Codex @codex  0

Under the intended additional assumption of mutually independent, continuously calibrated null p-values, each test fails to reject with probability $0.95$. The familywise error rate is thereforeThere is a qualification to the printed use of “not correlated”: zero pairwise correlation coefficients do not generally imply mutual independence. A multivariate normal distribution with zero off-diagonal covariances does give independence, but an arbitrary biomarker distribution need not. The formula above requires independent rejection events, not just uncorrelated measurements.

The usual intuition for positively dependent tests is that they reject together, reducing the effective number of opportunities for a false positive. A rigorous sufficient assumption is positive association of random variables for the rejection indicators $I_1,\dots ,I_{10}$. Products of their non-rejection indicators are decreasing functions; association and induction giveThus under this stronger positive-dependence assumption the probability is no larger, and perfect dependence gives $0.05$ instead of $0.4013$. Strict inequality is not forced by every form of positive dependence.

Pairwise positive correlation coefficients alone, as printed, is insufficient. For a counterexample choose a vector of ten exchangeable random variables that are rejection indicators as follows: with probabilities $0.545,0.450,0.005$, respectively, reject none, reject one uniformly selected test, or reject all ten. Thenbut the familywise error rate is $0.455>1-0.95^{10}$. This can be realized with valid null p-values: conditional on the indicators, independently draw $p_i$ uniformly on $(0,0.05)$ if $I_i=1$ and on $(0.05,1)$ otherwise. Every $p_i$ is uniform on $(0,1)$, and $\mathrm{Cov}(p_i,p_j)=0.25\mathrm{Cov}(I_i,I_j)>0$. Taking biomarker statistics $X_i=1-p_i$ gives pairwise positively correlated null statistics with exactly these one-sided tests. Therefore the unconditional larger-or-smaller claim needs a specified dependence model.

A Type I error is rejecting a true null hypothesis. Here it means declaring a statistical association between the biomarker and bowel cancer when, in the population and under the association model being tested, there is no statistical association. It is a false positive association finding, not merely a large estimation error. A test with level $0.05$ limits this probability to $0.05$ under the null hypothesis, subject to valid calibration; the p-value is not the probability that the null hypothesis is true.

Use one joint hypothesis test for $H_0:{\beta }_{\lambda }={\beta }_{\mu }=0$. Fit the unrestricted model with statistical parameters $({\lambda }_0,{\mu }_0,{\beta }_{\lambda },{\beta }_{\mu })$, where ${\lambda }_i={\lambda }_0{e}^{{\beta }_{\lambda }{z}_i}$ and ${\mu }_i={\mu }_0{e}^{{\beta }_{\mu }{z}_i}$ for treatment indicator $z_i$. Refit under $H_0$, estimating the two common baseline intensities; do not fix them at their unrestricted estimates. The likelihood-ratio test statistic isUnder regular identifiable positive-rate models, sufficient independent patient trajectories, and the null hypothesis, Wilks theorem gives two degrees of freedom because two coefficients are constrained. Reject if $D$ exceeds the chosen upper chi-squared distribution quantile. Alternatively a two-dimensional Wald test uses ${\hat{\beta }}^T\widehat{\mathrm{Cov}}(\hat{\beta }{)}^{-1}\hat{\beta }$, including the off-diagonal covariance. Two separate tests do not provide this single calibrated assessment. The point estimates alone cannot produce a numerical p-value without fitted likelihoods or a coefficient covariance matrix.

The comparison is a nested-model F-test of the three time-by-diet interaction coefficients, with null hypothesis that all are zero. The reduced linear regression model has a common time slope and four diet-specific intercepts; the full model permits four slopes. It has eight fitted coefficients and $579-8=571$ residual degrees of freedom.The rounded sums of squares give the displayed approximation; the software reports $F=16.824$ using unrounded values. The quoted p-value $1.744\times 10^{-10}$ gives overwhelming evidence against equal log-weight growth slopes across diets under the normal linear model assumptions. Thus reject the common-slope hypothesis. Because repeated measurements on the same chicken may be correlated, the exact reference distribution requires independent, homoscedastic Gaussian errors; the displayed ordinary regression output by itself does not establish that assumption.

