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A common-clock contribution to a maximum receiver horizon is a residual timing error that survives in every channel's inferred input-time horizon. Write $H_j=h_j+C+{\epsilon }_j$, with common error $C$ independent of the remaining channel errors. Then the identityshows that selecting the latest channel does not average away the common term. With finite variances,Here $C$ means a component remaining after each channel's subtraction $T_j-d_j$. A clock contribution that already cancels inside that subtraction must not be added again; see receiver-latency uncertainty in the inferred input-time horizon.

For two equal nominal horizons $h_j=\mu$, take independent channel errors ${\epsilon }_j\sim N(0,s^2)$ and $EC=0$. The independent maximum moments derived in channel switching under uncertainty in a combined receiver horizon giveIf the common error is also normal, each channel has variance ${\sigma }_C^2+s^2$ and their correlation is ${\sigma }_C^2/({\sigma }_C^2+s^2)$, recovering the equal-channel model in distribution of the maximum of correlated receiver horizons.

For an original example, choose ${\sigma }_C=0.8\mathrm{ns}$ and $s=0.3\mathrm{ns}$. The combined mean is about $0.1693\mathrm{ns}$ above $\mu$, and its standard deviation is about $0.8375\mathrm{ns}$. Yet the difference between channel horizons has standard deviation only $\sqrt{2}s\approx 0.4243\mathrm{ns}$, because $C$ cancels there. Good agreement between channels therefore need not establish a precise absolute horizon.

A timing experiment needs an independently justified reference for the common component, in addition to channel-to-channel alignment observations. Shared-input covariance is explicitly retained in NIST's uncertainty law.^[ref] Repeated reads sharing the same clock realization cannot estimate away its absolute displacement simply by choosing the largest result.

The distribution of the maximum of correlated receiver horizons must use the joint timing law. Let $H=\max (H_1,H_2)$. Then $\mathrm{Pr}(H\le h)=\mathrm{Pr}(H_1\le h,H_2\le h)$, which generally is not the product of the marginal probabilities. The independent special case is derived in channel switching under uncertainty in a combined receiver horizon.

For an explicit correlated model, suppose $(H_1,H_2)$ has a bivariate normal distribution with common mean $\mu$, common standard deviation $s>0$, and correlation coefficient $-1<\rho <1$. With $z=(h-\mu )/s$, the cumulative distribution function and density of the maximum areHere $\varphi ,\Phi$ are the standard normal distribution density and cdf, and ${\Phi }_2$ is the bivariate cdf. The density follows by differentiating along both boundaries and using the conditional normal law. The official SciPy reference supplies a cdf method parameterized by the mean vector and covariance matrix; it is a numerical evaluation, not an assumption of channel independence.^[ref]

The equal-variance model also gives exact moments. Write $H=(S+|D|)/2$, with $S=H_1+H_2$ and $D=H_1-H_2$. They are independent normal variables, with $\mathrm{Var}(D)=2s^2(1-\rho )$. The absolute-normal moment then givesFor an original shared-clock example, let $\mu =100\mathrm{ns}$, $s=1\mathrm{ns}$, and $\rho =1/2$. Then $EH\approx 100.399\mathrm{ns}$ and the standard deviation is about $0.917\mathrm{ns}$. Positive correlation reduces the expected gain from selecting the larger horizon, while increasing its variance relative to the independent equal-channel case.

A common-clock contribution to a maximum receiver horizon can be modeled for nonnegative $\rho$ by $H_j=\mu +s(\sqrt{\rho }{Z}_0+\sqrt{1-\rho }{Z}_j)$ with independent standard normal inputs. As $\rho$ tends to one, both channels share the same uncertain horizon: selection does not remove the common error. Actual receiver timing correlations must be justified by calibration and reference reuse; the example neither measures them nor guarantees that the receiver is invertible.