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Put $\hat{\Sigma }=X^{T}X/n$. The Debiased Lasso correction isFor the Nodewise Lasso, let $r_j=X_j-X_{-j}{\hat{\gamma }}^{(j)}$. Its Karush-Kuhn-Tucker conditions areso $z_{jk}=\mathrm{sgn}({\hat{\gamma }}_k^{(j)})$ at a nonzero coefficient and $|z_{jk}|\le 1$ at a zero coefficient. Multiplying by ${\hat{\gamma }}^{(j)}$ gives the nodewise-Lasso residual identityWe require nonzero design columns, a necessary qualification missing from the unrestricted printed formulation. Because ${\lambda }_j>0$, ${\hat{\tau }}_j^2=0$ would force both $r_j=0$ and ${\hat{\gamma }}^{(j)}=0$, and hence $X_j=0$. Thus nonzero columns suffice for ${\hat{\tau }}_j^2>0$ even when $p>n$ or the full Gram matrix is singular. A zero column instead gives ${\hat{\tau }}_j^2=0$ and its coefficient is unidentifiable from $Y$; the proposed normalization and confidence interval do not exist there.

For each $j$, define $c_j\in {\mathbb{R}}^p$ by $(c_j{)}_j=1$ and $(c_j{)}_{-j}=-{\hat{\gamma }}^{(j)}$. Set row $j$ of $\hat{\Theta }$ to $c_j^T/{\hat{\tau }}_j^2$. Since $X{c}_j=r_j$,This row construction is the nodewise normalization used by van de Geer and coauthors. The matrix $\hat{\Theta }$ need not be symmetric or be an actual inverse.

Substitute the linear model into the correction. The exact decomposition isIn the fixed-design normal linear model, or with $\epsilon \mid X\sim N_n(0,{\sigma }^{2}I)$ and nodewise estimates and tuning chosen from $X$ alone, a linear transformation of a multivariate normal distribution gives $W\mid X\sim N_p(0,{\sigma }^2\hat{\Omega })$. Marginal normality of $\epsilon$ without this conditional assumption would not suffice for a response-dependent random design. No independence between $W$ and $\Delta$ is asserted. The diagonal cancellation and the Holder inequality give the debiased-Lasso remainder bound

For a coordinate variance, ${\hat{\Theta }}_j{X}^T=r_j^T/{\hat{\tau }}_j^2$, whenceWriting $z_{1-\alpha /2}$ for the corresponding standard normal quantile, an approximate confidence interval isEquivalently its half-width is $z_{1-\alpha /2}\sigma \Vert X_j-X_{-j}{\hat{\gamma }}^{(j)}{\Vert }_2/(n{\hat{\tau }}_j^2)$. We assume $0<\alpha <1$ and $\sigma >0$. For asymptotic coverage, the standardized remainder ${\Delta }_j/(\sigma \sqrt{{\hat{\Omega }}_{jj}})$ must have convergence in probability to zero and the coordinate variance must be nondegenerate. Exact conditional normality of $W$ alone does not give finite-sample exact coverage for $\hat{b}$.

One sufficient random-design regime is the Gaussian identity design for debiased Lasso: take independent rows $X_i\sim N_p(0,I_p)$, independently of $\epsilon$, fixed $\sigma >0$, $p=p_n\to \infty$, $s=s_n\ge 1$, andChoose $\lambda =C\sigma \sqrt{\log p/n}$ and all ${\lambda }_j=D\sqrt{\log p/n}$ for sufficiently large fixed $C,D$. Here is why this regime has the requested remainder bound. Uniformly over columns, their squared norms divided by $n$ tend to one, while all off-diagonal sample cross-correlations have magnitude at most a constant times $\sqrt{\log p/n}$ with probability tending to one. Taking $D$ larger makes the zero vector satisfy all nodewise Karush-Kuhn-Tucker conditions strictly, so ${\hat{\gamma }}^{(j)}=0$ for every $j$ on that event, and ${\hat{\tau }}_j^2=\Vert X_j{\Vert }_2^2/n\ge 1/2$ eventually. This also gives ${\hat{\Omega }}_{jj}=1/{\hat{\tau }}_j^2\to 1$ uniformly.

