Optimal Quote Placement for a Market Maker
A market maker posts an ask quote at $m_0 + \delta$, where $m_0$ is the current mid-price and $\delta > 0$ is the spread she chooses. If a market order hits her quote, she earns edge $\delta$. Opposite-side market orders arrive as a Poisson process with intensity $\lambda(\delta) = Ae^{-k\delta}$, where $A, k > 0$. The exponential decay captures the idea that wider spreads attract fewer fills. Ignore price drift while waiting.
The market maker has a time horizon $T$ and cares only about whether she gets at least one fill (a binary outcome: either she captures edge $\delta$ or earns nothing).
1. Write the expected P&L as a function of $\delta$: $E[\pi(\delta)] = \delta\bigl(1 - e^{-\lambda(\delta)T}\bigr)$ and explain why this is the correct expression.
2. Differentiate to find the first-order condition for the optimal spread $\delta^{*}$, and show it reduces to: $1 - e^{-Ae^{-k\delta}T} = k\delta \cdot A T e^{-k\delta} \cdot e^{-Ae^{-k\delta}T}$
- Argue that a unique optimal $\delta^{*} > 0$ exists. Provide a Newton iteration to compute $\delta^{*}$ numerically, and discuss how $\delta^{*}$ scales with the horizon $T$ and the baseline intensity $A$.
Hints
- The number of fills in $[0, T]$ is Poisson with parameter $\lambda(\delta)T$, so think about the probability of getting at least one fill and how that depends on $\delta$.
- When you differentiate $E[\pi] = \delta(1 - e^{-\lambda T})$, use the chain rule with $\lambda'(\delta) = -k\lambda(\delta)$ to keep the algebra clean.
- For existence and uniqueness, check $E[\pi]$ at the boundaries ($\delta = 0$ and $\delta \to \infty$) and argue concavity of $\ln E[\pi]$ to rule out multiple local maxima.
Worked Solution
How to Think About It: This is the fundamental market-making tradeoff: post tight and you get filled often but earn little per fill; post wide and you earn a lot per fill but rarely get hit. Your job is to find the sweet spot. The Poisson arrival model with exponentially decaying intensity is the simplest tractable setup that captures this tradeoff. Before doing any calculus, think about limiting cases: if $T$ is huge, you will almost surely get filled regardless of $\delta$, so you should post wide. If $A$ is huge (tons of flow), fills are easy to get, so again you can afford to post wider. The optimal spread should increase with both $T$ and $A$.
Quick Estimate: Take $A = 10$, $k = 2$, $T = 1$. At $\delta = 0.5$: $\lambda = 10e^{-1} \approx 3.68$, fill probability $= 1 - e^{-3.68} \approx 0.975$, expected P&L $\approx 0.5 \times 0.975 = 0.49$. At $\delta = 1.0$: $\lambda = 10e^{-2} \approx 1.35$, fill probability $\approx 1 - e^{-1.35} \approx 0.74$, expected P&L $\approx 0.74$. At $\delta = 1.5$: $\lambda = 10e^{-3} \approx 0.50$, fill probability $\approx 0.39$, expected P&L $\approx 0.59$. At $\delta = 2.0$: $\lambda \approx 0.18$, fill probability $\approx 0.17$, expected P&L $\approx 0.33$. So the optimum is around $\delta^{*} \approx 1.0$, and the P&L curve is skewed right with a gentle peak.
---
Part (i): Expected P&L
The number of market-order arrivals in time $T$ follows a Poisson distribution with parameter $\lambda(\delta)T$. The probability of at least one fill is: $P(\text{fill}) = 1 - e^{-\lambda(\delta)T}$
Since the payoff is $\delta$ if filled and $0$ otherwise: $E[\pi(\delta)] = \delta \cdot P(\text{fill}) = \delta\bigl(1 - e^{-\lambda(\delta)T}\bigr)$
where $\lambda(\delta) = Ae^{-k\delta}$.
---
Part (ii): First-Order Condition
Write $\lambda = Ae^{-k\delta}$ so that $\lambda'(\delta) = -k\lambda$. Differentiate $E[\pi]$ using the product rule: $\frac{dE[\pi]}{d\delta} = \bigl(1 - e^{-\lambda T}\bigr) + \delta \cdot \frac{d}{d\delta}\bigl(1 - e^{-\lambda T}\bigr)$
For the second term, apply the chain rule. Since $\lambda'(\delta) = -k\lambda$: $\frac{d}{d\delta}e^{-\lambda T} = e^{-\lambda T} \cdot (-T)\lambda'(\delta) = e^{-\lambda T} \cdot (-T)(-k\lambda) = kT\lambda\, e^{-\lambda T}$
Therefore: $\frac{d}{d\delta}\bigl(1 - e^{-\lambda T}\bigr) = -kT\lambda\, e^{-\lambda T}$
Substituting into the product rule: $\frac{dE[\pi]}{d\delta} = \bigl(1 - e^{-\lambda T}\bigr) - k\delta T\lambda\, e^{-\lambda T}$
Setting this to zero and substituting $\lambda = Ae^{-k\delta}$: $\boxed{1 - e^{-Ae^{-k\delta}T} = k\delta \cdot AT\, e^{-k\delta}\cdot e^{-Ae^{-k\delta}T}}$
The left side is the fill probability (the "benefit" of tightening the spread), and the right side is proportional to $\delta$ times the marginal loss of fill probability from widening (the "cost"). At the optimum, these balance.
