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Here is an original reproducible finite set with 40 points and 512 coordinates. Let $e_1,\dots ,e_4$ be the first four vectors of the standard basis of ${\mathbb{R}}^{512}$. With point indices $i=0,\dots ,39$, generateThe four groups have different centers, and the standard normal random variables perturb all 512 coordinates. Generate the coordinates in row order using NumPy's PCG64 pseudorandom number generator with seed 20261013. Then keep this finite set fixed. A second PCG64 stream with seed 424242 generates a $32\times 512$ matrix of independent random variables with the standard normal distribution; divide it by $\sqrt{32}$ to obtain the Gaussian Johnson–Lindenstrauss transform.

For all 780 unordered pairs, compute the distance ratio $r_{ij}=\Vert T{x}_i-T{x}_j{\Vert }_2/\Vert x_i-x_j{\Vert }_2$. In this draw, its minimum is approximately $0.733010$ and its maximum $1.478785$, giving a maximum relative error in Euclidean distance of about $0.478785$. The squared-distance ratios $r_{ij}^2$ instead range from approximately $0.537304$ to $2.186806$: their maximum relative error is about $1.186806$. Thus this particular draw fails a 20% distance tolerance. The Johnson–Lindenstrauss lemma does not assert that every draw into 32 dimensions must succeed.

The two bands of original Euclidean distances distinguish pairs within a group from pairs between groups. Their spread after the linear map shows why checking an average distance is insufficient: the Johnson–Lindenstrauss lemma controls every pair. The dashed 20% distance band is an illustrative tolerance, not a confidence interval or a squared-distance band. The affine hull of 40 points has dimension at most 39; 512 is the ambient coordinate count, and no full-dimensionality assumption is required. For this draw the bi-Lipschitz distortion is approximately $1.478785/0.733010=2.017414$.

Now keep the same points and draw 48 fresh Gaussian Johnson–Lindenstrauss transforms for each target dimension $k\in \{8,16,32,64,128\}$. NumPy's SeedSequence(314159).spawn(240) supplies distinct PCG64 streams, ordered first by target dimension, then by draw. Each stream generates a new $k\times 512$ random matrix, normalized by $1/\sqrt{k}$; no map is a row prefix of another map. Each plotted value is

The observed median maximum relative errors in Euclidean distance are approximately $0.735$, $0.520$, $0.381$, $0.267$, and $0.184$, respectively. All pairs fall inside the 20% distance tolerance in $0$, $0$, $0$, $0$, and $35$ of the 48 draws, respectively. These are computations on one synthetic finite set, rather than optimal dimension bounds or measured physical data. The displayed quantiles describe the sampled maps; they do not certify a success probability for other sets. Even at 128 dimensions, 13 of the sampled maps fail this tolerance.

The practical benefit is that each point acquires fewer coordinates while its pairwise Euclidean distances can remain useful. The choice of target dimension balances compression, accuracy, and failure probability. For example, a conservative sufficient squared-distance tolerance $\epsilon =0.2$ with $n=40$ and $\delta =0.05$ gives $k\ge 1195$ by the explicit bound above. Here that exceeds the ambient dimension, so the identity map is better for a certified guarantee. The much smaller dimensions in the plots illustrate observed behavior without claiming that this sufficient bound is sharp.

In a parametric statistical model, the likelihood function compares parameter values, while sampling distributions calibrate tests and confidence intervals. A sufficient statistic retains the parameter information. Conditioning on an observed ancillary statistic restricts that calibration to experiments with the same ancillary configuration, instead of averaging over configurations whose distribution is unrelated to the parameter. Since the ancillary's marginal law is parameter-free, conditional and unconditional likelihoods have the same parameter-dependent factor. With a nuisance parameter, a conditional law or pivotal quantity must remove it before inference for the parameter of interest can use a nuisance-free reference law.

A forest plot shows each study's effect estimate and confidence interval, together with a pooled estimate when appropriate. A vertical null-effect line allows comparison on a common scale. For an odds ratio the null is one and a logarithmic axis treats reciprocal ratios symmetrically; marker areas commonly encode study weights.