In this regime, the entrywise entrywise concentration of a Gaussian Gram matrix and Lasso cone condition argument in Gaussian identity design for debiased Lasso gives a Compatibility condition for the Lasso bounded away from zero with probability tending to one. Together with the noise-score event and the Basic inequality for the Lasso, this yields, for a fixed $C_1$,For example, the cone inequality and compatibility bound give $\Vert \hat{\beta }-{\beta }^0{\Vert }_1\le 12\lambda s/{\varphi }^2$ on their joint event. Therefore, with a fixed $A\ge 2D{C}_1$,The stated sparsity condition makes this remainder vanish and supplies the preceding confidence-interval requirement. Both dimensions may grow: $p=n^2$ and $s=\lfloor n^{1/4}\rfloor$ satisfy these assumptions. The concentration and Gaussian-design compatibility results are sufficient standard conditions here; the question does not require their proofs. The displayed rate is a high-probability fixed-constant bound, rather than only a claim of boundedness in probability.

Write $y_{WP}$ and $y_{AP}$ for the estimated log odds ratios relative to the common control. Indirect treatment comparison uses the consistency relation ${\delta }_{WA}={\delta }_{WP}-{\delta }_{AP}$. For independent trials, subtraction adds their variances, givingThe substantive assumption is transitivity in network meta-analysis: the study populations, outcome definitions and follow-up are sufficiently comparable, particularly in their distributions of effect modifiers, that the two control comparisons estimate effects applicable to the same target population. The common control's name alone does not justify that assumption. This also assumes comparable marginal odds ratios; these need not equal covariate-adjusted odds ratios even without confounding.

Using the supplied rounded variances and the given standard normal quantile, the approximate confidence interval isOn the odds ratio scale this is $e^{-1.26}\simeq 0.284$, with approximate limits $(0.104,0.771)$. The estimated odds of recurrent stroke under treatment $W$ are about 28% of those under $A$, a reduction of about 72% in odds. The interval excludes equal odds, so this comparison provides evidence of lower odds under $W$ if the indirect treatment comparison assumptions hold. These are odds, not a 72% reduction in probability; a confidence interval describes the long-run coverage of the procedure, not a posterior probability for this particular interval.

A random-effects meta-analysis allows differences in true treatment effects across countries and eligibility criteria. For trial $i=1,\dots ,6$, let $y_i$ be its estimated log odds ratio and $v_i$ its estimated within-trial variance. A usual approximate statistical model isindependently across trials, with $b_i$ independent of sampling errors. Here $\mu$ is the mean true log odds ratio in the population of comparable trials, and ${\tau }^2\ge 0$ is between-trial heterogeneity, not additional sampling error. Therefore $y_i\sim N(\mu ,v_i+{\tau }^2)$, and for a fitted heterogeneity value,One can estimate ${\tau }^2$ by restricted maximum likelihood or another justified method; $({\sum }_i{w}_i{)}^{-1}$ is a plug-in conditional variance and does not fully account for estimating heterogeneity. With only six studies that uncertainty matters. Different eligibility rules motivate this random effect but do not by themselves establish exchangeability or remove bias; known effect modifiers may warrant stratification or regression.

Use leave-one-study-out influence analysis: fit all six trials, then omit Cohen 1989 and refit both $\mu$ and ${\tau }^2$. Report $\hat{\mu }-{\hat{\mu }}_{(-C)}$, the corresponding change in the odds ratio, and changes in the confidence interval and heterogeneity. As a diagnostic with heterogeneity held fixed, writing $W={\sum }_i{w}_i$ givesThe fitted Cohen weight would be $(0.15+{\hat{\tau }}^2{)}^{-1}$ and $y_C=-1.95$. A precise but discordant trial can have large influence; refitting assesses additional influence through heterogeneity. The other five trials' data are absent, so a numerical six-trial influence assessment is not identifiable from the displayed table.

