Kelly criterion and volatility targeting: how much to risk.

Kelly: how much to bet when you have an edge

Lesson 2.2.2 spread risk across markets with a hand-set volatility target, and 4.2.1 limits what is risked on each trade. The question above them is missing: how much total risk is worth taking with a given edge. John L. Kelly answered it in 1956 for a gambler who repeatedly bets a fraction $f$ of his capital: the one that maximizes the expected logarithmic growth, which is the rate at which capital grows in the long run. For a bet that wins $b$ per unit with probability $p$ and loses the unit with $q = 1 - p$:

A coin with a 55% hit rate and an even-money payout gives $f^* = 0.10$: bet 10% of the capital on each toss.

import numpy as np
import pandas as pd

p, b = 0.55, 1.0      # probability of winning and payout per unit
f_kelly = p - (1 - p) / b
print(f"Kelly fraction: {f_kelly:.2f}")


def g(f):
    """Expected logarithmic growth per bet."""
    return p * np.log1p(f * b) + (1 - p) * np.log1p(-f)


fracciones = [0.025, 0.05, 0.10, 0.15, 0.20, 0.30]
pd.Series({f: g(f) for f in fracciones}, name="g").round(5)

Kelly fraction: 0.10
0.025    0.00219
0.050    0.00375
0.100    0.00501
0.150    0.00374
0.200   -0.00014
0.300   -0.01620
Name: g, dtype: float64

Growth per bet is maximal at $f = 0.10$ (0.00501 in logarithms) and almost symmetric around it: 0.05 and 0.15 give the same (0.00375 and 0.00374). With 0.20, twice Kelly, growth is already negative (−0.00014): you play with an edge and, even so, capital tends to shrink. With 0.30 the capital vanishes. An average does not show it; a thousand repeated bets by 4,000 players do:

rng = np.random.default_rng(7)
gana = rng.random((4000, 1000)) < p     # 4,000 paths of 1,000

filas = {}
for f in (0.05, 0.10, 0.20, 0.30):
    paso = np.where(gana, np.log1p(f * b), np.log1p(-f))
    lw = np.cumsum(paso, axis=1)                # log of wealth
    pico = np.maximum.accumulate(np.maximum(lw, 0), axis=1)
    caida = np.exp(lw - pico) - 1
    filas[f] = {"mediana_final": np.exp(np.median(lw[:, -1])),
                "p_pierde": (lw[:, -1] < 0).mean(),
                "p_mitad_inicial": (lw.min(axis=1)
                                    <= np.log(0.5)).mean(),
                "p_caida_50": (caida.min(axis=1) <= -0.5).mean(),
                "p_caida_80": (caida.min(axis=1) <= -0.8).mean()}
pd.DataFrame(filas).T.round(3)

      mediana_final  p_pierde  p_mitad_inicial  p_caida_50  p_caida_80
0.05         42.632     0.009            0.116       0.854       0.064
0.10        149.660     0.061            0.483       1.000       0.850
0.20          0.871     0.514            0.911       1.000       1.000
0.30          0.000     0.954            0.997       1.000       1.000

With Kelly (0.10) the median player multiplies their capital by 150, but 6.1% end up losing, 48.3% touch half the initial capital at some point and all of them suffer a 50% drawdown from a peak. With half Kelly (0.05) the median multiplies by 43, only 0.9% lose, 11.6% touch half the initial capital and 6.4% suffer an 80% drawdown, versus 85% with Kelly. With twice Kelly (0.20) half of the players lose money with a coin in their favor.

In continuous time: the optimal volatility is the Sharpe ratio

For a strategy with expected excess return $\mu$ and volatility $\sigma$ (annual), leveraged $f$ times and rebalanced often, the logarithmic growth of the excess is $g(f) = f\mu - \tfrac{1}{2}f^2\sigma^2$, a parabola. Its maximum:

