Bayesian Update from a No-Trade Event
You are a market maker quoting a symmetric spread $[-s, +s]$ around zero for a one-shot game. The true value of the asset is $V \sim N(0, \sigma^2)$, unknown to you but observed perfectly by informed traders.
The arrival structure is as follows. With probability $p_0$, nobody arrives and no trade occurs. Conditional on arrival, with probability $\mu$ an informed trader shows up -- she buys if $V > +s$ and sells if $V < -s$, and does nothing if $|V| \leq s$. With probability $1 - \mu$, a noise trader arrives -- she buys or sells with equal probability $\frac{1}{2}$, independently of $V$, but declines to trade at all with probability $q$.
You observe that no trade occurs.
(a) Write down $P(\text{no trade} \mid V = v)$ as an explicit function of $v$, $s$, $\mu$, $p_0$, and $q$.
(b) Use Bayes' theorem to compute the posterior mean $E[V \mid \text{no trade}]$. Does observing no trade shift your mean estimate of $V$ away from zero? Why or why not?
(c) Even though the posterior mean may be unchanged, the posterior distribution is not the same as the prior. Describe qualitatively how the posterior variance $\text{Var}(V \mid \text{no trade})$ compares to $\sigma^2$, and explain what this implies for how you should requote your spread.
Hints
- Write down $P(\text{no trade} \mid V = v)$ explicitly by conditioning on who arrives -- no one, the informed trader, or the noise trader. Notice the result depends on $v$ only through $\mathbf{1}[|v| \leq s]$.
- To find $E[V \mid \text{no trade}]$, check whether the likelihood $P(\text{no trade} \mid V = v)$ is an even or odd function of $v$. The prior $N(0, \sigma^2)$ is symmetric -- what does multiplying an even likelihood by a symmetric prior tell you about the posterior mean?
- The posterior density is proportional to $\left[ L_{\text{out}} + \mu \cdot \mathbf{1}[|v| \leq s] \right] \cdot f(v)$. Split the integral into $|v| \leq s$ and $|v| > s$ and use the standard normal CDF $\Phi$ to compute $P(\text{no trade}) = L_{\text{out}} + \mu [2\Phi(s/\sigma) - 1]$.
Worked Solution
How to Think About It: You observed silence -- nobody traded. Who could have produced that silence? Either nobody arrived at all, or an informed trader arrived but found $V$ inside your spread (she trades if and only if $|V| > s$, so her silence is genuine information), or a noise trader arrived and happened to decline. The informed channel is the only one whose behavior depends on $V$, and it depends on $V$ only through $|V|$. So the no-trade likelihood is an *even* function of $v$: silence is exactly as consistent with $V = +v$ as with $V = -v$. Multiplying an even likelihood by a prior symmetric about zero gives a posterior symmetric about zero -- the mean cannot move. What does move is the spread of the distribution: large $|v|$ values would have triggered the informed trader, so silence downweights the tails and the posterior variance shrinks. One bookkeeping trap: the informed/noise split happens *conditional on arrival*, so those branches must carry the factor $(1 - p_0)$, otherwise the "likelihood" can exceed 1.
Quick Estimate: Take $\sigma = 2$, $s = 1$, $\mu = 0.3$, $p_0 = 0.1$, $q = 0.2$. Inside the spread ($|v| \le s$) nobody who arrives ever hits you: likelihood $= 0.1 + 0.9(0.3 + 0.7 \times 0.2) = 0.496$. Outside the spread the informed trader always trades: likelihood $= 0.1 + 0.9 \times 0.7 \times 0.2 = 0.226$. The inside region is roughly $2.2\times$ as consistent with silence as the tails, so the posterior piles up symmetrically near zero: mean stays $0$, variance drops (simulation: $2.84$ vs. prior $4$).
Approach: Condition on the arrival tree to get the likelihood, then exploit evenness in Bayes' rule.