Take the original synthetic model $T\sim N(0,0.6)$ and ${\epsilon }_1,{\epsilon }_2\sim N(0,0.4)$, mutually independent random variables, with the second argument denoting variance. The pair $(X,Y)$ has a multivariate normal distribution with mean zero, unit marginal variances, and Pearson correlation coefficient $0.6$. Its conditional multivariate normal distribution isFor a first reading $x=2$, the second reading has conditional expectation $1.2$ and conditional variance $0.64$, hence standard deviation $0.8$. There is still conditional probability $1-\Phi (1)\approx 0.1587$ of $Y>2$, a movement farther from zero, where $\Phi$ is the standard normal distribution function. The line $0.6x$ describes a conditional mean, not an individual trajectory.

A short derivation isolates $R=Y-0.6X$. Its covariance with $X$ is zero and its variance is $0.64$. Because $(X,R)$ has a multivariate normal distribution, zero covariance implies independence: $R\sim N(0,0.64)$ independently of $X$. This also explains why reversing the two measurements gives $\mathbb{E}[X\mid Y=y]=0.6y$. The statistical association does not identify a direction of causal effect.

For the Kaplan–Meier estimator, the estimated variance isThe formula follows by summing approximate variances of the logarithms of the conditional survival factors and applying the delta method to their product. Its square root is the estimated standard error. The same expression uses the actual delayed-entry risk sets for the Kaplan–Meier estimator with delayed entry. When $0<\hat{S}(t)<1$ and every contributing $r_j>d_j$, an approximate log-scale confidence interval is $\exp \{\log \hat{S}(t)\pm z_{1-\alpha /2}\widehat{\mathrm{se}}(\hat{S}(t))/\hat{S}(t)\}$, with the upper endpoint capped at one. This interval can be inaccurate with very small risk sets.

With $m$ groups of $q$ trials and common marginal failure probability $p$, each count $X_j$ has expectation $qp$, so the displayed estimate is unbiased without assuming independence within groups. Independent individual trials give variance $p(1-p)/(mq)$. If only groups are independent, estimate the variance instead by $s_X^2/(m{q}^2)$, retaining within-group dependence. Equal marginal probabilities alone do not justify a binomial confidence interval.

For a fixed positive sample size $n$ of independent and identically distributed Bernoulli random variables, letThe Hoeffding inequality makes $C_n$ a conservative confidence interval for $p$: ${\mathbb{P}}_p(p\in C_n)\ge 1-\delta$ for every $p\in [0,1]$. Indeed, exclusion from the closed interval implies $|{\hat{p}}_n-p|>r_n$, an event contained in the deviation event bounded by $\delta$. Clipping to $[0,1]$ retains coverage because $p$ already lies there. For $n=738$, $\delta =0.05$, and an observed frequency $369/738=1/2$, $r_n\approx 0.0499924$, giving approximately $[0.450008,0.549992]$.

Coverage concerns repeated independent samples: the unknown $p$ is fixed and the confidence interval is random. This fixed-$n$ guarantee does not justify repeatedly checking the same bound and stopping as soon as it looks favorable. The Hoeffding sample size for Bernoulli frequencies chooses $n$ in advance; guarantees across repeated inspection times require separate control of the additional events, for example through a union bound with an allocated failure probability at each time.

At a specified willingness-to-pay threshold $\lambda$, the incremental net monetary benefit is the value of the effect increment minus the cost increment. It avoids dividing by a potentially near-zero or negative effect estimate. For estimates with variances $v_E,v_C$ and covariance $v_{CE}$, its estimated variance is ${\lambda }^2{v}_E+v_C-2\lambda v_{CE}$. A normal approximation therefore gives a confidence interval directly on the net-benefit scale. The economic sign criterion does not require a positive effect increment.

The median survival estimate is the first event time at which the Kaplan–Meier estimator reaches or falls below one half. It is not identified beyond the observed follow-up if the curve never reaches that level. A few reported survival probabilities bound the crossing time but do not determine an exact median. Confidence intervals can be constructed by inverting confidence bands for the survivor function.

