Tracing Weight Growth Across Follow-Up

Why Track Weights Over Follow-Up?

A longitudinal treatment weight is a product of factors, one for each treatment decision. A moderate factor at each visit can therefore produce a large cumulative weight after many visits. Looking only at the final weight distribution can hide when that growth began: the early follow-up weights may be well behaved even though later treatment decisions create substantial dispersion. A time-indexed plot makes this pattern visible.

The goal here is to examine that growth, not to treat a plot as proof that the treatment model is correct. The derivations below describe weights formed from the true conditional treatment probabilities. In practice, probabilities are estimated, so finite-sample plots also reflect estimation error and sampling variation.

Definition: Let $A_k$ be treatment at visit $k$, $B$ baseline covariates, and $\bar L_k=(L_0,\ldots,L_k)$ the measured covariate history available before $A_k$. Write $\bar A_{k-1}=(A_0,\ldots,A_{k-1})$. Let $g_k(a_k\mid B,\bar L_k,\bar A_{k-1})$ be the observed conditional treatment distribution and $q_k(a_k\mid B,\bar A_{k-1})$ the chosen numerator distribution. The incremental factor and cumulative treatment weight through visit $k$ are $$ R_k=\frac{q_k(A_k\mid B,\bar A_{k-1})}{g_k(A_k\mid B,\bar L_k,\bar A_{k-1})}, \qquad W_k=\prod_{j=0}^{k}R_j, $$ with $W_{-1}=1$.

For discrete treatment, the support condition used here is that $q_k(a\mid B,\bar A_{k-1})=0$ whenever $g_k(a\mid B,\bar L_k,\bar A_{k-1})=0$, almost surely. For continuous treatment, the corresponding condition is absolute continuity of $q_k$ with respect to $g_k$. The denominator must be positive for the realized treatment whenever the numerator assigns it positive probability. Under these conditions, factors and cumulative weights are nonnegative, but need not be strictly positive: a realized treatment can have numerator probability zero.

Two Properties That Explain the Growth

The first property is that each incremental factor has conditional mean one under the observed treatment mechanism. Let $H_k=(B,\bar L_k,\bar A_{k-1})$ denote the information just before treatment at visit $k$. For discrete treatment,

E[R_k\mid H_k] =\sum_{a:g_k(a\mid H_k)>0}g_k(a\mid H_k) \frac{q_k(a\mid B,\bar A_{k-1})}{g_k(a\mid H_k)} =\sum_a q_k(a\mid B,\bar A_{k-1})=1.

The support condition justifies the cancellation: any treatment value assigned zero probability by $g_k$ also receives zero probability under $q_k$. In the continuous case the sum is replaced by an integral of the density ratio against $g_k$.

Result: If the conditional treatment probabilities are correctly specified and the support condition holds at every visit, then $E[W_k]=1$ for each $k$. Further, $E[W_k^2]\geq E[W_{k-1}^2]$, so the variance of the cumulative weight is nondecreasing with follow-up.

For the mean statement, $W_{k-1}$ is determined before $A_k$. Taking conditional expectations first given $H_k$, and then over $H_k$, gives

E[W_k] =E[W_{k-1}R_k] =E\{W_{k-1}E[R_k\mid H_k]\} =E[W_{k-1}].

Since $W_{-1}=1$, induction yields $E[W_k]=1$. This is a population statement, not a guarantee that the sample average of estimated weights equals one at every visit.

For the second moment, define the conditional second-moment factor

C_k(H_k)=E[R_k^2\mid H_k] =\sum_{a:g_k(a\mid H_k)>0} \frac{q_k(a\mid B,\bar A_{k-1})^2}{g_k(a\mid H_k)}.

Cauchy–Schwarz gives $C_k(H_k)\geq 1$. In detail, on the set where $q_k(a)>0$,

\left(\sum_a q_k(a)\right)^2 \leq \left(\sum_{a:q_k(a)>0}\frac{q_k(a)^2}{g_k(a)}\right) \left(\sum_{a:q_k(a)>0}g_k(a)\right) \leq C_k(H_k).

