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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.

For an exponential distribution with positive rate $\lambda$, the mean is $1/\lambda$. A confidence interval 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.

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.

Fitted values are the predictions at the predictor values used to fit a regression model. In linear regression with an explicit regression intercept,Here $X$ is the slope design matrix, and $r$ is the regression residual vector. The fitted values estimate the regression function at the observed predictors; they need not equal the observed responses, which may also contain random errors. Prediction at a new predictor vector $x_{\ast }$ uses $\hat{\alpha }+x_{\ast }^T\hat{\beta }$ with the same centering and scaling used in fitting. A prediction interval addresses uncertainty in a new response rather than just displaying this point prediction.

For the five constructed observations in linear regression, ordinary least squares gives fitted values $(4/5,9/5,14/5,19/5,24/5)$. The ridge regression fit at $\lambda =10$ instead gives $(9/5,23/10,14/5,33/10,19/5)$. Both fitted vectors have sample mean $14/5$ because the regression intercept is unpenalized; their regression residuals differ and their residual sums of squares are $4/5$ and $33/10$, respectively.

For ordinary least squares, the fitted values are the orthogonal projection of $Y$ onto the column space of the complete design matrix, including its regression intercept column. They remain unique even in rank-deficient ordinary least squares, although the regression coefficients can be nonunique. The regression residuals are then orthogonal to every column.^[ref] With centered slopes, ridge regression instead satisfies $X_c^{T}r=\lambda {\hat{\beta }}_{\lambda }$, so the regression residuals need not be orthogonal to the slope columns. Their sum is still zero because the regression intercept remains free.

For fixed predictors and scales and a fixed positive penalty, ridge regression gives a linear smoother,For $\lambda >0$ with at least one nonzero slope direction, this matrix is not an orthogonal projection matrix: applying it twice shrinks that direction again. Its effective degrees of freedom are $1+{\sum }_{j=1}^r{d}_j^2/(d_j^2+\lambda )$, where $d_j$ are the positive singular values of $X_c$; the extra $1$ counts the free regression intercept. If the penalty is selected using $Y$, the resulting fitting rule need not be a linear smoother, since its chosen $H_{\lambda }$ also depends on $Y$.

The descriptive statistics report 2466 observed ages and efficacy scores, with no missing values for either variable. Age ranges from 14 to 97 years, with sample mean 52.689 years and sample standard deviation 19.064 years. Efficacy ranges from 0 to 16, with sample mean 9.348 and sample standard deviation 3.176 scale points. The standard deviations use different units and should not be compared as though their numerical ratio measured relative substantive variability. The efficacy mean retains the conflict with Question 1's group summaries; that conflict is not resolved by this separate regression output.

A simple linear regression models $\mathbb{E}(Y\mid x)={\beta }_0+{\beta }_{1}x$, where $Y$ is efficacy and $x$ is age in years. Ordinary least squares chooses the coefficients to minimize the residual sum of squares. From the printed coefficients the fitted line isHolding to this model, ten additional years correspond to $-0.03192$ fitted scale points. For example, the fitted mean is about 9.453 at age 20 and 9.261 at age 80, a difference of about $-0.192$ points. These are fitted means, not exact predictions of individual scores. The line does not imply that every person's efficacy falls at that rate, nor that cross-sectional age differences identify within-person changes over time.

The intercept is the predicted mean at age zero, about 9.517. Since zero is outside the observed 14–97 age range, its direct interpretation requires extrapolation. The printed standard error of 0.188043 and $t\approx 50.608$ yield the extremely small displayed p-value for a zero intercept. That test does not test whether age is a useful predictor. Recentring age at its observed sample mean would put the intercept near $9.516515-0.003192(52.689)\approx 9.34833$, the fitted mean at a substantively relevant reference age. This agrees with the rounded overall efficacy mean.

The age standard error is 0.003356 points per year. The coefficient divided by this standard error gives $-0.003192/0.003356\approx -0.9511$, agreeing with the printed t value. Estimating an intercept and one regression slope from 2466 observations leaves $2466-2=2464$ residual statistical degrees of freedom. Under a correctly specified normal linear model, the Student t-distribution provides the stated two-sided p-value. At 5%, or 1%, it does not reject $H_0:{\beta }_1=0$. The p-value is not a 34.2% probability that age has no effect, and nonrejection does not establish a practically negligible effect size.^[ref]

A 95% confidence interval calculated from the printed estimate and standard error is approximatelyor about $[-0.09773,0.03389]$ points per decade. It includes zero and gives the scale of slopes compatible with these summaries and assumptions. Assessing a practically negligible effect would require an explicit substantive tolerance and an appropriate equivalence test, not simply the fact that $p>0.05$. A nonlinear association could also coexist with a near-zero overall linear slope.