Formal Solution:
(a) Likelihood of no trade. Three disjoint events produce no trade. With probability $p_0$, nobody arrives. With probability $(1-p_0)\mu$, an informed trader arrives and stays silent only when $|V| \le s$. With probability $(1-p_0)(1-\mu)$, a noise trader arrives and declines with probability $q$, independent of $V$. Summing:
$$\boxed{P(\text{no trade} \mid V = v) = p_0 + (1-p_0)\,\mu\,\mathbf{1}[|v| \le s] + (1-p_0)(1-\mu)\,q}$$
Note the $(1-p_0)$ factors: the $\mu$ vs. $1-\mu$ split is *conditional on arrival*. Dropping them (i.e., writing $p_0 + \mu\mathbf{1}[|v|\le s] + (1-\mu)q$) is not a probability -- with $p_0 = 0.5$, $\mu = 0.8$, $q = 0.5$ it gives $1.4$. Write $L_{\text{in}} = p_0 + (1-p_0)[\mu + (1-\mu)q]$ for $|v| \le s$ and $L_{\text{out}} = p_0 + (1-p_0)(1-\mu)q$ for $|v| > s$; always $L_{\text{in}} > L_{\text{out}}$.
(b) Posterior mean. By Bayes' theorem, with $f$ the $N(0, \sigma^2)$ density,
$$f(v \mid \text{no trade}) \propto \Big[L_{\text{out}} + (1-p_0)\,\mu\,\mathbf{1}[|v| \le s]\Big] f(v).$$
Both $f(v)$ and $\mathbf{1}[|v| \le s]$ are even in $v$, so the posterior density is even, and therefore
$$\boxed{E[V \mid \text{no trade}] = 0}$$
No trade does not shift your mean estimate: silence is directionless evidence. It tells you $V$ is more likely to be *small*, not more likely to be positive or negative. The normalizing constant is
$$P(\text{no trade}) = L_{\text{out}} + (1-p_0)\,\mu\left[2\Phi\!\left(\tfrac{s}{\sigma}\right) - 1\right].$$
(Check with the Quick Estimate numbers: $0.226 + 0.9 \times 0.3 \times 0.383 = 0.329$, matching simulation.)
(c) Posterior variance. Since the posterior mean is zero, $\text{Var}(V \mid \text{no trade}) = E[V^2 \mid \text{no trade}]$:
$$\text{Var}(V \mid \text{no trade}) = \frac{L_{\text{out}}\,\sigma^2 + (1-p_0)\,\mu\,E\!\left[V^2\,\mathbf{1}[|V| \le s]\right]}{P(\text{no trade})},$$
where $E[V^2\mathbf{1}[|V| \le s]] = \sigma^2\left[2\Phi(\tfrac{s}{\sigma}) - 1\right] - 2s\sigma\,\phi(\tfrac{s}{\sigma})$. The no-trade posterior reweights mass toward the inside region $|v| \le s$ (weight boosted by the informed trader's silence), and values there have smaller $V^2$ than the prior average. Hence
$$\boxed{\text{Var}(V \mid \text{no trade}) < \sigma^2}$$
(With the example parameters: $2.85 < 4$.) Economically: silence lowers the probability that an informed trader is lurking with $|V| > s$, so adverse-selection risk has fallen and you should tighten your spread on the next quote.
Answer: $P(\text{no trade} \mid V = v) = p_0 + (1-p_0)\mu\,\mathbf{1}[|v| \le s] + (1-p_0)(1-\mu)q$. The posterior mean is $E[V \mid \text{no trade}] = 0$ by symmetry -- silence carries no directional information. The posterior variance satisfies $\text{Var}(V \mid \text{no trade}) < \sigma^2$: silence concentrates belief near zero, so you should tighten the spread.
Intuition
The punchline here is that silence is informative about variance, not about direction. You started with $V \sim N(0, \sigma^2)$ -- symmetric, centered at zero. The no-trade event is equally likely whether $V$ is mildly positive or mildly negative. So it cannot break the symmetry, and the posterior mean stays at zero. But silence is not neutral about the magnitude of $V$. The informed trader would have acted if $V$ were large in absolute value. Her inaction is evidence that $V$ is probably sitting quietly inside the spread. The posterior is a truncated-toward-zero version of the prior -- same center, smaller spread.
This is a foundational result in market microstructure, closely related to the Glosten-Milgrom model. Market makers facing adverse selection use the order flow -- or lack thereof -- to update their beliefs about the asset's true value. After a period of no trading, a market maker rationally tightens her quotes, not because she knows which direction $V$ moved, but because the absence of informed flow signals that $V$ is probably not far from fair value. The practical implication is that inactivity reduces adverse selection risk and justifies narrower spreads. Confusing mean-updating with variance-updating is the classic error -- no trade gives you information about how uncertain you should be, not about which side to lean.