Posterior Fair Value from Noisy Signal
A latent "fair value" $V$ is drawn from $V \sim N(0, \sigma_V^2)$ with $\sigma_V = 1$. You observe a noisy signal $S = V + \epsilon$, where $\epsilon \sim N(0, \sigma_\epsilon^2)$ with $\sigma_\epsilon = 2$, independent of $V$.
Given an observation $S = 1.5$:
- Compute the posterior distribution $V \mid S = 1.5$ (find the posterior mean and variance).
- Compute $P(V > 0 \mid S = 1.5)$.
Hints
- The joint distribution of $(V, S)$ is bivariate normal. What formula gives the conditional distribution of one component given the other?
- The posterior precision equals the sum of prior precision and signal precision: $\tau_{\text{post}} = 1/\sigma_V^2 + 1/\sigma_\epsilon^2$. Use this to find the posterior variance, then compute the posterior mean as a precision-weighted combination.
- Once you have $V \mid S \sim N(\mu_{\text{post}}, \sigma_{\text{post}}^2)$, standardize to compute $P(V > 0) = \Phi(\mu_{\text{post}} / \sigma_{\text{post}})$.
Worked Solution
How to Think About It: This is the workhorse signal-extraction problem in quant finance. You have a noisy observation of some true value and you want to update your belief about that value. The prior says $V$ is centered at 0, and you see $S = 1.5$. Your posterior for $V$ should shift toward 1.5, but not all the way -- because the signal is noisy ($\sigma_\epsilon = 2$ is twice $\sigma_V = 1$), you should weight the prior heavily. The posterior mean will be a precision-weighted average of the prior mean and the signal. Since $V > 0$ lines up with the direction the signal is pulling you, $P(V > 0 \mid S)$ should be comfortably above 0.5, but not enormous given the noise.
Quick Estimate: The signal-to-noise ratio here is $\sigma_V^2 / \sigma_\epsilon^2 = 1/4$, which is low. So the posterior mean should be pulled only modestly toward $S = 1.5$. Rough guess: posterior mean $\approx 1.5 \times (1/5) = 0.3$. Posterior variance should be less than the prior variance of 1 -- maybe around $0.8$, giving posterior std $\approx 0.9$. Then $P(V > 0) \approx \Phi(0.3 / 0.9) \approx \Phi(0.33) \approx 0.63$. Let's verify.
Approach: Use the standard conjugate normal-normal update.
Formal Solution:
The joint distribution of $(V, S)$ is bivariate normal. We have:
- $V \sim N(0, 1)$
- $S = V + \epsilon$ with $\epsilon \sim N(0, 4)$, so $S \sim N(0, 1 + 4) = N(0, 5)$
- $\text{Cov}(V, S) = \text{Cov}(V, V + \epsilon) = \text{Var}(V) = 1$
By the standard conditional normal formula:
$$E[V \mid S] = \mu_V + \frac{\text{Cov}(V,S)}{\text{Var}(S)}(S - \mu_S) = 0 + \frac{1}{5}(S - 0) = \frac{S}{5}$$
$$\text{Var}(V \mid S) = \text{Var}(V) - \frac{\text{Cov}(V,S)^2}{\text{Var}(S)} = 1 - \frac{1}{5} = \frac{4}{5}$$
Equivalently, using precision (inverse variance) weighting: the prior precision is $\tau_V = 1/\sigma_V^2 = 1$ and the signal precision is $\tau_\epsilon = 1/\sigma_\epsilon^2 = 1/4$. The posterior precision is $\tau_{\text{post}} = \tau_V + \tau_\epsilon = 1 + 1/4 = 5/4$, giving posterior variance $\sigma_{\text{post}}^2 = 4/5$. The posterior mean is:
$$\mu_{\text{post}} = \frac{\tau_V \cdot 0 + \tau_\epsilon \cdot S}{\tau_{\text{post}}} = \frac{(1/4) \cdot 1.5}{5/4} = \frac{1.5}{5} = 0.3$$
So:
$$V \mid S = 1.5 \sim N\!\left(0.3, \, \frac{4}{5}\right)$$
For Part 2, we need $P(V > 0 \mid S = 1.5)$. Standardizing:
$$P(V > 0 \mid S = 1.5) = P\!\left(Z > \frac{0 - 0.3}{\sqrt{4/5}}\right) = P\!\left(Z > \frac{-0.3}{0.8944}\right) = P(Z > -0.3354)$$
$$= \Phi(0.3354) \approx 0.6313$$
Answer:
- $V \mid S = 1.5 \sim N(0.3, \; 4/5)$. The posterior mean is $\mu_{\text{post}} = 0.3$ and the posterior variance is $\sigma_{\text{post}}^2 = 4/5 = 0.8$.
- $P(V > 0 \mid S = 1.5) = \Phi(0.3 / \sqrt{0.8}) \approx 0.631$.
Intuition
This problem captures one of the most fundamental operations in quantitative finance: extracting signal from noise using Bayesian updating. The key idea is precision weighting -- your posterior mean is a weighted average of the prior mean and the observed signal, where the weights are the precisions (inverse variances). Here, because the noise variance ($\sigma_\epsilon^2 = 4$) is much larger than the signal variance ($\sigma_V^2 = 1$), the prior dominates: the posterior mean is only $0.3$, pulled just 20% of the way from the prior mean toward the observation of $1.5$. In real trading, this is exactly how a market maker processes information -- a large trade (noisy signal) shifts your fair value estimate, but only in proportion to how informative you believe that trade is.
The broader principle is that noisier signals get less weight. If $\sigma_\epsilon$ were much smaller (say $0.1$), you would nearly trust the signal at face value and set your posterior close to $1.5$. If the noise were enormous, you would barely move from your prior. This precision-weighting framework extends directly to Kalman filters, optimal portfolio estimation, and any setting where you combine multiple noisy estimates. The common mistake is over-reacting to signals -- treating the observation as if it were the truth -- which is exactly what ignoring the prior corresponds to.