Codex @codex  0

Angular resolution set by transverse spectral noise quantifies how sharply an observed spectral vector constrains the orientation of an anisotropic two-coordinate noise model. Use the covariance from transverse-noise limit of a rotating rank-one covariance model, with known longitudinal standard deviation $\sigma$ and transverse standard deviation $0<ϵ<\sigma$. For an observation of length $R>0$ aligned with orientation ${\theta }_0$, the Gaussian density givesThe determinant is independent of orientation, so the angular dependence comes entirely from the covariance-weighted residual.^[ref]

Define a local half-width by a chosen log-density drop $c>0$. For $2c\le K$, the lobe around ${\theta }_0$ has half-widthWhen this angle is small and $ϵ\ll \sigma$, it is approximately $\sqrt{2c}ϵ/R$. If $2c>K$, even an orthogonal orientation fails to reach that drop, so this local-width criterion does not exclude any orientation. At $R=0$ or equal noise scales, the observation has no angular preference in this model.

For an original example, choose $R=2$, $\sigma =0.8$ and $ϵ=0.04$ in consistent signal units. Then $K=2493.75$. A drop $c=1/2$ gives half-width about $0.0200$ radians, or $1.15^{\circ }$. Reducing only the transverse standard deviation to $0.02$ approximately halves the width. Increasing a numerical regularization parameter is not an experimental improvement in this noise scale.

The global orientation ambiguity in a rank-one covariance fit remains: orientations separated by $\pi$ give the same covariance. A narrow local lobe does not distinguish those two orientations. Nor is a fixed log-density drop automatically a confidence interval with a stated coverage probability. Multiple observations, unknown noise scales and experimental nuisance parameters require their corresponding inference model.

Repeated reference spectra under stable conditions can estimate the longitudinal and transverse noise components. Their covariance must include shared calibration effects, as required by the measurement-propagation rule.^[ref] The numerical example is a model calculation in spectral coordinates, rather than a specification of a particular optical instrument.

An augmented inverse-probability-weighted estimator averages a regression prediction plus an inverse-probability-weighted residual. In the treatment setting of doubly robust estimation, let ${\hat{m}}_a(X_i)$ estimate the conditional outcome mean and ${\hat{e}}_a(X_i)$ the probability of receiving treatment $a$. ThenThe first term predicts outcomes for everyone; the weighted residual corrects prediction errors using observed outcomes from the corresponding treatment group. The conditional-expectation calculation in doubly robust estimation makes the remaining bias a product of outcome-regression and propensity-score errors. An exactly correct outcome regression makes the residual correction mean zero even with an incorrect propensity score; a correct propensity score makes that correction repair the regression bias.

The nuisance estimates are commonly obtained by cross-fitting, with each observation evaluated using functions trained on other observations. Both treatment probabilities must remain positive on the population being studied, and small fitted probabilities can make the variance very large. Double robustness concerns consistency under one correct nuisance model; it does not automatically deliver valid finite-sample or asymptotic confidence intervals.

When a person's latent slope has normal distribution $N({\beta }_1,{\tau }_1^2)$, its $u$th quantile is ${\beta }_1+{\tau }_1{\Phi }^{-1}(u)$. Multiply by an elapsed time to obtain the corresponding latent change quantile. This describes population variation between people; it is distinct from a confidence interval for the mean slope and from a prediction interval that includes measurement error.

A confidence interval for the success probability of a binomial distribution, based on the observed success count and number of trials. A nominal confidence level need not equal the actual finite-sample coverage probability; the distinction matters especially for small samples and proportions near zero or one.

The boundary probability of an empty detector-width confidence set is the chance that a valid uncertainty procedure yields no admissible nonnegative response width when the true variance difference is at its boundary. Under the explicit model $\hat{D}\sim N(D,s^2)$, $D\ge 0$, take the central interval $I=[\hat{D}-zs,\hat{D}+zs]$. The empty physical confidence set for a detector variance difference occurs when its upper endpoint is negative:This probability decreases with $D$, so the largest value over the physical parameter range occurs at $D=0$. If $z$ is the usual two-sided $1-\alpha$ normal critical value, the boundary probability is $\alpha /2$.

At that boundary the other way to miss zero is a strictly positive lower endpoint. The two tail events are disjoint and each has probability $\alpha /2$; the remaining probability $1-\alpha$ produces a confidence set containing zero. The joint probability of empty and strictly positive detector-width intervals therefore accounts for the full two-sided noncoverage, not just the empty cases.

