A strategy has excess return mu and volatility sigma, and at leverage f its expected log growth is f*mu - f^2*sigma^2/2. What share of the maximum growth rate survives at 1.5 times full Kelly, and at what multiple does the growth rate reach zero?

A strategy has excess return mu and volatility sigma, and at leverage f its expected log growth is f*mu - f^2*sigma^2/2. What share of the maximum growth rate survives at 1.5 times full Kelly, and at what multiple does the growth rate reach zero?

Approach: Maximise the quadratic in f, then rewrite the growth rate as a function of the ratio k = f/f* so the answer is independent of mu and sigma.

75%. Maximising f*mu - f^2*sigma^2/2 gives f* = mu/sigma^2 and a maximum log growth rate of mu^2/(2*sigma^2). Writing f = k*f*, the growth becomes (2k - k^2) times that maximum, so k = 1.5 keeps 3 - 2.25 = 0.75 of it and k = 2 keeps 4 - 4 = 0, where the leverage is twice full Kelly and every unit of the growth has been paid away in variance drag. Past k = 2 the log growth is negative and the account goes to zero with probability one. The curve is symmetric about k = 1, so half Kelly also keeps 75% while running half the volatility, which is the whole case for sizing below the optimum.

Follow-up: How does the 2 times Kelly ruin threshold change when returns are fat tailed rather than normal?

Key concepts: kelly fraction, log growth rate, leverage, half kelly.