Kelly Criterion with Proportional Transaction Costs

Optimization · Hard · Free problem

You have a repeated even-money bet: each round you wager a fraction $f \in [0, 1)$ of your current wealth. With probability $p > \tfrac{1}{2}$ you win $+f$ times your wealth, and with probability

-p$ you lose $-f$ times your wealth. However, each round you also pay a proportional fee $c \in [0, 1)$ on the dollar amount wagered, regardless of outcome.

Your per-round gross wealth multiplier is:

$G(f) = \begin{cases} 1 + f(1-c) & \text{with probability } p \\ 1 - f(1+c) & \text{with probability } 1-p \end{cases}$

You want to maximize the expected log-growth rate $E[\log G(f)]$.

  1. Derive the first-order condition for the optimal wager $f^{*}$ and give the closed-form solution (assuming the optimum is interior).
  1. State the condition on $(p, c)$ under which the optimal strategy is to not bet at all ($f^{*} = 0$).
  1. For small $c$, linearize the shrinkage of $f^{*}$ relative to the frictionless Kelly fraction $f_0 = 2p - 1$. How much does the optimal bet shrink per unit of transaction cost?

Hints

  1. Write out $E[\log G(f)]$ explicitly and take the derivative with respect to $f$. The log terms make this clean.
  2. After cross-multiplying the first-order condition, notice that the $f$-coefficient simplifies to
    - c^2 = (1-c)(1+c)$, giving a neat closed form.
  3. For the no-bet condition, check the sign of $h'(0)$. For the linearization, write $f^{*} = (f_0 - c)/(1 - c^2)$ and expand for small $c$.

Worked Solution

How to Think About It: This is the classic Kelly problem with a twist -- every round, the house skims a cut $c$ off the top of whatever you wager, win or lose. Without costs, Kelly says bet $f_0 = 2p - 1$. The fee effectively makes winning less profitable and losing more painful, so you should bet less. The real question is: how much less? And when should you stop betting entirely?

Before doing any algebra, your gut should say: if the fee eats up your entire edge (i.e., $c \ge 2p - 1$), the game has negative or zero expected log-growth and you should walk away. That turns out to be exactly right.

Quick Estimate: Take $p = 0.6$, so the frictionless Kelly is $f_0 = 0.2$. With a 5% fee ($c = 0.05$), you'd guess the optimal bet drops by roughly $c = 0.05$, giving $f^{*} \approx 0.15$. The exact formula gives $f^{*} = (0.2 - 0.05)/(1 - 0.0025) = 0.15/0.9975 \approx 0.1504$, so the linear approximation $f^{*} \approx f_0 - c$ is excellent for small $c$. If instead $c = 0.2$, you are right at the break-even point: $f^{*} = 0$ and you should not bet.

Approach: Maximize $E[\log G(f)]$ by taking the derivative, setting it to zero, and solving for $f$.

Formal Solution:

*Part (i): First-order condition and closed-form*

The objective function is:

$h(f) = p \log(1 + f(1-c)) + (1-p) \log(1 - f(1+c))$

Differentiating with respect to $f$:

$h'(f) = \frac{p(1-c)}{1 + f(1-c)} - \frac{(1-p)(1+c)}{1 - f(1+c)}$

Setting $h'(f) = 0$ and cross-multiplying:

$p(1-c)\bigl(1 - f(1+c)\bigr) = (1-p)(1+c)\bigl(1 + f(1-c)\bigr)$

Expanding the left side: $p(1-c) - pf(1-c)(1+c)$.

Expanding the right side: $(1-p)(1+c) + (1-p)(1+c)f(1-c)$.

Collecting the $f$ terms on one side:

$p(1-c) - (1-p)(1+c) = f(1-c)(1+c)\bigl[p + (1-p)\bigr] = f(1-c^2)$

The left side simplifies as $p - pc - 1 + p - c + pc = 2p - 1 - c$. Therefore:

$\boxed{f^{*} = \frac{2p - 1 - c}{1 - c^2}}$

Note that

- c^2 = (1-c)(1+c)$, so this can also be written as $f^{*} = \frac{2p - 1 - c}{(1-c)(1+c)}$.

The second-order condition $h''(f) < 0$ is automatically satisfied because $h$ is strictly concave in $f$ (it is a sum of concave log terms with negative second derivatives).

*Part (ii): When is $f^{*} = 0$ optimal?*

The interior solution $f^{*} > 0$ requires the numerator

p - 1 - c > 0$. When this fails, the derivative $h'(0)$ is non-positive, so the optimal fraction is at the boundary:

$f^{*} = 0 \quad \text{when} \quad c \ge 2p - 1$

In words, you should not bet at all when the proportional fee exceeds or equals your edge p - 1$. For example, if $p = 0.55$ (edge $= 0.10$), any fee $c \ge 0.10$ wipes out the incentive to bet.

*Part (iii): Linearized shrinkage for small $c$*

The frictionless Kelly fraction is $f_0 = 2p - 1$. Write the optimal fraction as:

$f^{*} = \frac{f_0 - c}{1 - c^2}$

For small $c$, use $\frac{1}{1-c^2} = 1 + c^2 + O(c^4) \approx 1$:

$f^{*} \approx f_0 - c + O(c^2)$

So the absolute shrinkage is:

$f_0 - f^{*} \approx c$

The optimal bet shrinks by approximately $c$ -- one dollar of bet size per dollar of cost rate. The fractional shrinkage relative to the frictionless Kelly is:

$\frac{f^{*}}{f_0} \approx 1 - \frac{c}{2p - 1}$

The smaller your edge, the more sensitive you are to costs. An edge of $0.04$ with a fee of $0.02$ already loses half the optimal bet size.

Answer: The optimal Kelly fraction with proportional cost $c$ is $f^{*} = \frac{2p - 1 - c}{1 - c^2}$. You should not bet when $c \ge 2p - 1$. For small costs, the bet shrinks linearly: $f^{*} \approx (2p-1) - c$, losing roughly one unit of bet fraction per unit of fee rate.

Intuition

The Kelly criterion tells you to size your bets in proportion to your edge divided by the odds. Transaction costs eat directly into that edge. What makes this problem clean is that proportional costs enter the wealth multiplier asymmetrically -- they make wins smaller and losses bigger -- but the log-growth maximization still yields a closed-form fraction. The key structural insight is that costs effectively reduce your edge from p - 1$ to p - 1 - c$ while also slightly adjusting the denominator by a factor of

- c^2$.

In practice, this is one of the most important corrections in systematic trading. Strategies that look great on paper (positive edge, Kelly-optimal sizing) can become worthless once you account for real-world execution costs -- commissions, spread, slippage. The linearized result $f^{*} \approx f_0 - c$ is the workhorse approximation: it says each percentage point of round-trip cost shrinks your optimal position by roughly one percentage point, and the damage is proportionally worse for low-edge strategies. This is why high-frequency strategies need tiny costs to survive, and why cost estimation is arguably as important as alpha estimation.

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