The coefficient of determination is $R^2=0.000367$, meaning that the fitted line explains about $100R^2=0.0367\%$ of the sample's centred efficacy sum of squares compared with an intercept-only fit. This is not 36.7%, a fraction of people whose efficacy is predicted correctly, or the proportion of efficacy caused by age. The adjusted coefficient of determination, approximately $-0.00003873$, is slightly negative because its penalty for estimating the extra coefficient exceeds the tiny improvement in fit. It need not be constrained to be nonnegative and does not mean a negative amount of explained variation.

The overall $F\approx 0.9045$ on 1 and 2464 statistical degrees of freedom, with $p\approx 0.3417$, tests the same zero-slope hypothesis as the age t test because there is only one predictor. The simple-regression output identities give $F=t_{\mathrm{slope}}^2$ and $F=(2464)R^2/(1-R^2)$; the printed numbers agree allowing their displayed precision. The signed Pearson correlation coefficient is consequently about $-\sqrt{R^2}\approx -0.0192$. These are related summaries of the same weak linear association, not independent corroborating tests.

The residual standard error of 3.177 points estimates the typical unexplained error scale under the fitted model. It is close to the efficacy sample standard deviation, as expected when the line explains almost none of the variation. It is not a standard error of the slope or a guarantee that each individual's prediction error lies within 3.177 points. The residual summary spans roughly $-9.46$ to $6.75$, with sample median about 0.573 and central quartiles about $-1.437$ and $2.579$. An ordinary least-squares fit with an intercept has zero mean residual, not necessarily zero median residual. These few quantiles cannot establish normality, constant conditional spread, absence of outliers or linearity. The apparent last-decimal discrepancies among rounded error scales and other summary fields do not force an additional contradiction: their rounding intervals admit compatible regression summaries.

The conventional uncertainty calculations assume independent, mean-zero errors conditional on age, an appropriate linear mean, and homoscedasticity; normal errors give the usual exact finite-sample t and F calibration. Large samples can relax some distributional requirements but do not remove confounding, selection bias or dependence. A residual-versus-fitted plot could reveal curvature or changing spread, and a normal probability plot could assess error shape; neither plot can be reconstructed from the printed residual quantiles. A prediction interval for an individual includes residual variation and is wider than a confidence interval for the mean response. The supplied output supports a very weak, imprecisely signed linear association with little predictive gain from age alone, while leaving nonlinear, causal and design-based questions unresolved.

The response is a device count, while the predictor is a skill score. A simple linear regression models $\mathbb{E}(Y\mid x)={\beta }_0+{\beta }_{1}x$, where $Y$ is the device count and $x$ the skill score. Ordinary least squares chooses the intercept and regression slope to minimize the residual sum of squares. The regression slope describes a conditional association, not the causal effect of training or acquiring a device.

The descriptive statistics put the device count between $0$ and $8$, with sample mean about $4.494$ and sample standard deviation about $2.128$ among $2053$ records. Skill scores range from $0$ to $24$, with sample mean about $13.546$ and sample standard deviation about $5.989$ among $2051$ records. Two missing skill values account for the stated two deleted records; because device counts have no missing values, the regression complete-case sample has $2051$ observations. Estimating two regression coefficients leaves $2051-2=2049$ residual statistical degrees of freedom, matching that part of the output. The units of the two standard deviations differ, so comparing $5.989$ with $2.128$ does not establish that skill is more variable in a substantive sense.

If the coefficient estimates are trusted, the fitted values follow $\hat{y}=0.392122+0.209240x$. The intercept predicts about $0.392$ devices at a skill score of zero; zero is within the stated score range, but precision and the number of observations near zero are not shown. A five-point increase in score corresponds to about $1.0462$ additional fitted devices and a ten-point increase to about $2.0924$. At a score of ten the fitted value is about $2.4845$, and at 24 it is about $5.4139$. Expected counts may be fractional even though individual counts are integers. These are conditional interpretations of the coefficient fields, not reconstructed raw observations or validated individual predictions.

