The Kelly Criterion: Comparison with Expected Values

This is the final post in a four-part series exploring the Kelly criterion:

In a previous post, we looked at the Kelly formula, which maximizes earnings in a series of gambles where winnings are constantly re-invested. Is this equivalent to maximizing the expected return in each game? It turns out that the answer is "no". In this post we'll look into the reasons for this and discover the pitfalls of expected values.

We will look at the same game as in the previous post:

$$ \frac{V_1}{V_0} = (1+lr_W)^W(1+lr_L)^{(1-W)} $$

with the variables:

  • \(V_0, V_1\): the available money before and after the first round
  • \(l\): fraction of available money to bet in each round (the variable to optimize)
  • \(r_W, r_L\): return on win and loss, 0.4 and -1 in our example (i.e. 40% of wager awarded on win, otherwise 100% of wager lost)
  • \(W\): Random variable describing our chances to win; valued \(1\) with \(p=0.8\), \(0\) with \(p=0.2\)

The Kelly formula obtained from maximizing \(\log V_1/V_0\) tells us to invest 30% of our capital in such a gamble. Let's see what the result is if we maximize the expected value \(E[V_1/V_0]\) instead.

This is trivial by hand, but we'll use SymPy, because we can:

import sympy as sp
import sympy.stats as ss
sp.init_printing()

l = sp.symbols('l')             # Define the symbol/variable l
W = ss.Bernoulli('W', 0.8)      # Random variable, 1 with p=0.8, else 0
def f1(W):                      # Define f1 = V_1/V_0
    return (1 + 0.4*l)**W * (1 - l)**(1-W)
ss.E(f1(W))                     # Calculate the expected value

Evaluating this gives:

$$ E[\frac{V_1}{V_0}] = 1 + 0.12l $$

Uh-hum... so the expected return has no maximum, but grows linearly with increasing \(l\). Essentially, this approach advises that you should bet all your money, and more if you can borrow it for negligible interest rates.

Could the problem be that we only look at a single round? Let's examine the expected return after playing 10 rounds:

# Note: We can not do f10 = f1(W)**10, since we need independent samples
W_list = [ss.Bernoulli('W_%d' % i, 0.8) for i in range(10)]
f10 = sp.prod([f1(Wi) for Wi in W_list])
sp.expand(ss.E(f10))
$$ E[\frac{V_{10}}{V_0}] = 6 \cdot 10^{-10} l^{10} + 5 \cdot 10^{-8} l^9 + ... + 1.2 l + 1.0 $$

All the coefficients of the polynomial are positive, there are no maxima for \(l >= 0\). What's going on?

Time to dig deeper. Let's say we bet all of our money each round. If we lose just once, all of our money is gone. After 10 rounds playing this strategy, the probability of total loss is:

$$ p({\text{one loss in 10 games}}) = 1 - 0.8^{10} = 0.89 $$

So 89% of the time we would lose all our money. However, the expected return after 10 rounds at \(l=1\) is:

$$ E[\frac{V_{10}}{V_0}]_{l=1} = 3.1 $$

So on average we'd have \$3.10 after 10 rounds for every Dollar initially bet, but 90% of the time we'd lose everything. Strange.

The big reveal

Things become clearer when we look at in more detail at the calculation of \(E[V_{10}/V_0]\):

\begin{align*} E[\frac{V_{10}}{V_0}]_{l=1} && = && 1.4^0 0^{10} && \times && 0.8^0 0.2^{10} && + \\ && && 1.4^1 0^9 && \times && 0.8^1 0.2^{9} && + \\ && && 1.4^2 0^8 && \times && 0.8^2 0.2^{8} && + \\ && && ... && && \\ && && 1.4^9 0^1 && \times && 0.8^9 0.2^{1} && + \\ && && 1.4^{10} 0^0 && \times && 0.8^{10} 0.2^{0} \tag{*} \\ \end{align*}

The expected value is the sum of probability-weighted outcomes(\(1.4\) and \(0\) are the per-round outcomes for win and loss). Since a single loss results in loss of all money, the only non-zero term in the sum is the starred one, that occurs with about 11% probability, at a value gain of \(1.4^{10} = 28\). This high gain is enough to drag the expected return up to 3.1. When more then 10 rounds are played, these numbers become more extreme: the winning probability plummets, but the winning payoff skyrockets, dragging the expected return further upwards.

