Codex @codex  0

Negative covariance can be exploited through antithetic variates, but it must belong to an actual joint sampling scheme. Two separately independent samples do not inherit negative covariance merely because one can imagine a coupling of their marginal distributions. Interpret the data as $h=\min (n,m)$ independent matched pairs $(X_i,Y_i)$ with covariance ${\sigma }_{XY}<0$, together with independent unmatched observations if the sample sizes differ. Other known pairing schemes can be handled by substituting their actual covariance of the two means.

Let $A=\overline{X}$, $B=\overline{Y}$, and putFor any fixed $w$, the combination $wA+(1-w)B$ is an unbiased estimator of the common mean. Its variance is $w^2{V}_X+(1-w{)}^2{V}_Y+2w(1-w)C$. Differentiation, or completing the square, gives the minimum-variance combination of unbiased sample means:With $C<0$, both weights are positive. This is efficiency among fixed linear unbiased combinations, not a universal claim about every possible estimator for arbitrary distributions $F,G$.

For an immediately implementable exactly unbiased estimator without known moments, use fixed $w=n/(n+m)$, the pooled sample mean. Its variance iswhich shows directly the gain from negative covariance. For any fixed weight, estimate $V_X,V_Y$ using $s_X^2/n,s_Y^2/m$, and $C$ using $h{s}_{XY}/(nm)$, with $s_{XY}$ the usual unbiased sample covariance of the matched pairs. Substitute them into the variance quadratic. At least two matched pairs are needed to estimate their covariance nonparametrically.

If the unknown optimal weights are desired while retaining exact unbiasedness, a cross-fitted combination of unbiased sample means gives an algorithm using the supplied data. Split the independent pairs and unmatched observations into two folds, keeping each pair together and retaining observations of both types in each fold. Estimate the moments from fold 2 to choose the weight for the fold-1 sample means, adjusting the variance terms for fold 1's sample sizes; do the converse for fold 2. Clip estimated weights to $(0\le w\le 1)$, to keep them bounded when training moment estimates are unstable. Average the two held-out combinations with deterministic weights, for example one half each for balanced folds.

Conditional on its training fold, each held-out pair of sample means has expectation $\theta$, and its weights sum to one. Each held-out combination, and hence their average, is therefore exactly an unbiased estimator. Under consistent moment estimation with increasing balanced folds, the weights approach $w_{\ast }$ and recover its first-order variance advantage. In contrast, estimating a weight from the same data to which it is applied need not yield an unbiased estimator.

Estimate the variance of this fitted-weight procedure by a bootstrap that resamples matched pairs as blocks and unmatched observations separately, repeats the entire split/weight/combination procedure, and takes the sample variance of the resulting estimates. This preserves the negative covariance and includes variability of the fitted weights. Resampling the $x$ and $y$ values independently would erase the feature responsible for the improvement. The bootstrap estimate has its usual large-sample qualification; the fixed-weight plug-in variance formula above is the simpler alternative when weights are predetermined.

A random vector $X$ has multivariate normal distribution $N_d(\mu ,\Sigma )$ if every real linear combination $t^{T}X$ is normally distributed, with mean $t^T\mu$ and variance $t^T\Sigma t$. The covariance matrix is symmetric positive semidefinite; zero variances permit degenerate normals. Its moment-generating function is $M_X(t)=\exp (t^T\mu +t^T\Sigma t/2)$, including the degenerate cases.

For the given $X$, let $t_j\in {\mathbb{R}}^{d_j}$. The joint moment-generating function of the $Y_j=A_{j}X$ isThe last equality uses the assumed orthogonality of the row spaces. Its factors are exactly the individual moment-generating functions, so the hint implies the random vectors $Y_j$ are independent. Also every linear combination of the stacked vector $Y$ is a linear combination of $X$, hence normally distributed. ConsequentlyThis establishes both claims without excluding singular block covariances.

