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.

  • 3
    $\begingroup$ 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. $\endgroup$ Commented Apr 19, 2021 at 14:37
  • 1
    $\begingroup$ @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. $\endgroup$ Commented Apr 19, 2021 at 16:33

1 Answer 1


I’ve already made a comment about how to find the Kelly bet size, f, but I guess that misunderstood the question. I come by it honestly though, because it is so vague. There are an infinite number of possible discrete distributions with payoffs between -1 and 1 for certain probabilities. You know best what question you want answering, and if you know the tools, which “of course” you do, you can solve it.

Numerical methods can find an exact answer to the question too. I don’t understand why they are not applicable. You need numerical methods to find basic stuff like IRR. That function is super close to the derivative of your Kelly bet. Yet there’s no motivation for a closed form IRR formula because numerical methods are so accurate.

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.

  • $\begingroup$ 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. $\endgroup$ Commented Apr 21, 2021 at 6:28
  • $\begingroup$ 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 $k=3,p=0.8$). $\endgroup$ Commented Apr 22, 2021 at 10:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.