This is a bit reminiscent of the St. Petersburg paradox, in that arbitrarily small probabilities can drag the expected return to completely different (read "unrealistic") values.

Different kinds of playing

The Kelly approach builds on the assumption that you play with all your available wealth as base capital, and tells you what fraction of that amount to invest. It requires reinvestment of your winnings. Obviously, investing everything in one game (\(l=1\)) is insane, since a single loss would brankrupt you. However, following Kelly's strategy is the fastest way to grow total wealth.

The expected-value approach of "invest everything you have" is applicable in a different kind of situation. Let's say you can play only one game per day, have a fixed gambling budget each day, and thus are barred from reinvesting your wins. If you invest your full daily gambling budet, you may win or lose, but over the long run you will average a daily return of \(1.4 \cdot 0.8 = 1.12\) for every dollar invested. The more you can invest per day, the higher your wins, thus the pressure towards large \(l\) values.

In a way, Kelly optimizes for the highest probability of large returns when re-investing winnings, while the expected value strategy optimizes for large returns, even if the probability is very low.

A dubious game

Should you play a game where the winning probability \(p\) is \(10^{-6}\), but the winning return \(r_W\) is \(2\cdot 10^6\)? Mathematically it seems like a solid bet with a 100% return on investment in the long run. The question is whether you can reach "the long run". Can you afford to play the game a million times? If not, you'll most likely lose money. If you can afford to play a few million times, it becomes a nice investment indeed.

Kelly would tell you to invest only a very small fraction of your total wealth into such a game. The expected-value formalism advises to invest as much as possible, which for most people is bad advice even when playing with a fixed daily budget and no reinvestment (i.e. the expected-value play style).

This is an interesting example for two reasons:

  • It demonstrates one of the ways how "rich become richer" - the game has high returns, but also a significant barrier to entry.
  • It demonstrates a downside of both the Kelly- and the expected-value approach. The two strategies are optimal in their use cases in the limit of infinitely many games, however for finitely many games they may give bad advice, especially regarding to very low-probability winning scenarios.

Conclusion

So, a brief discussion of the relationship between the Kelly strategy and the expected return. For me, it was striking how two seemingly similar approaches ("maximize the moneys") lead to so different results and how unintuitively the expectation value can be in the face of outliers.

If you're interested in a Jupyter Notebook containing several interactive plots that really helped me understand this material, you can find it here.

The Kelly criterion

Over the course of this blog post series, we looked at the classical Kelly criterion in the first post, and how it can be extended to situations such as stock buying, with multiple parallel investment opportunities, in the second post. Next, we investigated the origin of the logarithm in the Kelly formula in the third post, before finishing up with the current discussion about expected values.

Surely, there's more to say about the Kelly criterion. If you want to leave your thoughts, please do so in the comments below!


The Kelly Criterion: Where does the logarithm come from?

This is the third post in a four-part series exploring the Kelly criterion:

The neat thing about the derivations in the last two posts is that they give a motivation for "optimizing the logarithm of wealth". The logarithm is not put in by decree, but is a mathematical technicality that arises from the repeated betting process! Kelly mentions this in his original paper:

At every bet he maximizes the expected value of the logarithm of his capital. The reason has nothing to do with the value function which he attached to his money, but merely with the fact that it is the logarithm which is additive in repeated bets and to which the law of large numbers applies.

