# Kelly criterion for normally distributed returns

If the returns of my strategy are distributed like 𝒩[μ,σ], what is the optimal fraction of capital to invest in each single trade, as a function μ and σ? Help!

PS. I know that normally distributed returns are an abstraction. But I'd like to grasp the concept in an ideal world, before exploring the implications of fat tails on the formula...

• Do you assume you close each trade at the beginning of the next investment period ? – Ezy Jan 4 '19 at 19:09
• yes, I do 1 trade per period. There are never 2 trades open at the same time. – elemolotiv Jan 4 '19 at 19:14
• Have you perused Thorp’s papers, for instance, eecs.harvard.edu/cs286r/courses/fall12/papers/… , for one derivation? Another can be done via Itô’s lemma, iirc. – Nap D. Lover Jan 4 '19 at 22:59
• @LoveTooNap29 thanks for the suggestion! – elemolotiv Jan 5 '19 at 7:24

This problem can be expressed as the original Merton's portfolio problem.

Consider wealth process defined by SDE

$$d X _ { t } = \frac { X _ { t } \alpha _ { t } } { S _ { t } } d S _ { t } + \frac { X _ { t } \left( 1 - \alpha _ { t } \right) } { S _ { t } ^ { 0 } } d S _ { t } ^ { 0 }$$

where $$\alpha_t$$ is proportion of the investment in the risky asset $$S_t$$, and $$S_t^0$$ is the risk-free asset.

Optimality criterion may depend on the risk aversion of the investor, and the problem is to maximize expected utility of the investor for appropriate utility function $$U$$:

$$E \left[ U \left( X _ { T } \right) \right] \rightarrow \max$$

Classical choice of the utility function is CRRA:

$$u ( x ) = \frac { x ^ { 1 - \gamma } } { 1 - \gamma }$$

where $$\gamma$$ is constant and corresponds to the risk-aversion of the investor.

If the asset $$S_t$$ follows Black-Scholes dynamics (in conformance with your assumption of log-normal returns)

\begin{aligned} d S _ { t } ^ { 0 } & = r S _ { t } ^ { 0 } d t \\ d S _ { t } & = \mu S _ { t } d t + \sigma S _ { t } d W _ { t } \end{aligned}

remarkably there is a closed-form solution which it is to invest a constant proportion of wealth in the risky asset

$$\alpha_t = \frac { \mu - r } { \gamma \sigma ^ { 2 } }$$

Notice that the solution can be interpreted as the mean-variance trade-off.

• Note also that for log-utility (the Kelly case) $\gamma=1$ – noob2 Jan 5 '19 at 20:50
• @nakajuice thanks for the answer! I can't explain this case though. If we suppose γ=1 (kelly case), r=0, μ=0.01€ and σ=0.05€ we get ⍺=4. How can ⍺>1? – elemolotiv Jan 6 '19 at 8:31
• We assume short-selling is allowed. $\alpha=4$ implies that for each 4 shares you hold, you need to short-sell 3 bonds (read: borrow money). – starovoitovs Jan 6 '19 at 12:29

ok I found it 🙂and this works for any distribution, not just the normal distribution

$$f^*=\frac μ {σ^2 + μ^2} \approx \frac μ {σ^2} \space if μ\llσ$$

here the steps: https://www.dropbox.com/s/4nqd5yfk2xcuag5/kelly.pdf

• This answer is close to, but not identical, to the other well-known one. It would be interesting to understand why the difference. – noob2 Jan 5 '19 at 20:54
• Why don't they just do $$E(\prod_i (1 + f r_i)) = \prod_i E(1 + f r_i) = (1 + f \mu)^t$$ since the $$r_i$$ are IID? I am probably doing something dumb here. – mathtick Apr 5 '19 at 11:30
• @mathtick I am trying to figure that out myself too. Can it be that: $\frac d {df} E\left[\prod_{i=1}^t(1+f\space r_i)\right] \ne \frac d {df}(1+f\space μ)^t$ that is, for some property of the $E[ ... ]$ operator, you must first differentiate and then apply the $E[ ... ]$ operator? – elemolotiv Apr 13 '19 at 20:55
• @elemolotiv I read a lot more after that post ... so basically Kelly is exactly what I wrote above, and then then exp/log and end up maximizing the mean of the $\log(1 + f u)$ in the exponential. So basically, in the most general sense, "Kelly" just means use a log-utility when balancing risks. A key thing that they miss, is that the log is only defined if you survive. i.e. you don't go bust. So there is a term missing in most of this stuff to do with $P(1 + f r < 0)$ that one should do something with. – mathtick Apr 14 '19 at 9:04
• Thanks! @mathtick Your comment helped me realise that maximising 𝐸[log(𝐺)] is not a mathematically convenient way to maximise 𝐸[𝐺]. That would be mathematically incorrect. 𝐸[log(𝐺)] is a whole different utility function that incorporates risk aversion. – elemolotiv Apr 14 '19 at 12:51