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Kelly is calculated as mu / sigma^2. If we remove our highest performing returns from our calculations this actually increases our Kelly leverage, which does not make sense to me. A less profitable return history means we should be lowering our Kelly factor, not increasing it

I've seen Kelly derivations that account for higher moments (skew, kurtosis) but never this. Can anyone help?

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The underlying assumption to your mu/simga^2 formula is that the pricing process follows geometric Brownian motion, so your returns are therefore symmetric and normal. The existence of a very high positive return implies the possibility of a very low negative return, even if you haven’t realized it yet in the time series you used to calibrate your sigma. If the very low negative return wasn’t as likely as the very high positive return, your return distribution wouldn’t be normal (because it’s not symmetric) and you shouldn’t be using that formula anyway. Then, the log utility function you are maximizing punishes losses more than it does gains, like most utility functions. Depending on your time series, the chance of the big loss isn’t worth the chance of another big gain, so your total allocation decreases.

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  • $\begingroup$ I understand your point re: symmetry and non-normality, as well the implication of symmetrical fat tails (leverage —>0 as it increases). However, the issue also applies when removing slightly positive returns (directly right of mu) - Kelly will increase even in this case. I dont think the symmetry explanation works in that case $\endgroup$
    – redpowertie
    May 19, 2021 at 3:53
  • $\begingroup$ To clarify, I personally use a Kelly formula adapted for non-normality (where the 3rd and 4th moments are left intact in the Taylor series derivation, unlike Ed Thorp’s version I listed above). Perhaps a rephrase would be more accurate - in my Kelly variant (where skew adjusts the formula for ‘symmetry’), why can I not use semi variance? I assume the skewness term is sufficient to adjust the formula for upside/downside vol? $\endgroup$
    – redpowertie
    May 19, 2021 at 3:53
  • $\begingroup$ Your first comment isn't correct. Say you have a return time series: -.5 and .5. Kelly allocation is 0, as is mu. Take another time series: -.5, .5, .01. The last term is to the right of mu, but it will increase your Kelly allocation to .02. Removing relatively small returns to the right of mu does NOT increase your leverage. Your return has to be "large" enough, and that is what the original question was concerned with: the highest performers. The symmetry explanation still holds. $\endgroup$
    – Mild_Thornberry
    May 19, 2021 at 13:54
  • $\begingroup$ To your second comment, you can use semi-variance, but it does not define the full distribution of possible outcomes. If you are not defining the full distribution properly, you won't be maximizing your wealth given the distribution, so it won't be Kelly betting anymore. If your skewness adjustment is given by the Kelly fraction papers.ssrn.com/sol3/papers.cfm?abstract_id=2956161: (mu/(mu^2+sigma^2), then it does not necessarily address the issue you're concerned with in the original post. You can still remove the best return and get higher leverage. $\endgroup$
    – Mild_Thornberry
    May 19, 2021 at 14:08
  • $\begingroup$ That is actually not the formula I use, although the derivation is indeed based on Sinclair's paper. The formula I use derives Kelly from the Taylor series truncated after the 4th term (0 = µ - f (µ² + σ²) + f²λ3 - f³λ4) and uses Cardano's formula to solve for f. It is too difficult to format here, but my point is that there is indeed a distinct skewness term in my formula. Maybe that will make my question clearer $\endgroup$
    – redpowertie
    May 19, 2021 at 19:29
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The Kelly's derivation is the maximization of the expectation of a function (log) of a random variable, which is usually approximate by the Taylor expansion of this function. For small $E[X]$, the variance is a good proxy of $E[X^2]$ which appears in the expectation of the Taylor expansion.

You could substitute the variance by the semivariance, for sure, but it wouldn't maximize the expected log-return. Maybe maximize some type of utility function, though.

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