
  
  0
  
  Just the clarify my question, I understand eq 1 is eq3 if a=1. But my understanding is that the original kelly was focused on complete loss. At some point the partial loss was introduced. Trying to understand how that partial loss derivative was incorporated given that original kelly "b" was all or nothing.
Walter

0

@Blue: Please respond to Dforootan comment above.
Blue

0

The Kelly Criterion is more general than the complete loss case. You start with the partial loss and derive the complete loss formula, not the other way around.
- Dforootan
  
  0
  
  My understanding is that the original Kelly Criterion, from the paper (pdf below) only has full loss per, αs, not partial, is that incorrect? https://www.princeton.edu/~wbialek/rome/refs/kelly_56.pdf
- Blue
  
  0
  
  Sure, but there exists another proof for the more general result, why you can find sketched and cited on the Wikipedia page you linked earlier. The historical order of the results need not affect which one is the more general criterion.
- Dforootan
  
  0
  
  Oh, I see. Could you please share that proof? there's a lot in the wikipedia and Thorton paper.
Blue

0

https://wayback.archive-it.org/all/20090320125959/http://www.edwardothorp.com/sitebuildercontent/sitebuilderfiles/KellyCriterion2007.pdf
Dforootan

0

Thank you for your help!

$Reconciling Kelly Criterion derivatives$

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