For two independent binomial distributions with event and nonevent counts $a,b$ and $c,d$, the odds ratio is $ad/(bc)$. The delta method applied to $\log (p/(1-p))$, whose derivative is $1/[p(1-p)]$, gives a group variance $1/[Np(1-p)]$. Substitution of the fitted proportion gives $1/a+1/b$ in the first group and $1/c+1/d$ in the second. Their independence gives the displayed sum. A normal approximation then gives a confidence interval on the log odds ratio scale; exponentiation produces an odds-ratio interval. Zero cells require another approximation or model.

For the same normal distribution model, a two-sided test compares the null hypothesis $H_0:\mu ={\mu }_0$ with the alternative hypothesis $H_1:\mu \ne {\mu }_0$. Under the null hypothesis, the standardized test statistic has standard normal distribution. Write $\Phi$ for the standard normal distribution function. With $\bar{x}=1.2$, $\sigma =2$, $n=16$ and ${\mu }_0=0$, the observed test statistic is $z=1.2/0.5=2.4$. Its p-value isThus the p-value is a tail probability computed under the null hypothesis, including departures in both directions. It is not $\mathbb{P}(H_0\mid x)$, the posterior probability of the null hypothesis, nor the probability that the observations arose "by chance". The normal distribution test statistic and its calibration appear in NIST's teaching reference for testing a process mean; the tail interpretation is stated in Penn State's hypothesis-testing notes.

Choose the significance level $\alpha =0.05$ before observing the sample. The rejection region is $|Z|>z_{0.975}$, whose probability under $H_0$ is exactly 0.05. Here $2.4>1.9599639845\dots$, so reject $H_0$. This agrees with zero lying outside $[0.22,2.18]$. More generally, a proposed ${\mu }_0$ belongs to the unrounded 95% confidence interval precisely when its two-sided test does not reject under this strict cutoff. The equality boundary has probability zero in this continuous statistical model.

The significance level and statistical power describe different repeated-experiment probabilities. At a specified alternative $\mu$, put $\delta =(\mu -{\mu }_0)/(\sigma /\sqrt{n})$ and $q=z_{1-\alpha /2}$. ThenThe first is the Type I error probability; the second is the Type II error probability at that alternative. For a planning alternative $\mu =1$, keeping ${\mu }_0=0$, $\sigma =2$ and $n=16$ gives $\delta =2$, $\beta (1)\approx 0.483995$ and statistical power $0.516005$. At $n=64$, $\delta =4$ and statistical power rises to $0.979327$. These are model calculations for specified alternatives, not posterior probabilities inferred from a nonsignificant result. They explain why failure to reject cannot establish equality and why choosing sample size before an experiment is useful.

Historically, Ronald Fisher's 1925 book discusses the normal cutoff $1.96$ and tail probability 0.05 as a convenient convention on page 47, rather than a mathematically compulsory threshold: ^[ref].

Suppose $X_1,\dots ,X_n$ are independent and identically distributed with normal distribution $N(\mu ,{\sigma }^2)$, where the mean $\mu$ is unknown and the standard deviation $\sigma >0$ is known. The sample mean $\bar{X}=n^{-1}{\sum }_i{X}_i$ has normal distribution $N(\mu ,{\sigma }^2/n)$ and standard error $\sigma /\sqrt{n}$. With $z_q$ the standard normal quantile at $q$, the displayed confidence interval has exact coverage probability $1-\alpha$ for every $\mu$ and every positive integer $n$. No central limit theorem approximation is needed.

Take $\sigma =2$, $n=16$ and nominal confidence 95%. The standard error is $2/\sqrt{16}=0.5$, and $z_{0.975}=1.9599639845\dots$. ThereforeRounded for reporting, this is $\bar{x}\pm 0.98$. If $\bar{x}=1.2$, the realized confidence interval is approximately $[0.22,2.18]$. Exact 95% calibration refers to the unrounded standard normal quantile; using $1.96$ literally gives coverage probability $0.9500042097\dots$, almost but not exactly 0.95.

The confidence interval's width scales as $n^{-1/2}$: increasing sample size from 16 to 64 halves its half-width to $0.4899909961\dots$. If $\sigma$ is unknown, merely substituting the square root of a sample variance loses the displayed exact calibration; for the same normal distribution model, use a Student t confidence interval.