Under the null hypothesis that treatment and recovery are independent, estimate each expected cell count byand form the Pearson chi-squared test of independence statistic ${\chi }^2=\sum (O-E{)}^2/E$. For a $2\times 2$ table its asymptotic null law is a chi-squared distribution with one degree of freedom. A second-order Taylor expansion of the generalized likelihood-ratio test statistic $-2\log \Lambda$ around the fitted counts gives this Pearson statistic, which explains its asymptotic null distribution.

Writing the four residuals over their fitted counts and simplifying givesFor $(a,b,c,d)=(50,10,15,5)$,whose one-degree-of-freedom p-value is about $0.41$. There is no evidence of an effect. One fitted count is only $3.75$, however, so the chi-squared approximation is questionable; Fisher's exact test would be preferable.

Estimate the signal by one-dimensional total variation denoising, which penalizes adjacent differences rather than the constant level:This is a fused Lasso objective. Its L1 norm penalty makes many adjacent differences exactly zero, producing a piecewise-constant fit. For change-point detection, estimate the locations by $\hat{S}=\{i:{\hat{\mu }}_i\ne {\hat{\mu }}_{i+1}\}$. The tuning parameter controls the strength of the penalty; it can be chosen, for example, using a validation criterion appropriate to the noise model.

For a general family of elementary nulls $H_1,\dots ,H_m$, the closed testing procedure forms every nonempty intersection $H_B={\bigcap }_{j\in B}{H}_j$ and gives it a local test of level $\alpha$. Reject $H_j$ precisely when all intersections with $j\in B$ are locally rejected. This gives strong control of the familywise error rate: let $T_0$ be the indices of the true elementary nulls. If $T_0$ is empty there is no false rejection. Otherwise, rejecting any $H_j$ with $j\in T_0$ requires rejection of the true intersection $H_{T_0}$. Its local test has rejection probability at most $\alpha$, henceThis proof of closed-testing control of the familywise error rate needs no independence of the tests.

For the interval procedure, a true tested interval is contained in one maximal constant block $B\in \mathcal{T}$. If $I\subseteq B$, the intervals over which $p_B$ takes its maximum are a subset of those used for $p_I$, so $p_I\ge p_B$. Therefore rejection of a true $I$ forces rejection of its containing block. Conversely, a rejected block is itself a rejected true null.

There is a small indexing issue in the PDF: its testing family excludes singleton intervals, while $\mathcal{T}$ may contain singleton blocks. Put ${\mathcal{T}}_{\ast }=\{B\in \mathcal{T}:|B|\ge 2\}$. Every true tested interval lies in one of these blocks, so the exact statement, using only hypotheses defined in the paper, isSingleton blocks contain no eligible tested interval and can be ignored. Equivalently, one may define their nulls as never rejected.

For each $B\in {\mathcal{T}}_{\ast }$, the term $J=B$ occurs in the maximum defining $p_B$, giving $p_B\ge n{q}_B/|B|$. A valid p-value is a super-uniform random variable under its null, so for $0\le \alpha \le 1$,The union bound and disjointness of the constant blocks now giveThis establishes weighted interval testing for a piecewise-constant mean even when the interval p-values are dependent. Values $p_I>1$ do not affect rejection at levels at most one.

There are two defects. The mixed fit supplies a restricted maximum likelihood, whereas the ordinary fit supplies an ordinary log-likelihood, so their difference is not a likelihood-ratio test statistic. Also, $H_0:{\tau }^2=0$ lies on the boundary of ${\tau }^2\ge 0$; the usual Wilks theorem does not give a ${\chi }_1^2$ null law.

A valid approach first refits the mixed model by ordinary maximum likelihood estimation, setting REML to false, and fits the same fixed effects under $H_0$. For $V={\tau }^{2}Z{Z}^T+{\sigma }^{2}I$, maximizeover $\beta ,{\sigma }^2>0,{\tau }^2\ge 0$, and separately over $\beta ,{\sigma }^2>0$ with ${\tau }^2=0$. Set $T=2({\hat{\ell }}_1-{\hat{\ell }}_0)\ge 0$.

