Channel switching under uncertainty in a combined receiver horizon

Channel switching under uncertainty in a combined receiver horizon occurs when uncertainty changes which receiver has the latest accessible input time. For channel $j$, let $H_j=T_j-d_j$ be its last output time minus its effective input-to-output offset. The combined horizon is $H={\max }_j{H}_j$, as in receiver-latency uncertainty in the inferred input-time horizon. Away from ties a small perturbation may preserve the maximizing channel; near a tie it need not. Propagating only that channel's variance omits the switching operation.

An original two-channel model makes the effect exact. Suppose $H_1$ and $H_2$ are independent normal random variables with common mean $\mu$ and standard deviation $s>0$. Independence giveswhere $\Phi$ is the standard normal distribution cumulative distribution function. This square is already enough to show that $H$ is not normal: $\mathrm{Pr}(H\le \mu )=1/4$.

To calculate the resulting bias relative to the common nominal horizon, write $H=(H_1+H_2+|H_1-H_2|)/2$. The difference has normal distribution with variance $2s^2$, hence $E|H_1-H_2|=2s/\sqrt{\pi }$. The sum and difference are independent in this model, givingFor $\mu =100\mathrm{ns}$ and $s=1\mathrm{ns}$, the combined mean is about $100.564\mathrm{ns}$ and its standard deviation is about $0.826\mathrm{ns}$. The formulas concern the uncertainty distribution of the horizon, not a promise of extra observable future illumination. The normal cdf/ppf routines provide direct evaluation of the probabilities.^[ref]

Shared clock or calibration errors invalidate the independence product. The distribution of the maximum of correlated receiver horizons must retain the joint timing model; NIST's general uncertainty guidance retains input covariance and requires distributional justification for coverage.^[ref] For hard bounds instead, $H_1\in [99,101]\mathrm{ns}$ and $H_2\in [100,102]\mathrm{ns}$ imply $H\in [100,102]\mathrm{ns}$ without selecting a fixed winning channel. Neither calculation establishes invertibility of the physical receiver.

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.

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.