Rahul Shah, Jean-Stanislas Denain, Jacob Steinhardt

In this lecture, we will describe Kelly Betting, which is a reasonable heuristic to decide how much of your money to invest in an asset that is profitable in expectation, but risky.

As a first example, consider the following situation:

Coin Flip

You pay \(n\) dollars, and a fair coin is flipped:

if the outcome is Heads, you receive \(3n\) dollars: your net earnings are \(2n\) dollars

if the outcome is Tails, you receive nothing: your net earnings are \(-n\) dollars

Question: What value of \(n\) do you choose?

This is a good bet in expectation: your expected wealth is \(\frac 1 2 \times (3n) + \frac 1 2 \times (-n) = +n\) dollars. Therefore, if you wanted to maximize your expected amount of dollars, you should choose the highest possible value of \(n\) and bet all of your money.

However, this is very risky: if you bet all of your money, you will lose everything with probability \(\frac 1 2\), if the coin comes up Tails. Thus, most of you are unlikely to bet all of your money, but rather some fraction of your money. The Kelly Criterion provides a useful rule of thumb to choose this fraction.

Consider the following iterated version of the Coin Flip example above:

Iterated Coin Flip

There are 100 days. Every day, you get the same offer:

You pay some fraction \(\alpha\) of your current wealth

A fair coin is flipped

if the outcome is Heads, you receive \(3\) times the amount you paid

if the outcome is Tails, you receive nothing

We make the simplifying assumption that \(\alpha\) is the same for every day: you have to choose it once and for all.

Question: What fraction \(\alpha\) of your current wealth should you bet on each day?

Let \(x\) be your wealth at the beginning of the first day. After the first coin flip, your wealth becomes:

\(x - \alpha x + 3 \alpha x = (1 + 2 \alpha) x\) if the coin comes up Heads

\(x - \alpha x = (1 - \alpha) x\) if the coin comes up Tails

After 100 days, your wealth is

\[
(1 + 2 \alpha)^k (1 - \alpha)^{100 - k} x, \text{ with } k \sim \text{ Binomial}(n = 100, p = 1/2).
\]

As in the previous example, the expected value of this quantity is greatest when \(\alpha = 1\): to maximize your expected wealth, you should bet all of your money every day. However, this means that every day you have a probability of one half of losing all of your money. Even though you have a tiny chance of becoming enourmously rich, you will be ruined with overwhelming probability\(1 - (1 / 2)^{100}\).

Instead of looking at the expectation of your wealth after 100 days, you could focus on its value in the typical case, by considering the modal or median value of your wealth. This corresponds to taking \(k = 50\), hence a typical wealth of

\[
[(1 + 2 \alpha)(1 - \alpha)]^{50}.
\]

You can then maximize this modal value over \(\alpha\): this is tantamount to finding the maximum of the second degree polynomial \((1 + 2 \alpha)(1 - \alpha)\), which is reached at \(\alpha = \frac 1 4\), leading to a median wealth of \(\left(\frac 9 8\right)^{50}\).

Let us now consider a different bet:

Iterated Biased Coin Flip

Once again, there are 100 days, and every day you get the same offer:

You pay some fraction \(\alpha\) of your current wealth

With probability \(\frac 2 3\), you receive twice the amount you paid

With probability \(\frac 1 3\), you receive nothing

Question: What fraction \(\alpha\) of your current wealth should you bet on each day?

The same reasoning as before shows that your wealth after 100 days is

\[
(1 + \alpha)^k (1 - \alpha)^{100 - k}, \text{ with } k \sim \text{ Binomial}(n = 100, p = 2/3).
\]

The median and most likely value of this random variable is \((1 + \alpha)^{\frac 2 3 \cdot 100} (1 - \alpha)^{\frac 1 3 \cdot 100}\). So we want to find \(\alpha \in [0, 1]\) that maximizes the cubic \((1 + \alpha)^2 (1 - \alpha)\): this corresponds to \(\alpha = \frac 1 3\). This makes us approximately 290x our wealth!

In both these examples, we saw how maximizing the modal or median value of your wealth leads to more intuitive levels of risk-taking than maximizing the expected value. In the next section, we will introduce the Kelly Criterion in a more general setting.

In general, we understand the limiting behavior of sums of random variables better than the limiting behavior of products. This motivates taking the logarithm, and forming the average:

This fraction increases when when your probability of winning \(p\) increases, and when \(b\) increases. It vanishes when \(b = q / p\): in this case, the bet has zero expectation and you have no edge, so the Kelly criterion recommends betting nothing.

As an example, consider the Iterated Biased Coin Flip example from the previous section. In that case, \(p = 2/3\), \(q = 1/3\) and \(b = 1\), thus the Kelly criterion suggests \(\alpha = 1/3\). This is the same as the result obtained by maximizing the median value of the wealth.

20.3 Choosing between multiple bets

So far, we have considered cases where there is only one possible bet: you can either invest your money in this bet, or keep it. However, a more realistic situation is that at any point in time you have to choose between multiple bets. In these cases, how should you split your money between these bets?

For example, suppose you could invest a fraction \(\alpha_1\) of your wealth in a first bet with return \(X^{(1)} \sim p_1\), and a fraction \(\alpha_2\) of your wealth in a second bet with return \(X^{(2)} \sim p_2\). Using the same reasoning as the previous section, you could maximize the following quantity over \((\alpha_1, \alpha_2)\):

However, this optimization problem often becomes intractable as the number of possible bets becomes large. For example, think about the enormous number of possible investments a trader can make at any point in time.

One idea could be to use the Taylor approximation \(\log(1 + z) \simeq z - \frac {z^2} 2\). For a single bet, this yields:

which is maximized by \(\alpha = \frac {\mathbb{E}(X)} {\mathbb{E}(X^2)}.\) Intuitively, this approximation favors bets whose returns have a high expected value but a low variance.

When choosing between \(n\) bets with returns \(X^{(i)} \sim p_i\), a first simpler case is when these bets are all independent. Then, the same reasoning applies to each bet independently, and

which is maximized by \(\alpha = \mathbb{E}(XX^\top)^{-1} \mathbb{E}(X)\) (assuming that \(\mathbb{E}(XX^\top)\) is invertible). Qualitatively, this approximation favors diversification, by assigning greater fractions \(\alpha_i\) to bets that are less dependent on the others.