For finite-sample calibration, use the location-scale invariant simulation test for a Gaussian variance component. With the actual $X,Z$ fixed, simulate $B$ independent vectors $Y^{(b)}\sim N(0,I_{176})$, refit both models by ordinary maximum likelihood estimation to each, and calculate $T_b$ in exactly the same way. Under $H_0$, $Y=X\beta +\sigma z$. Translating by a vector in the column space of $X$ and multiplying by a positive scalar preserves the statistic: both maximized log-likelihoods acquire the same scale constant. Thus this simulation has the correct null law without knowing $\beta$ or $\sigma$. A conservative Monte Carlo p-value isIn the usual regular limit with increasing independent groups, a variance-component likelihood-ratio test at a boundary instead uses $\frac{1}{2}{\delta }_0+\frac{1}{2}{\chi }_1^2$: for $T>0$, its approximate tail probability is $\frac{1}{2}\mathrm{Pr}({\chi }_1^2\ge T)$, and at $T=0$ the nonrandomized p-value is one. This approximation is not an exact guarantee for sixteen rats. Alternatively one can compare consistently defined restricted likelihoods with the same $X$ and simulate their restricted-ratio null distribution, as in the RLRsim documentation.

A procedure controls the familywise error rate at level $\alpha$ when, for every configuration of true nulls $I$,Let $i_{\ast }=\min I$. The step-down procedure can reject any true null only if it reaches and rejects $H_{i_{\ast }}$, which requires $p_{i_{\ast }}\le \alpha$. Since a valid p-value under its null satisfies $\mathbb{P}(p_{i_{\ast }}\le \alpha )\le \alpha$, the procedure controls FWER without any assumption on dependence among the p-values.

A distribution $P$ has the global Markov property for a directed acyclic graph $G$ when every D-separation statement $A{\perp }_{G}B\mid C$ implies the corresponding conditional independence $Z_A\perp \perp Z_B\mid Z_C$ under $P$.

If $G\in \mathcal{S}$ has nonadjacent vertices $j,k$, choose a topological ordering. Suppose $j$ precedes $k$. Then $j$ is a non-descendant and non-parent of $k$, so the directed local Markov property givesThe reversed ordering case is analogous. HenceUse the intersection-union test: reject $H_0$ exactly when $p_S\le \alpha$ for every $S$. Under $H_0$, at least one $H_{S_{\ast }}$ is true, soThis is nontrivial because it rejects whenever every tested conditional independence is rejected.

The null hypothesis is that price and score are independent, equivalently that the independence log-linear model for a two-way contingency table is correct. Against the saturated alternative, the reported Poisson deviance is the likelihood-ratio test statisticUnder the null and the usual large-sample regularity assumptions, Wilks theorem givesThe observations must be independent, the cell probabilities must not lie on the boundary of the parameter space, and the expected counts must be large enough for the chi-squared asymptotic approximation. The fitted counts are either $31/3$ or $28/3$, so the customary expected-count check is comfortably satisfied. Since(equivalently, the p-value is about $0.329$), we do not reject the null at the $1\%$ significance level. The data provide no significant lack of fit for independence.

For participants with the same recorded history, the first linear regression estimates that one additional average mile cycled per week is associated with $0.04917$ fewer asthma episodes. The corresponding $p$-value is $0.00761$, giving evidence against a zero conditional association.

The second model estimates that, among participants with the same history, receiving the incentive is associated with $0.91387$ more episodes. The third estimates that receiving the incentive is associated with $1.73282$ more miles cycled per week. Both incentive coefficients have very small p-values. Because the incentive was assigned in a randomized controlled trial, these last two comparisons estimate a causal effect of the incentive.

The familywise error rate isThe Bonferroni correction rejects $H_i$ when $p_i\le \alpha /m$. Since a valid p-value is super-uniform under its null, the union bound givesNo independence assumption is needed.

