Optimal Tick Improvement for a Market Maker

Worked Solution

How to Think About It: This is a classic rate-times-margin optimization. Improving your quote makes you more likely to get filled (higher intensity), but each fill is worth less (smaller spread capture). The expected profit per unit time is the product of the fill rate and the per-fill profit. You want to find the sweet spot where the marginal gain in fill rate exactly offsets the marginal loss in spread. This is the same tradeoff a retailer faces between price and volume.

Quick Estimate: The profit function is $\Pi(k) = e^{\lambda k}(s - k\tau)$. At $k = 0$, profit is $s$ (times the base intensity, which we can normalize to $1$). At $k = 1$, profit is $e^{\lambda}(s - \tau)$. Improving by one tick is worthwhile if $e^{\lambda}(s - \tau) > s$, i.e., $e^{\lambda} > s/(s - \tau)$. For small $\lambda$ and $\tau$, this approximates to $\lambda > \tau/s$ -- the fill-rate elasticity must exceed the fractional spread loss.

Approach: Treat $k$ as continuous, optimize with calculus, then round to the nearest integer.

Formal Solution:

The expected profit per unit time (up to the base intensity constant) is:

$$\Pi(k) = e^{\lambda k}(s - k\tau)$$

where $k \geq 0$ and $s - k\tau > 0$ (i.e., $k < s/\tau$).

*Continuous relaxation:* Taking the derivative and setting it to zero:

$$\Pi'(k) = \lambda e^{\lambda k}(s - k\tau) + e^{\lambda k}(-\tau) = e^{\lambda k}[\lambda(s - k\tau) - \tau] = 0$$

Since $e^{\lambda k} > 0$, we need:

$$\lambda(s - k\tau) - \tau = 0$$

$$k^{*} = \frac{s}{\tau} - \frac{1}{\lambda}$$

Since $k$ must be a non-negative integer, the optimal discrete solution is:

$$k^{*} = \max\left(0, \left\lfloor \frac{s}{\tau} - \frac{1}{\lambda} + 1 \right\rfloor \text{ or } \left\lceil \frac{s}{\tau} - \frac{1}{\lambda} \right\rceil\right)$$

In practice, evaluate $\Pi$ at $\lfloor k^{*}_{\text{cont}} \rfloor$ and $\lceil k^{*}_{\text{cont}} \rceil$ and pick the better one.

*Second-order check:*

$$\Pi''(k) = e^{\lambda k}[\lambda^2(s - k\tau) - 2\lambda\tau]$$

At the optimum $s - k^{*}\tau = \tau/\lambda$, so $\Pi''(k^{*}) = e^{\lambda k^{*}}[\lambda^2 \cdot \tau/\lambda - 2\lambda\tau] = e^{\lambda k^{*}}[-\lambda\tau] < 0$. This confirms a maximum.

*Condition for improving at least one tick ($k^{*} \geq 1$):*

From $k^{*}_{\text{cont}} = s/\tau - 1/\lambda \geq 1$:

$$\frac{s}{\tau} - \frac{1}{\lambda} \geq 1 \implies \lambda \geq \frac{\tau}{s - \tau}$$

Equivalently, $\lambda(s - \tau) \geq \tau$, or $\lambda s \geq \tau(1 + \lambda)$.

For the exact discrete condition, it is optimal to improve by at least one tick when $\Pi(1) > \Pi(0)$:

$$e^{\lambda}(s - \tau) > s \implies e^{\lambda} > \frac{s}{s - \tau}$$

Answer: The continuous optimum is $k^{*} = s/\tau - 1/\lambda$. It is optimal to improve by at least one tick when $e^{\lambda} > s/(s - \tau)$, i.e., the exponential boost in fill rate from one tick improvement exceeds the ratio of current spread to reduced spread. Equivalently, for small $\lambda$ and $\tau/s$, the condition is approximately $\lambda > \tau/(s - \tau)$.