This argument is very general. Let's say we model the wealth \(V_n\) after \(n\) rounds of betting based on the initial wealth \(V_0\) in terms of a function \(f(R, l) = V_{n+1} / V_n\) as

$$ V_n = V_0 f(R, l)^n \tag 1 $$

where \(R\) is a random variable describing the possible outcomes of the game, and \(l\) is the fraction of available money to invest in each round. Then we can derive the following formula for the best \(l\) value:

$$ l_{opt} = \operatorname*{argmax}_l \prod_j f(r_j, l)^{p_j n} = \operatorname*{argmax}_l \sum_j p_j \log f(r_j, l) = \operatorname*{argmax}_l E \left[ \log \frac{V_{n+1}}{V_n} \right] \tag 2 $$

where \(r_j\) are the possible investment outcomes and \(p_j\) are the associated probabilities. The crucial change from random variable \(R\) to outcomes and probabilities \(r_j\) and \(p_j\) is justified by the law of large numbers. Based on the exponential nature of the formula, switching to a logarithmic view feels very natural.

Consequently, neither the details of the game — represented by the random variable \(R\) — nor the exact form of the per-round return \(f\) matter. Any iterative scheme with reinvestment of profits should be representable in the form of equation (1), leading to the logarithm in solution (2). Beyond the origin of the logarithm, this analysis also shows the universality of the Kelly derivation.

Unfortunately, this argument is mostly skipped in online discussions. Often the logarithm is not justified at all, or it is treated as "genius from the 1950s says: use \(\log\)". Sometimes the result is also linked to utility theory, which posits that having twice the money is not twice as useful. While utility theory may be true, reasonable people can disagree on their utility function — exactly how useful more or less money is to them. However, Kelly's result is not grounded in utility, and the \(\log\) does not represent logarithmic utility of money. Consequently, even people who disagree on their utility function should agree that the Kelly criterion is the fastest way to gain wealth.

I hope this post shed some light on the reasoning behind the Kelly decision scheme. If you're interested in a Jupyter Notebook containing several interactive plots that really helped me understand this material, you can find it here.

In the next post, we'll take a closer look at the relation of the Kelly criterion to expected values.


The Kelly Criterion: Multiple Investment Opportunities

This is the second post in a four-part series exploring the Kelly criterion:

After the last post introduced the Kelly criterion and its application in deciding how much to invest in a single gamble, we'll investigate whether Kelly can help us choosing between multiple investment opportunities.

We'll start with a mathematical model:

$$ V_1 = V_0 \left( (1+r_0)(1 - \sum_i l_i) + \sum_i l_i(1 + r_i) \right) $$

Simple, right? :-)

  • \(V_0, V_1\) are the available money before and after the first round, as before.
  • \(r_0\) is the risk-free return rate. This allows us to model opportunity costs (could put the money into a bank account instead of investing, \(r_0 = 0.004\)), or inflation (non-invested money loses value over time \(r_0 = -0.02\)).
  • \(l_i\) is the fraction of the available money invested in stock \(i\).
  • \(r_i\) is a random variable describing returns for stock \(i\).

The first term \((1+r_0) (1 - \sum_i l_i)\) represents all the money not invested in any stock, being invested at the risk-free return rate \(r_0\). The second term \(\sum_i l_i(1 + r_i)\) represents the outcomes of the investments in individual stocks \(i\).

We will re-formulate a bit to simplify the expression:

$$ \frac{V_1}{V_0} = (1+r_0) - \sum_i l_i (1+r_0) + \sum_i l_i(1 + r_i) = (1+r_0) + \sum_i l_i(r_i - r_0) $$

We can further simplify by merging the \(l_i\) and \(r_i\) into the vectors \(\vec l\) and \(\vec r\), with \(\vec 1\) being a vector where all elements are \(1\):

$$ \frac{V_1}{V_0} = (1+r_0) + \vec l \cdot (\vec r - r_0 \vec 1) $$

Let's move from a single game round to \(n\) rounds:

$$ \frac{V_n}{V_0} = \left( (1+r_0) + \vec l \cdot (\vec r - r_0 \vec 1) \right)^n $$

Now we are again in a position where we can follow Kelly's prescription: invoking the law of large numbers and finding the \(l\) which maximizes the gain \(V_n/V_0\).