For the sample mean and sample variance result, assume $n\ge 2$ and set $W=(Z_1-\mu ,\dots ,Z_n-\mu {)}^T$. Independence of the observations gives $W\sim N_n(0,{\sigma }^{2}I)$. Let $u=(1,\dots ,1{)}^T$, $A_1=u^T/n$, and $A_2=I-u{u}^T/n$. Then $A_1{A}_2^T=0$. The first transformed vector is the scalar $\overline{Z}-\mu$, and the second is the residual vector $(Z_i-\overline{Z}{)}_i$. They are independent by the just-proved result. Applying the measurable function $v\mapsto \Vert v{\Vert }^2/(n-1)$ to the residual vector preserves independence from the sample mean. Thus

An unordered bootstrap sample is determined by the multiplicities $N_i\ge 0$ of the $n$ distinct observations, with ${\sum }_i{N}_i=n$. Conversely each such vector specifies exactly one sample up to rearrangement. Place $n$ identical marks into $n$ boxes using $n-1$ separators. The stars and bars argument therefore givesThese bootstrap count vectors are not equally likely: the probability of a vector is $n!/(n^n{\prod }_i{N}_i!)$, counting its possible ordered sequences. The ordered index sequences themselves number $n^n$.

The first simulation uses independent standard Cauchy distributions and returnsThe tail probability and estimator moments areThe indicator count has a binomial distribution with parameters $n,\theta$; the infinite variance of a Cauchy observation is irrelevant because this estimator averages bounded indicators.

For the next code block, each row is an independent sample of size $n$ from the empirical distribution of the original $x_i$. Its entry in the vector $v$ is the corresponding bootstrap indicator mean ${\hat{\theta }}_b^{\ast }$. The last expression is the variance of these $B=199$ replicate estimates, with denominator $B-1$:Condition on the original data and write a star on bootstrap expectations. Every resampled indicator has a Bernoulli distribution with success probability $\hat{\theta }$, and the $n$ draws within a row are independent. Thus the conditional bootstrap variance of a sample mean givesIt estimates the sampling variance of the tail-probability estimator, not the variance of the raw Cauchy data. Its square root is an estimated standard error. Finite $B$ leaves bootstrap simulation noise. Averaging over the original data, its expectation is $(n-1)\theta (1-\theta )/n^2$, so it is not exactly unbiased for the true variance $\theta (1-\theta )/n$ at finite $n$.

In the importance-sampling block, the transformation $Y=2/(1-U)$ gives, for $y\ge 2$,The variable $Y$ is a proposal draw supported entirely in the integration region. The deterministic value computed from it is the importance sampling weightConsequentlyThe code uses ordinary importance sampling of a Cauchy tail, not self-normalization; $Y$ itself is neither an observation from the target Cauchy law nor the estimator's summand. The weight compensates for sampling from its proposal law.

The final block bootstraps these same proposal observations $Y_i$ and recomputes the weight average in each row. Since $W$ is a deterministic function of $Y$, this is equivalent to resampling the observed weights $W_i$. If ${\tilde{\theta }}_b^{\ast }$ denotes a replicate, the final output is its sample variance over the $B$ replicates. Its conditional expectation isIts expectation over the original proposal sample is $(n-1)\mathrm{Var}(W)/n^2$.

The variance reduction can be calculated, rather than inferred just from the appearance of the weights. Write $A=\arctan (1/2)$. Thenbecause an antiderivative of the integrand without its prefactor is $\frac{1}{2}(\arctan y-y/(1+y^2))$. HenceBy comparison, $\theta (1-\theta )\approx 0.125803$. For $n=100$ the true estimator variances are approximately $9.55253\times 10^{-7}$ and $1.25803\times 10^{-3}$ respectively. The importance estimator has about 1/1,317 of the indicator estimator's variance, because all proposals land in the tail and the weights remain in the narrow interval $[2/(5\pi ),1/(2\pi ))$.

Thus the last bootstrap output should usually be much smaller than the first, and its expected value is smaller by the same variance ratio. This is not a deterministic ordering of two finite-run outputs: if all the original Cauchy draws happen to be at most two, the first bootstrap variance is zero, while unequal importance weights can still give a positive second variance.

