Make-or-Take Decision with Drift Forecasts
You are a market maker with a one-step forecast $\hat{\mu}$ for the mid-price move over horizon $T$. The mid price is currently $m$, and you model the future mid as $m_T \sim N(\hat{\mu}, \sigma^2 T)$.
You have two choices:
- Take (cross the spread now): You buy immediately at the ask $m + s/2$, locking in a cost of $s/2$ relative to mid.
2. Make (post a passive ask): You post an ask at $m + \delta$. This order fills with probability $p(\delta) = 1 - e^{-A e^{-k \delta T}}$ where $A, k > 0$ are constants governing the fill rate. If filled, your realized P&L is $\delta - \hat{\mu}$ (you sold at $m + \delta$, but mid drifts by $\hat{\mu}$ against you on average).
Derive the inequality in $\delta$ under which the expected P&L from making dominates taking. Then explain how the optimal threshold $\delta^{*}$ depends on the drift forecast $\hat{\mu}$.
Hints
- Compare the expected P&L of each strategy: taking gives you a certain cost of $s/2$ offset by the drift, while making gives you $\delta$ per fill but only with probability $p(\delta)$.
- Write out $\text{E}[\text{PnL}_{\text{make}}] = p(\delta)(\delta - \hat{\mu})$ and set it greater than $\text{E}[\text{PnL}_{\text{take}}] = \hat{\mu} - s/2$. The inequality is implicit in $\delta$ because $p$ depends on $\delta$.
- Notice that as $\hat{\mu}$ grows, the right-hand side increases while the left-hand side shrinks (since $\delta - \hat{\mu}$ falls). This means $\delta^{*}$ must increase with $\hat{\mu}$ -- stronger signals favor taking.
Worked Solution
How to Think About It: This is the classic make-or-take tradeoff that every electronic market maker faces dozens of times per second. Taking is expensive but certain -- you pay half the spread and you are done. Making is cheaper (you earn the spread if filled) but uncertain -- you might not get filled, and if the market moves against you while you wait, your expected P&L deteriorates. The drift forecast $\hat{\mu}$ is the key lever: if you are confident the price is about to move in your favor, you want to take now; if the drift is small or adverse, you can afford to be patient and make.
Quick Estimate: Suppose $s = 2$ ticks, so taking costs $s/2 = 1$ tick. If you post an ask at $\delta = 1.5$ ticks above mid and get filled with probability $p = 0.4$, your expected making P&L is $0.4 \times (1.5 - \hat{\mu})$. For making to beat taking (expected cost $-1$ tick after drift, i.e., $\hat{\mu} - s/2$), you need $0.4(1.5 - \hat{\mu}) > \hat{\mu} - 1$. If $\hat{\mu} = 0.5$, the left side is $0.4(1.0) = 0.4$ and the right side is $-0.5$, so making wins easily. If $\hat{\mu} = 2.0$, the left side is $0.4(-0.5) = -0.2$ and the right side is
Approach: Compare expected P&L from each strategy and solve for the inequality.
Formal Solution:
The expected P&L from taking is: $\text{E}[\text{PnL}_{\text{take}}] = \hat{\mu} - \frac{s}{2}$ You pay $s/2$ to cross the spread, then the mid moves by $\hat{\mu}$ in your favor on average.
The expected P&L from making at level $\delta$ is: $\text{E}[\text{PnL}_{\text{make}}] = p(\delta) \cdot (\delta - \hat{\mu})$ where $p(\delta) = 1 - e^{-A e^{-k \delta T}}$. You collect $\delta$ when filled, but the mid drifts by $\hat{\mu}$ against your position.
Making dominates taking when: $p(\delta)(\delta - \hat{\mu}) > \hat{\mu} - \frac{s}{2}$
Expanding with the fill probability: $\left(1 - e^{-A e^{-k \delta T}}\right)(\delta - \hat{\mu}) > \hat{\mu} - \frac{s}{2}$
Rearranging to isolate the condition on $\delta$: $\delta > \hat{\mu} + \frac{\hat{\mu} - s/2}{p(\delta)}$
This is implicit in $\delta$ (since $p$ depends on $\delta$), so in practice you solve it numerically. But the structure is clear:
- The threshold $\delta^{*}$ is the value where the two sides are equal.
- As $\hat{\mu}$ increases (stronger bullish forecast), $\delta^{*}$ rises because making requires a larger edge per fill to compensate for the opportunity cost of not taking immediately.
- When $\hat{\mu} < s/2$, the right-hand side is negative, so making dominates for a wide range of $\delta$ -- the drift is too weak to justify paying the spread.
- When $\hat{\mu} \gg s/2$, taking dominates for almost any $\delta$ because the drift is so large that waiting to get filled is too costly.
**Dependence of $\delta^{*}$ on $\hat{\mu}$:**
$\delta^{*}$ is increasing in $\hat{\mu}$. Intuitively: - Small $\hat{\mu}$ (weak signal): You expect little price movement, so crossing the spread is wasteful. Post passively at a modest $\delta$ and wait. - Large $\hat{\mu}$ (strong signal): The price is about to move. Every moment you wait unexecuted, you lose expected edge. You need a very large $\delta$ to compensate -- or just take. - At the extreme, there exists a $\hat{\mu}^{*}$ above which no $\delta$ makes making profitable, and you should always take.
The fill probability function $p(\delta)$ decays as $\delta$ increases (posting further from mid reduces fill likelihood), which creates the tension: higher $\delta$ earns more per fill but fills less often.
Answer: Making dominates taking when $p(\delta)(\delta - \hat{\mu}) > \hat{\mu} - s/2$, or equivalently $\left(1 - e^{-A e^{-k \delta T}}\right)(\delta - \hat{\mu}) > \hat{\mu} - s/2$. The threshold $\delta^{*}$ is increasing in $\hat{\mu}$: stronger drift forecasts raise the bar for passive orders because the opportunity cost of waiting grows. When $\hat{\mu} < s/2$, making is broadly favored; when $\hat{\mu}$ is large enough, taking always dominates.
Intuition
This problem captures the fundamental tension in electronic market making: every order is a choice between certainty and optionality. Taking is expensive but guaranteed -- you pay the spread and move on. Making is cheap but risky -- you earn the spread only if someone hits your order, and while you wait, the market can move away from you. The drift forecast is the bridge between the two: it quantifies how much you expect to gain (or lose) by waiting.
In practice, this is why market makers care so much about short-term alpha signals. A strong directional forecast compresses the time window in which passive orders are profitable -- if you know the price is about to jump, you cannot afford to sit on the offer hoping for a fill. Conversely, in quiet markets with weak signals, the spread you earn from making is essentially free money. The fill probability function adds another layer: posting far from mid earns more per fill but fills rarely, creating a classic risk-reward curve. The optimal $\delta^{*}$ balances all three forces -- spread capture, fill probability, and drift cost -- and shifts with every new forecast update.