How deep are Kelly drawdowns?

Question

A full-Kelly bettor grows their bankroll optimally in the long run, yet the ride is wild.

Using the continuous approximation, find the probability that a full-Kelly bettor ever sees their bankroll fall to a fraction $a$ of its starting value. Then repeat for a half-Kelly bettor, and interpret.

Accepted Answer

Model log-wealth Xt = \ln(Wt / W0) as a Brownian motion with upward drift g (the growth rate) and variance rate V. The classic hitting result: the probability that X ever falls to a level -L (with L 0) is \exp(-2gL/V). Reaching a fraction a < 1 of bankroll means X hits \ln a, so L = \ln(1/a). Full Kelly. In the continuous model, full Kelly makes the growth rate exactly half the variance rate: g = V/2. Then P(	ext{ever reach fraction } a) = \exp\!\left(-\frac{2g}{V}\ln\frac{1}{a}ight) = \exp\!\left(-\ln\frac{1}{a}ight) = a. So the probability you ever fall to fraction a is simply a. You w…

How deep are Kelly drawdowns?

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