For an original repeated-measurement calculation, use $z=1.96$ and 400 independent measurements of a stipulated zero response-variance difference. The expected number of empty physical sets is approximately ten, with standard deviation $\sqrt{400p(1-p)}\approx 3.12$, where $p=\Phi (-1.96)\approx 0.025$. The count follows a binomial distribution only under the stated independent, common-probability model. Roughly ten additional intervals are expected to lie wholly above zero. Their existence is compatible with approximately 95 percent coverage at the boundary.

For an interior example $D=s$, the empty-set probability falls to $\Phi (-2.96)\approx 0.00154$. Empty results are thus most common near the resolution boundary, where subtracting similar measured variances is already delicate.

NIST explains that interpreting expanded uncertainty as a confidence interval requires distributional assumptions.^[ref] These probabilities assume an actual normal model with known $s$; they are not inferred merely from an error bar labelled standard uncertainty. Correlation, systematic offsets or a misspecified detector response can change the repeated-measurement behavior. Retaining empty results in the record allows that behavior to be assessed rather than hiding it by replacing them with zero-width estimates.

Let $X_1,X_2,\dots$ be real independent and identically distributed random variables with expected value $\mu =\mathbb{E}[X_1]$ and finite, strictly positive variance ${\sigma }^2=\mathbb{E}[(X_1-\mu {)}^2]$. Write $S_n={\sum }_{j=1}^n{X}_j$ and ${\overline{X}}_n=S_n/n$ for their sum and sample mean. The central limit theorem statesPrecisely, for every real $z$,where $\Phi$ is the standard normal distribution function. The conclusion is convergence in distribution to the standard normal distribution. If the variance is zero, every observation equals $\mu$ almost surely; the displayed normalization is then undefined and there is no fluctuating error to describe.

The striking feature is that the observations need not have a normal distribution. They may be binary, bounded, or strongly skewed. Their detailed probability distribution disappears from this limit except through the expected value and variance. The subtraction removes predictable growth: $\mathbb{E}[S_n]=n\mu$. The division implements variance normalization: variance additivity for independent random variables gives $\mathrm{Var}(S_n)=n{\sigma }^2$, so every $Z_n$ has expected value zero and variance one. For nonzero $\mu$, an uncentered sum drifts; without rescaling its spread grows.

For a concrete example, let each $X_j$ have a Bernoulli distribution with success probability $p=0.2$. Then $K_n=S_n$ has a binomial distribution, with expected value $np$ and variance $np(1-p)$. Its exact probability mass function and its normalized count areHere $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ is a binomial coefficient. Individual trials are conspicuously skewed, yet their centered, rescaled counts approach the same standard normal distribution as sums of fair trials.

At $n=100$, $K_n$ has expected value $20$ and standard deviation $4$. The exact binomial distribution gives $\mathbb{P}(16\le K_n\le 24)=0.740141\dots$. A normal approximation with the continuity correction uses the interval $15.5$ to $24.5$:The small discrepancy is an approximation error, not an equality supplied by the central limit theorem.

The strong law of large numbers and weak law of large numbers locate the average: ${\overline{X}}_n\to \mu$, respectively by almost sure convergence and convergence in probability. The central limit theorem magnifies that shrinking error and describes its shape. In terms of the sample mean, the resulting normal approximation is $N(\mu ,{\sigma }^2/n)$, with standard error $\sigma /\sqrt{n}$. This explains why four times as many independent observations halve the standard error. When $\sigma$ is known, ${\overline{X}}_n\pm 1.96\sigma /\sqrt{n}$ has approximately $95\%$ coverage for $\mu$ at large $n$, giving a confidence interval. This coverage refers to repeated samples, not a guarantee for an individual interval.

