Bayesian Updating with Sequential Trade Direction
A market maker classifies each arriving customer as either informed (probability 0.30) or noise (probability 0.70). An informed trader buys with probability 0.80 and sells with probability 0.20. A noise trader buys or sells with equal probability 0.50.
You observe the next two customer trades are both buys. Compute $P(\text{informed} \mid \text{buy, buy})$.
Hints
- Two consecutive buys from an informed trader are far more likely than from a noise trader -- use this likelihood ratio to update your prior.
- Compute $P(\text{buy, buy} \mid \text{informed}) = 0.8^2$ and $P(\text{buy, buy} \mid \text{noise}) = 0.5^2$, then apply Bayes' theorem.
- The marginal probability is $P(\text{buy, buy}) = 0.30 \times 0.64 + 0.70 \times 0.25$; plug into $P(I \mid \text{data}) = P(\text{data} \mid I) P(I) / P(\text{data})$.
Worked Solution
How to Think About It: This is a classic adverse selection problem -- a market maker's bread and butter. Two consecutive buys are a signal of information, but you need to update how strong that signal is. Your prior says 30% of customers are informed; two buys should push that up meaningfully because informed buyers are far more likely to buy twice than noise traders. Get a quick feel for the answer before grinding the algebra: noise traders buy twice with probability $0.5^2 = 0.25$, while informed traders buy twice with probability $0.8^2 = 0.64$. That is more than twice the rate, so expect the posterior to be well above 0.30.
Quick Estimate: Likelihood ratio: $(0.64) / (0.25) = 2.56$. Prior odds: $0.30/0.70 \approx 0.43$. Posterior odds: $0.43 \times 2.56 \approx 1.10$. Posterior:
Approach: Apply Bayes' theorem directly, computing the joint likelihood of two buys under each hypothesis.
Formal Solution:
Let $I$ denote the informed type. The joint likelihood of observing (buy, buy) under each type:
$P(\text{buy, buy} \mid I) = 0.80^2 = 0.64$
$P(\text{buy, buy} \mid \text{noise}) = 0.50^2 = 0.25$
The marginal probability of observing (buy, buy):
$P(\text{buy, buy}) = 0.30 \times 0.64 + 0.70 \times 0.25 = 0.192 + 0.175 = 0.367$
Applying Bayes' theorem:
$P(I \mid \text{buy, buy}) = \frac{P(\text{buy, buy} \mid I) \cdot P(I)}{P(\text{buy, buy})} = \frac{0.192}{0.367} \approx 0.523$
Answer: $P(\text{informed} \mid \text{buy, buy}) = \dfrac{0.192}{0.367} \approx 52.3\%$.
Intuition
This problem captures the core of adverse selection in market making. Every time you see directional flow -- especially repeated buys or sells -- you should be asking: is this informed or noise? The likelihood ratio here (0.64 vs 0.25) is the key object. After two buys, your probability that the customer is informed nearly doubles from 30% to 52%. If you were quoting a bid-ask spread, you would widen it in response to this signal.
The broader lesson: Bayesian updating is multiplicative in log-odds space. Each observation multiplies your odds by the likelihood ratio. Two buys multiply by $(0.8/0.5)^2 = 2.56$. Three buys would multiply by $(0.8/0.5)^3 = 4.10$. This exponential growth in evidence is why sequential trades are so informative -- and why a smart market maker adjusts quotes dynamically after each trade, not just at the start.