The left side is one, and the final inequality uses $\sum_{a:q_k(a)>0}g_k(a)\leq 1$. Therefore

E[W_k^2] =E[W_{k-1}^2 C_k(H_k)] \geq E[W_{k-1}^2].

Because each cumulative weight has mean one, $\operatorname{Var}(W_k)=E[W_k^2]-1$, so its variance cannot decrease. Growth is strict if $P\{W_{k-1}>0,\ C_k(H_k)>1\}>0$. If $q_k=g_k$ at a visit, then $R_k=1$ and that visit adds no dispersion. These arguments prove a population property under the stated model; estimated weights in a finite sample need not show a smooth or strictly increasing empirical spread. $\square$

Worked Examples

Worked Example: Following Four Cumulative Trajectories

Suppose treatment is binary and the numerator probability is $q(A_k=1)=q(A_k=0)=0.5$ at both visits. For illustration, consider four observed treatment paths. The observed probabilities of the realized treatments are shown in the table; the incremental factor is $q(A_k)/g(A_k)$.

Path	Visit 0: $A_0,\ g_0(A_0)$	$R_0$	Visit 1: $A_1,\ g_1(A_1)$	$R_1$	$W_0$	$W_1$
1	1, 0.8	0.5/0.8 = 0.625	1, 0.8	0.5/0.8 = 0.625	0.625	0.390625
2	1, 0.8	0.625	0, 0.2	0.5/0.2 = 2.5	0.625	1.5625
3	0, 0.2	2.5	1, 0.8	0.625	2.5	1.5625
4	0, 0.2	2.5	0, 0.2	2.5	2.5	6.25

Each cumulative value is the product of the incremental factors; for example, path 4 has $W_1=2.5\times2.5=6.25$. A plot connecting each path’s $W_0$ and $W_1$ would show that some trajectories rise while others fall. The increasing population variance result does not mean every individual weight rises, nor does this four-path illustration claim that these paths form a representative random sample.

Worked Example: Calculating the Increase in Second Moment

At each of two visits, suppose treatment is binary, $g(A_k=1)=0.9$, $g(A_k=0)=0.1$, and $q(A_k=1)=q(A_k=0)=0.5$. Assume the same conditional probabilities apply at each visit. At one visit, the incremental factors are $R=0.5/0.9=5/9$ when $A=1$ and $R=0.5/0.1=5$ when $A=0$.

The conditional mean is $$ E[R]=0.9(5/9)+0.1(5)=0.5+0.5=1. $$ The second moment is $$ E[R^2]=0.9(25/81)+0.1(25) =\frac{25}{90}+2.5 =\frac{25}{9}\approx2.778. $$ Thus $\operatorname{Var}(R)=25/9-1=16/9\approx1.778$.

If the two visit-specific treatment draws are independent with these same probabilities, then $W_1=R_0R_1$, and independence gives $$ E[W_1^2]=E[R_0^2]E[R_1^2]=(25/9)^2=625/81\approx7.716. $$ Hence $\operatorname{Var}(W_1)=625/81-1=544/81\approx6.716$, compared with the one-visit variance $16/9\approx1.778$. The calculation illustrates how repeated nonconstant ratios can compound dispersion.

Worked Example: A Zero Weight and an Invalid Log Plot

Let $q(A=1)=1,\ q(A=0)=0$, while $g(A=1)=g(A=0)=0.5$. The support condition holds because $q$ assigns no mass outside the support of $g$. The incremental weight is $R=1/0.5=2$ for $A=1$, and $R=0/0.5=0$ for $A=0$.

The mean remains one: $E[R]=0.5(2)+0.5(0)=1$. Its second moment is $E[R^2]=0.5(4)+0.5(0)=2$, so its variance is one. However, for an individual with $A=0$, $\log R=\log 0$ is undefined. A log-weight plot must not silently discard or alter that value. One can show the zero weights on the original scale and use a log scale only for strictly positive weights, with the excluded or separately marked zero values clearly reported.