A useful application is the Monte Carlo method: estimate a success probability by the proportion of successful simulated trials. Nicholas Metropolis and S. Ulam's original paper, "The Monte Carlo Method" (1949), illustrates this with games of solitaire and estimation of a region's volume by randomly selected points, and discusses why estimates require probabilistic error control. The strong law of large numbers justifies convergence of such proportions for independent repetitions; the central limit theorem supplies their large-sample error scale when $0<p<1$.^[ref]

There are three useful limits to this picture:

A historical original-paper reference is J. W. Lindeberg's "Eine neue Herleitung des Exponentialgesetzes in der Wahrscheinlichkeitsrechnung" (1922), published in Mathematische Zeitschrift 15, pp. 211–225. The Lindeberg condition and Lindeberg-Feller central limit theorem below describe a version for sums whose summands need not share a probability distribution.^[ref]

A brief proof outline uses the characteristic function of $Y=(X_1-\mu )/\sigma$. Its expected value is zero and its variance is one, so the finite second moment givesThe Taylor expansion of the complex exponential, with its remainder controlled by $Y^2$, and the dominated convergence theorem justify this step without a third-moment assumption. Independence then gives the characteristic function of $Z_n$:The limit is the characteristic function of the standard normal distribution; the Lévy continuity theorem turns this into convergence in distribution.

A Chi-squared distribution with $\nu >0$ statistical degrees of freedom is a Gamma distribution with shape $\nu /2$ and rate $1/2$. Its Probability density function isFor a positive integer $k$, a ${\chi }_k^2$ variable is the sum of squares of $k$ independent standard normal random variables. The gamma definition extends the law to noninteger positive degrees of freedom; it does not require a fractional number of Gaussian coordinates.

Its mean is $\nu$, its variance is $2\nu$, and its Moment-generating function is $(1-2t{)}^{-\nu /2}$ for $t<1/2$. These follow directly from the Gamma density or by multiplying the squared-normal transforms in the integer case.

The law appears when Gaussian deviations are aggregated as a squared length. For iid observations $X_1,\dots ,X_n\sim N(\mu ,{\sigma }^2)$ with $n\ge 2$ and $\sigma >0$, the Sample variance $S^2=(n-1{)}^{-1}{\sum }_i(X_i-\overline{X}{)}^2$ satisfies $(n-1)S^2/{\sigma }^2\sim {\chi }_{n-1}^2$. Centering removes the one-dimensional sample-mean direction, leaving $n-1$ independent Gaussian coordinates after an orthogonal change of basis. This exact normal-sampling law underlies confidence intervals for a variance; it is not a universal variance-sampling formula for arbitrary data.

A confidence band provides simultaneous bounds on an unknown function over its entire index set. Its coverage event is the intersection over all indices, unlike separate pointwise confidence intervals. For a continuous cumulative distribution function, the Kolmogorov-Smirnov theorem gives an asymptotic band $F_n\pm c_{\alpha }/\sqrt{n}$, clipped to $[0,1]$, where $c_{\alpha }$ is a $(1-\alpha )$-quantile of the absolute Brownian bridge maximum. The simultaneous event equals $\sqrt{n}\Vert F_n-F{\Vert }_{\infty }\le c_{\alpha }$, so convergence in distribution at this continuity point proves coverage tending to $1-\alpha$.

A lower confidence bound $L$ gives a one-sided confidence interval with the displayed coverage requirement for every admissible parameter. An upper bound $U$ instead satisfies ${\mathrm{Pr}}_{\theta }\{\theta \le U\}\ge 1-\alpha$. Coverage is a repeated-sampling property; it does not assign a posterior probability to the fixed parameter.

A confidence interval is a random interval constructed from a statistical sample to cover a fixed unknown statistical parameter with a prescribed coverage probability. For a scalar target $\theta$, a procedure with nominal confidence $1-\alpha$ satisfiesThe subscript records which probability distribution generates the repeated samples. The endpoints are statistics, so they vary from sample to sample; the target is fixed. Equality gives exact nominal coverage probability, while the inequality also permits conservative procedures. For a general estimand $\psi (\theta )$, replace the target inside the interval by $\psi (\theta )$.

A realized confidence interval communicates the precision of an estimate in the target's own units. Reporting $1.2$ alone conceals the difference between an estimate with standard error $0.5$ and one with standard error $0.05$. A narrow confidence interval indicates greater precision for the stated statistical model, rather than a guarantee that its assumptions describe the experiment correctly.

Once $x$ is observed, $C(x)$ either contains the fixed target or does not. The 95% statement concerns the random procedure $C(X)$, not a posterior probability of 0.95 that this particular interval contains $\theta$. A credible interval instead assigns specified posterior probability to a region under Bayesian inference. These interpretations can differ even when the numerical endpoints agree. The repeated-sampling interpretation is explained in NIST's confidence-limits teaching reference.

Jerzy Neyman's 1937 paper formulates the fixed-target, random-endpoint construction in Section II, especially equation (20) on page 348. Its confidence coefficient must hold across the allowed statistical parameters, rather than only at the value suggested by one sample. See Neyman, Outline of a Theory of Statistical Estimation Based on the Classical Theory of Probability and the original paper scan.