Let $Y_i$ be iid with density $f$, set $H_i=h(Y_i)$, and put $I=E_{f}h(Y)$ and ${\tau }^2={\mathrm{Var}}_f(h(Y))$. The finite-second-moment assumption gives, for ${\tau }^2>0$,The sample variance $s_N^2=(N-1{)}^{-1}{\sum }_i(H_i-{\hat{I}}_N{)}^2$ consistently estimates ${\tau }^2$. The central limit theorem and Slutsky theorem therefore give the asymptotic confidence intervalIts confidence probability is approximately $1-\alpha$, conventionally written $100(1-\alpha )\%$; the printed $(1-\alpha )\%$ omits the factor $100$. For ${\tau }^2=0$, $h(Y)=I$ almost surely and the estimator has zero error. This interval concerns iid Monte Carlo estimators; correlated simulation output requires a long-run variance of a stationary process estimate instead of the iid sample variance.

Under fit.2, observations in each period are independent Poisson random variables with a common mean, so each sample variance should fluctuate around its sample mean. The plot instead has variances roughly 130–300 for means roughly 80–120, well above the line variance equals mean. This is overdispersion.

Day-to-day traffic variation can produce a Poisson mixture: a random daily intensity increases the marginal variance and can correlate observations from the same day. A useful replacement is a negative binomial regression, with variance $\mu +{\mu }^2/\kappa$, or a Poisson generalized linear mixed model with a day random intercept. A day-factor effect can also be included if inference is restricted to the observed days. A Quasi-Poisson regression estimates a dispersion multiplier for uncertainty, but does not itself explain the shared-day dependence and has no ordinary likelihood-based AIC. The means can retain the period-factor or selected jump structure.

An unbiased estimator $T$ of a parameter $\theta$ satisfies ${\mathbb{E}}_{\theta }[T]=\theta$ for every permissible value of $\theta$. Since the sample mean is linear,so $\hat{\mu }$ is unbiased.

Use the identityThe two expectations on the right are $n{\sigma }^2$ and $n\mathrm{Var}(\overline{X})={\sigma }^2$, respectively. HenceThusThe unbiased sample variance instead divides the same sum of squares by $n-1$.

Because every column of $X$ has sample mean zero, every column $u_i$ of $U=XV$ also has sample mean zero. The spectral theorem for real symmetric matrices givesHence for $i\ne j$ the sample covariance is $n^{-1}{u}_i^T{u}_j=0$, while the sample variance of $u_i$ isThus the $u_i$ are pairwise uncorrelated sample principal components, ordered by decreasing sample variance.

The given identities implyThus, writingwe have the exact identityThe sample variance satisfies $s_n^2\stackrel{P}{\to }1$, soThe Slutsky theorem and part (c) now give

At sample size 100, keep essentially the same numerical confidence interval, but treat its calibration as approximate. For independent heights with finite, nonzero variance, the central limit theorem makes the sample mean approximately normally distributed, and the sample variance consistently estimates the population variance. Hence the studentized sample mean is approximately standard normal even without a normal distribution for individual heights.

Using the permitted cutoff 2 givesA Student t-distribution cutoff gives a similar approximation for large sample size, but its justification at fixed sample size is no longer exact. Extreme skewness or influential outliers could make 100 observations insufficient for accurate coverage probability; abandoning normality alone supplies no specific numerical correction. The sampling with replacement assumption still precludes a finite population correction.

A $100(1-\alpha )\%$ confidence interval procedure produces random endpoints $L,U$ such that ${\mathbb{P}}_{\mu }(L\le \mu \le U)=1-\alpha$, under the model. In everyday terms, repeating the same sampling and interval construction many times would give intervals containing the true mean in about $100(1-\alpha )\%$ of repetitions.