A coefficient's standard error estimates its sampling uncertainty, whereas the residual standard error describes unexplained response variation. For the intercept, $0.392122/0.114570\approx 3.42255$, agreeing with the printed $t=3.423$ and rounded $p=0.000633$. Conditional on the model assumptions, this would reject a zero intercept at conventional significance levels. The intercept test is not a test of whether skills predict devices. For the slope, the printed $p=0.013672$ would reject $H_0:{\beta }_1=0$ at 5% but not 1%; it does not assign a probability to the null hypothesis or measure the usefulness of prediction. The significance stars simply encode thresholds and add no independent evidence.^[ref]

The slope row fails its own arithmetic check:At $2049$ statistical degrees of freedom the ratio gives a two-sided p-value about $0.00845$, while the printed $t=2.882$ gives about $0.003993$; neither is $0.013672$. Thus the choice of which source field is trusted changes the apparent strength of evidence, including the 1% decision. A 95% confidence interval built conditionally from the printed estimate and standard error would be about $[0.0536,0.3649]$, but it is not a verified interval from a coherent fitted model. In fact, using the printed residual standard error $1.793$, the complete predictor sample size and predictor standard deviation gives a slope standard error near $0.006612$, not $0.079377$.

The printed coefficient of determination $R^2=0.2893$ would mean that the fitted line explains $28.93\%$ of the complete-case sample's centred device-count sum of squares relative to an intercept-only fit; $71.07\%$ remains unexplained by that comparison. It is not the percentage of devices caused by skill, the fraction of correctly predicted households, or a measure of future predictive accuracy. The adjusted coefficient of determination $0.289$ is consistent with that $R^2$ and the given sample size, allowing rounding. The small adjustment reflects fitting one slope among many cases; it does not validate the model.

For a one-predictor ordinary least squares fit with an intercept and the usual common-variance standard errors, the simple-regression output identities requireHere the printed slope t gives $F\approx 8.305924$, the printed F field says $83.2$, and the printed $R^2$ instead implies $F\approx 834.073$. These cannot all hold. Moreover, $F=83.2$ on $1$ and $2049$ statistical degrees of freedom has p-value about $1.72\times 10^{-19}$, not $0.001978$. Reading the printed F p-value alone would give a 1% significant test of the zero-slope model, but its reported calibration is incompatible with its statistic. No significance label resolves these identities.

There is also a contradiction that does not depend on tail probabilities. Ordinary least squares with an intercept makes the arithmetic mean of fitted values equal the arithmetic mean of observed responses on the analysed records. The printed coefficients and predictor mean give $0.392122+0.209240(13.546)\approx 3.22649$. Deleting two device counts anywhere in the printed range $[0,8]$, and allowing the reported overall mean to be rounded, would leave the complete-case device mean between approximately $4.49008$ and $4.49888$. The mean bounds after deleting bounded observations therefore rule out reconciling $3.22649$ with the response summary merely by citing those two missing predictor values. This is not the mistaken assumption that full-sample and complete-case sample means must always be identical.

If its own field were trusted, residual standard error $1.793$ estimates the unexplained response standard deviation, in device-count units. It is neither the standard error of the slope nor a guarantee that every prediction is accurate to $1.793$ devices. The residual five-number summary has extremes near $-5.206$ and $5.086$ and median near $-0.060$, with central quartiles near $-1.372$ and $1.252$. The roughly central median does not establish normally distributed errors, homoscedasticity, independence or a linear conditional expectation. A residual-versus-fitted plot could check curvature and spread; a normal probability plot could assess an error-shape approximation. Neither diagnostic can be recovered from five residual quantiles alone. Individual prediction intervals include response variation as well as coefficient uncertainty and are wider than confidence intervals for the conditional expectation.

For conventional finite-sample t and F calibration, a normal linear model assumes independent mean-zero normal errors of common variance with an appropriate predictor specification. Mean-zero errors conditional on skill and adequate sampling remain essential even when large-sample approximations relax normality. A bounded count response can have nonconstant conditional variance, so residual diagnostics and a scientifically suitable response model matter; count data do not become normally distributed just because the sample is large. Poisson regression is one possible alternative to investigate, but its own conditional mean–variance assumptions and the bounded device range require assessment, not automatic substitution. Complete-case selection can also bias the fitted association if it depends on the response in a relevant way. The only safe summary is conditional: the printed slope is positive, but the numerical effect, its standard error, t/F evidence and explanatory fit require a corrected, coherent output before a single definitive regression claim is made.