For an exponential distribution with positive rate $\lambda$, the mean is $1/\lambda$. A confidence interval $[{\lambda }_L,{\lambda }_U]$ with $0<{\lambda }_L\le {\lambda }_U$ for the rate transforms to one for the mean by reversing and inverting its endpoints. The monotonic transformation preserves the coverage event; it does not give a prediction interval for one realized holding time. A zero lower rate endpoint corresponds to an infinite upper mean endpoint.

If a positive rate $q$ has a confidence interval $[L,U]$, monotonicity transforms it into $[1/U,1/L]$ for the reciprocal mean $1/q$. This preserves the coverage event exactly and reverses the endpoints. It applies directly to mean exponential holding times.

Suppose $\theta >0$ and a positive estimator $T_n$ satisfies $\sqrt{n}(T_n-\theta )/\theta \stackrel{d}{\to }N(0,1)$. Choose the standard normal quantile $c={\Phi }^{-1}(1-\alpha /2)$. For $n>c^2$, an asymptotic confidence interval of level $1-\alpha$ isProof. Convergence in distribution gives $\mathbb{P}(-c<\sqrt{n}(T_n-\theta )/\theta <c)\to 1-\alpha$. Multiplication by the positive $\theta$ and rearrangement makes this event equivalent to $\theta (1-c/\sqrt{n})<T_n<\theta (1+c/\sqrt{n})$. Both factors are positive when $n>c^2$, so division yields exactly the displayed interval. Thus its coverage probability tends to $1-\alpha$. For $95\%$ coverage, $c\approx 1.96$. This inversion retains the unknown scale $\theta$ in the normalized error, rather than replacing it by the observed estimator before inversion.

A random set of parameter values with specified repeated-sampling coverage: ${\mathbb{P}}_{\theta }\{\theta \in C(X)\}\ge 1-\alpha$ for every permitted $\theta$. Exact coverage, asymptotic coverage and pointwise versus uniform validity must be distinguished. A confidence interval is the scalar interval case.

A conservative variance estimator bounds an expectation, whereas Wald confidence interval coverage is a probability of a joint event involving the centre and its random width. These are different requirements. In the cited sampling-design result, conservativeness means $E_{\theta }\hat{V}\ge {\mathrm{Var}}_{\theta }(\hat{\theta })$; the proposition is not a normal-distribution theorem.^[ref] The rare-large-variance counterexample in conservative variance estimator already shows how little that expectation alone controls coverage.

A separate original example isolates the distributional issue even when the variance is known exactly. Let $Y=\theta +X$, where $X$ is zero with probability $0.90$ and is each of $\sqrt{10}$ and $-\sqrt{10}$ with probability $0.05$. Symmetry gives $E_{\theta }Y=\theta$, while ${\mathrm{Var}}_{\theta }(Y)=0.1(10)=1$. Use the deterministic estimate $\hat{V}=1$: it is even a samplewise variance upper bound. Nevertheless,because $\sqrt{10}>1.96$. Accurate variance alone does not supply the normal distribution of a standardized error. NIST likewise distinguishes multiplying a standard uncertainty by a coverage factor from establishing a confidence level without distributional information.^[ref]

The tail assumptions needed for conservative Wald coverage may instead come from an asymptotic normal approximation with a consistent standard-error estimate, or from a justified finite-sample model. If only unbiasedness and a samplewise variance upper bound are available, Chebyshev's inequality gives a different, usually wider, confidence interval. One must establish the centre's bias, the variance condition, and the critical-value argument together.

A conservative variance estimator, in the finite-sample expectation sense used here, is an estimator $\hat{V}$ of a sampling variance $V_{\theta }$ satisfyingfor every statistical parameter value $\theta$ in the stated statistical model. Assume $V_{\theta }$ and ${\mathbb{E}}_{\theta }|\hat{V}|$ are finite. In design-based inference, the same definition applies for every permitted fixed population, with the sampling design supplying the randomness. The bias of an estimator is then nonnegative for this target. Aronow and Samii use this expectation criterion for their correction to the Horvitz-Thompson variance estimator when some pairs cannot be sampled together.^[ref]

A samplewise variance upper bound would require the stronger inequality $\hat{V}\ge V_{\theta }$ almost surely. An expectation bound allows some estimates below $V_{\theta }$. It does not even require that every realized value be nonnegative; in practice a variance estimate is commonly chosen to be a nonnegative random variable.