---
Part (iii): Existence, Uniqueness, and Newton Iteration
Existence: At $\delta = 0$, $E[\pi(0)] = 0$. As $\delta \to \infty$, $\lambda \to 0$ so $E[\pi] \to 0$. For any small $\delta > 0$, $E[\pi(\delta)] > 0$ (e.g., our estimate showed $E[\pi(0.5)] \approx 0.49$). Since $E[\pi]$ is continuous, starts at zero, goes positive, and returns to zero, it attains a maximum at some interior $\delta^{*} > 0$.
Uniqueness: Define $F(\delta) = \frac{dE[\pi]}{d\delta}$. At $\delta = 0$: $F(0) = 1 - e^{-AT} > 0$. As $\delta \to \infty$: $F(\delta) \to 0^{-}$ (the fill probability term vanishes faster than the marginal cost term because of the factor $\delta$). One can verify that $F'(\delta) < 0$ at any root of $F$ by computing the second derivative. Equivalently, define $u = \lambda T = ATe^{-k\delta}$ and rewrite the FOC as: $1 - e^{-u} = k\delta \cdot u \cdot e^{-u}$
Since $u$ is a monotonically decreasing function of $\delta$, and the LHS $(1 - e^{-u})/u$ is strictly decreasing in $\delta$ while $k\delta e^{-u}$ eventually decreases (after initial increase), a single crossing is guaranteed. More directly: the log of the P&L, $\ln \delta + \ln(1 - e^{-\lambda T})$, is strictly concave in $\delta$ (since $\ln \delta$ is concave and the second term is concave in $\delta$ given the log-linear structure of $\lambda$). Strict concavity of $\ln E[\pi]$ implies quasi-concavity of $E[\pi]$, which gives a unique maximum.
Newton Iteration: Define: $g(\delta) = \bigl(1 - e^{-\lambda T}\bigr) - k\delta T\lambda\, e^{-\lambda T}, \quad \lambda = Ae^{-k\delta}$
We need $g(\delta^{*}) = 0$. The Newton update is: $\delta_{n+1} = \delta_n - \frac{g(\delta_n)}{g'(\delta_n)}$
To compute $g'$, differentiate: $g'(\delta) = -kT\lambda e^{-\lambda T} - kT\lambda e^{-\lambda T}\bigl(1 - k\delta - k\delta\lambda T\bigr)$
Simplifying with $\lambda' = -k\lambda$: $g'(\delta) = -kT\lambda e^{-\lambda T}\bigl(2 - k\delta(1 + \lambda T)\bigr)$
A good starting point is $\delta_0 = 1/k$ (a natural scale for the problem). Convergence is typically achieved in 3-5 iterations.
**Scaling of $\delta^{*}$:**
- With $T$: As $T$ increases, fills become more likely at any spread, so the market maker can afford to post wider. $\delta^{*}$ increases with $T$, roughly as $\frac{1}{k}\ln(AT)$ for large $T$. Intuitively, when the horizon is long, you push your quote out until the marginal fill probability per unit time just balances the extra edge.
- With $A$: Higher $A$ means more baseline flow, so again fills are easier to get and the optimal spread widens. $\delta^{*}$ increases with $A$, again roughly logarithmically: $\delta^{*} \sim \frac{1}{k}\ln(AT)$ for large $A$. The logarithmic dependence comes from the exponential decay of $\lambda$ in $\delta$ -- you need to increase $\delta$ by /k$ to offset a factor-of-$e$ increase in flow.
Answer: The optimal spread $\delta^{*}$ is the unique root of
- e^{-\lambda T} = k\delta \lambda T e^{-\lambda T}$ with $\lambda = Ae^{-k\delta}$. It increases logarithmically in both $T$ and $A$, scaling as $\delta^{*} \approx \frac{1}{k}\ln(AT)$ for large $AT$. Compute it via Newton's method starting from $\delta_0 = 1/k$.Intuition
This problem captures the central tension in market making: you want to earn as much as possible per trade, but wider quotes mean fewer trades. The exponential intensity model $\lambda(\delta) = Ae^{-k\delta}$ is a workhorse in microstructure because it makes the tradeoff analytically tractable while being empirically reasonable -- order flow really does decay roughly exponentially with distance from the mid. The optimal spread balances the marginal gain from a wider quote (more edge if filled) against the marginal loss (lower fill probability). The logarithmic scaling $\delta^{*} \sim \frac{1}{k}\ln(AT)$ is the key practical insight: doubling your time horizon or the flow intensity only nudges your optimal quote out by $\frac{\ln 2}{k}$. This means optimal spreads are surprisingly stable across different market conditions -- a fact that real market makers exploit by not adjusting quotes aggressively with every tick of volatility in flow.
The Newton iteration part is also practically relevant. In production, you cannot solve this FOC in closed form, so numerical methods are essential. Starting from $\delta_0 = 1/k$ (the natural length scale of the intensity decay) gives fast convergence. This kind of "set up the FOC, solve numerically, interpret the comparative statics" workflow is exactly how quant traders calibrate their quoting strategies in practice.