Fixed fraction betting - the Kelly criterion

Let $X$ be a real random variable with support $[-1,\infty[$ . Following Kelly [1] consider the reinvested growth of a fixed fraction investment in an instrument that has payoff $X$ .

$$W_n = W_{n-1}(1+\alpha\,X_n), \quad X_n \sim X, \alpha\in[0,1]\quad.$$

(If the support of $X$ is bounded the restriction on $\alpha$ can be relaxed.) The long run growth rate is

$$G_\alpha = \lim_{N\rightarrow\infty} \frac{1}{N} \log\frac{W_N}{W_0}\quad .$$

By the law of large numbers

$$G_\alpha = {\mathsf{E}}\log(1+\alpha\,X), \quad\mathrm{a.s.}\quad .$$ (1)

For a given $X$ the function $G_\alpha$ is concave in $\alpha$ , and, writing $G'(\alpha)=dG/d\alpha$ we have $G'(0)={\mathsf{E}}X$ . Define the maximal growth rate as

$$G[X] = \max_\alpha\,{\mathsf{E}}\log(1+\alpha\,X)\quad.$$ (2)

$G[X]$ and ${\mathsf{E}}X$ :
Since the slope of $G(\alpha)$ at $\alpha=0$ , i.e. $G'(0)={\mathsf{E}}X$ , determines the location of the maximum we have, for $\alpha\in[0,1]$ ,

$${\mathsf{E}}X\leq0 \Leftrightarrow G[X]=0, \alpha_\mathrm{opt}=0$$ (3)

$${\mathsf{E}}X>0 \Leftrightarrow G[X]>0, \alpha_\mathrm{opt}>0$$ (4)

Upper bound for $G[X]$ :
Jensen's inequality yields $G_\alpha \leq \log(1+\alpha {\mathsf{E}}X)$ and because of $\alpha<1$ and $\log(1+x) \leq x$ it is

$$G[X] \leq \log(1+{\mathsf{E}}X) \leq {\mathsf{E}}X \quad .$$ (5)

Mixtures:
Consider a mixture $p(X)=\int dY p(X\vert Y)p(Y)$ . We have

$$G[X] = \max_\alpha{\mathsf{E}}_X\log(1+\alpha X) = \max_\alpha{\mathsf{E... ...mathsf{E}}_{X\vert Y} \underbrace{\log(1+\alpha X\vert Y)}_{\leq G[X\vert Y]} =$$

$$\leq {\mathsf{E}}_Y G[X\vert Y] \leq {\mathsf{E}}_Y \max_Y G[X\vert Y],$$

hence

$$G[X] \leq \max_Y G[X\vert Y] \quad ,$$ (6)

i.e. the gain of a mixture cannot be larger than the gain of the best variable in the mixture.

Gambles that are offered with lower fequency $\rho$ :
Suppose a gamble is offered at a reduced rate, i.e. at every step with probability $\rho$ the payoff $X$ is available and with probability $1-\rho$ no investment is offered, i.e. the payoff is 0 . The pdf for payoff $x$ is $\widetilde p(x)=\rho\,p(x)+(1-\rho) \delta(x)$ . Then $\max_\alpha\widetilde{\mathsf{E}}\log(1+\alpha X) = \max_\alpha\left[\rho{\mathsf{E}}\log(1+\alpha X) + (1-\rho)\log(1)\right] = \rho G[X]$ . Denoting the reduced rate gain, i.e. the maximal gain under $\widetilde{\mathsf{E}}$ , as $G^\rho[X]$ , it is

$$G^\rho[X] = \rho\, G[X] \quad .$$ (7)

When several gambles $\{X_i\}_{i=1}^{N}$ are offered with occurrence probabilities $\{\rho_i\}_{i=1}^{N}$ the individual gains are $\rho_iG[X_i]$ . Amongst the offered gambles the one with largest $\rho_iG[X_i]$ is favourable.

