Common-clock contribution to a maximum receiver horizon

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.