Plotting Weight Growth in Practice

For each follow-up time, calculate $W_{ik}$ for every individual still represented in the treatment-weight calculation. Plot time on the horizontal axis and weight on the vertical axis. Useful displays include a set of individual trajectories when the sample is small, or time-specific summaries—such as the median, upper quantiles, and maximum—when trajectories would overlap too heavily. State how the plotted sample is defined at each time; otherwise a changing set of individuals can look like changing weights.

The ordinary scale preserves zeros and makes absolute differences visible. A log-scale companion can make multiplicative growth easier to see, but only when the relevant weights are strictly positive. Under that additional condition,

\log W_{ik}=\sum_{j=0}^{k}\log R_{ij}.

The difference $\log W_{ik}-\log W_{i,k-1}=\log R_{ik}$ then identifies the contribution of visit $k$ to that individual’s cumulative log weight. If an incremental numerator probability is zero for a realized treatment, this identity is unavailable for that individual at that time and later cumulative weights remain zero.

Read the graph for the time at which the distribution starts to spread, whether upper quantiles rise gradually or abruptly, and whether a small number of trajectories account for the visible extremes. Compare time-specific plots on a consistent scale when possible. A changing scale can conceal growth or exaggerate small differences.

A sharp rise after a particular visit points to examining that visit’s incremental ratios and fitted conditional probabilities.
A widening upper tail with a relatively stable median suggests that growth is concentrated among some individuals, rather than shared by the whole sample.
A sudden flattening or collapse may reflect factors near one, zero numerator probabilities, a changed analysis sample, or a data-processing error; the plot alone cannot distinguish these explanations.
Small empirical fluctuations in quantiles do not contradict the nondecreasing population variance result. Quantiles are not variances, and sample summaries vary randomly.

Common Pitfalls and Limits

A common mistake is to infer that every cumulative weight must increase because it is a product. Products can decrease whenever a factor is between zero and one. The result proved above concerns the population second moment under the specified treatment mechanism, not the direction of each person’s trajectory.

Another mistake is to read increasing empirical spread as proof of positivity failure or a misspecified model. It is a diagnostic signal, not a diagnosis: real treatment probabilities, estimated probabilities, sampling variation, and changing follow-up composition can all affect the display. Conversely, a stable-looking plot does not establish correct model specification or adequate support. In particular, finite plots may not reveal rare treatment histories.

A time-indexed plot complements rather than replaces a single-time weight distribution, log-scale inspection for positive weights, or checks of model construction. Compared with a final-time-only display, it localizes when cumulative growth occurs; compared with plotting incremental factors alone, it shows the accumulated consequence of their sequence. Neither view by itself establishes that the weighted analysis is unbiased. The variance result also assumes true conditional distributions and the support condition; estimated probabilities can produce empirical weight means and spreads that depart from the population identities.

Check Your Understanding

Answer each question, then compare with the solutions below.

Under what condition does $E[R_k\mid H_k]=1$ hold for a discrete treatment?
Does nondecreasing population variance imply that every person’s cumulative weight is nondecreasing? Explain.
With $q(1)=0.4,\ q(0)=0.6,\ g(1)=0.8,\ g(0)=0.2$, calculate the two possible incremental weights and verify their mean under $g$.
When is the log cumulative-weight identity valid, and what should be done with zero weights?

Solutions. (1) The numerator must be a probability distribution supported wherever the observed treatment distribution has positive mass; then summing $g(a)q(a)/g(a)$ over the observed support gives $\sum_a q(a)=1$. (2) No. A factor below one reduces an individual’s product, and another factor above one increases it; the theorem concerns the population second moment and variance. (3) For $A=1$, $R=0.4/0.8=0.5$; for $A=0$, $R=0.6/0.2=3$. Thus $E_g[R]=0.8(0.5)+0.2(3)=0.4+0.6=1$. (4) All relevant incremental factors must be strictly positive, so the cumulative weights are strictly positive. Show zero weights on the original scale and do not take their logarithm.