Average estimates from several ties into one

Combines the ties' estimates cell by cell, as a weighted average.

Usage

pool_ties(
  ...,
  estimator = c("ind", "agg"),
  weights = c("inverse-variance", "equal", "exposure"),
  method = c("auto", "replicate", "analytic")
)

Arguments

...: estimator results, named by tie
estimator: "ind" or "agg"
weights: how to weight the ties: "inverse-variance" (the default), "equal", or "exposure" (each tie's denom.hat)
method: "replicate" to pool within bootstrap replicates, "analytic" for the inverse-variance formula, or "auto" (the default) to use replicates when every input has them

Value

a tie_pool: a tibble with one row per cell

The independence problem, and how to avoid it

The obvious way to pool is inverse-variance weighting, which assumes the estimates being combined are independent. In a multi-tie survey they are not: every tie is reported by the same respondents, so a respondent weighted up perturbs every tie at once.

Where that correlation is positive — which is what shared respondents usually produce — the independence formula gives an interval that is too narrow. It is not guaranteed to err that way, though: with negatively correlated ties it errs the other, and the size of the discrepancy is a property of the data rather than something that can be reasoned out in advance. The point is not the direction. It is that the formula is answering a question about ties that do not exist.

There is a way round it that costs nothing extra if the estimates were bootstrapped with the same replicate weights. Pool within each replicate, then take the spread across replicates: whatever correlation the ties have is already in there, because each replicate perturbs all of them together. That is method = "replicate", and it is the default wherever the inputs allow it.

method = "analytic" is the inverse-variance formula, available for when replicate estimates are not to hand. It warns, because the interval it produces is not one you should quote without saying how it was made.

Pooling is not the same as union

This treats each tie as a separate estimate of one quantity and averages them. It does not add the ties' reports together — that is union, it double-counts any alter reachable through more than one tie, and it needs information this package does not carry. See ties_union_check().

Examples