Posterior After Aggressive Buy in Binary Market

Market Microstructure · Hard · Free problem

A unit digital contract pays $\

$ if event $E$ occurs. A market maker holds a prior $p_0 = P(E)$ and quotes a symmetric bid-ask

$(b,\, a) = \left(p_0 - \frac{s}{2},\; p_0 + \frac{s}{2}\right).$

A single counterparty arrives. With probability $\eta$ the counterparty is fully informed -- they know whether $E$ will occur and act myopically (buy if $E$, sell if not $E$). With probability

- \eta$ the counterparty is a noise trader who buys or sells each with probability $\tfrac{1}{2}$, independent of $E$.

You observe a buy at the ask price $a$.

Compute the posterior $p_1 = P(E \mid \text{buy at } a)$.

You must immediately requote a new symmetric market $(b',\, a') = (m - s'/2,\; m + s'/2)$. Against the same informed/noise mixture, find the midpoint $m$ that gives zero unconditional expected P&L, and find the minimal spread $s'$ that makes your expected P&L nonnegative.

Hints

Decompose $P(\text{buy})$ by conditioning on whether the counterparty is informed or noise, and on whether $E$ is true. The likelihood ratio $(1+\eta)/(1-\eta)$ tells you how much a buy shifts your belief.
After the update, your new mid should be $p_1$ -- that is just fair value. The real question is how wide to quote around it.
Write out the market maker's P&L in all four cases (informed buy, informed sell, noise buy, noise sell). The noise terms collapse to a clean $s'/2$ profit, and the adverse selection cost is

Worked Solution

How to Think About It: This is textbook adverse selection -- the bread and butter of electronic market making. Someone just lifted your offer on a binary contract. Should you be worried? An informed trader only buys when they know $E$ will happen, so that buy is bullish evidence. But it might just be noise. Bayes' theorem lets you weigh these two explanations. Once you update your belief, the practical question is: how do you requote? Your new mid should be your posterior (that is just fair value), and your spread has to be wide enough that the edge you earn from noise flow covers the losses you take when an informed trader picks you off. This is the Glosten-Milgrom model in its simplest form -- order flow is informative, and prices must move to reflect it.

Quick Estimate: Take $p_0 = 0.5$ and $\eta = 0.2$. The buy tilts us toward $E$. How much?

Likelihood of a buy given $E$: the informed trader (20% of flow) always buys, the noise trader (80%) buys half the time. So $P(\text{buy} \mid E) = 0.2 + 0.8 \times 0.5 = 0.6$.

Likelihood of a buy given $\neg E$: the informed trader never buys, noise buys half the time. So $P(\text{buy} \mid \neg E) = 0 + 0.8 \times 0.5 = 0.4$.

Likelihood ratio: $0.6 / 0.4 = 1.5$. With equal prior odds (1:1), posterior odds become

.5:1$, giving $p_1 = 1.5 / 2.5 = 0.6$.

So the buy moves us from 50% to 60%. That is a meaningful but not dramatic shift -- most of the flow is noise.

For the requote: center the new market at $m = p_1 = 0.6$. The minimal spread needs to cover adverse selection costs. Plugging in: $s' = 4 \eta \, p_1(1 - p_1) = 4(0.2)(0.6)(0.4) = 0.192$. So you would requote roughly $0.504$ bid, $0.696$ offer -- a 19.2-cent spread centered at $0.6$. That is a wide market, reflecting the toxicity risk from informed flow.

Approach: Apply Bayes' theorem to update $p_0$ after the observed buy. Then write out expected P&L across all four trade scenarios (informed buy, informed sell, noise buy, noise sell) and solve for the break-even spread.

---

Formal Derivation

Part (i): Posterior probability

Compute the likelihood of a buy conditional on $E$ and on $\neg E$.

If $E$ is true: - Informed trader (probability $\eta$): buys with certainty. - Noise trader (probability

- \eta$): buys with probability

/2$.

$P(\text{buy} \mid E) = \eta \cdot 1 + (1 - \eta) \cdot \tfrac{1}{2} = \frac{1 + \eta}{2}.$

If $E$ is false: - Informed trader: does not buy (would sell or abstain). - Noise trader: buys with probability

/2$.

