Two-Trade Bayesian Updating and Microprice Reversion
Consider a market-making mixture model with parameters $(p, q, r)$: with probability $p$ the true value is high ($V = H$), with probability
You are quoting a bid and an ask that satisfy the conditional zero-profit rule (i.e., each quote equals the expected value conditional on being filled at that quote).
Two consecutive trades occur: first your ask is lifted (a buy), then your bid is hit (a sell).
- Compute $\pi_1 = P(V = H \mid \text{first trade is Buy})$.
- Compute $\pi_2 = P(V = H \mid \text{Buy then Sell})$.
- Derive the implied microprice after each fill and show conditions on $(p, q, r)$ under which the second fill tends to revert the microprice move caused by the first fill.
Hints
- Think about what a Buy signal tells you about $V$ and write out the likelihoods $P(\text{Buy} \mid V = H)$ vs. $P(\text{Buy} \mid V = L)$ in terms of $q$ and $r$.
- Apply Bayes' theorem sequentially: after updating on the Buy to get $\pi_1$, use $\pi_1$ as the new prior and update on the Sell to get $\pi_2$.
- For the reversion condition, note that conditional on $V$ the two trades are independent. Write the joint likelihoods $P(\text{Buy},\text{Sell}\mid V=H)=(q+r/2)(r/2)$ and $P(\text{Buy},\text{Sell}\mid V=L)=(r/2)(q+r/2)$ -- they are equal, so the likelihood ratio is $ and $\pi_2 = p$ exactly.
Worked Solution
How to Think About It: You are a market maker and you just saw two consecutive trades in opposite directions -- first a buy, then a sell. After the buy, you updated your belief that the asset is worth $H$ upward (buys are more likely from informed traders when $V = H$). Now a sell comes in. If informedness is high relative to noise, this sell is strong evidence that $V = L$, which should push the microprice back down. The question is: does the sell fully reverse the buy's impact, overshoot, or only partially revert? This depends on the mix of informed vs. noise flow.
Quick Estimate: Take $p = 0.5$, $q = 0.8$, $r = 0.3$. A buy arrives from an informed trader with probability $pq = 0.4$ (informed, $V = H$) plus noise buy probability $r/2 = 0.15$, total buy probability from a high-value state is $pq + p \cdot r/2$. After the buy, $\pi_1$ jumps above 0.5. Then a sell arrives -- informed sells happen when $V = L$, so this pushes $\pi_2$ back toward 0.5. With symmetric parameters ($p = 0.5$), a buy then a sell are equally explained by either state, so $\pi_2 \approx p = 0.5$. In fact, as we will see, $\pi_2 = p$ exactly for any $p$: the second trade exactly undoes the first, and the microprice reverts fully to the prior (no overshoot).
Approach: Apply Bayes' theorem sequentially, first conditioning on the buy, then conditioning on the sell given the buy. Note that conditional on the true value $V$, the two trades are independent.
Formal Solution:
Step 1: Posterior after the first trade (Buy)
The probability of observing a Buy in each state: - $P(\text{Buy} \mid V = H) = q + (1 - q) \cdot \frac{r}{2} \cdot \frac{1}{1} = q + \frac{r}{2}(1 - q)$
More precisely, using the standard mixture model: an informed trader buys when $V = H$ with probability $q$, and a noise trader buys with probability $r/2$ regardless of state. So:
$P(\text{Buy} \mid V = H) = q + \frac{r}{2}$
$P(\text{Buy} \mid V = L) = (1 - q) \cdot 0 + \frac{r}{2} = \frac{r}{2}$
(Here we assume an informed trader never buys when $V = L$, only sells.)
By Bayes' theorem:
$\pi_1 = P(V = H \mid \text{Buy}) = \frac{p(q + r/2)}{p(q + r/2) + (1-p)(r/2)}$
Simplifying:
$\pi_1 = \frac{p(q + r/2)}{pq + r/2}$
Since $q > 0$, we have $\pi_1 > p$ -- the buy pushes the posterior toward $V = H$.