Conditional predictions for observed birds and years add their fitted random intercepts to the fixed part of the Gaussian linear mixed model:These are predicted expected values, so they need not be integers. The first is likely more reliable: age three is within the displayed age range, whereas age seven requires extrapolation beyond ages one to six. A quantitative comparison would need the prediction intervals, including uncertainty in both fixed coefficients and random effects; the displayed coefficients alone do not supply them.

Assume $n\ge 2$ and ${\sigma }^2>0$. Up to a constant, the log-likelihood of the normal sample isFor each $s$, the square is minimized at $\mu =\bar{X}$; optimizing the remaining one-variable likelihood gives the maximum-likelihood estimatorsThey are the global maximum when the residual sum is positive, which occurs with probability one. On the exceptional all-equal sample the likelihood increases without bound as $s\downarrow 0$.

To derive the distributions and independence, write $Z_i=(X_i-\mu )/\sigma$ and perform an orthogonal change of coordinates whose first row is $n^{-1/2}(1,\dots ,1)$. The transformed components $Z_1^{\prime },\dots ,Z_n^{\prime }$ remain independent standard normal random variables, since their joint density depends only on the squared Euclidean norm. ThenThe first expression uses only $Z_1^{\prime }$, the second only the other components. ConsequentlyThis also exhibits why the maximum-likelihood variance uses divisor $n$, even though its chi-squared degrees of freedom are $n-1$.

The independent future observation satisfies $X_0-\bar{X}\sim N(0,{\sigma }^2(1+1/n))$. It is independent of the residual sum of squares, because both $X_0$ and $\bar{X}$ are independent of the residual vector. Combining this normal variable with the independent chi-squared variable givesThe event in the proposed prediction interval is equivalent to $|T|<\gamma \sqrt{(n-1)/n}$. By symmetry and continuity of Student's t-distribution, exact joint coverage $1-\alpha$ therefore requiresThis is the normal prediction interval with maximum-likelihood variance. The factor $\sqrt{n/(n-1)}$ cannot be dropped: the interval uses $\hat{\sigma }$, not the unbiased residual standard deviation. Coverage here is over repeated joint sampling of the old and new observations, not a claim of parameter-free conditional coverage for every realized sample.

For a homogeneous continuous-time Markov chain, the holding time in a living state $r$ before the next transition has survival probability $e^{-{\lambda }_{r}t}$, where ${\lambda }_r={\sum }_{s\ne r}{q}_{rs}=-q_{rr}$. Indeed, conditional on still occupying that state, the probability of leaving in the next $dt$ is ${\lambda }_{r}dt+o(dt)$, so its survival function satisfies $S^{\prime }(t)=-{\lambda }_{r}S(t)$ and $S(0)=1$. Integrating gives the mean holding time from a transition intensity matrix:This is the mean time until any allowed exit from that state, not necessarily the patient's remaining lifetime.

For confidence limits, use the supplied standard error $s_r$ of the corresponding diagonal generator estimate. Changing its sign does not change its standard error, so it is also the standard error of ${\hat{\lambda }}_r$. The delta method, with derivative $d(1/\lambda )/d\lambda =-1/{\lambda }^2$, givesThe resulting approximate 95% Wald confidence intervals areThe diagonal standard errors already incorporate the covariance among outgoing rates. They should not be replaced by the square root of a sum of squared off-diagonal standard errors, which would silently assume those rate estimates independent.

Alternatively construct a positive rate confidence interval and apply the confidence interval for an inverse rate transformation. Here every normal rate interval has a positive lower endpoint, givingThese approximate intervals are, respectively, $(10.863,12.830)$, $(4.497,7.161)$, $(2.397,3.117)$ and $(1.548,2.631)$ years. They are asymmetric because inversion is nonlinear; positive log-rate or likelihood-based intervals could also be inverted. The symmetric delta intervals and the inverted-rate intervals are two legitimate first-order approaches and need not have numerically identical endpoints.

The death state has zero exit rate, so its holding time is infinite. There is no finite mean-sojourn estimate or ordinary finite confidence interval for this structurally absorbing state. All intervals above concern the mean parameter, not a prediction interval for one patient's realized holding time.