The figure holds the true mean at $\mu =0$ and independently generates 40 samples of 16 observations from the same normal distribution. NumPy's PCG64 generator with seed 42 gives 39 covering confidence intervals and one miss, at sample 32: observed coverage is $39/40=97.5\%$. Every bar uses the unrounded standard normal quantile. This finite collection illustrates the moving endpoints; it does not establish the analytic coverage probability or require exactly two misses.

The beauty of the construction is that a familiar standard normal distribution calibrates every possible unknown mean at once. For a short proof, the pivotal quantityhas a probability distribution independent of $\mu$. The event $|Z|\le z_{1-\alpha /2}$ has probability $1-\alpha$; rearranging it gives $\mu \in C(X)$. The right panel shows the equivalent event $|\bar{X}|\le 0.9799819923\dots$ when $\mu =0$.

Ordinary kriging assumes $\mathbb{E}[Z(s)]=\mu$ for an unknown constant $\mu$. A predictor $w^{T}Z$ has zero bias for every $\mu$ precisely when $1^{T}w=1$, where $1$ is the all-ones vector. Ordinary kriging minimizes the error variance $Q(w)$ subject to this constraint, giving the best linear unbiased predictor. Its explicit system isThe sign of the last term follows from the displayed convention $\Sigma w+\eta 1=c$. To derive the system, differentiate $Q(w)+2\eta (1^{T}w-1)$. To prove the minimum, let $w_{\ast }$ solve it and let $w=w_{\ast }+d$ be another feasible choice. Then $1^{T}d=0$, soA positive-definite matrix $\Sigma$ makes equality possible only for $d=0$. This argument uses second moments, not a normal distribution assumption. Ordinary kriging is the special case of universal kriging with $X=1$ and $x_0=1$.

For an original one-dimensional teaching example, observe $z=(1,3,2{)}^T$ at $s=(0,1,3)$, and predict at $s_0=2$. Choose unit marginal variance, exact observations without measurement noise, an unknown constant mean, and the exponential covariance functionThe exponential covariance function corresponds to a stationary Ornstein-Uhlenbeck process, so it gives a valid covariance model. Its semivariogram is $1-2^{-|h|}$. The resulting matrix and vector areSolving the displayed system givesThese values can be checked directly: the weights sum to one and $\Sigma w=(41/116,35/58,35/58{)}^T=c+(3/29)1$. Substitution into $Q(w)$ gives $18/29$. Nearest-neighbor interpolation must choose between the two equally close samples at one and three, returning either three or two; their error variances under this model are both $1+1-2(1/2)=1$. Thus the benefit is quantified under the chosen model, without knowing the unobserved target value.

The upper panel applies the same ordinary kriging calculation at each target location. The lower panel shows the square root of its error variance, in the units of the observations. It is zero at the exact observation locations and rises between and beyond them. No confidence interval is inferred from a variance alone; a probability distribution assumption is needed to attach a coverage level.

The weights are not required to be positive. Keep the same observations and target, but replace the covariogram by the smoother Gaussian kernel $C(h)=2^{-h^2}$. ThenThe negative coefficient compensates for dependence between samples; the constraint eliminating bias still holds. Kriging is not generally a convex combination, and its prediction need not stay between the observed values. A smaller model-based error variance does not establish that one covariance function is better supported by these three observations.

The survival estimates favor the new treatment at the two reported times, with absolute survival differences $0.199$ at two years and $0.157$ at five years. These point differences alone do not establish efficacy. Randomization supports a causal comparison of the treatment assignments, but sampling variance, censoring, follow-up, protocol adherence and the prespecified endpoint still matter. One should not infer a statistically established benefit merely from two larger estimates.

A Kaplan–Meier estimator iswhere $r_j$ is the risk set and $d_j$ the event count. Its uncertainty depends on the actual risk sets, not just the initial group size. The Greenwood formula estimatesWithout the missing survival tables one cannot calculate these standard errors or the uncertainty of the group difference. In particular, replacing the estimator variance by the variance of an uncensored binomial proportion would generally be wrong.

