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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.

Let ${\mathcal{G}}_n=\sigma (S_n,S_{n+1},\dots )$. Since $X_{n+j}=S_{n+j}-S_{n+j-1}$,Permuting $X_1,\dots ,X_n$ preserves their joint probability distribution and leaves every generator of ${\mathcal{G}}_n$ unchanged. Therefore, for each bounded ${\mathcal{G}}_n$-measurable $H$,This is the symmetry of exchangeable random variables. By the defining identity of conditional expectation, all $\mathbb{E}[X_j\mid {\mathcal{G}}_n]$ are equal almost surely. Summing these conditional expectations and using that $S_n$ is ${\mathcal{G}}_n$-measurable givesHence the conditional expectation of a summand given future partial sums isThe integrability of each $X_j$ justifies every conditional expectation above.