Monte Carlo Backtest of Kelly and Fractional Kelly Strategies

Worked Solution

How to Think About It: This problem connects the Kelly Criterion theory to practical implementation. Part (a) is the standard Kelly derivation. The interesting parts are (b) and (c): in theory, full Kelly ($c = 1$) maximizes long-run growth, but over any finite number of bets, the path can be extremely volatile. Fractional Kelly ($c < 1$) sacrifices some growth for dramatically lower variance, which matters when you have a finite bankroll and a finite horizon. The Monte Carlo simulation is how you quantify this tradeoff empirically.

Part (a): Kelly-Optimal Fraction

The expected log-growth rate per round when betting fraction $f$ is: $$G(f) = p \ln(1 + bf) + (1-p) \ln(1 - f)$$

Setting $G'(f) = 0$: $$\frac{pb}{1 + bf} = \frac{1-p}{1 - f}$$

Cross-multiplying and solving: $$f^{*} = \frac{pb - (1-p)}{b} = p - \frac{1-p}{b}$$

This is positive when $pb > 1-p$, i.e., when the expected return per dollar bet is positive.

The maximum growth rate is: $$G^{*} = G(f^{*}) = p \ln\left(1 + bf^{*}\right) + (1-p) \ln\left(1 - f^{*}\right)$$

Part (b): Fractional Kelly -- Growth Rate and Volatility as Functions of $c$

When you bet $f = cf^{*}$ with $c \in (0, 1)$, the log-growth rate per round is: $$G(c) = p \ln(1 + bcf^{*}) + (1-p) \ln(1 - cf^{*})$$

Growth rate: $G(c)$ is a concave function of $c$ that peaks at $c = 1$ (full Kelly). For small deviations from Kelly, the growth rate is approximately: $$G(c) \approx G^{*} - \frac{1}{2}|G''(f^{*})|(f^{*})^2(1 - c)^2$$

So the growth loss from fractional Kelly is quadratic in $(1-c)$: halving to $c = 0.5$ costs only about $25\%$ of the maximum growth rate, not $50\%$.

Volatility of log-wealth: The variance of the single-round log-return when betting $cf^{*}$ is: $$\sigma^2(c) = p(1-p)\left[\ln\left(\frac{1 + bcf^{*}}{1 - cf^{*}}\right)\right]^2$$

This is approximately proportional to $c^2$ for small $cf^{*}$: $$\sigma^2(c) \approx c^2 (f^{*})^2 (b+1)^2 p(1-p)$$

So the standard deviation scales linearly with $c$. Over $T$ rounds (independent bets), the total log-wealth has: - Mean: $T \cdot G(c)$ (roughly linear in $c$ near $c = 0$) - Standard deviation: $\sqrt{T} \cdot \sigma(c)$ (linear in $c$)

The Sharpe ratio of log-returns (growth rate divided by volatility) is approximately: $$\text{SR}(c) \approx \frac{G(c)}{\sigma(c)}$$

Since $G(c)$ is roughly linear in $c$ for small $c$ and $\sigma(c)$ is also linear in $c$, the Sharpe ratio is approximately constant for small $c$. But for larger $c$, the concavity of $G(c)$ means the Sharpe ratio actually decreases -- full Kelly has the highest growth but not the highest risk-adjusted return over finite horizons.

Part (c): Monte Carlo Simulation Design

Simulation procedure:

1. Fix parameters $p$, $b$, initial wealth $W_0 = 1$, horizon $T$ (e.g., $T = 1000$ rounds). 2. Choose a grid of $c$ values: e.g., $c \in \{0.1, 0.2, 0.3, \ldots, 1.0, 1.5, 2.0\}$. 3. For each $c$, run $M$ independent simulations (e.g., $M = 10{,}000$): - In each simulation, generate $T$ independent Bernoulli($p$) outcomes. - Track wealth: $W_{t+1} = W_t \cdot (1 + bcf^{*})$ on a win, $W_{t+1} = W_t \cdot (1 - cf^{*})$ on a loss. - Record terminal wealth $W_T$. 4. For each $c$, compute: - Median terminal wealth (more informative than mean for multiplicative processes). - 5th and 95th percentiles of terminal wealth. - Probability of drawdown exceeding some threshold (e.g., $P(W_T < 0.5 W_0)$). - Mean and standard deviation of $\log W_T$.

```python import numpy as np

def kelly_mc(p, b, T, c_values, M=10000): f_star = (p * b - (1 - p)) / b results = {} for c in c_values: f = c * f_star # Generate all outcomes at once: shape (M, T) wins = np.random.binomial(1, p, size=(M, T)) log_returns = np.where( wins, np.log(1 + b * f), np.log(1 - f) ) log_wealth = np.sum(log_returns, axis=1) wealth = np.exp(log_wealth) results[c] = { 'median': np.median(wealth), 'p5': np.percentile(wealth, 5), 'p95': np.percentile(wealth, 95), 'mean_log': np.mean(log_wealth), 'std_log': np.std(log_wealth), 'prob_ruin': np.mean(wealth < 0.5), } return results ```

Why the tradeoff matters:

Full Kelly ($c = 1$) maximizes the median terminal wealth as $T \to \infty$. But over finite horizons: - The distribution of terminal wealth under full Kelly is extremely right-skewed (log-normal with high variance). The mean is enormous, but the median can still be disappointing for moderate $T$. - Drawdowns under full Kelly can be severe: losing 50-80% of wealth before recovering is common. - Fractional Kelly ($c = 0.5$) gives up only about 25% of the growth rate but cuts the volatility of log-wealth in half, dramatically reducing the probability of large drawdowns. - Over-Kelly ($c > 1$) is catastrophic: expected log-growth decreases, and the probability of severe drawdown increases. At $c = 2$, expected log-growth is approximately zero.

Interviewers emphasize this because it tests whether you understand that theoretical optimality (infinite-horizon, log-utility) does not equal practical optimality (finite horizon, real constraints, career risk). A trader who bets full Kelly will eventually go through drawdowns that get them fired, even if the strategy is positive-EV.

Answer: (a) $f^{*} = (pb - q)/b$. (b) Growth rate $G(c)$ is concave in $c$, peaking at $c = 1$; volatility $\sigma(c)$ scales linearly with $c$. Half-Kelly sacrifices about 25% growth but halves volatility. (c) Simulate $M$ paths of $T$ rounds for each $c$, compare median wealth, percentiles, and drawdown probabilities. The key lesson: full Kelly is theoretically optimal but practically too aggressive for finite horizons and real-world constraints.