Inverse-Probability Weighting for Panel Attrition

Why Weight Panel Respondents?

When panel members stop responding, the observed records can represent the original cohort unevenly. If response is more likely for some measured histories than others, an unweighted respondent average gives those histories more influence than their share of the target population warrants. Inverse-probability weighting corrects this imbalance by giving each observed record weight inversely proportional to its probability of being observed.

The method does not recover outcomes for particular nonrespondents. Instead, under explicit assumptions, it uses respondents to estimate population means or moments that would have been calculated had the full panel been observed. The target remains the original population or baseline cohort, not merely the set of people who replied.

Definition: For unit $i$ at wave $t$, let $R_{it}=1$ if the outcome is observed and $R_{it}=0$ otherwise. Let $H_{it}$ denote the measured history available before the response at wave $t$, and define the response probability $$ \pi_{it}=P(R_{it}=1\mid H_{it}). $$ For an observed record with $R_{it}=1$, its inverse-probability weight is $w_{it}=1/\pi_{it}$. A unit with a lower response probability receives a larger weight.

The history $H_{it}$ can include baseline covariates and previously observed outcomes or covariates. It must be available for the units whose response probabilities are being modeled, including nonrespondents at wave $t$. This tutorial considers weighting at a specified wave using that wave’s conditional response probability; it does not construct products of probabilities across successive waves.

Assumptions for Recovering a Panel Mean

Write $Y_{it}$ for the wave-$t$ outcome and consider the target mean $\mu_t=E[Y_{it}]$ in the population represented by the baseline panel. The key response assumption is that, after conditioning on measured history, response carries no additional information about the current outcome:

Assumptions for the mean identity: (1) $H_{it}$ is measured for the target units; (2) $\pi_{it}$ is correctly specified, or known; (3) $Y_{it}$ is conditionally mean independent of response given history, so $E[Y_{it}\mid H_{it},R_{it}=1]=E[Y_{it}\mid H_{it}]$; and (4) $\pi_{it}>0$ for histories occurring in the target population. Conditional independence $R_{it}\perp Y_{it}\mid H_{it}$ is a sufficient, stronger version of the third condition.

The positivity condition means that every relevant history has some chance of being observed. If a history has response probability zero, no respondent with that history can reveal its outcome distribution; inverse-probability weighting cannot fill that gap. The condition does not require all histories to have the same response probability.

Under these assumptions, the defining identity is

E\!\left[\frac{R_{it}Y_{it}}{\pi_{it}}\right]=E[Y_{it}].

When $R_{it}=0$, the numerator is zero and the unobserved $Y_{it}$ is not needed to calculate the weighted contribution. When $R_{it}=1$, the contribution is $Y_{it}/\pi_{it}$. The derivation makes clear why that reciprocal is the appropriate correction.

Condition on measured history.
Given $H_{it}$, the response indicator has conditional expectation $E[R_{it}\mid H_{it}]=\pi_{it}$.

Use conditional independence.
Conditional independence of response and outcome gives $E[R_{it}Y_{it}\mid H_{it}]=\pi_{it}E[Y_{it}\mid H_{it}]$.

Divide by the response probability.
Since $\pi_{it}$ is determined by history and is positive, $E[R_{it}Y_{it}/\pi_{it}\mid H_{it}]=E[Y_{it}\mid H_{it}]$.

Average over histories.
Taking expectations over $H_{it}$ on both sides yields $E[R_{it}Y_{it}/\pi_{it}]=E[Y_{it}]$.

The same argument works if conditional mean independence holds for $Y_{it}$, even when the full conditional distributions of $Y_{it}$ differ in other respects. The identity is about the mean; it does not imply that every feature of the outcome distribution is recovered by this mean calculation.

Estimating a Full-Population Mean

For a baseline cohort of $N$ units with response probabilities treated as known, an unnormalized inverse-probability estimator of the cohort mean is

\widehat{\mu}_{t,\mathrm{HT}}=\frac{1}{N}\sum_{i=1}^{N}\frac{R_{it}Y_{it}}{\pi_{it}}.

This form retains the baseline cohort size in the denominator. A commonly used normalized estimator instead divides the weighted outcome sum by the sum of respondent weights:

\widehat{\mu}_{t,\mathrm{H}}= \frac{\sum_{i=1}^{N}R_{it}Y_{it}/\pi_{it}} {\sum_{i=1}^{N}R_{it}/\pi_{it}}.

The first is an inverse-probability weighted Horvitz–Thompson form; the second is a normalized, or Hájek, form. In a finite sample their values need not agree. The first directly estimates the mean by weighting observed outcomes and dividing by $N$; the second rescales the weights to sum to the observed weighted population size. Under the stated assumptions and suitable large-sample conditions, both target the same population mean. A reported estimate should make clear which form was used.

Worked Example: A Weighted Mean With Two Histories

Suppose a hypothetical baseline panel has six units: three with history $A$ and three with history $B$. The response probabilities at the target wave are $\pi(A)=0.5$ and $\pi(B)=1$. Two $A$ units respond, with outcomes 10 and 14; the third does not respond. All three $B$ units respond, with outcomes 18, 20, and 22. Assume response is conditionally mean independent of the outcome given history, and that the stated response probabilities are correct.

The inverse probabilities are $1/0.5=2$ for history $A$ and $1/1=1$ for history $B$. The weighted outcome sum is

$$ 2(10+14)+1(18+20+22)=48+60=108. $$

The Horvitz–Thompson estimate divides by the six baseline units:

\widehat{\mu}_{t,\mathrm{HT}}=\frac{108}{6}=18.