Step 2: Posterior after the second trade (Sell given Buy)
Now condition on a Sell arriving given current posterior $\pi_1$:
$P(\text{Sell} \mid V = H) = \frac{r}{2}$
$P(\text{Sell} \mid V = L) = q + \frac{r}{2}$
Updating from $\pi_1$:
$\pi_2 = \frac{\pi_1 \cdot (r/2)}{\pi_1 \cdot (r/2) + (1 - \pi_1)(q + r/2)}$
Substituting $\pi_1 = \frac{p(q+r/2)}{D}$ and
- \pi_1 = \frac{(1-p)(r/2)}{D}$ with common denominator $D = p(q + r/2) + (1-p)(r/2)$, the $D$ cancels and we get:$\pi_2 = \frac{p(q + r/2) \cdot (r/2)}{p(q + r/2)(r/2) + (1-p)(r/2)(q + r/2)}$
Both terms in the denominator carry the common factor $(r/2)(q + r/2)$, which also appears in the numerator. Cancelling it:
$\pi_2 = \frac{p}{p + (1-p)} = p$
This is no accident. Conditional on $V$, the two trades are independent, so the Buy-then-Sell likelihoods are
$P(\text{Buy}, \text{Sell} \mid V = H) = \left(q + \tfrac{r}{2}\right)\left(\tfrac{r}{2}\right), \qquad P(\text{Buy}, \text{Sell} \mid V = L) = \left(\tfrac{r}{2}\right)\left(q + \tfrac{r}{2}\right),$
which are equal. The likelihood ratio is exactly
$, so the joint event "Buy then Sell" is equally probable under $V = H$ and $V = L$ and carries no net information. Hence $\pi_2 = p$.Step 3: Microprice and reversion conditions
The microprice after each fill is:
$M_0 = pH + (1-p)L, \quad M_1 = \pi_1 H + (1 - \pi_1)L, \quad M_2 = \pi_2 H + (1 - \pi_2)L$
The first trade moves the microprice up: $M_1 > M_0$ since $\pi_1 > p$.
Reversion occurs when $M_2 < M_1$, i.e., $\pi_2 < \pi_1$. Since $\pi_2 = p < \pi_1$ (the latter because $q > 0$), the sell always reverts the microprice move caused by the buy, for any $(p, q, r)$ with $q > 0$.
In fact the reversion is exact, not partial and not an overshoot: because $\pi_2 = p$, we have $M_2 = M_0$. The microprice returns precisely to its prior level $M_0$ -- it does not undershoot past the prior. The strength of the round-trip swing (how far $M_1$ rose before snapping back to $M_0$) grows with the informativeness of a single trade, i.e. with $q$ relative to $r$: when $q$ is large and $r$ is small, $\pi_1$ is close to
$, so $M_1 - M_0$ is large and the subsequent reversion back to $M_0$ is correspondingly large.The intuition for $\pi_2 = p$: an informed trader only ever trades in one direction (buy when $V = H$, sell when $V = L$), so a Buy-then-Sell sequence cannot come from a single informed trader on both legs. Conditional on either state, one leg is "informed-compatible" and the other is forced to be noise, and the two states are perfectly symmetric in how they explain the pair. The sell exactly cancels the information in the buy, leaving the posterior at the prior.
Answer: $\pi_1 = \dfrac{p(q + r/2)}{pq + r/2}$ and $\boxed{\pi_2 = p}$. Because the two trades are conditionally independent given $V$, the Buy-then-Sell pair is equally likely under $V = H$ and $V = L$ (likelihood ratio
$), so the posterior reverts exactly to the prior: $M_2 = M_0$. For any $q > 0$ the second fill fully reverses the first fill's microprice move (no overshoot), and the magnitude of the round trip grows as $q$ rises relative to $r$.Intuition
This problem illustrates the fundamental tension in market microstructure between information and noise. When you see a buy followed by a sell, there are two stories: either these are noise traders on both sides (no information content), or two different informed traders disagree (impossible in a single-state model). In the standard mixture model, an informed trader only trades in one direction, so a buy-sell sequence is more likely to be noise. The reversion effect arises because the sell partially "cancels" the information content of the buy -- the combination is less informative than either trade alone. This is why market makers widen spreads after seeing volatile order flow: alternating buys and sells suggest noise dominance, but each individual trade still moves the microprice, creating oscillation around the true value. In practice, this is the microstructure foundation of bid-ask bounce and short-horizon mean reversion in transaction prices.