The association is strongly statistically significant under the reported regression assumptions, but age alone has only moderate predictive ability. The reported confidence interval for the age regression coefficient excludes zero by a wide margin. If it is the usual two-sided 95% interval from a normal linear model with an intercept and one predictor, there are $206-2=204$ degrees of freedom. Taking $t_{0.975,204}\simeq 1.97$ givesThus the two-sided p-value against zero slope is far below $0.001$. This conclusion concerns a statistical association in the sampled population; it does not establish an individual growth rate. Clustering by practitioner, nonconstant variance or nonindependent observations would require an appropriate standard error rather than blind reliance on this calculation.

The scatter plot has substantial vertical spread at each age. A narrow slope confidence interval describes uncertainty about the average trend, whereas a prediction interval for a new person's ear length must also include the residual variance. The reported interval even permits an approximate numerical assessment: write $S_{xx}={\sum }_i(x_i-\bar{x}{)}^2$ and let $s_e^2=\mathrm{RSS}/(n-2)$. In simple linear regression, $t^2={\hat{\beta }}^2{S}_{xx}/s_e^2$, while the fitted and residual sums of squares are ${\hat{\beta }}^2{S}_{xx}$ and $(n-2)s_e^2$. Hence the coefficient of determination isUnder these assumptions, age explains only about 27% of the sample variation; roughly 73% remains unexplained. The calculation is approximate because the printed slope and interval are rounded. For a new individual aged $x_0$, the usual prediction interval is based onThe leading one represents individual residual variation. The provided summaries do not determine $s_e$ and $S_{xx}$ separately, so they do not supply a numerical prediction interval.

Record sex, height or another measure of body size, ancestry, family characteristics and practitioner or measurement method. Use these covariates in a multiple linear regression, allowing scientifically justified nonlinear age effects or interactions, and assess prediction error on held-out observations. A guessed overall sex proportion cannot replace individual sex measurements or identify sex-adjusted effects.

There is a source inconsistency: the PDF's numerical ear-length summary is ten times the scale of its plotted measurements and fitted line. Indeed, the fitted line at the reported mean age gives about $67.7$ mm, whereas the printed mean is about ten times larger. Missing decimal points in the summary are a plausible explanation. The significance and approximate $R^2$ above use the mutually consistent slope, interval and graph; they do not treat the inconsistent summary as valid data.

library(mgcv) loads the fitting package; gam fits a Gaussian generalised additive model with the printed identity link. The smooth specification s(x, bs="cr") uses a penalized cubic regression spline. This is a finite-dimensional smooth basis, rather than a full interpolating spline with a free value at all $401$ observations. The fitted model isHere $f(x)={\sum }_{r=1}^q{\theta }_r{B}_r(x)$ in the centred cubic regression-spline basis, with its curvature penalized through $\lambda \int (f^{\prime \prime }{)}^2$. The intercept is estimated separately. In the displayed fit the smooth test space has rank $9$, consistent with nine smooth coefficients after the centring constraint, while penalization lowers its effective dimension.

The intercept estimate is $-0.13241$, with standard error $0.05426$, $t=-2.44$ and $p=0.0151$. Because the smooth is centred over the observations, the intercept is the overall fitted average, not necessarily the fitted response at $x=0$. Its test compares this average with zero. The significance-code legend maps progressively smaller $p$-values to stars; stars are an annotation, not an effect-size measure.

The smooth has $8.237$ effective degrees of freedom, displaying substantial nonlinearity: a centred purely linear term would use one. Est.rank=9 reports the rank used for its approximate significance calculation. The $p$-value below $2\times 10^{-16}$ is very strong evidence that the mean depends on $x$, against a zero smooth term. It does not mean the fitted curve is the exact generating function, or that lack of fit has been tested. The total effective dimension including the intercept is $1+8.237=9.237$.

The adjusted coefficient of determination $R^2$ of $0.983$ and deviance explained of $98.4\%$ indicate that nearly all variation about the response mean is explained. For this Gaussian model, deviance explained is $1-\mathrm{RSS}/\mathrm{SST}$. The scale estimate $0.96737$ estimates residual variance, corresponding to a residual standard deviation about $0.984$. The GCV score $0.99018$ is the prediction-oriented criterion used for smoothing selection, not a probability. It is consistent with the reported scale and effective dimension: $0.96737\times 401/(401-9.237)\simeq 0.99018$.

