Optimal Kelly Bet Sizing for a Mean-Reversion Signal
You have an opportunity each day to trade an ETF mean-reversion signal. Each day the trade either wins or loses:
- With probability $p = 0.54$, you earn $+1\%$ on the capital you deploy.
- With probability $q = 0.46$, you lose $-1\%$ on the capital you deploy.
You choose a fraction $f \in [0, 1]$ of your total bankroll to invest in the trade each day. The remaining fraction $1 - f$ sits in cash and earns nothing.
- Derive the fraction $f^{*}$ that maximizes the expected log-growth of your bankroll over one day, $E[\log(W_1 / W_0)]$.
- What is the maximum expected log-growth at $f^{*}$?
Hints
- Think about what objective the Kelly criterion maximizes and how it applies to a binary bet with small percentage gains and losses.
- Write out $E[\log(W_1/W_0)]$ as a function of $f$ and note that the per-trade return is only $\pm 0.01 f$, not $\pm f$. The Kelly fraction scales inversely with the return magnitude.
- Set $g'(f) = 0$ to find $f_{\text{unconstrained}}^{*} = (p - q)/r = 0.08/0.01 = 8$. Since this exceeds the constraint $f \leq 1$, argue by concavity that $f^{*} = 1$.
Worked Solution
How to Think About It: This is a textbook Kelly criterion setup. You have a small edge (54% vs. 46%) on a bet that pays 1:1 in percentage terms. The question is: how much of your bankroll should you deploy each day to maximize long-run compounded growth? The Kelly framework says bet $f^{*} = (p - q)/b$ where $b$ is the gain-per-unit-bet. But here the gain-per-unit-bet is only $0.01$ (you earn 1% on deployed capital), so the Kelly fraction blows up well past 1. Since you are constrained to $f \in [0,1]$, the answer is to go all in -- the per-trade risk is so small relative to the edge that even full deployment barely scratches the bankroll in a bad outcome.
Quick Estimate: If you bet everything ($f = 1$), your bankroll grows by $+1\%$ with probability 0.54 and shrinks by $-1\%$ with probability 0.46. Expected log-growth per day:
$$g(1) = 0.54 \ln(1.01) + 0.46 \ln(0.99)$$
Using $\ln(1+x) \approx x - x^2/2$ for small $x$:
$$g(1) \approx 0.54(0.01 - 0.00005) + 0.46(-0.01 - 0.00005) = 0.005397 - 0.004623 = 0.000774$$
So roughly $7.5$ basis points of log-growth per day. Over 252 trading days that compounds to about $0.000750 \times 252 \approx 0.189$, or roughly $19\%$ annual log-growth. Not bad for a tiny daily edge.
Approach: Write the expected log-growth as a function of $f$, take the derivative, solve for the unconstrained optimum, then apply the constraint $f \leq 1$.
Formal Solution:
After one day, your wealth is:
$$W_1 = W_0 \cdot \begin{cases} 1 + 0.01 f & \text{with probability } 0.54 \\ 1 - 0.01 f & \text{with probability } 0.46 \end{cases}$$
The expected log-growth rate is:
$$g(f) = 0.54 \ln(1 + 0.01 f) + 0.46 \ln(1 - 0.01 f)$$
Differentiate with respect to $f$:
$$g'(f) = \frac{0.54 \times 0.01}{1 + 0.01 f} - \frac{0.46 \times 0.01}{1 - 0.01 f} = \frac{0.0054}{1 + 0.01 f} - \frac{0.0046}{1 - 0.01 f}$$
Setting $g'(f) = 0$:
$$0.54(1 - 0.01 f) = 0.46(1 + 0.01 f)$$
$$0.54 - 0.0054 f = 0.46 + 0.0046 f$$
$$0.08 = 0.01 f$$
$$f_{\text{unconstrained}}^{*} = 8$$
Since $g(f)$ is strictly concave (you can verify $g''(f) < 0$) and the unconstrained maximizer $f = 8$ lies far above the feasible region $[0,1]$, the function $g(f)$ is strictly increasing on $[0,1]$. Therefore the constrained optimum is:
$$f^{*} = 1$$
This makes intuitive sense: the unconstrained Kelly fraction is $f_{\text{Kelly}} = (p - q) / r = 0.08 / 0.01 = 8$, meaning you would want to lever up 8x if you could. Capped at $f = 1$, you invest everything.
The maximum expected log-growth at $f^{*} = 1$:
$$g(1) = 0.54 \ln(1.01) + 0.46 \ln(0.99)$$
Computing precisely: $\ln(1.01) = 0.0099503\ldots$ and $\ln(0.99) = -0.0100503\ldots$
$$g(1) = 0.54 \times 0.0099503 + 0.46 \times (-0.0100503)$$
$$= 0.0053732 - 0.0046231 = 0.0007501$$
Using the second-order approximation as a check: $g \approx E[r] - \text{Var}(r)/2$ where $r$ is the portfolio return. With $f = 1$: $E[r] = 0.01(0.54 - 0.46) = 0.0008$ and $\text{Var}(r) = (0.01)^2 \cdot 0.54 \cdot 0.46 \cdot 4 = 0.00009936$, giving $g \approx 0.0008 - 0.0000497 = 0.000750$. Consistent.
Answer: The optimal fraction is $f^{*} = 1$ (invest your entire bankroll). The unconstrained Kelly fraction is $f = (p - q)/r = 0.08/0.01 = 8$, which exceeds the constraint, so the boundary solution applies. The maximum expected log-growth per day is:
$$g(f^{*}) = 0.54 \ln(1.01) + 0.46 \ln(0.99) \approx 7.50 \times 10^{-4}$$
or about $0.075\%$ (0.75 basis points) per day.
Intuition
The Kelly criterion tells you to bet a fraction proportional to your edge divided by the variance of the bet's return. When returns per trade are tiny (here just 1%), the variance per dollar deployed is minuscule, so the optimal Kelly fraction is enormous -- you would lever up 8x if you could. Capped at 100% deployment, you simply go all in every day. The expected log-growth is only about 0.75 basis points per day, but that compounds to roughly 19% annually, which is the power of a small persistent edge applied consistently.
This is exactly the situation many systematic trading desks face: each individual trade has a tiny edge and tiny risk, so you want to put on as much size as your constraints allow. The danger comes when the edge estimate is wrong -- if the true win rate is 50% instead of 54%, the Kelly math says bet nothing. In practice, firms size well below the Kelly optimum ("half-Kelly" is common) to guard against edge estimation error, which is why this problem is a starting point, not the final word, on real position sizing.