The third equality is the most useful one: at Kelly, account volatility equals the Sharpe ratio. A strategy with a Sharpe of 0.5 calls for 50% annual volatility, whatever its own volatility. With a fraction $c$ of Kelly, growth is $g^*(2c - c^2)$ and volatility $c\cdot\mathrm{SR}$. The logarithm of capital is then a Brownian motion with drift $\nu = \mathrm{SR}^2(c - c^2/2)$ and variance $s^2 = c^2\mathrm{SR}^2$, and the probability that it ever falls to a fraction $x$ of the initial capital is $x^{2\nu/s^2} = x^{2/c - 1}$ (the probability of touching a lower barrier, with known parameters and an infinite horizon):

c = np.array([0.25, 0.5, 0.75, 1.0, 1.5, 2.0])
expo = 2 / c - 1          # P(ever falling to x) = x ** expo
pd.DataFrame({"crecimiento_vs_max": 2 * c - c**2,
              "vol_en_sharpes": c,
              "p_caer_a_la_mitad": 0.5 ** expo,
              "p_caer_al_20": 0.2 ** expo},
             index=pd.Index(c, name="fraccion")).round(3)

          crecimiento_vs_max  vol_en_sharpes  p_caer_a_la_mitad  p_caer_al_20
fraccion
0.25                   0.438            0.25              0.008         0.000
0.50                   0.750            0.50              0.125         0.008
0.75                   0.938            0.75              0.315         0.068
1.00                   1.000            1.00              0.500         0.200
1.50                   0.750            1.50              0.794         0.585
2.00                   0.000            2.00              1.000         1.000

The header figure draws those two curves. Growth is flat near the maximum and the drawdown probability is not: going from Kelly to half Kelly costs 25% of the growth and divides by four the probability of losing half (from 50% to 12.5%); a quarter Kelly keeps 44% of the growth with 0.8%. Above Kelly everything gets worse at once: with 1.5 times Kelly growth is the same as with half Kelly (0.75) and the probability of losing half, 79%. The coin simulation agrees with the formula (0.116 and 0.483 versus 0.125 and 0.5). The error is not symmetric: underbetting costs little; overbetting, a lot.

Kelly on the level's strategies

The ten strategies in estrategias4_diarias.parquet are simple daily returns in excess of the risk-free rate, 2006-2025, with a known true mean (exceso_anual_esperado in estrategias4_verdad.csv). The lesson leaves out degradada, which stops working in 2019 and is the case of lesson 4.2.5. In addition to $\mu/\sigma^2$, the cell computes the “exact” Kelly of the data (the $f$ that maximizes the mean of $\ln(1 + f r_t)$ with the actual daily returns, tails included) and the leverage that ruins the account on the worst day of the sample:

from scipy.optimize import minimize_scalar

est = pd.read_parquet("estrategias4_diarias.parquet")
est = est.set_index("fecha").drop(columns="degradada")
verdad = pd.read_csv("estrategias4_verdad.csv",
                     index_col="estrategia").reindex(est.columns)


def kelly_log(r):
    """f that maximizes the mean of log(1 + f·r) in the data."""
    tope = 0.999 / abs(r.min())          # beyond this, ruin
    res = minimize_scalar(lambda f: -np.log1p(f * r).mean(),
                          bounds=(0, tope), method="bounded")
    return res.x


mu, var = est.mean() * 252, est.var() * 252
tabla = pd.DataFrame({
    "mu": mu, "vol": np.sqrt(var),
    "kelly": mu / var,
    "kelly_verdad": verdad["exceso_anual_esperado"] / var,
    "kelly_log": est.apply(kelly_log),
    "ruina": 1 / est.min().abs()})
tabla.round(2)

                     mu   vol  kelly  kelly_verdad  kelly_log  ruina
tendencia_futuros  0.06  0.11   4.97          4.30       4.94  20.25
carry_futuros      0.03  0.08   5.06          5.63       5.02  31.35
carry_divisas      0.05  0.09   5.81          3.81       5.73  21.78
venta_puts         0.04  0.12   2.56          2.82       2.54  15.25
pares              0.02  0.06   6.72          8.43       6.61  25.65
momentum_acciones  0.05  0.12   3.19          2.83       3.12  12.44
valor_acciones     0.02  0.09   3.01          3.06       3.01  31.49
venta_volatilidad  0.04  0.16   1.55          2.61       1.47   5.19
reversion_corto    0.03  0.07   6.41          5.61       6.36  24.95

Three readings:

• The Kelly fractions are large: from 1.5 to 6.7 times the capital for single strategies with volatilities from 6 to 16%. It is the consequence of $f^*\sigma = \mathrm{SR}$: with a Sharpe of 0.4, Kelly calls for 40% volatility.
• The sample mean drifts away from the true one even with 20 years: in carry_divisas the sample Kelly is 5.81 and the true one 3.81; in venta_volatilidad, 1.55 versus 2.61. With the sample mean you bet 50% too much on the first and 40% too little on the second.
• Tails dominate in volatility selling. Its exact Kelly (1.47) is smaller than the formula's, and the ruin leverage, 5.19 times, is only twice the true Kelly: the worst day of the sample loses 19.3%. In the rest, ruin lies at 12-31 times the capital and the formula and the exact Kelly almost coincide.