The supplementary excerpt uses an adjusted observational study and a two-sample Mendelian randomization analysis.

The observational study examines co-occurrence of hypercalcaemia and migraine diagnoses in electronic health records, adjusting for age, sex, and ancestry. Its adjusted odds ratio is 1.58, with $P=4.75\times 10^{-13}$.

The two-sample Mendelian randomization analysis combines the associations of eight genetic variants with serum calcium and with migraine, using a weighted genetic risk score. It estimates an odds ratio of 1.80 per genetically predicted 1 mg/dL higher serum calcium, with 95% confidence interval $(1.31,2.46)$. The weighted-median Mendelian randomization estimator and MR-Egger regression provide sensitivity analyses.

The measured and instrumented exposure is serum calcium, rather than dietary calcium intake itself. Calcium intake and serum calcium are not interchangeable interventions, and an odds ratio of 1.80 is an odds contrast rather than necessarily an 80% increase in probability. The shortened excerpt supplies two approaches; a further analysis in the full research paper is outside this excerpt.

The multivariate Wald statistic for a proposed value $\theta$ isBecause ${\hat{i}}_n\to I_p$ andSlutsky theorem gives

If $q_{p,1-\alpha }$ is the $(1-\alpha )$-quantile of the chi-squared distribution with $p$ degrees of freedom, an asymptotic $(1-\alpha )$ confidence region isIn this known-identity-covariance model one may replace ${\hat{i}}_n$ by $I_p$; then the confidence ball is exact. For an individual coordinate, the corresponding exact confidence interval is

Let $z_p$ denote the $p$-quantile of the standard normal distribution. Since the variance is known,so a $100(1-\alpha )\%$ confidence interval is

Because $0\le U\le 1$, for fixed $t$ and sufficiently large $n$,ThereforeEquivalently, $n(1-\hat{\theta }/\theta )$ converges in distribution to an exponential distribution of rate one.

Put $c=\log (1/\alpha )$. The limiting probabilityand the certain inequality $\hat{\theta }\le \theta$ suggest the approximate confidence intervalSince $\mathbb{E}[\hat{\theta }]=n\theta /(n+1)$, its expected length isas required.

The reported 95% credible interval for ${\beta }_1$ isConditional on the model, prior, and observed data, its posterior probability is 0.95. A frequentist 95% confidence interval instead has 95% coverage under repeated sampling before the data are observed; it does not assign a sampling probability to the fixed parameter after observing this dataset.

Apply the central limit theorem to the independent and identically distributed random variables $X_i^2$, whose mean and variance are $\theta$ and ${\theta }^2$. It gives the approximate standard normal distributionThe supplied standard normal quantile $c=1.96$ leaves probability $0.025$ in each tail. HenceSince $\theta >0$, this event is equivalent toFor $n>1.96^2$, equivalently integer $n\ge 4$, both denominators below are positive. Inverting the inequalities gives the approximate $95\%$ confidence intervalThe approximation is justified for large $n$; positivity at $n\ge 4$ by itself does not guarantee accurate coverage. This construction is a confidence interval from a relative normal pivot: the unknown scale remains inside the pivotal ratio until the inequalities are inverted. Its meaning is that the random interval covers the fixed statistical parameter $\theta$ with probability approximately $0.95$ under repeated sampling.

The pivotal quantity $U=M/\theta$ lies in $[0,1]$ and satisfies $\mathbb{P}(U\le t)=t^n$. ThusRearranging these inequalities yields the exact confidence intervalFor every $\theta >0$, its coverage probability is exactly $1-\alpha$; the result uses the finite-sample probability distribution, with no normal approximation.

The central limit theorem givesWrite $z=z_{1-\alpha /2}$ for the indicated standard normal quantile. Inverting the approximate pivot inequality and substituting $Y/n=1/\hat{\theta }$ gives the approximate confidence intervalIts coverage is approximately $1-\alpha$ for large $n$. The intersection respects the positive rate parameter; asymptotically the lower endpoint is positive. This is an asymptotic interval, not an exact finite-sample confidence interval.

