Posterior Return Probability with Binary Regime Indicators
A binary latent regime $Z \in \{0, 1\}$ drives the sign of tomorrow's FX carry return $R \in \{+1, -1\}$. Your prior is $P(Z = 1) = \theta$. Conditional on the regime, the return probabilities are:
$$P(R = +1 \mid Z = 1) = a, \quad P(R = +1 \mid Z = 0) = b$$
where $0 < b < a < 1$ (regime 1 is the "good" regime for positive returns).
You also observe two binary indicators $X_1, X_2 \in \{0, 1\}$ that are conditionally independent given $Z$, with:
$$P(X_i = 1 \mid Z = 1) = p, \quad P(X_i = 1 \mid Z = 0) = q$$
where $0 < q < p < 1$ (each indicator is more likely to fire in regime 1).
You observe that neither indicator fired: $X_1 + X_2 = 0$. Compute $P(R = +1 \mid X_1 + X_2 = 0)$ in closed form.
Hints
- The indicators tell you about the regime, not the return directly. Update your belief about $Z$ first, then use that to get the return probability.
- Both indicators being zero means neither fired. Use conditional independence to write the joint likelihood as $(1 - p)^2$ under regime 1 and $(1 - q)^2$ under regime 0, then apply Bayes' rule.
- Once you have $P(Z = 1 \mid X_1 + X_2 = 0)$, the return probability is just the posterior-weighted mixture: $a \cdot P(Z=1 \mid \cdot) + b \cdot P(Z=0 \mid \cdot)$.
Worked Solution
How to Think About It: You have a two-regime model where regime 1 is favorable for positive returns. Two noisy sensors each "fire" more often in regime 1. Neither sensor fired, so that is evidence against regime 1 -- your posterior on $Z = 1$ should drop below the prior $\theta$. The question is just: how much does that shift your return probability from the prior-weighted average $a\theta + b(1 - \theta)$ toward $b$ (the bad-regime return probability)? The mechanics are standard Bayes, applied in two layers: first update your belief about $Z$, then marginalize over the updated belief to get the return probability.
Quick Estimate: Take $\theta = 0.5$, $a = 0.7$, $b = 0.3$, $p = 0.8$, $q = 0.2$. Before any data, $P(R = +1) = 0.5(0.7) + 0.5(0.3) = 0.50$. Now condition on both indicators being zero. The likelihood of $X_1 + X_2 = 0$ under regime 1 is $(1 - 0.8)^2 = 0.04$. Under regime 0 it is $(1 - 0.2)^2 = 0.64$. The posterior on $Z = 1$ drops to $0.5 \times 0.04 / (0.5 \times 0.04 + 0.5 \times 0.64) = 0.02 / 0.34 \approx 0.059$. So the updated return probability is roughly $0.059 \times 0.7 + 0.941 \times 0.3 \approx 0.041 + 0.282 = 0.32$. That is a big move from 0.50 to 0.32 -- both indicators being silent is strong evidence for the bad regime, which drags the return probability well below the unconditional average.
Approach: Apply Bayes' rule to update the regime posterior given $X_1 + X_2 = 0$, then use the law of total probability to compute the return probability.
Formal Solution:
Since $X_1$ and $X_2$ are conditionally independent given $Z$, the likelihood of both being zero is:
$$P(X_1 + X_2 = 0 \mid Z = 1) = (1 - p)^2, \quad P(X_1 + X_2 = 0 \mid Z = 0) = (1 - q)^2$$
The marginal probability of the observation is:
$$P(X_1 + X_2 = 0) = \theta(1 - p)^2 + (1 - \theta)(1 - q)^2$$
By Bayes' rule, the posterior regime probability is:
$$P(Z = 1 \mid X_1 + X_2 = 0) = \frac{\theta(1 - p)^2}{\theta(1 - p)^2 + (1 - \theta)(1 - q)^2}$$
Now apply the law of total probability for the return, conditioning on $Z$. Note that $R$ depends on $Z$ but is conditionally independent of $X_1, X_2$ given $Z$, so:
$$P(R = +1 \mid X_1 + X_2 = 0) = a \cdot P(Z = 1 \mid X_1 + X_2 = 0) + b \cdot P(Z = 0 \mid X_1 + X_2 = 0)$$
Substituting:
$$P(R = +1 \mid X_1 + X_2 = 0) = \frac{a\theta(1 - p)^2 + b(1 - \theta)(1 - q)^2}{\theta(1 - p)^2 + (1 - \theta)(1 - q)^2}$$
Verification with the numerical example: Plugging in $\theta = 0.5$, $a = 0.7$, $b = 0.3$, $p = 0.8$, $q = 0.2$:
$$\frac{0.7 \times 0.5 \times 0.04 + 0.3 \times 0.5 \times 0.64}{0.5 \times 0.04 + 0.5 \times 0.64} = \frac{0.014 + 0.096}{0.02 + 0.32} = \frac{0.110}{0.34} \approx 0.324$$
This matches our quick estimate.
Answer:
$$P(R = +1 \mid X_1 + X_2 = 0) = \frac{a\theta(1 - p)^2 + b(1 - \theta)(1 - q)^2}{\theta(1 - p)^2 + (1 - \theta)(1 - q)^2}$$
Intuition
This is the basic engine behind any signal-based regime model in trading. You never observe the regime directly -- you observe noisy proxies (carry signals, volatility levels, flow indicators) and use Bayes' rule to form a posterior belief. The posterior then feeds into your return forecast. Here, both signals being silent is informative precisely because silence is more likely in regime 0 than regime 1. The more asymmetric $p$ and $q$ are, the stronger the evidence from silence.
The structure generalizes cleanly: with $n$ conditionally independent indicators, observing $k$ of them fire gives likelihoods $p^k(1-p)^{n-k}$ vs. $q^k(1-q)^{n-k}$, and the log-likelihood ratio is $k \log(p/q) + (n-k)\log((1-p)/(1-q))$. Each indicator contributes additively to the log-odds, which is why regime models with multiple weak signals aggregate evidence so naturally. The common mistake is to condition on the indicators without going through the regime -- that throws away the conditional independence structure and makes the problem much harder than it needs to be.