With nine weakly correlated strategies, the equal-weight portfolio has a higher Sharpe and a much lower volatility, and Kelly shoots up:

cartera = est.mean(axis=1)          # equal-weight, simple returns
k_cartera = cartera.mean() / cartera.var()
print(f"Portfolio Kelly: {k_cartera:.1f} times the capital")
print(f"Volatility at Kelly: {k_cartera * cartera.std() * 252**.5:.0%}")


def resultado(r):
    w = (1 + r).cumprod()
    años = len(r) / 252
    return {"cagr": w.iloc[-1] ** (1 / años) - 1
            if w.iloc[-1] > 0 else -1.0,
            "max_caida": (w / w.cummax() - 1).min(),
            "peor_dia": r.min(),
            "ruina": bool((w <= 0).any())}


fr = {c: resultado(c * k_cartera * cartera)
      for c in (0.125, 0.25, 0.5, 1.0, 2.0)}
print("Worst day:", cartera.idxmin().date())
pd.DataFrame(fr).T.astype(float).round(3)

Portfolio Kelly: 18.8 times the capital
Volatility at Kelly: 84%
Worst day: 2018-02-05
        cagr  max_caida  peor_dia  ruina
0.125  0.086     -0.315    -0.053    0.0
0.250  0.167     -0.544    -0.106    0.0
0.500  0.303     -0.816    -0.213    0.0
1.000  0.417     -0.979    -0.426    0.0
2.000 -0.081     -1.000    -0.851    0.0

Kelly calls for 18.8 times the capital and 84% volatility. With the full-sample Kelly (which uses the 20-year mean from the first day: look-ahead) the result is the theoretical one with a nuance: full Kelly gives the highest CAGR (41.7%), but with a maximum drawdown of −97.9% and one day, February 5, 2018, on which it loses 42.6% of the capital. One eighth of Kelly (2.3 times) gives 8.6% with a drawdown of −31.5%. Twice Kelly loses 8.1% per year and ends up with a 100% drawdown (rounded). No fraction reaches strict ruin (the account does not go below zero because the portfolio's worst day loses 2.3%), but a 98% drawdown is ruin for practical purposes.

With what was known each day (mean and variance since the start, with a three-year warm-up and no short positions), the portfolio Kelly starts at zero in 2009, because the 2008 crisis left the sample mean negative, and rises to 5.9 in 2013, 14.6 in 2017 and 18.7 in 2025:

mu_t = cartera.expanding(756).mean().shift(1)
var_t = cartera.expanding(756).var().shift(1)
k_t = (mu_t / var_t).clip(lower=0)       # no short positions
print(k_t.resample("YE").last().loc["2009":"2025":4].round(1))
pd.DataFrame({c: resultado((c * k_t * cartera).dropna())
              for c in (0.25, 0.5, 1.0)}).T.astype(float).round(3)

fecha
2009-12-31     0.0
2013-12-31     5.9
2017-12-31    14.6
2021-12-31    16.2
2025-12-31    18.7
Freq: 4YE-DEC, dtype: float64
       cagr  max_caida  peor_dia  ruina
0.25  0.129     -0.346    -0.084    0.0
0.50  0.251     -0.581    -0.168    0.0
1.00  0.450     -0.839    -0.337    0.0

The CAGR of full Kelly comes out even higher (45.0%) because it reached the strategy late and with little leverage in the 2009-2012 drawdowns; the maximum drawdown, −83.9%, and the worst day, −33.7%, are still those of an account that nobody would keep. Calendar luck decides more here than the rule.

Estimation error

Kelly assumes that $\mu$ and $\sigma$ are known. Volatility is estimated well with little data; the mean is not. With $T$ years of data, the sample mean has a standard error of $\sigma/\sqrt{T}$, so the estimated Kelly is $\hat f = f^*\,u$ with $u \sim N\!\left(1, \tfrac{1}{\mathrm{SR}^2 T}\right)$. Betting a fraction $c$ of the estimated Kelly gives an expected growth, relative to that of the true Kelly, of