# Intro

In this post I am going to talk a bit about Kelly betting, since today I was flabbergasted that I did not write anything about it yet. Kelly betting is a strategy how to allocate money to bets.

## Problem

Assume we are playing a game. I am going to flip a coin. You can bet an amount $$x$$$on the result of the coinflip. If you are right, you win $$x$$$. If you are wrong, you lose your $$x$$\$. Let us say that you know that my coin is nonrandom and lands on head with probability $$p>0.5$$. To further ease the problem assume that you are always betting the same fraction $$f$$ of your capital. Now, how should you choose $$f$$ to get returns bigger than any other strategy? Another way of formulating the problem would be to maximize the expected return.

Get out and try it before reading on. And I hope you will come back with the right solution or at least utterly frustrated and nearly crying.

## Solution

The expected return is $$(1+f)$$ each time we win and $$(1-f)$$ each time we lose. Now putting those nasty probabilities in we get a return of $$(1+f)^p+(1-f)^{1-p}$$. We need to find the maximal value of that expression. To make things easier we can also find the maximal value of the logarithm of that expression

\begin{align*} \log ((1+f)^p+(1-f)^{1-p})= p\log(1+f) + (1-p) \log(1-f) \end{align*}

since the logarithm is increasing monotonically. So we have to set teh derivative to zero. Observe that

\begin{align*} \frac{\partial}{\partial f} p\log(1+f) + (1-p) \log(1-f) &= 0\\ \Leftrightarrow \frac{p}{1+f}+\frac{p-1}{1-f} &= 0\\ \Leftrightarrow \frac{p}{1+f} &= \frac{1-p}{1-f}\\ \Leftrightarrow p(1-f) &= (1-p)(1+f)\\ \Leftrightarrow f &= 2p-1 \end{align*}

You can plug some sensible values for $$p$$ in it to check correctness.

Now, let us generalize this problem because of mathematical elitism. Let us say if you win with probability $$p$$, you win a proportion $$a$$ of your initial bet and if you loose, then you win a proportion $$b$$, where $$b$$ is possibly negative. So our returns are $$(1+fa)^p+(1+fb)^{1-p}$$. Note that this is essentially the same as before if we take $$a=1$$ and $$b=-1$$. Now let us again maximize this, more exactly let us maximize the logarithm. So we again set the derivative to zero.

\begin{align*} \frac{\partial}{\partial f} \log[(1+fa)^p+(1+fb)^{1-p}]\\ = \frac{\partial}{\partial f} p\log(1+fa)+(1-p)\log(1+fb)\\ = \frac{pa}{1+fa}+\frac{(1-p)b}{(1+fb)} \end{align*}

This is equal to zero if and only if

\begin{align*} \frac{pa}{1+fa}+\frac{(1-p)b}{(1+fb)} &= 0\\ \Leftrightarrow pa+pafb+(1-p)b+(1-p)bfa &= 0\\ \Leftrightarrow pa+(1-p)b+bfa &= 0\\ \Leftrightarrow fab=-pa+(p-1)b &= 0\\ \Leftrightarrow f=\frac{-pa+(p-1)b}{ab} &= 0 \end{align*}

If we would now plug in the values for our first exercise $$a=1$$ and $$b=-1$$, we would get the same result.

20 May 2015

Brainteaser