In the PDF's plot, the solid curve is the estimated centred smooth $f(x)$, and the dashed curves show pointwise uncertainty around that smooth, rather than prediction intervals for new noisy responses. The label s(x,8.24) gives its rounded effective degrees of freedom. The rug marks the observed covariate locations. Reading the endpoints givesThe smooth is increasing with an almost flat middle and one inflection, resembling a cubic. A simpler model to investigate is ordinary Gaussian polynomial regressionIt uses four mean coefficients rather than about nine effective ones. The apparent near-odd shape also makes ${\beta }_0+{\beta }_3{x}^3$ a plausible further reduction, but the linear and quadratic terms should be checked rather than discarded solely from the picture. The figure suggests this alternative; it does not supply coefficients or establish the true simulation model.

Add the new individual's latent mean and log variance to the same population hierarchy. Enter only that individual's first two measurements as observations; use the historical data and these two values to fit the model. The third measurement must be held out, or the predictive check reuses the value it is meant to test. For example, augment the unchanged population hyperpriors with this WinBUGS structure:

new.mean ~ dnorm(mu, invtau2)
new.logvar ~ dnorm(phi, invpsi2)
new.precision <- exp(-new.logvar)
for (k in 1:2) {
    new.Y[k] ~ dnorm(new.mean, new.precision)
}
future.Y ~ dnorm(new.mean, new.precision)
upper.indicator <- step(future.Y - held.out.third)

future.Y is unobserved. Its simulated values give the posterior predictive distribution, including uncertainty in the population parameters and personal variance. A central 99.9% prediction interval uses its $0.0005$ and $0.9995$ empirical quantiles. Flag the third value if it is outside that interval. Equivalently, estimate its predictive tail probability by the mean of upper.indicator and flag if that mean is below $0.0005$ or above $0.9995$. This is two-sided, unlike the earlier one-sided upper limit.

Put $M=X^{T}X$, which is positive definite because $X$ has full column rank. The log likelihood differs from $-\Vert Y-X\beta {\Vert }^2/(2{\sigma }^2)$ by a term independent of $\beta$. Differentiation gives the normal equations $M\hat{\beta }=X^{T}Y$, henceThe distribution follows by writing $\hat{\beta }=\beta +M^{-1}{X}^T\epsilon$ and applying the linear transformation rule for a multivariate normal distribution.

Let $s={\sigma }^2$ and $R=\Vert Y-X\hat{\beta }{\Vert }^2$. Orthogonality of the residual to the columns of $X$ gives the exact decompositionThe prior is density proportional to $s^{-1}$ with respect to $d\beta ds$. Thus the joint posterior distribution isCompleting the quadratic identifiesThe sampling distribution treats the true $\beta$ as fixed and varies the data, whereas the conditional posterior fixes the observed data and distributes $\beta$. Their covariance formulas agree, but their centres and their interpretations differ.

For a new row vector $x^{\ast }$, let $h^{\ast }=x^{\ast }{M}^{-1}(x^{\ast }{)}^T$ and suppose its new error is independent. With known ${\sigma }^2$, the maximum-likelihood plug-in prediction uses mean $x^{\ast }\hat{\beta }$ and noise variance ${\sigma }^2$, while a calibrated frequentist prediction must account for the estimator: $y^{\ast }-x^{\ast }\hat{\beta }\sim N(0,{\sigma }^2(1+h^{\ast }))$. If variance is estimated, the maximum-likelihood estimator is $R/n$, whereas the exact frequentist prediction pivot uses $\hat{s}=R/(n-p)$ and has Student t-distribution with $n-p$ degrees of freedom.

In the Bayesian calculation, integrating $\beta$ first gives $y^{\ast }\mid s,X,Y\sim N(x^{\ast }\hat{\beta },s(1+h^{\ast }))$. Integrating $\beta$ out of the joint posterior gives $s\mid X,Y\sim \mathrm{InvGamma}((n-p)/2,R/2)$. For $\nu =n-p$, integration of the predictive normal against this inverse-gamma distribution has density proportional toThe substitution $t=[R+(y^{\ast }-x^{\ast }\hat{\beta }{)}^2/(1+h^{\ast })]/(2s)$ reduces the integral to a Gamma function. Normalizing identifies the Student t-distribution, givingwhere the second parameter is a scale, not its variance. These Bayesian intervals coincide with the usual exact Gaussian-regression prediction intervals for this prior, although their probability interpretations differ. Posterior propriety requires $R>0$ and $n>p$; the former holds almost surely under the nondegenerate Gaussian model. Prediction of the noiseless conditional mean omits the added one in $1+h^{\ast }$.

