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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.

The Gram matrix representation and $Y_{ii}=1$ give $\Vert v_i{\Vert }_2=1$. For a standard Gaussian random vector $Z$, each $⟨v_i,Z⟩$ is a standard normal scalar, so the zero event has probability zero. Choose either sign convention at zero; the resulting vector is almost surely a feasible sign vector. Consequently $y^{T}Ay\le v^{\ast }$ almost surely.

The Gaussian sign-correlation identity givesThe last equality uses symmetry of the real matrices. HenceThe inverse sine is applied entrywise, not through spectral matrix functional calculus.

For completeness, the correlation coefficient identity has a geometric proof. If the angle between two unit vectors is $\theta$, the isotropic Gaussian random vector direction in their two-dimensional span gives opposite signs on angular sectors with probability $\theta /\pi$. Thus the sign product has expectation $1-2\theta /\pi =(2/\pi )\arcsin (\cos \theta )$. Parallel and antiparallel pairs give the endpoint values $1$ and $-1$ directly. This is Gaussian hyperplane rounding, and needs no independence between the rounded coordinates.

Assume the usual positive initial stock price and strike, so the logarithm is defined, and interpret $U$ as a classical solution of the displayed partial differential equation. For a fixed complex $z$, put $F_t(z)=S_t^{z}U(t,{\sigma }_t,z)$, defining $S_t^z=\exp (z\log S_t)$. The Itô formula for the two Brownian motions with correlation coefficient $\rho$ givesThe partial differential equation cancels the entire drift. By the permission to treat the resulting local martingales as true martingales, and the terminal condition, $F_t(z)=\mathbb{E}[S_T^z\mid {\mathcal{F}}_t]$.

The bounded spot volatility ensures $\mathbb{E}{S}_T^{x_0}<\infty$: the Itô formula for a real power and localization bound its moment by $S_0^{x_0}\exp (x_0(x_0-1)\Vert \sigma {\Vert }_{\infty }^{2}T/2)$. Rewrite the proposed integrand asNow $|F_t(x_0+iy)|\le \mathbb{E}[S_T^{x_0}\mid {\mathcal{F}}_t]$, an integrable bound independent of $y$. Conditional Fubini's theorem and part (b) therefore implyThis route justifies the contour exchange without assuming bounds on $U$ uniform in all complex $z$. Cash is constant, $S$ is a true martingale under the stated allowance, and the displayed $C$ is a true martingale. The original measure is thus an equivalent martingale measure for all three assets. The fundamental theorem of asset pricing gives the market has no arbitrage under the usual admissible trading convention.

The two predictors have correlation coefficient $\rho =0.9813846$, so there is severe multicollinearity. Their centered slope Gram matrix isThus either partial slope has variance ${\sigma }^2/[n(1-{\rho }^2)]$. Relative to a univariate regression with the same error variance, the two-predictor variance inflation factor and standard-error factor areThe small eigenvalue $n(1-\rho )$ makes the contrast between the two slope effects particularly poorly estimated. The univariate regressions test marginal association with the response; the bivariate tests ask whether one predictor adds an effect while holding its nearly identical companion fixed. Their much larger standard errors therefore make the partial tests insignificant without contradicting the marginal significance. The two fits also estimate slightly different residual variances, explaining why the reported standard-error ratio is not exactly 5.21.

For distinct sibling supernovae $s$ and $s^{\prime }$ in the same galaxy, the shared fluctuation is the only common random term. Independence and linearity of covariance therefore giveTheir correlation coefficient isThis is the intraclass correlation coefficient of the galaxy-level random intercept model.

Take all components to be centered and let their marginal variances be $V_A,V_C,V_E$. A linear structural equation model iswhere $C_i,E_{i1},E_{i2}$ are mutually independent and the $E_{ij}$ are identically distributed. The causal directed acyclic graph has arrows

For monozygotic twins, set $A_{i1}=A_{i2}=G_i$ with $\mathrm{Var}(G_i)=V_A$; the genetic cause is completely shared. For dizygotic twins, one explicit construction iswhere $G_i,G_{i1},G_{i2}$ are independent with variance $V_A/2$. Then both additive genetic terms have variance $V_A$, while their covariance is $V_A/2$ and their correlation coefficient is $1/2$. In a fuller graph, the shared $G_i$ points to both genetic components and the unique $G_{ij}$ points only to its own component.

Provided $X$ and $Y$ have finite positive variances, their correlation coefficient isIt lies in $[-1,1]$ by the Cauchy--Schwarz inequality.

A random vector $X$ is positively associated if $\mathrm{Cov}(f(X),g(X))\ge 0$ for every pair of coordinatewise nondecreasing bounded functions $f,g$. The same inequality holds when both are nonincreasing, by negating them. Repeatedly applying it to products of nonnegative increasing indicators gives $\mathbb{P}({\bigcap }_i{E}_i)\ge {\prod }_i\mathbb{P}(E_i)$ for increasing events; the same is true for decreasing events. Thus associated rejection indicators have familywise error rate no greater than independent indicators with the same marginals. Pairwise positive correlation coefficients alone do not give this stronger property. The classical definition is in Esary, Proschan and Walkup (1967).

Responses observed on the same process or individual at different times may have nonzero correlation coefficients. Such serial correlation can arise from a shared random effect or from history-dependent conditional means. A scalar dispersion parameter adjustment alone does not represent this temporal dependence.

For two centered predictors with squared norms $n$ and correlation coefficient $\rho$, the ordinary least squares estimate of either partial slope has variance ${\sigma }^2/[n(1-{\rho }^2)]$. Compared with the same predictor alone and the same error variance, the inflation factor is $1/(1-{\rho }^2)$. The eigenvalues of the slope Gram matrix are $n(1+\rho )$ and $n(1-\rho )$.

Independent normal arm means with variances $v_0,v_1,v_2$ give treatment-contrast Wald test statistics whose correlation coefficient is $v_0/\sqrt{(v_0+v_1)(v_0+v_2)}$. The common control mean creates positive dependence between comparisons.