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Gaussian input uncertainty at a clipped acquisition-time threshold produces a mixed output distribution rather than an ordinary Gaussian error bar. Hold the decay time $\tau >0$ fixed and assume the logarithmic input ratio $X=\log (A\tau /\epsilon )$ has distribution $N(\mu ,s^2)$, with $s>0$. The required extension isThe directional derivative of a clipped logarithmic acquisition time explains the kink; here the full input distribution is propagated through it.

Writing $\Phi$ and $\phi$ for the standard normal cumulative distribution function and density, respectively,For $\ell >0$ the continuous part has density $\phi ((\ell /\tau -\mu )/s)/(\tau s)$. It integrates to less than one because the remaining probability is an atom at zero. Treating this density alone as the entire distribution loses that atom.

At the nominal threshold $\mu =0$, half the outcomes are zero and the conditional positive part has a half-normal distribution. Direct integration of the normal density givesFor an original example, $\tau =20\mathrm{ns}$ and logarithmic standard deviation $s=0.02$ give mean extension about $0.1596\mathrm{ns}$ and standard deviation $0.2335\mathrm{ns}$, although inserting the nominal input into the clipped formula gives zero. The probability of an extension above $0.80\mathrm{ns}$ is $1-\Phi (2)\approx 0.0228$.

The quantiles of a clipped Gaussian acquisition duration must likewise include the zero atom. A symmetric interval around the nominal zero duration cannot represent these probabilities. NIST's uncertainty discussion stresses the need to know and combine input distributions when assigning an interval probability.^[ref]

This is a stated uncertainty model for a conditional recording requirement, not a worst-case optical-energy guarantee. Uncertain decay time, correlation among the original envelope parameters or a non-Gaussian logarithmic ratio requires the corresponding joint propagation instead. Establishing the optical envelope remains a separate experimental task.

Logarithmic tail-duration change at the zero-extension boundary describes a recording requirement whose input ratio crosses the point where no further acquisition is needed. For fixed decay time $\tau >0$, put $R=A\tau /\epsilon$ and $x=\log R$. The physical extension isHere the logarithm applies to the input ratio, not to a zero output duration. Logarithmic condition numbers of a tail-acquisition requirement applies only where $L>0$.

At $x=0$, the left derivative is zero and the right derivative is $\tau$. The directional derivative of a clipped logarithmic acquisition time along a signed input change $h$ isThus no single ordinary derivative describes threshold-crossing perturbations. Nevertheless, the duration is stable in absolute units:This follows because the positive-part map has Lipschitz continuity with constant one.

For an original interval example, take $\tau =25\mathrm{ns}$ and a justified input-ratio range $0.99\le R\le 1.02$. The corresponding duration range is $0\le L\le 25\log 1.02\approx 0.4951\mathrm{ns}$. Its lower part is clipped at zero. Linearizing only on the zero-duration side would incorrectly predict no possible extension even though the allowed set crosses the threshold. A relative error divided by a nominal $L=0$ is undefined, while this absolute bound remains meaningful.

If $\tau$ is uncertain too, optimize the full expression over the justified joint parameter set rather than treating this fixed-$\tau$ inequality as a complete uncertainty analysis. The tail model must bound optical power throughout the missed interval. NIST's calibration apparatus measured detector/cable/oscilloscope response separately, illustrating why a recorded electrical decay cannot automatically certify such an optical envelope.^[ref] The clipping analysis is an original property of the stated conditional acquisition model.