Integrate each supplied predictive density to get its cumulative distribution function $F_t$, then form the sequential probability integral transformFor a correct continuous one-step conditional predictive model, $\mathbb{P}(U_t\le u\mid y_1,\dots ,y_{t-1})=u$. Iterating this identity shows that the $U_t$ are independent uniform variables. A histogram or quantile plot checks uniformity; serial plots and autocorrelations check for temporal structure left unexplained by the forecasts.

Also inspect empirical coverage of central prediction intervals, tail exceedances and interval widths, assessing calibration together with sharpness. These checks need only the supplied forecasts and observations. Compare chosen predictive discrepancy statistics with simulated uniform reference sequences. A total log score alone is a relative reward and does not provide a universal absolute goodness-of-fit threshold.

For a new independent tumour grown for ten days in B, use $x_0=(1,10,10,0{)}^T$. Its fitted mean isWrite $h_0=x_0^T(X^{T}X{)}^{-1}{x}_0$. The prediction interval in a normal linear model iswhere $s^2=\mathrm{RSS}/26$ and $t_{26,1-\alpha /2}$ is the indicated Student's t-distribution quantile. The design information needed to evaluate $h_0$ is not supplied, so a fully numerical interval cannot be recovered from the coefficient standard errors alone: the coefficient covariances are also needed.

To prove coverage, write $Y_0=x_0^T\beta +{\epsilon }_0$, with ${\epsilon }_0\sim N(0,{\sigma }^2)$ independent of the original sample. The error $Y_0-x_0^T\hat{\beta }$ has a normal distribution with mean zero and variance ${\sigma }^2(1+h_0)$. By the joint distribution of least-squares and variance estimators, $26s^2/{\sigma }^2\sim {\chi }_{26}^2$ and is independent of $\hat{\beta }$; it is also independent of ${\epsilon }_0$. Hence the pivotal quantityConsequently $\mathrm{Pr}(-t_{26,1-\alpha /2}\le T\le t_{26,1-\alpha /2})=1-\alpha$, which is exactly the claimed prediction interval coverage. This is repeated-sampling coverage conditional on the design, assuming the model; it need not hold conditional on the particular observed training responses. The extra $1$ represents new-observation noise and must not be omitted as in a mean confidence interval.

Write $z=(Z_{s_1},\dots ,Z_{s_n}{)}^T$, let ${\Sigma }_{ij}=C(\Vert s_i-s_j\Vert )$, let $c_i=C(\Vert s_i-s_0\Vert )$, and put $v_0=C(0)$. Assume the covariance matrix $\Sigma$ is invertible. Because the mean is known to be zero, every linear predictor $w^{T}z$ is unbiased. Its mean squared prediction error isCompleting the square yieldsThe positive-definite matrix property makes the last term nonnegative, proving the simple kriging formulasThe residual is uncorrelated with $z$. Since the joint law is multivariate normal, it is independent of $z$, so the same predictor and variance describe the conditional distribution. Consequently a 95-percent prediction interval isThere is no constraint that the weights sum to one: that is required for an unknown constant mean in ordinary kriging, not for the stated known-zero-mean problem. If $\Sigma$ is singular, use a Moore-Penrose inverse with the corresponding covariance compatibility condition. At an already observed location prediction of that very same field value has zero error; predicting an independent noisy replicate is a different target. Estimated covariance parameters make this a plug-in interval and generally add uncertainty that the formula does not include.

A $(1-\alpha )$ prediction interval is a pair of statistics $L(Y),U(Y)$ such that, for every $\beta$ and ${\sigma }^2>0$, an independent future response $Y^{\ast }$ at the specified input satisfies

In the normal linear model, assume $X$ has full column rank and writeFor a new observationindependent of the training sample, the prediction error isThe two terms on the right are independent centered normal random variables, whileConsequently, with the new-point regression leveragewe have

The assumption in the question givesThis variable is independent of $\hat{\beta }$; it is also independent of the new error ${\epsilon }^{\ast }$. Therefore it is independent of the standardized prediction error, and the defining normal-over-chi-square representation of Student's t-distribution givesIf $t_{n-p,1-\alpha /2}$ is the corresponding upper quantile, symmetry yieldsSolving these inequalities for $Y^{\ast }$ proves that the prediction interval in a normal linear model isIts center and endpoints depend only on the observed data and $x^{\ast }$, and the pivotal calculation proves the required coverage uniformly in the unknown $\beta$ and ${\sigma }^2$.

Because model2 uses a logarithmic transformation, the logarithm of the payment ratio is the difference of two predicted log payments. The cars have equal Brand and Bonus values and differ by one Kilometres level, so its fitted mean is the displayed mileage regression coefficient, ${\hat{\beta }}_K=0.057132$.