For large \(n\), we can approximate:

$$ \frac{V_n}{V_0} \approx \prod_j \left( (1+r_0) + \vec l \cdot (\vec r_j - r_0 \vec 1) \right)^{p_j n} $$

In this step we switched from the random variable \(\vec r\) to its outcomes \(\vec r_j\). Each possible outcome \(\vec r_j\) occurs with probability \(p_j\). We justify this switch with the law of large numbers: the outcome \(\vec r_j\) will be observed proportionally to its probability, \(p_j n\) times (for large \(n\)). Consequently, we will have \(p_j n\) factors involving \(\vec r_j\) in the overall product, with \(j\) iterating over all potential outcomes of \(\vec r\).

Note that since \(\vec r_j\) is vector-valued, \(p_j\) is a joint probability distribution.

At this point we are nearly finished. We are looking for the vector \(\vec l_{opt}\) that maximizes \({V_n}/{V_0}\). In analogy to last post, we arrive at:

$$ \vec l_{opt} = \operatorname*{argmax}_{\vec l} \frac{V_n}{V_0} = \operatorname*{argmax}_{\vec l} \log \frac{V_n}{V_0} = \operatorname*{argmax}_{\vec l} \sum_j p_j \log \left( (1+r_0) + \vec l \cdot (\vec r_j - r_0 \vec 1) \right) $$

This equation is not analytically solvable, but may be approximated as a quadratic programming problem as described in a paper by Vasily Nekrasov.

Conclusion

It should be obvious that the Kelly criterion is applicable in a wide range of scenarios, from gambling over investment decisions to whether to buy insurance. If you're interested in a Jupyter Notebook containing several interactive plots that really helped me understand this material, you can find it here.

In the next post we'll discuss the origins of the logarithm in the Kelly formula.


The Kelly Criterion: Introduction

This is the first post in a four-part series exploring the Kelly criterion:

The Kelly criterion is a formula used to determine the optimal size of a series of bets in order to maximize wealth. It is often described as optimizing the logarithm of wealth, and will do better than any other strategy in the long run.

Let's look at a simple gamble: when you play, you win with probability \(p\) and lose with probability \(q = 1-p\). If you win, you get back your initial bet plus a fraction \(r\) of that bet. If you lose, your bet is forfeited. The Kelly formula calculating the optimal fraction of your available wealth to bet is:

$$ l_{opt} = \frac{pr - q}{r} = \frac{pr - (1-p)}{r} = p - \frac{1-p}{r} $$

Here is a small demo to explore the formula:

Kelly calculation demo

\(p\){{(p/100).toFixed(2)}}
\(r\){{(r/100).toFixed(2)}}

Probability of winning: {{p}} %
Profit on win, as percentage of placed bet: {{r}} %
Optimal fraction of wealth bet in this game: {{ (100*computeKelly(p/100, r/100)).toFixed(1) }} %

The main takeaways are:

  • In each game you invest a fraction \(l\) of the total amount of money you have. If you play the game from the demo (\(p=0.8, r=0.4\)) and you have $100 available, you should invest $30 (since \(l=0.3\)). If you win, you will have $112 (\(100 + 30 \times 0.4\)) and invest $33.60 (\(112 \times 0.3\)) in the second round. If you lost the first round you'd have $70, and invest $21 in the second round. If you can not reinvest your earnings, Kelly does not apply.
  • This formula holds when a large number of games are played. If you play only few rounds of a game this may not apply.
  • When it applies, the Kelly strategy will generate the most wealth out of your initial starting capital.

