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Finite ratio moments under a positive denominator cutoff follow from a simple pointwise bound: dividing by a denominator known to be at least $c>0$ cannot amplify an accepted numerator by more than $1/c$. The conclusion still needs a numerator-moment assumption and a nonzero acceptance probability. It applies to the pulse-contrast denominator acceptance rule, including dependent numerator and denominator noise.

Let $A=\{C\ge c\}$, $p_A=\mathbb{P}(A)>0$, and let $p>0$. If $\mathbb{E}|S{|}^p<\infty$, thenTo prove it, on $A$ one has $|S/C{|}^p\le |S{|}^p/c^p$. Integrate over $A$ and divide by $p_A$:Here $1_A$ is an indicator function. The accepted ratio is defined on $A$; any indicator-weighted expression is assigned zero outside $A$, so this proof does not divide by an unaccepted zero denominator. No independence assumption was used.

For $p=2$, the accepted ratio has finite second moment and hence finite variance. The Cauchy-Schwarz inequality also gives a finite first absolute moment:This makes the accepted expected value well-defined. Standard sampling conclusions, such as a law of large numbers for independent accepted ratios with the same conditional distribution, still require their own sampling assumptions.

For an original example matching the two-state model in selection bias from a noisy denominator threshold, take $S=0.02$, $C\in \{4,6\}$ with equal probabilities and $c=5$. Then $p_A=1/2$ and $\mathbb{E}{S}^2=0.0004$, so the displayed second-moment bound is0.000032. The actual accepted ratio is constantly1/300, whose second moment is $1/90000\approx 0.00001111$. The bound is conservative. Using the conditional numerator moment directly gives the sharper bound $0.0004/25=0.000016$.

A cutoff alone is insufficient if the numerator has no required moment. If $C=2$ always and $S$ has a standard Cauchy distribution, accepting $C\ge 1$ retains every shot, but $S/2$ still has infinite first absolute moment and infinite second moment. Conversely, if $|S|\le M$ on accepted shots, the stronger deterministic bound $|S/C|\le M/c$ gives all positive moments without the factor $1/p_A$.

The cutoff and probability must remain explicit. A positive cutoff for each analysis does not supply a uniform bound if cutoffs approach zero or accepted numerator distributions change without control. Nor does moment finiteness identify an unbiased measurement: the accepted conditional expectation can differ from the desired pulse-population quantity, as selection bias from a noisy denominator threshold demonstrates.

Pulse-contrast ratio denominator noise is the distortion introduced when a noisy central pulse signal is used to normalize a side-delay signal. A shot-resolved pulse-contrast estimator uses $r=S/C$ after background correction, so errors in $C$ affect both precision and the distribution of the ratio.

For central values ${\mu }_S,{\mu }_C$ with ${\mu }_C>0$, a second-order local expansion givesThis approximation requires sufficiently small fluctuations and well-behaved ratio moments. It shows why an unbiased central-signal measurement need not give an unbiased normalized ratio. First-order uncertainty propagation alone describes local spread, not this mean shift; the GUM warns that nonlinear models can require higher-order terms.^[ref]

An original bounded example takes constant side signal $S=0.1$ and equally likely central readings $C=9$ or 11. The mean central reading is 10, so the ratio of means is 0.01. ButThe second-order prediction is 0.0101 because $\mathrm{Var}(C)=1$. Increasing the number of such shots estimates the shifted mean more precisely; it does not remove the nonlinear effect.

Correlation can instead make normalization exact. If $S=r_{0}C$ for every accepted shot with nonzero $C$, their ratio is always $r_0$, however variable the common multiplier. This is a different error structure from adding independent noise to the two channels. A shared additive background is also not generally equivalent to a shared multiplicative gain.

Near zero, even existence of a mean can fail. If $S=s_0>0$ is fixed and $C$ is uniform on $(0,1)$, every ratio is positive butThis mathematical example explains why “the denominator was never exactly zero” is insufficient. Detector noise is not asserted to follow this distribution.

A pulse-contrast denominator acceptance rule such as $C\ge C_{\min }$ changes the sampled population. Report the rule, rejected-shot count and dependence between brightness and pulse shape. The primary single-shot correlator's nonlinear gate and CCD readout motivate acquiring the relevant signals together, but do not guarantee that all ratios are statistically well conditioned.^[ref] Resolve the denominator and background experimentally before interpreting a long average of ratios.