For the normalized estimate, the weighted respondent count is $2(2)+1(3)=7$. Therefore

\widehat{\mu}_{t,\mathrm{H}}=\frac{108}{7}\approx15.43.

Both calculations use the same respondent outcomes and response probabilities, but the finite-sample estimates differ because their denominators differ. Neither calculation asserts that the missing outcome is known. The assumptions justify using the weighted respondents to estimate the full-cohort mean.

Inverse-Probability Weighting for Regression Moments

The same identity applies to regression moments. Let $X_{it}$ be a vector of regressors, including an intercept if desired, and define the full-population linear-projection coefficient $\beta_t$ as the value minimizing $E[(Y_{it}-X_{it}'b)^2]$ over $b$. If the required second moments exist and the regressor moment matrix is nonsingular, its normal equations are

E\!\left[X_{it}(Y_{it}-X_{it}'\beta_t)\right]=0.

For the observed records to identify these moments, assume the response probability is correctly specified from measured history, is positive for relevant histories, and response is conditionally independent of the variables in the moment given that history. A sufficient condition is $R_{it}\perp(Y_{it},X_{it})\mid H_{it}$. If the regressors are already known for all units, a suitably weaker condition on the outcome moment may suffice. Under the sufficient condition, apply the mean identity component by component to $X_{it}(Y_{it}-X_{it}'b)$:

E\!\left[ \frac{R_{it}}{\pi_{it}}X_{it}(Y_{it}-X_{it}'b) \right] = E\!\left[X_{it}(Y_{it}-X_{it}'b)\right].

Thus an inverse-probability weighted regression estimates the full-population projection by solving the sample version of the weighted normal equations. Equivalently, it minimizes the respondent-weighted sum of squared residuals, with each respondent weighted by $1/\pi_{it}$. These are population regression moments, not a claim that weighting makes an incorrectly specified conditional mean model correct.

Worked Example: A Weighted Regression From Respondent Records

Consider three hypothetical observed records $(X,Y)$: $(0,1)$, $(1,2)$, and $(2,2)$. Their response probabilities, conditional on measured histories, are $0.5$, $0.5$, and $1$, so their weights are $2$, $2$, and $1$. Assume the conditional independence and positivity conditions above hold for the regression moments. Fit a weighted line with an intercept.

The weighted sums are

\sum_i w_i=5,\quad \sum_i w_iX_i=4,\quad \sum_i w_iY_i=8,\quad \sum_i w_iX_i^2=6,\quad \sum_i w_iX_iY_i=8.

For example, the weighted outcome sum is $2(1)+2(2)+1(2)=8$, and the weighted cross-product sum is $2(0)(1)+2(1)(2)+1(2)(2)=8$. The weighted least-squares slope is

\widehat{\beta}_1= \frac{\sum_i w_iX_iY_i-(\sum_iw_iX_i)(\sum_iw_iY_i)/\sum_iw_i} {\sum_iw_iX_i^2-(\sum_iw_iX_i)^2/\sum_iw_i} = \frac{8-(4)(8)/5}{6-4^2/5} = \frac{1.6}{2.8} = \frac{4}{7} \approx0.5714.

The weighted means are $\bar X_w=4/5=0.8$ and $\bar Y_w=8/5=1.6$. The intercept is

\widehat{\beta}_0=\bar Y_w-\widehat{\beta}_1\bar X_w =1.6-\frac{4}{7}(0.8) =\frac{8}{7} \approx1.1429.

The fitted weighted line is approximately $1.1429+0.5714X$. The slope has units of $Y$ per unit of $X$; its interpretation concerns the full-population linear projection under the weighting assumptions. It is not, by itself, evidence of a causal effect.

What the Weights Do—and Do Not Do

Inverse-probability weights re-balance observed records according to measured response histories. A respondent with a history shared by many target units but a low response probability represents more of those units than a respondent from a history that almost always responds. The correction relies on information in $H_{it}$: if response depends on unmeasured features of the outcome even after conditioning on that history, the identity need not hold.

Weights also do not guarantee a precise estimate. A small response probability produces a large reciprocal weight, so a small number of records can have substantial influence. This is a reason to examine the response model and the behavior of estimated weights in applied work. How to stabilize weights and assess positivity and attrition diagnostics requires additional tools beyond the basic identification argument here.

Common Mistake

A common error is to weight respondents by their response probabilities rather than by their reciprocals. That gives less influence to people who were unlikely to respond, even though each such respondent stands in for more members of the target population. For example, a response probability of $0.5$ implies weight $1/0.5=2$, not $0.5$. Another error is to interpret the weighted estimate as automatically unbiased in any data set: the correction depends on correct response probabilities, measured history, conditional independence for the target quantity, and positivity.

Key takeaway: Under conditional independence given measured history and positive, correctly specified response probabilities, multiplying each observed outcome or regression moment by $1/\pi_{it}$ recovers its full-population expectation. The assumption and target must be stated before interpreting a weighted estimate.

Check Your Understanding

Use the definitions and assumptions above to answer these questions.

If a unit’s response probability given its history is $0.25$, what inverse-probability weight does its observed record receive, and why?
State the conditional mean-independence assumption that supports inverse-probability estimation of a panel outcome mean.
Why can the Horvitz–Thompson and normalized weighted mean differ in a finite sample?
What additional variables must satisfy an appropriate response-independence condition when using weighted regression moments?
What does positivity rule out, and why can weighting not recover outcomes for a history with zero response probability?