Derivation of the Kelly criterion

We will take a short dive into mathematics to understand how Kelly came up with this formula. Let's compute the wealth \(V_1\) after the first round of gaming, when starting from an initial wealth \(V_0\):

$$ V_{1} = V_{0} (1 + Wlr_W + (1-W)lr_L) $$

Here, \(l\) is the fraction of money bet, \(r_W\) is the profit on win, \(r_L\) is the fraction of the bet forfeited on loss (in the above scenario, \(r_L=-1\)). \(W\) is a random variable that's \(1\) with probability \(p\) and \(0\) with \(1-p\). So with probability \(p\) we'll have \(W=1\), giving \(V_{1,win} = V_0 (1+lr_W)\), with probability \(1-p\) we'll get \(V_{1,loss} = V_0 (1+lr_L)\).

We can reformulate the expression in a way that will show an important generalization:

$$ V_1 = V_0 (1+lr_W)^W(1+lr_L)^{(1-W)} $$

The two expressions look very different, but they are equivalent since \(W\) can only take on the values \(0\) and \(1\). If \(W=1\) the second term will vanish since \(x^0 = 1\), and we will get the same result for \(V_{1,win}\). The same consideration holds for \(V_{1,loss}\).

If we repeat this game \(n\) times, we arrive at

$$ V_n = V_0 (1+lr_W)^{W_n}(1+lr_L)^{(n-W_n)} $$

with \(W_n\) being the number of wins in those \(n\) games. Essentially, each won round adds a factor \((1+lr_W)\) to the expression, each lost round \((1+lr_L)\).

Now comes a crucial step: how many wins do we expect after playing many games? Asked mathematically, what value will \(W_n\) have for large \(n\)? Because of the law of large numbers we expect \(W = pn\). So for \(p=0.8\) and \(n=10000\) we would expect to win 8000 out of the 10000 games.

Let's plug this into our formula:

$$ V_n = V_0 (1+lr_W)^{pn}(1+lr_L)^{(1-p)n} $$

The law of large numbers allowed us to eliminate the random variables \(W\), \(W_n\) and instead obtain an explicit formula in terms of \(p, n, r_W, r_L\). Since we are interested in maximizing our profit, we are looking for \(l_opt\) — the \(l\) value that maximizes \({V_n}/{V_0}\).

$$ l_{opt} = \operatorname*{argmax}_l \frac{v_n}{v_0} = \operatorname*{argmax}_l \log \frac{v_n}{v_0} = \operatorname*{argmax}_l p \log(1 + l r_W) + (1-p) \log(1 + l r_L) $$

We are free to introduce the logarithm and drop the \(n\) factor because this does not change the value of \(l_{opt}\).1 Differentiating the logarithmic expression and solving for \(l\) gives:

$$ l_{opt} = -\frac{p}{r_L} - \frac{1-p}{r_W} $$

This is identical with the formula given in the introduction after substituting \(r_L = -1\).

Extension to games with more than two possible outcomes

We can extend the Kelly argument to games that have a greater number of possible outcomes. Such games could be:

  • A game with 6 different rewards selected by roll of a 6-sided die, where rewards may be negative (money lost)
  • An investment in a stock, here we have a continuous range of outcomes.

If we have multiple outcomes \(r_i\), each happening with probability \(p_i\), we obtain:

$$ l_{opt} = \operatorname*{argmax}_l \sum_i p_i \log(1 + l r_i) $$

While this equation is not easily solvable analytically, it is trivial to solve numerically. This allows us to extend the Kelly approach to a new kind of game with some interesting practical applications.

If you're interested in a Jupyter Notebook containing several interactive plots that really helped me understand the material outlined in this post, you can find it here.

The Kelly formalism can be generalized further still, taking into account multiple parallel investment opportunities. We'll look at this in the next post.


1 The logarithm is a strictly monotonically increasing function, so \(\alpha \gt \beta \Leftrightarrow \log \alpha \gt \log \beta\). Thus it doesn't change the location of any maxima: \(\operatorname*{argmax} f(x) = \operatorname*{argmax} \log f(x)\). See here for more information about properties of \(\operatorname*{argmax}\).