For two independent future payments, the prediction interval must include both the standard error $0.032654$ of ${\hat{\beta }}_K$ and two independent residual errors. The estimated variance of the predicted log ratio is thereforeUsing the residual degrees of freedom shown in the output, a 95% prediction interval for a ratio of log-normal responses isWith $t_{284,0.975}\simeq 1.97$, this is approximately $[0.120,9.33]$. The interval is wide because it predicts the ratio of two individual future payments; an interval for the ratio of their fitted mean responses would omit the $2{\hat{\sigma }}^2$ term.

A return level plot graphs $T$ horizontally and $z_T$ vertically, often using a logarithmic scale for $T$. To read it, choose a return period, move up to the curve, and read the corresponding return level. Conversely, a threshold's horizontal coordinate gives the reciprocal of its annual exceedance probability. The usual plot of fitted quantiles against return periods, and optional pointwise confidence intervals, are documented in Mathieu Ribatet, POT package: return level plot.

The near-straight shape has a mathematical explanation: since $-\log (1-1/T)\sim 1/T$ as $T\to \infty$, the Gumbel distribution gives $z_T=\mu +b\log T+o(1)$. Multiplying a large return period by ten therefore increases the return level by approximately $b\log 10$, about $4.605$ units for this toy curve. A curve indexed by rarity puts several thresholds on one common scale while retaining their measurement units.

For a fitted curve, a pointwise confidence interval concerns uncertainty in the estimated quantile at a specified $T$; it is distinct from a prediction interval for a future annual sample maximum, and pointwise intervals need not form a simultaneous confidence band. A far-right curve value can also be an extrapolation beyond the observation record: extending the line adds no observations to support the chosen tail model. The diagram above uses fixed toy parameters, so it shows neither estimation uncertainty nor an observational record. Its interpretation throughout assumes the stated annual block and stationarity.

Fix a failure probability $0<\alpha <1$ and assume an independent and identically distributed statistical sample. A finite-sample confidence band for its population cumulative distribution function isHere $\log$ is the natural logarithm. The DKW inequality impliesThis is simultaneous coverage over the whole real line, valid at each fixed sample size.^[ref]

To obtain the band, solve $2e^{-2n{\epsilon }^2}=\alpha$ for the positive $\epsilon$. Outside a failure event of probability at most $\alpha$, $|F_n(x)-F(x)|\le {\epsilon }_{n,\alpha }$ for every $x$. Since a cumulative distribution function already lies in $[0,1]$, clipping the endpoints to this interval preserves the coverage event. The guarantee is at least $1-\alpha$, and need not equal it.

Choose $\alpha =0.1$, giving at least $90\%$ simultaneous coverage under the stated model. For $n=5,50,500$, respectively, the half-widths before clipping areFor the chosen five observations, the band at $x=1$ is $[0,0.94733]$, while at $x=4$ it is approximately $[0.25267,1]$. At $x=3$ the unclipped band would be $[0.05267,1.14733]$, so clipping gives $[0.05267,1]$. The broad uncertainty is informative: five observations give a precise description of themselves, but a coarse bound for an unknown population cumulative distribution function.

A collection of $90\%$ pointwise confidence intervals only promises coverage separately at each preselected threshold; it does not promise $90\%$ coverage of an entire cumulative distribution function. Here the one simultaneous event permits inspecting any threshold after seeing the data. The probability refers to repeated sampling of the entire statistical sample, with $F$ fixed. It does not assign $90\%$ probability to a fixed $F$ after observing this particular band, and it is not a prediction interval for a future observation. The displayed $n=5,50,500$ comparison gives a guarantee for each fixed $n$ separately, not simultaneous validity over all sample sizes or an optional stopping rule. The chosen observations are an illustration; the coverage claim follows from the DKW inequality, not from their appearance or a simulation. Dependence or observations with different sampling laws require another argument.

At fixed $\alpha$, multiplying $n$ by $100$ divides the half-width by $10$. Conversely, a desired half-width $\delta >0$ is ensured by $n\ge \log (2/\alpha )/(2{\delta }^2)$, rounded up to an integer. For $\alpha =0.1$ and $\delta =0.1$, this sufficient sample size is $150$. This makes the confidence band useful for planning how many observations are needed before estimating threshold probabilities to a chosen simultaneous accuracy.