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.

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Published

20 May 2015

Category

Recreational Math

Tags

• brainteaser 2
• stochastic 14
• trading 13
• math 28