$P(\text{buy} \mid \neg E) = \eta \cdot 0 + (1 - \eta) \cdot \tfrac{1}{2} = \frac{1 - \eta}{2}.$

The total probability of a buy:

$P(\text{buy}) = p_0 \cdot \frac{1+\eta}{2} + (1-p_0) \cdot \frac{1-\eta}{2} = \frac{1 + \eta(2p_0 - 1)}{2}.$

By Bayes' theorem:

$\boxed{p_1 = P(E \mid \text{buy}) = \frac{p_0(1+\eta)}{p_0(1+\eta) + (1-p_0)(1-\eta)}}$

Sanity checks: - $\eta = 0$ (all noise): $p_1 = p_0$. The buy carries no information. - $\eta = 1$ (all informed): $p_1 = 1$. The buy guarantees $E$. - For any $\eta > 0$: $p_1 > p_0$, as expected -- a buy is bullish.

---

Part (ii): Requoting

After updating to $p_1$, the market maker posts $(m - s'/2,\; m + s'/2)$. A fresh counterparty arrives from the same mixture. There are four scenarios:

| Scenario | Probability | MM P&L | |----------|-------------|--------| | Informed, $E$ true: buys at ask | $\eta p_1$ | $(m + s'/2) - 1$ | | Informed, $E$ false: sells at bid | $\eta(1-p_1)$ | $0 - (m - s'/2)$ | | Noise buys at ask | $(1-\eta)/2$ | $(m + s'/2) - p_1$ | | Noise sells at bid | $(1-\eta)/2$ | $p_1 - (m - s'/2)$ |

The noise trader legs combine cleanly:

$\frac{1-\eta}{2}\bigl[(m + s'/2 - p_1) + (p_1 - m + s'/2)\bigr] = \frac{(1-\eta)\,s'}{2}.$

The informed trader legs:

$\eta\bigl[p_1(m + s'/2 - 1) + (1-p_1)(s'/2 - m)\bigr] = \eta\bigl[m(2p_1 - 1) - p_1 + s'/2\bigr].$

Total expected P&L:

$\mathbb{E}[\text{P\&L}] = \eta\bigl[m(2p_1-1) - p_1 + s'/2\bigr] + \frac{(1-\eta)\,s'}{2}.$

Finding the midpoint $m$: The term involving $m$ is $\eta \cdot m(2p_1 - 1)$. For the market maker to have zero directional exposure, the midpoint must equal the posterior fair value:

$m = p_1.$

Substituting $m = p_1$:

$\mathbb{E}[\text{P\&L}] = \eta\bigl[p_1(2p_1 - 1) - p_1 + s'/2\bigr] + \frac{(1-\eta)\,s'}{2}$

$= \eta\bigl[-2p_1(1-p_1) + s'/2\bigr] + \frac{(1-\eta)\,s'}{2}$

$= -2\eta\, p_1(1-p_1) + \frac{s'}{2}.$

Finding minimal $s'$: Setting $\mathbb{E}[\text{P\&L}] \geq 0$:

$\frac{s'}{2} \geq 2\eta\, p_1(1-p_1)$

$\boxed{s'^{\,\star} = 4\eta\, p_1(1 - p_1), \quad m = p_1}$

Interpretation: The spread formula $s' = 4\eta\, p_1(1-p_1)$ encodes the entire economics of adverse selection. Wider spreads are needed when informed flow is more prevalent (higher $\eta$) and when uncertainty about the payoff is greatest ($p_1(1-p_1)$ is maximized at $p_1 = 1/2$). The midpoint $m = p_1$ is the Glosten-Milgrom result: the market maker's quote midpoint equals the expected value conditional on all available information, including the information revealed by the order itself. In practice, this is why bid-ask spreads widen around binary events like earnings or central bank decisions -- both the fraction of informed participants and the payoff uncertainty spike at the same time.

Answer:

(i) $p_1 = \dfrac{p_0(1+\eta)}{p_0(1+\eta) + (1-p_0)(1-\eta)}$

(ii) $m = p_1$, $\;s'^{\,\star} = 4\eta\, p_1(1 - p_1)$

Intuition

This problem distills the core economics of market making under adverse selection. Every time someone trades against you, it could be because they know something or because they are just noise. Bayes' theorem lets you partially sort this out, but you can never fully distinguish the two -- and that residual uncertainty is exactly why spreads exist. The posterior update $p_1 > p_0$ after a buy is the simplest version of the Glosten-Milgrom model: order flow is informative, and prices must move to reflect it.

The factor of 4 in the spread formula deserves a moment. Adverse selection hits both sides of the book -- a buy when $E$ is true costs you, and a sell when $E$ is false costs you -- and the informed trader's edge is the distance from your quote to the true value, which scales with $p_1(1-p_1)$. This is why you see bid-ask spreads blow out around binary events like earnings announcements or FOMC decisions: both the fraction of informed participants and the payoff uncertainty spike simultaneously. Understanding this tradeoff between capturing noise-trader spread and losing to informed flow is the single most important idea in market microstructure.