# Kelly fraction for discrete distributions

The Kelly fraction is $$f^\star$$ maximizing $$\mathbb E[\log(1+f X)]$$. For instance, if $$X\sim\begin{cases} 1 & w.p. p\\ -1 & w.p. 1-p \end{cases},$$ we get that $$f^\star=2p-1$$. I'm curious about closed-forms of $$f^\star$$ for discrete distributions with $$X\in [-1,1]$$. I wonder if such a closed-form is known in the economic literature.

I can derive a simple closed-form if $$supp(x) = \{-1,0,1\}$$ by rescaling $$p$$, but I'm interested in a larger (finite) support, say $$\{-1,-\frac{1}{2},0,\frac{1}{2},1\}$$.

Any ideas?

Edit Given the comment, I'll clarify. The function $$\mathbb E[\log(1+f X)]$$ is strictly concave in $$f$$ for $$f\in[0,1]$$; hence, there exists only one maximum. Since it is also differentiable, that maximum is the root of its derivative.

Taylor series implies that $$\log(1+y)=\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {y^{n}}{n}}=y-{\frac {y^{2}}{2}}+{\frac {y^{3}}{3}}-\cdots$$ Hence, $$\frac{d}{df}\mathbb E[\log(1+f X)]=\frac{d}{df}\mathbb E\left[\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {(fX)^{n}}{n}}\right].$$ Invoking linearity of expectation, we have $$\frac{d}{df}\mathbb E[\log(1+f X)]=\sum _{n=1}^{\infty }(-1)^{n+1}{f^{n-1} \mathbb E[ X^n]}.$$ The Kelly criterion is $$f^\star$$ is the root of the above equation. I'm wondering whether it has a nice closed-form for some discrete, non-Bernoulli distributions.

• As you have already stated, Kelly maximizes log-utility. It’s not confined to Bernoulli bets. To maximize the wealth of a discrete distribution, you just need the probabilities of each outcome, along with the payoffs. Then you can solve for expected log wealth and maximize it by finding where the derivative of the wealth function equals zero with respect to bet size. Apr 19 at 14:37
• @Mild_Thornberry Of course it is not restricted to Bernoulli, otherwise I won't be asking this question :) The issue is that the root(s) of the derivative doesn't always have a closed-form like in the Bernoulli case. My question is about "nice" distributions for which we have closed-form solutions. Apr 19 at 16:33

I also don’t understand why a Taylor Series expansion won’t work. Take it out to the fourth term. Then you have $$O(x^5)$$ error. Is that not precise enough, given any forward probability estimate in finance probably very likely contains WAY more error? One should not lose the forest through the trees.
• In my application, I'm interested in the value of $f$. Even if two functions are arbitrarily close, we can say nothing about their argmaxes being close. Despite we can approximate the utility function up to the factor you mentioned, I don't see how this translates to an approximation of $f$. If you do, I'd appreciate it if you could add it to your answer. Apr 21 at 6:28
• To further illustrate why the approach you suggest is wrong: if $X$ is a scaled Binomial, namely $X=\frac{2B}{k}-1$ for $B\sim Bin(k,p)$, taking out the third term can result in $f$ greater than 1 (bounded away by a constant, say for $k=3,p=0.8$). Apr 22 at 10:27