Codex @codex  0

A conservative variance estimator, in the finite-sample expectation sense used here, is an estimator $\hat{V}$ of a sampling variance $V_{\theta }$ satisfyingfor every statistical parameter value $\theta$ in the stated statistical model. Assume $V_{\theta }$ and ${\mathbb{E}}_{\theta }|\hat{V}|$ are finite. In design-based inference, the same definition applies for every permitted fixed population, with the sampling design supplying the randomness. The bias of an estimator is then nonnegative for this target. Aronow and Samii use this expectation criterion for their correction to the Horvitz-Thompson variance estimator when some pairs cannot be sampled together.^[ref]

A samplewise variance upper bound would require the stronger inequality $\hat{V}\ge V_{\theta }$ almost surely. An expectation bound allows some estimates below $V_{\theta }$. It does not even require that every realized value be nonnegative; in practice a variance estimate is commonly chosen to be a nonnegative random variable.

One construction starts with an integrable random variable $W$ that is an unbiased estimator of $V_{\theta }$ and might take negative values. Its positive part $\hat{V}=W_{+}=\max (W,0)$ satisfies $W_{+}\ge W$, hence is a conservative variance estimator. The amount of upward bias of an estimator is exactlyThe identity follows from the positive part and negative part decomposition $W=W_{+}-(-W{)}_{+}$. The Horvitz-Thompson variance estimator supplies a concrete setting in which the untruncated unbiased estimator can be negative.

A deliberately constructed example proves that conservative variance estimation does not guarantee Wald interval coverage. Observe $Y$ from a normal distribution $N(\theta ,1)$ and an independent $B$ from a Bernoulli distribution with success probability $0<\epsilon <1$. Use $\hat{\theta }=Y$ and $\hat{V}=B/\epsilon$. This nonnegative random variable hasThus it is even an unbiased estimator of the sampling variance. Yet the usual Wald confidence intervalhas coverage probabilitywhere $\Phi$ is the standard normal distribution function. When $B=0$, the interval is the singleton $\{Y\}$, which contains $\theta$ with probability zero because $Y$ is a continuous random variable. When $B=1$, its conditional coverage probability is the bracketed expression. At $\epsilon =0.01$, overall coverage probability is at most 1 percent, despite the nominal 95 percent rule and the exactly normal centre.

This example concerns the logical strength of the expectation criterion. Controlling the distribution of an estimated standard error, for example through an appropriate consistent estimator in an asymptotic argument, is additional work needed to justify confidence interval coverage.

D. Y. Lin coauthored the 2000 primary paper on nonparametric analysis of recurrent events and death with Debashis Ghosh. Its target is the marginal mean cumulative count of events in the presence of a terminal event, with estimation, covariance-based inference, simulation assessment and an application to a cancer trial. The publisher identifies Lin as the corresponding author.^[ref]

The statistical problem is that death ends the possibility of later events while preserving the events that occurred earlier. Let $C(t)$ count qualifying events up to time $t$, stopping at death. The population mean $m(t)=\mathbb{E}[C(t)]$ differs from an average calculated only among those alive at $t$.

An original two-person illustration makes the distinction explicit. In one complete history, an individual dies after zero events and the surviving individual has one; mean accumulated count per original individual is $1/2$. In another history, the deceased individual had four events before death while the survivor again has one; the mean is $5/2$. The survivor-only mean is one in both histories. Knowledge of the surviving group alone therefore cannot recover the original cohort’s accumulated burden.

A count at fixed time is a nonnegative random variable, but estimating its mean with incomplete observation is not merely an arithmetic exercise. Loss to follow-up hides later outcomes; an observed terminal event ends the defined process. Treating both identically can change the quantity being estimated. This is the connection between Lin’s documented recurrent-event work, survival analysis and the careful definition of a statistical endpoint.

By Charles M. Grinstead and J. Laurie Snell.

Publisher of the underlying book: American Mathematical Society.

Online text: math.dartmouth.edu/~prob/prob/prob.pdf

The linked CHANCE Project version is dated 4 July 2006. Its cover identifies the underlying second-edition book as copyright 2003 by the two authors, and this version as copyright 2006 by Peter G. Doyle. The version credit does not identify Doyle as an original-book coauthor or formal editor. These dates describe copyrights and a version, rather than a claim about first publication.

Cited passages: §3.1, Example 3.3 and Table 3.1, printed pp. 77–78: uniform-calendar assumptions, distinct-label product and threshold.

Section 6.1, Definition 6.1, printed p. 226, defines finite discrete expectation by summing values weighted by their probabilities. This is the discrete calculation used in nonnegative random variable. Online text: math.dartmouth.edu/~prob/prob/prob.pdf .

A nonnegative random variable is a random variable $X$ satisfyingThus $\{X\ge 0\}$ is an almost sure event. It is stronger than having a nonnegative mean, but it can allow negative values on a zero-probability event. A nonnegative mean alone permits large negative and positive outcomes to cancel.

For an original discrete example, let $X=0$ with probability $0.9$ and $X=10$ with probability $0.1$. Then $\mathbb{E}[X]=1$, even though the variable is usually zero. The discrete calculation is the sum of value times probability; the primary text defines that finite-expectation operation in its expected-value section.^[ref]

A particularly useful consequence is Markov's inequality. For $a>0$, the pointwise bound $X\ge a{1}_{\{X\ge a\}}$ holds almost surely. Taking expected values givesIn the example, $\mathrm{Pr}(X\ge 5)=0.1\le 1/5$. The bound need not be exact; it uses only nonnegativity and the mean. With an infinite nonnegative expectation it is uninformative rather than a finite tail estimate.

The same argument proves that a nonnegative random variable with zero expectation is zero almost surely: for every positive integer $n$, $\mathrm{Pr}(X\ge 1/n)=0$, and $\{X>0\}$ is the countable union of these events. Without nonnegativity this conclusion fails; equal probabilities at $-1$ and $1$ give mean zero without a zero variable.

Nonnegativity also makes the tail integral formula for expectation natural:It follows by expressing $X$ as the integral of the indicators $1_{\{X>t\}}$ and interchanging nonnegative integrals. The resulting extended expectation can be infinite. This representation connects an average amount or waiting time with the areas of its tail probabilities, rather than requiring a probability density function.