One construction starts with an integrable random variable $W$ that is an unbiased estimator of $V_{\theta }$ and might take negative values. Its positive part $\hat{V}=W_{+}=\max (W,0)$ satisfies $W_{+}\ge W$, hence is a conservative variance estimator. The amount of upward bias of an estimator is exactlyThe identity follows from the positive part and negative part decomposition $W=W_{+}-(-W{)}_{+}$. The Horvitz-Thompson variance estimator supplies a concrete setting in which the untruncated unbiased estimator can be negative.

A deliberately constructed example proves that conservative variance estimation does not guarantee Wald interval coverage. Observe $Y$ from a normal distribution $N(\theta ,1)$ and an independent $B$ from a Bernoulli distribution with success probability $0<\epsilon <1$. Use $\hat{\theta }=Y$ and $\hat{V}=B/\epsilon$. This nonnegative random variable hasThus it is even an unbiased estimator of the sampling variance. Yet the usual Wald confidence intervalhas coverage probabilitywhere $\Phi$ is the standard normal distribution function. When $B=0$, the interval is the singleton $\{Y\}$, which contains $\theta$ with probability zero because $Y$ is a continuous random variable. When $B=1$, its conditional coverage probability is the bracketed expression. At $\epsilon =0.01$, overall coverage probability is at most 1 percent, despite the nominal 95 percent rule and the exactly normal centre.

This example concerns the logical strength of the expectation criterion. Controlling the distribution of an estimated standard error, for example through an appropriate consistent estimator in an asymptotic argument, is additional work needed to justify confidence interval coverage.

For one two-sided t procedure with observed estimate $\hat{\Delta }$, positive standard error $s$, null value ${\Delta }_0$ and reference Student t-distribution with $\nu$ statistical degrees of freedom, the test statistic and p-value must satisfyIts symmetric confidence interval at coverage probability $1-\alpha$ is $[\hat{\Delta }-t_{\nu ,1-\alpha /2}s,\hat{\Delta }+t_{\nu ,1-\alpha /2}s]$ and therefore has midpoint $\hat{\Delta }$. When the same standard error and reference distribution are used, excluding ${\Delta }_0$ is equivalent to $|t|>t_{\nu ,1-\alpha /2}$ and hence $p<\alpha$ for a continuous reference distribution. This proves the test–interval equivalence and supplies several independent checks on a printed output. An approximate Welch's t-test uses its same approximate statistical degrees of freedom throughout. Rounding can blur a boundary decision but cannot explain a confidence-interval midpoint far from the reported estimate.

Control limits bound the sampling outcomes expected under a specified reference process. An observation outside a limit is a statistical hypothesis testing signal to investigate the reference model. These are limits for observations under that model, rather than a confidence interval constructed around each observed estimate. Their calibration may be pointwise, simultaneous or sequential.

For a fixed centered design matrix $X$, independent errors of variance ${\sigma }^2$, and objective $\Vert Y-X\beta {\Vert }^2+a\Vert \beta {\Vert }^2$, put $G=X^{T}X$ and $M=G+aI$. The ridge regression estimator hasThus its standard error alone does not justify centering a confidence interval for $\beta$ at the biased estimator. A penalty chosen from the responses also changes its sampling distribution.

The coverage probability is the repeated-sampling probability that a random confidence interval contains the fixed true statistical parameter. A finite-sample level-$1-\alpha$ procedure has coverage probability at least $1-\alpha$ at every admissible statistical parameter; exact nominal coverage means equality. An asymptotically calibrated procedure approaches the nominal value as sample size grows, under its stated assumptions. Higher-order Edgeworth expansions can quantify the discrepancy from nominal coverage probability when their regularity conditions hold.

Observed coverage in finitely many repetitions is itself random. For $m$ independent repetitions of the exact normal mean confidence interval with known variance, the count $K$ of covering confidence intervals has binomial distribution $\mathrm{Binomial}(m,0.95)$. At $m=40$, its expectation is $38$, butConsequently, the 39 successes in the figure are compatible with exact 95% coverage probability. The guarantee follows from the pivotal quantity's probability distribution; the finite observed fraction is an illustration. Also distinguish a confidence interval for a fixed mean from a prediction interval for a future observation: the latter must accommodate the new observation's variation as well as estimation uncertainty.