The sample mean is exactly $N(\mu ,{\sigma }^2/n)$, so $(\overline{X}-\mu )/(\sigma /\sqrt{n})$ has the standard normal distribution. The given quantile gives probability $0.95$ to the interval $[-1.96,1.96]$. Inverting this pivot yields the exact $95\%$ confidence interval

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.

The independent uniform choices from the student list describe sampling with replacement. Under the stated independent normal distribution model for the heights, the population variance is unknown and the given $8.5$ cm is the sample standard deviation $s$. The exact pivotal quantity isThus the exact 95% Student t confidence interval isusing $t_{99,0.975}\approx 1.9842$. The supplied approximation $\Phi (2)=0.975$ instead gives a normal approximation for large sample sizeThis latter confidence interval is approximate even under normality because it substitutes the sample standard deviation for the unknown population value. There is no finite population correction under the specified independent selections. Interpreting the finite student list literally as a discrete population makes normality an idealized model; it does not justify changing the stated sampling design.

For this rate parametrization of the Laplace distribution, $\mu$ is its mean and median. By symmetry, $\mathbb{E}(X-\mu )=0$, while $|X-\mu |$ has exponential distribution with rate $\lambda$, soThe integral follows by two applications of integration by parts. The central limit theorem therefore givesWith known $\lambda$ and the supplied cutoff 2, an approximate 95% confidence interval isThis known-rate Laplace location confidence interval uses the sample mean; the later maximum-likelihood estimator of location need not be used to construct this valid asymptotic confidence interval.

Independence of the slope estimate and residual variance, together with their normal and chi-square laws, gives the Student pivotal quantityThus an exact Gaussian-model two-sided 95% confidence interval isUsing a normal quantile without estimating the variance uncertainty would not give this exact finite-sample coverage.

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

A frequentist confidence interval is a pair of data-dependent endpoints whose random interval covers the fixed parameter with probability at least $1-\alpha$ under every allowed parameter value. In repeated uses of the same sampling procedure, at least the advertised fraction of intervals cover the true value. It is not a posterior probability assigned to the parameter after observing the data.

For a sample of size $n$ from a Poisson law with mean $\mu$, the sum $S$ is Poisson with mean $n\mu$. At observed $s>0$, define the lower and upper limits byAt $s=0$ set $L=0$ and use the same upper equation. This inverts the two exact one-sided tail tests. Discreteness means the resulting central interval is generally conservative: its coverage is at least the nominal value, not identically equal to it.

Here one observes a four-year total of 39, so the lower endpoint satisfies ${\varphi }_L(38)=0.975$ and the upper satisfies ${\varphi }_U(39)=0.025$. Selecting the corresponding entries supplied in the source givesThis interval concerns the expected four-year total, not an annual rate. The other two supplied numbers correspond to the wrong inclusive tail for one of these endpoints.

Since the true mean is known to be zero, the chi-squared distribution gives $Q/\theta \sim {\chi }_n^2$. Let $q_{\beta }$ be the $\beta$ quantile of this law. The probabilityrearranges, using positivity, to the exact confidence intervalThere are $n$ degrees of freedom, because no unknown mean was fitted.

Let $z=z_{1-\alpha /2}={\Phi }^{-1}(1-\alpha /2)$, where $0<\alpha <1$. By the preceding asymptotic normality and Slutsky's theorem, $\sqrt{n}(\hat{\theta }-\theta )/\hat{\theta }$ is approximately standard normal. Inverting $|\sqrt{n}(\hat{\theta }-\theta )/\hat{\theta }|\le z$ gives the Wald confidence intervalIts coverage approaches $1-\alpha$ as $n\to \infty$. The interpretation of the confidence interval is repeated-sample coverage of the fixed parameter; it does not assign a posterior probability to $\theta$.