 Finite ratio moments under a positive denominator cutoff

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.