For $n\ge 2$, define the sample mean $\bar{X}=n^{-1}{\sum }_i{X}_i$ and the sample variance $S^2=(n-1{)}^{-1}{\sum }_i(X_i-\bar{X}{)}^2$. For independent observations from a normal distribution with ${\sigma }^2>0$, the standard sampling results areThe first follows from the independent normal sample mean and sample variance. Let $q=t_{n-1,1-\alpha /2}$ be the positive Student quantile. Symmetry gives the central confidence intervalIts width is $W=2qS/\sqrt{n}$, so

Each before/after pair concerns the same individual, so subject-specific baseline blood pressure can affect both measurements. Analyzing the within-person difference retains this pairing and removes a common baseline contribution. Under the stated independent normal-difference model,The normal-sample variance identity givesMoreover $\overline{D}$ and $S_D^2$ are independent, because the sample-mean projection and centered residual projection of a Gaussian vector are orthogonal and hence independent. Thus the standardized normal variable $N=\sqrt{n}(\overline{D}-\mu )/\sigma$ is independent of $U=(n-1)S_D^2/{\sigma }^2\sim {\chi }_{n-1}^2$. ConsequentlyThis distribution is free of both unknown parameters.

For each normal sample, its sample mean is independent of its sample variance, and the scaled residual sum of squares is chi-squared. Independence of the samples also makes their two residual sums independent. Adding their chi-squared distributions givesConsequentlyThe pooled variance is also independent of the difference of sample means.

For each normal sample, its sample mean is independent of its sample variance, and the scaled residual sum of squares is chi-squared. Independence of the samples also makes their two residual sums independent. Adding their chi-squared distributions givesConsequentlyThe pooled variance is also independent of the difference of sample means.

Use the sample mean to estimate each magnetic susceptibility, and the unbiased sample variance with denominator $n-1=4$ to estimate the repeatability of individual readings. In the numerical scale supplied by the paper,If the repeated measurements have independent random errors, their mean estimates have standard errorsThe difference and its estimated standard uncertainty are thereforeThe separation is about $2.53$ standard errors. Under an approximately normal distribution model with no experimenter-dependent bias, Welch's t-test gives about $7.34$ statistical degrees of freedom and a two-sided p-value of $0.038$. Thus there is modest evidence against equal underlying susceptibilities at a conventional $5\%$ significance level; overlap of the ranges of individual readings is not an adequate comparison of the means. Five readings per sample provide limited information about the error distribution.

Crucially, material and student exhibit complete confounding: each student measured only one sample. An experimenter-dependent systematic error could mimic the difference. Paramagnetism also depends on conditions such as temperature, so equal material alone does not guarantee equal readings without controlling those conditions. The defensible physical conclusion is the data suggest a difference, but do not establish different materials. Both students should measure both samples in randomized or alternating order, with calibration and temperature controlled, to separate a sample effect from an operator effect. No unit or scale factor absent from the original table has been invented.

The variables $(m-1)S_X^2/{\sigma }^2$ and $(n-1)S_Y^2/{\sigma }^2$ are independent chi-squared variables with $m-1$ and $n-1$ degrees of freedom. Independence follows because the two whole samples are independent. Their sum is chi-squared with the sum of the degrees of freedom, soWithin each normal sample the sample mean is independent of its sample variance, and independence of the samples makes the pair of means jointly independent of the pair of variances. Hence this pooled chi-squared variable is independent of the standardized mean difference, which is the independence required for the Student pivot.

For a sample size $m\ge 2$ define $\overline{X}=m^{-1}{\sum }_i{X}_i$ and $S_X^2=(m-1{)}^{-1}{\sum }_i(X_i-\overline{X}{)}^2$. The normal-sample theorem givesIt also states that the mean and sample variance are independent. Combining (i) and (ii), and the independence across samples, the correctly centred pivot isThe source reverses the mean difference inside the subtraction. Its literal statistic is generally noncentral when ${\mu }_X\ne {\mu }_Y$, rather than having a central Student distribution. Under the null hypothesis both differences are zero, so this misprint does not change the requested null test.

For a two-sided level-$\alpha$ test reject when $|T|>t_{m+n-2,1-\alpha /2}$. In the numerical example the pooled variance is $0.038$, the standard error is $\sqrt{0.038(2/5)}\approx 0.12329$, and $T\approx -2.190$. There are eight degrees of freedom and the supplied critical value is $2.31$. Thus do not reject equal mean cell volume at the 5% level. This does not establish equality of the means; it states that the observed difference does not cross this test's two-sided threshold.

