= 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 $\mathbb E|S|^p<\infty$, then
$$
\boxed{\mathbb E[|S/C|^p\mid A]
\le\frac{\mathbb E|S|^p}{p_Ac^p}<\infty.}
$$
To prove it, on $A$ one has $|S/C|^p\le |S|^p/c^p$. Integrate over $A$ and divide by $p_A$:
$$
\mathbb E[|S/C|^p\mid A]
\le\frac{\mathbb E[\mathbf1_A|S|^p]}{p_Ac^p}
\le\frac{\mathbb E|S|^p}{p_Ac^p}.
$$
Here $\mathbf1_A$ 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>:
$$
\mathbb E[|S/C|\mid A]\le\sqrt{\mathbb E[(S/C)^2\mid A]}.
$$
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 $1/90000\approx0.00001111$. 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 $C\ge1$ 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.
Back to article page
