Posted on July 22, 2018

A utility function can be used to model the value of something. Kelly's Criterion involves finding the maximum of the expected value of the logarithmic utility function. In probability theory, expected value represents the 'center of gravity' of the probability mass function (or probability density function in case our random variable is continous). Morever, when our probability mass function is symmetric, we often refer to the expected value as the 'mean'.

Back to Kelly, let $X_i$ denote a random variable that represents our wealth after $i$-trials and define $X_0 = $ initial capital. We want to maximize the expected value after $n$-trials, that is $E[X_n]$. Consider a basic game of chance where the probability of winning is $p$ and the probability of losing is $q$, and the pay-off is even (we win or lose the same amount) . Our wealth can be written down in the following way,

$\begin{align*} X_i = X_{i-1} + S_i B_i \end{align*}$

such that by recursion we arrive to,

$\begin{align*} X_n & = X_0 + \sum\limits_{i=1}^n S_i B_i \end{align*}$

where $B_i$ represents our bankroll at trial $i$ and $ S_i = \begin{cases} 1 & \text{if we win, with probability} \quad p \\ -1 & \text{if we lose, with probability} \quad q = 1 - p \end{cases}$

Then, the expected value of our random variable would be,

$\begin{align*} E[X_n] & = E[X_0] + \sum\limits_{i=1}^n E[S_i B_i] \\ E[X_n] & = X_0 + \sum\limits_{i=1}^n E[S_i] E[B_i] \end{align*}$

Lets remember that expected value of a constant is the constant and if events A and B are independent then $E[AB] = E[A]E[B]$. The expected value of $E[S_i] = 1p + (-1)q = p - q$.

It would be reasonable to bet all of our wealth if $p > q$ such that we maximize our gain. Likewise, if we bet all our wealth we also run into the posibility of ruin, where we lose our entire wealth. Kelly's argument begins by considering a bankroll that represents a fraction of your wealth, say, a midpoint between maximum gain and ruin. We define our subsequent bankroll as a fraction of our current wealth.

$\begin{align*} B_i = f X_{i-1} \end{align*}$

After $n$-trials, our wealth would be,

$\begin{align*} X_n = X_0 (1+f)^W (1-f)^L, \text{where } W+L=n \end{align*}$

Lets understand this equation through an example. At trial 0, our wealth is equal to 0. Now lets suppose we have 2 Wins followed by 1 Loss, WWL,

$\begin{align*} \text{At trial 1} \quad X_1 & = X_0 + X_0*f = X_0(1+f) \\ \text{At trial 2} \quad X_2 & = X_1 + X_1*f = X_0(1+f) + X_0(1+f)f = X_0 + X_0 f + X_0 f + X_0 f^2 \\ & = X_0 + 2 X_0 f + X_0 f^2 = X_0(1 + 2f + f^2) = X_0(1+f)^2 \\ \text{At trial 3} \quad X_3 & = X_2 - X_2*f = X_0(1+f)^2 (1-f) \quad \text{left as exercise} \end{align*}$

Our rate of wealth increase is equal to

$\begin{align*} \frac{X_n}{X_0} = (1+f)^W (1-f)^L \end{align*}$

and this fracion can be rewritten as

$\begin{align*} \frac{X_n}{X_0} = e^{\log (\frac{X_n}{X_0}^\frac{n}{n})} = e^{n \log (\frac{X_n}{X_0}^\frac{1}{n})} \end{align*}$

such that $\log (\frac{X_n}{X_0}^\frac{1}{n})$ represents the measure of exponential rate of increase per trial (there are $n$-trials total). Call this function $G(f)$. We define

$\begin{align*} G_n(f) & = \log (\frac{X_n}{X_0}^\frac{1}{n}) \\ & = \frac{1}{n} \log (\frac{X_n}{X_0}) \\ & = \frac{1}{n} \log [(1+f)^W (1-f)^L] \\ & = \frac{1}{n} [ \log (1+f)^W + \log (1-f)^L ] \\ & = \frac{W}{n} \log (1+f) + \frac{L}{n} \log (1-f) \end{align*}$

We want to maximize the expected value of of the exponential growth of wealth. We first find the expected value,

$\begin{align*} E[G_n(f)] & = E[\frac{W}{n} \log (1+f) + \frac{L}{n} \log (1-f) ] \\ & = E[\frac{W}{n} \log (1+f)] + E[\frac{L}{n} \log (1-f)] \\ & = \frac{1}{n}E[W] \log (1+f) + \frac{1}{n}E[L] \log (1-f) \\ & = \frac{1}{n} \cdot p \cdot n \cdot \log (1+f) + \frac{1}{n} \cdot q \cdot n \cdot \log (1-f) \\ & = p \log (1+f) + q \log (1-f) \end{align*}$

In calculus, one learns to find local maximums and minimums by first finding the critical points, or inflection points of a function. We should know that $\log$ function is smooth and sum of two smooth functions is smooth. First we find the critical points by taking the derivative of $E[G_n(f)]$ and finding those points $f$ such that it equals 0,

$\begin{align*} 0 = \frac{d E[G_n(f)]}{df} & = \frac{p}{1+f} + \frac{q}{1-f} \cdot -1 \\ & = \frac{p-q-f}{(1+f)(1-f)} \end{align*}$

This equation is satisfied with $f^* = p-q$. It is left to the reader to show this is a local maximum (hint: second derivative test).

Thus we have found the fraction of the bankroll we must utilize to maximize the expected value of the exponential growth of our wealth. However, remember that for the example above, our pay-off is even (that is if we win, we profit the bankroll we bet, and if we lost we lose the bankroll we bet)

Finally, we can derive Kelly's Criterion for when the pay-off is uneven. That is, suppose we bet 1 unit at odds of 1.5, then we stand to profit 0.5 units and lose 1 unit (uneven pay-off). Let $b$ represents the odds and if we win, we increase our wealth by $1+bf$, and if we lose we decreae our wealth by $1-f$. The expected value of the exponential growth of our wealth is now the following,

$\begin{align*} E[G_n(f)] & = p \log (1+bf) + q \log (1-f) \end{align*}$

such that to maximize this function we find the critical points,

$\begin{align*} 0 & = \frac{d E[G_n(f)]}{df} = \frac{p}{1+bf}\cdot b + \frac{q}{1-f} \cdot -1 \\ 0 & = \frac{pb(1-f) - q(1+bf)}{(1+bf)(1-f)} \\ f(pb + qb) & = pb-q \\ f & = \frac{pb - q}{b(p+q)} \\ f & = \frac{pb-q}{b} \quad \text{remember} \quad p+q=1 \end{align*}$