Age matching introduces a shared block factor: the two observations in a pair can be associated, and their age-dependent means also undermine the simple unblocked independent-samples model. A paired t-test on the within-pair drug-minus-placebo differences accounts for that design, assuming independent pairs and the corresponding normal-difference model. Shared age variation cancels in those differences; positive within-pair association reduces their variance and can increase power, although the reduction in degrees of freedom means improvement is not automatic. The paired test cannot be computed from the two marginal means and variances alone: the within-pair differences or covariance are needed.

Let $m=\mathbb{E}X$ and $v=\mathrm{Var}(X)$, evaluated from the given polylogarithm formulas. Generate a large sample of independent and identically distributed random variables with r_sample_f, and compute its sample mean ${\overline{X}}_n$ and sample variance $S_n^2$ using mean and var. The central limit theorem gives the first standardized check:For the second check, the correct asymptotic variance of the sample variance involves the fourth central moment ${\mu }_4=\mathbb{E}[(X-m{)}^4]$:The denominator is not $\sqrt{2v^2}$ unless the probability distribution has the requisite fourth central moment, as does the normal distribution. Here the probability density function decays exponentially, so all required moments are finite.

To compute ${\mu }_4$, expand the denominator of the probability density function as a geometric series and integrate each nonnegative term. The gamma function integral gives the raw momentsThus $v=m_2-m_1^2$ and ${\mu }_4=m_4-4m_1{m}_3+6m_1^2{m}_2-3m_1^4$. The following R code implements both checks for chosen interior parameters and large n:

p <- 0.4
beta <- 1.3
n <- 100000
raw_m <- vapply(1:4, function(j) {
  -factorial(j) * polylog(j + 1, 1 - p) / (beta^j * log(p))
}, numeric(1))
m <- raw_m[1]
v <- raw_m[2] - m^2
mu4 <- raw_m[4] - 4 * m * raw_m[3] +
  6 * m^2 * raw_m[2] - 3 * m^4
x <- r_sample_f(n, p, beta)
c(mean_check = sqrt(n) * (mean(x) - m) / sqrt(v),
  variance_check = sqrt(n) * (var(x) - v) / sqrt(mu4 - v^2))

For each check separately, an absolute value below $1.96$ is consistent with its approximate $95\%$ confidence interval. Repeat with many independent samples: the two standardized statistics should each resemble a standard normal distribution, with about $5\%$ outside that range. Also vary $p$, $\beta$ and n; a systematic discrepancy that grows with $\sqrt{n}$ suggests a bias. The two checks from the same sample are generally dependent, so passing both is not itself a joint $95\%$ statement, nor can a finite test prove that a generator is correct.

The second central limit theorem follows from the central limit theorem for the sample variance: $\sqrt{n}(S_n^2-v)$ differs by a quantity tending to zero in probability from $n^{-1/2}{\sum }_i[(X_i-m{)}^2-v]$.

TODO: Attachment C is absent from the held release. Recover it to check any supplied critical-value table, significance convention or additional instructions. This leaf remains incomplete against that missing input. The printed summary statistics nevertheless allow the following conditional numerical answer, using a two-sided $5\%$ Student's t-test and independent observations with common variance.

Test $H_0:{\mu }_W={\mu }_S$ against $H_1:{\mu }_W\ne {\mu }_S$. The pooled sample variance isThe test statistic and statistical degrees of freedom areUsing the Student t-distribution directly gives a two-sided p-value $p\simeq 0.00250$ and a $5\%$ critical value $1.96076$. Thus reject equal seasonal means at $5\%$ and $1\%$; the winter mean is higher by $0.36$ concentration units. This does not establish a large physical effect or its cause.

The two variances are similar, and Welch's t-test gives essentially the same result: $t=3.02344$, $\nu \simeq 2958.36$, $p\simeq 0.00252$. Both calculations assume the counts represent independent observations. Serial dependence, repeated measurements and strong non-normality can invalidate nominal p-values; the summary table alone cannot diagnose them. The unequal means are a conclusion under these statistical assumptions, pending recovery of Attachment C.