I would advise presenting both curves with numbers at risk and confidence intervals, checking independent censoring, and reporting a prespecified overall comparison plus an effect estimate and interval. For an unadjusted log-rank test, write $O_1={\sum }_j{d}_{1j}$ and $E_1={\sum }_j{d}_j{r}_{1j}/r_j$. Its hypergeometric null variance isTerms with no possible variation contribute zero. This compares the entire event history rather than selecting two arbitrary times, uses the available risk sets, and avoids choosing favorable time points after examining the curves. It is especially effective under proportional hazards. Crossing curves can make a log-rank comparison insensitive; a prespecified restricted mean survival time difference or another suitable time-varying analysis may then be more informative. An overall comparison does not make a clinically important prespecified time-point contrast invalid.

The Kaplan–Meier median survival time is the first event time at which $\hat{S}(t)\le 1/2$. Monotonicity of the reported estimates impliesIf the new-treatment curve never reaches one half during observed follow-up, its median is not estimable from that curve. The exact crossing times cannot be reconstructed from two survival probabilities: the standard curve could first cross at three or at four years and still have the same values at two and five years. The point estimates suggest benefit, but the evidence and exact medians require the missing event/censoring tables. Neither a conclusive agreement nor a conclusive rejection of benefit is justified from these numbers alone.

These are separate data, so results from the randomized trial must not be carried over. First inspect the construction of the survival response: a command creating a survival time object must pair follow-up in months with an event indicator, whose death/censoring coding should be checked explicitly. A censoring time is not an observed death time. Stratified product-limit fits would then summarize Kaplan–Meier estimators for the dose groups before adjusted regression.

If the fitted command is a Cox proportional-hazards model with dose, sex and age, its specification isThere is no separate intercept estimate because an intercept is absorbed into the unspecified baseline hazard $h_0$. With the stated coding, $e^{{\beta }_G}$ compares high-dose with low-dose death hazards at equal sex and age; $e^{{\beta }_F}$ compares female with male hazards at equal dose and age; $e^{{\beta }_A}$ is the hazard multiplier for a one-year age increase. A negative treatment coefficient would correspond to a lower fitted hazard under high dose. A hazard ratio is not a ratio of survival probabilities, nor directly a ratio of survival times. Under this model the survivor function is $S(t\mid x)=S_0(t{)}^{\exp (x^T\beta )}$.

The coefficient standard errors, Wald ratios and exponentiated confidence intervals should be read together. The common large-sample interval for a coefficient's hazard ratio isGlobal likelihood-ratio, score and Wald tests assess the included slopes jointly; for exactly these three one-coefficient terms their reference degrees of freedom are three. They need not coincide in a small sample. The Cox partial likelihood uses the event risk sets; the fitted log-likelihood is not a full likelihood for an estimated baseline survival curve. The method for tied event times should be recorded.

Check the proportional hazards assumption using scaled Schoenfeld residuals and time trends; check the functional form of age, influential subjects and independent censoring. Dose-by-age or dose-by-sex interaction terms are scientifically possible but require enough information and a stated model comparison. Adjustment can alter a crude dose association, and the allocation mechanism here is not specified as randomized. No coefficients, confidence limits, significance results, exact fitted calls or diagnostics for this dataset are recoverable without the output sheet. The displayed model is an interpretation template, not a claim that it was the exact missing command.

The posterior mean is a weighted average of the sample proportion and prior mean:Thus it is pulled from 10% toward the prior's 5%, giving about 7.17%. This is an affine shrinkage estimator for a binomial proportion. The prior carries substantial concentration relative to the 90 observations, so the posterior is also more precise under this model: its standard deviation is about $0.01785$, compared with the binomial plug-in standard error $\sqrt{0.1(0.9)/90}=0.03162$. The posterior 95% credible interval $[0.04076,0.11035]$ is narrower and shifted down compared with the simple Wald interval $[0.03802,0.16198]$ based on $9/90$ alone.

The point estimate shrinks toward the historical mean, and the interval is narrower under the informative prior. A credible interval and a frequentist confidence interval have different probability interpretations; the comparison does not make the prior's relevance automatic.