For the closed testing procedure, form the intersection hypothesis $H_I={\bigcap }_{i\in I}{H}_i$ for every nonempty $I\subseteq \{1,\dots ,m\}$ and choose a level-$\alpha$ local test for each $H_I$. Reject $H_i$ exactly when every $H_I$ with $i\in I$ is rejected by its local test. If any true $H_i$ is rejected, then the intersection $H_{I_0}$ of all true nulls is rejected. Since its local test has significance level $\alpha$,This is the closed-testing control of the familywise error rate.

Write $W={\sum }_{j=1}^m{w}_j$. Procedure (A) is the Weighted Bonferroni correction: it rejects $H_i$ when $p_i\le \alpha w_i/W$. Therefore

Procedure (B) is the Weighted Holm step-down procedure. Let $k$ be the first rank in the ordering $q_{(1)}<\cdots <q_{(m)}$ whose hypothesis is a true null, and put $W_0={\sum }_{i\in I_0}{w}_i$. If any true null is rejected, the procedure reaches step $k$ andbecause every true null remains among ranks $k,\dots ,m$. Henceand another union bound gives FWER at most $\alpha$.

At step $k$, procedure (B) divides by the total weight still under consideration, which is no larger than $W$. Its critical values therefore increase as hypotheses are rejected. Moreover, if procedure (A) would reject a hypothesis, every earlier ordered $q$ also passes the initial threshold, so procedure (B) reaches and rejects it. Thus (B) contains every rejection of (A) and can make strictly more rejections while retaining the same strong FWER control.

Under $H_1$, the slope in this through-origin linear regression is $1$. Its observed Wald statistic isThe corresponding two-sided p-value is about $0.07$, so the fit does not give evidence against $H_1$ at the $5\%$ significance level.

The usual least-squares standard error assumes constant conditional variance. Goal counts are counts, and under the natural Poisson distribution model their variance equals their mean $x_i$, so observations with different expected goals are heteroscedastic. The reported homoscedastic standard error is therefore doubtful; a heteroscedasticity-robust calculation or a suitable count model would be more credible.

If $B=\varnothing$, then for every $j$ at least one of $H_j^{\beta }$ and $H_j^{\eta }$ is true. Sincevalidity of the p-value for whichever component null is true implies $\mathbb{P}(q_j\le t)\le t$. Therefore the union bound givesThis is the Bonferroni correction for the composite intersection alternatives.

Let $I_0=B^c$ and $m_0=|I_0|$. Each $q_j$ with $j\in I_0$ is a valid p-value by the argument in part c. If the Holm step-down procedure selects any index from $I_0$, let $r$ be the rank of the first such index. All $r-1$ earlier selections belong to $B$, soSelection through rank $r$ impliesConsequentlyThus the procedure controls the familywise error rate without requiring independence among the p-values.

A p-value $p_i$ for $H_i$ is super-uniform under that null: ${\mathbb{P}}_{H_i}(p_i\le u)\le u$ for every $u\in [0,1]$. For the Benjamini-Hochberg procedure, order $p_{(1)}\le \cdots \le p_{(m)}$, setwith $\hat{k}=0$ if the set is empty, and reject the hypotheses whose p-values are at most $p_{(\hat{k})}$.

The significance level is the chosen upper bound on a test's probability of rejecting the null hypothesis when it is true.

For $m\ge 1$ independent valid null p-values, the displayed inequality follows by applying the Benjamini-Hochberg procedure with all nulls true. Its false discovery rate then equals the probability of at least one rejection. Valid means $\mathrm{Pr}(p_i\le t)\le t$ for $0\le t\le 1$; independence is a sufficient hypothesis, and arbitrary dependence is not covered by this proof.

For the intersection of $m$ null hypotheses, reject when ${\min }_{i}m{p}_{(i)}/i\le \alpha$. The Simes inequality establishes a level-$\alpha$ test under independence of valid true-null p-values. The displayed quantity is its combined p-value.