The potential scale reduction factor compares variation within simulated chains to a variance estimate also incorporating differences between chains. In its classical form, $\hat{R}=\sqrt{\hat{V}/W}$, where $W$ is the average within-chain sample variance. Values appreciably above one indicate disagreement; values close to one do not prove convergence. Splitting chains helps expose changes over time, and modern rank-based variants improve robustness.^[ref]

Fix $N>1$ real population values $y_1,\dots ,y_N$. Define the finite population mean and two normalizations of the finite population variance byThe values are fixed; only the selected subset $A$ is random. For its sample mean ${\bar{y}}_A=n^{-1}{\sum }_{i\in A}{y}_i$, the exact design expectation and variance areThe sample mean is an unbiased estimator of the finite population mean. At $n=N$, every unit is observed and ${\bar{y}}_A=\bar{Y}$ exactly, so the variance is zero. The $N=1$ case is deterministic and is handled directly, since $S^2$ is undefined then.^[ref]

The finite population correction depends on the baseline. With the $N-1$ normalization it is $1-n/N$ multiplying $S^2/n$. With the $N$ normalization it is $(N-n)/(N-1)$ multiplying ${\sigma }_N^2/n$, the exact variance for independent uniform sampling with replacement. The standard error uses the square root of the relevant variance factor. Mixing the two normalizations gives a wrong answer even at $n=1$, when replacement cannot matter. For the six-value example, $\bar{Y}=7$, ${\sum }_i(y_i-7{)}^2=70$, $S^2=14$ and ${\sigma }_N^2=35/3$, reproducing $14/3$ without replacement and $35/6$ with replacement.

A short proof exposes the statistical dependence. Let $X_r$ be the value at position $r$ in uniform sequential sampling without replacement, so each $X_r$ has expected value $\bar{Y}$ and variance ${\sigma }_N^2$. Write $d_i=y_i-\bar{Y}$. Every ordered pair of distinct labels is equally likely, and ${\sum }_i{d}_i=0$, hence for $r\ne s$,Linearity of expectation proves unbiasedness without requiring independence. The variance of a sum includes all $n(n-1)$ ordered covariance terms, givingFor the original population, each pair of distinct draw positions has covariance $-7/3$. Drawing a high value leaves a lower mean among the remaining values, balancing subsequent draws. Equal marginal probability distributions therefore coexist with negative covariance.

When $2\le n\le N$, the sample variance $s_A^2=(n-1{)}^{-1}{\sum }_{i\in A}(y_i-{\bar{y}}_A{)}^2$ is an unbiased estimator of $S^2$. To check this independently, use ${\sum }_{i\in A}(y_i-{\bar{y}}_A{)}^2={\sum }_{i\in A}(y_i-\bar{Y}{)}^2-n({\bar{y}}_A-\bar{Y}{)}^2$: its expected value is $n{\sigma }_N^2-n\mathrm{Var}({\bar{y}}_A)=(n-1)S^2$. Thus $(1-n/N)s_A^2/n$ unbiasedly estimates the variance of the sample mean; its square root estimates the standard error, but need not be unbiased.^[ref]

The diagram compares variance with the expected fraction of distinct population labels observed. Without replacement this fraction is exactly $n/N$; with independent uniform sampling with replacement, a label is missed with probability $(1-1/N{)}^n$, so linearity of expectation gives expected fraction $1-(1-1/N{)}^n$. At six draws, replacement covers only about $66.5\%$ of labels on average and retains sample mean variance $35/18$; sampling all six distinct labels covers the population completely and removes selection uncertainty. This population coverage is distinct from confidence interval coverage probability.

The correction matters when the sampling fraction is appreciable. It calibrates uncertainty about this fixed population under the specified selection mechanism. It does not correct unequal selection weights, missing observations or measurement errors; observing every label also does not eliminate uncertainty about a larger unobserved population. Moreover, an exact variance alone does not guarantee the coverage probability of a proposed confidence interval: a small discrete sampling distribution can make a normal approximation inaccurate.