Small edge ${\mathsf{E}}X\rightarrow 0$ :
For small ${\mathsf{E}}X$ we expand around $\alpha=0$

$$\log(1+\alpha X) = \alpha X - \frac{1}{2}\alpha^2 X^2 + O(\alpha^3)$$

and the quadratic approximation gives $\alpha_\mathrm{opt} = \frac{{\mathsf{E}}X}{{\mathsf{E}}X^2}$ and then

$$G[X] = \frac{1}{2} \frac{({\mathsf{E}}X)^2}{{\mathsf{E}}X^2},\quad \mathrm{for\ } {\mathsf{E}}X\rightarrow 0$$ (8)

Gain of averages $\sum_i w_i X_i$ :
A lower bound to $G[\sum_i w_i X_i]$ with $\sum_i w_i=1$ can easily be found:

$$G\left[\sum_i w_i X_i\right] = \max_\alpha{\mathsf{E}}\log\left(1+\alpha\sum_i w_i X_i\right) \geq$$

$$\geq \max_\alpha{\mathsf{E}}\sum_i w_i\log(1+\alpha X_i) = \max_\alpha\sum_i w_i\,G_i(\alpha)$$

By convexity of the $\max$ function it is $\max_\alpha\sum_i w_i\,G_i(\alpha) \geq\sum_i w_i\, \max_\alpha G_i(\alpha) = \sum_i w_i G[X_i]$ , so we have

$$G\left[\sum_i w_i X_i\right] \geq \sum_i w_i G[X_i] \geq \min_i G[X_i] \quad$$ (9)

i.e. the gain of an average of payoffs is not smaller than the smallest gain.

Representation in terms of the cumulative distribution function:
Partial integration of the integral $\int_{-1}^\infty p(x)\log(1+\alpha x)d x$ results in a useful representation for $G_\alpha$ . Denote the cdf $\Phi(x)=\int_{-1}^x p(x) dx$ . Starting with $G_\alpha\leftarrow G_\alpha^M=\int_{-1}^M\frac{d \Phi(x)}{d x}\frac{\alpha}{1+\alpha x} dx$ partial integration gives (use $\Phi(-1)=0$ )

$$\int_{-1}^M p(x) \log\left(1+\alpha x\right) dx = \Phi(M)\log\left(1+\alpha M\right) - \int_{-1}^M \frac{\alpha\,\Phi(x)} {1+\alpha x} dx$$

observe that both terms separately diverge with $M\rightarrow\infty$ . To alleviate this problem use $\int_{-1}^M\frac{\alpha}{1+\alpha x}dx=\log(1+\alpha M)-\log(1-\alpha)$ ,

$$G_\alpha^M = \Phi(M)\left[\int_{-1}^M\frac{\alpha}{1+\alpha x} dx+\log(1-\alpha)\right] - \int_{-1}^M \frac{\alpha\,\Phi(x)} {1+\alpha x}\, dx$$

and after rearranging and letting $M\rightarrow\infty$ we have the desired result

$$G_\alpha = \log(1-\alpha) + \alpha\int_{-1}^\infty\frac{1-\Phi(x)}{1+\alpha x}\, dx\quad.$$ (10)

Representation in terms of the Laplace transform:
We will need a result for the Laplace transform of a (positive) random variable $X$ , $(\mathcal{L}X)(s) = {\mathsf{E}}e^{-s X}$ . Doing partial integration of $(\mathcal{L}X)(s) = \int_0^\infty p(x) e^{-s x}dx$ analogous to the partial integration leading to eq. (10) gives the following result that expresses the Laplace transform in terms of the cdf:

$$(\mathcal{L}X)(s) = 1-s \int_0^\infty e^{-s x}\left(1-\Phi(x)\right)\, dx\quad.$$ (11)

Starting with the shifted version of the cdf representation eq. (10)

$$G_\alpha = \log(1-\alpha) + \alpha\int_0^\infty\frac{1-\Phi(x)}{1+\alpha (x-1)}\, dx$$

and using

$$\frac{1}{1+\alpha (x-1)} = \frac{1}{\alpha}\int_0^\infty e^{-s x} \exp\left(-s\frac{1-\alpha}{\alpha}\right)d s$$

we get

$$G_\alpha = \log(1-\alpha)+\int_0^\infty dp\, e^{-s\frac{1-\alpha}{\alpha}} \int_0^\infty\, dx\, e^{-s x} (1-\Phi(x))$$

and we can express the integral $\int_0^\infty\dots dx$ via eq. (11) in terms of the cdf $\Phi(x)$ to obtain

$$G_\alpha = \log(1-\alpha) + \int_0^\infty \frac{1-(\mathcal{L}X)(s)}{s} \, \exp\left(-s\frac{1-\alpha}{\alpha}\right)\, ds, \quad X\ge 0\quad .$$ (12)

The restriction $X\ge0$ in the equation above is used as a reminder that the relation holds for the unshifted, i.e. positive random variable $X$ .

It is tempting to use $\log x = \int_0^\infty\!d\zeta\,\frac{1}{\zeta}\left(e^{-\zeta}-e^{-x\zeta}\right), \Re(x)>0$ to derive a relation, but issues with uniform continuity arise.