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Estimation uncertainty and Bayesian updating

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Start with the idea

An interval makes the uncertainty in an estimate visible. Frequentist confidence intervals and Bayesian posterior intervals answer different questions; neither corrects an unrepresentative sample automatically. Priors are especially influential when data are sparse.

Symbols, units & horizon

x̄: sample mean
s: sample SD
n: observation count
SE=s/√n: IID standard error
μ: population mean
T: t-distributed statistic
α: probability outside the confidence interval, e.g. .05 for 95% confidence
t_(n−1,1−α/2), also written t*: positive critical value for the chosen confidence and n−1 degrees of freedom
a, b: positive Beta prior shape parameters
W, L: win and loss counts
p: unknown stable win probability
∝: proportional to, omitting the normalising constant
∼: distributed according to
Beta: probability distribution on [0,1]

When and why to use this

Use intervals when deciding whether to continue research, and posterior sensitivity when small samples tempt you to size aggressively. Report the assumptions alongside the estimate.

\overline{x} \pm t_{n - 1,1 - \frac{α}{2}} \frac{s}{\sqrt{n}}

Algebra and arithmetic

Invert a t-statistic to get an interval

Under IID normal sampling, $T = \frac{(\overline{X} - μ)}{(\frac{s}{\sqrt{n}})}$ has a t distribution with n−1 degrees of freedom.
Start from $- t^{*} \leq \frac{(\overline{x} - μ)}{S E} \leq t^{*}$ . Multiply by positive SE and isolate μ: $\overline{x} - t^{*} S E \leq μ \leq \overline{x} + t^{*} S E$ .

Work it by hand

n=25, mean .20%, s=1%, t*=2.064 at 95%: SE=.20%; interval=.20%±.4128%=[−.2128%, .6128%].

This confidence interval for an IID normal sample mean uses sample standard deviation s and a Student-t critical value. Its repeated-sampling coverage is about the procedure; it is not a posterior probability that a fixed mean lies in this realised interval. Dependence and selection require a different uncertainty calculation.

p | W, L \sim Beta (a + W, b + L), 𝔼 [p | W, L] = \frac{a + W}{a + b + W + L}

Probability-density algebra

Multiply the prior by the likelihood

A Beta prior has kernel $p^{a - 1} (1 - p)^{b - 1}$ . W wins and L losses have likelihood proportional to $p^{W} (1 - p)^{L}$ .
Add exponents: posterior kernel is $p^{a + W - 1} (1 - p)^{b + L - 1}$ , identifying Beta(a+W,b+L). Its mean is first shape parameter divided by their sum.

Work it by hand

Beta(1,1) plus 12 wins and 8 losses → Beta(13,9), mean 13/22=59.09%. The raw win fraction was 60%; the prior pulls it toward 50%.

A Beta(a,b) prior with conditionally independent Bernoulli wins updates by adding wins W and losses L. Starting from Beta(1,1), 12 wins and 8 losses produce Beta(13,9), with posterior mean 13/22 ≈ 59.1%. Payoff magnitudes and changing market conditions need additional modelling; win probability alone does not establish an edge.

Research sources, review dates and limitations

Extend the research question

Use a binary contract to separate a forecast probability, its sampling uncertainty and the price you could pay. Assess calibration on events withheld from model selection.

Continue with the connected research module →

Connect the ideas: Information and decision time

Retrieve: Use only information available when the decision is made.

Check the change: Observation dates, release delays, revisions and label maturity require different availability checks.

Research & backtests → Point-in-time data → Time series → Research & robust tuning → Financial machine learning → Putting it all together

Explain it yourself: Does shifting a feature by one row guarantee that it was available?

Self-assessed. Write your explanation before opening this comparison.

No. A revised value or delayed release may still contain unavailable information. Audit actual availability timestamps and fit preprocessing inside each training window.

Connect the ideas: Dependence and diversification

Retrieve: Joint behavior matters when combining uncertain outcomes.

Check the change: Correlation, cointegration, covariance and event dependence answer different questions.

Linear algebra → Time series → Portfolio construction → Portfolio management theory → Prediction foundations

Explain it yourself: Why does a highly correlated pair not automatically provide a converging spread?

Self-assessed. Write your explanation before opening this comparison.

Correlation measures co-movement under a chosen sample and horizon. It does not establish a stationary combination or contractual convergence; those require separate definitions, tests and implementation checks.

Python implementation

Self-contained teaching example. Python 3.10+; dependencies and input conventions are shown in the code and notation. Run in your own Python environment.

def mean_interval(mean_value, sample_sd, n, t_critical):
    """Supply the t-table critical value for confidence and n-1 df."""
    radius = t_critical*sample_sd/n**.5
    return mean_value-radius, mean_value+radius

def beta_update(a, b, wins, losses):
    if min(a,b) <= 0 or min(wins,losses) < 0:
        raise ValueError("Positive shapes and nonnegative counts required")
    posterior_a, posterior_b = a+wins, b+losses
    return posterior_a, posterior_b, posterior_a/(posterior_a+posterior_b)

print(mean_interval(.002, .01, 25, 2.064))
print(beta_update(1, 1, 12, 8))

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