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Pattern-Mixture Models for Dropout · Tutorial 214 of 499

Skewed Continuous Outcomes by Dropout Pattern

Learn how lognormal and gamma models describe skewed outcomes within dropout patterns and how their fitted parameters translate into means, medians, and other quantiles.

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What You'll Learn

  • Specify lognormal and gamma distributions for positive outcomes within each dropout pattern and visit.
  • Convert lognormal location and scale parameters into arithmetic means and quantiles.
  • Interpret gamma shape and scale parameters through the mean, variance, and quantile function.
  • Average pattern-specific means using dropout-pattern weights to obtain a population mean.
  • Recognize why pattern-specific models do not by themselves specify longitudinal dependence or identify unobserved outcomes after dropout.

Why Model Skewed Outcomes by Pattern?

A positive continuous outcome can have a long right tail: most measurements may be modest, while a few are much larger. Modeling such data with a symmetric Gaussian distribution on the original scale can imply impossible negative values or poorly represent the upper tail. In a pattern-mixture model, the distribution can instead be specified separately within each dropout pattern. This allows both the typical outcome and the degree of skewness to differ across patterns.

The focus here is the conditional distribution of a continuous outcome at a visit within a dropout pattern, and how its parameters translate into quantities reported on the original measurement scale. A distribution at each visit is not, by itself, a full joint model for a participant’s repeated measurements. In particular, it does not specify their within-person dependence.

Definition: Let \(D=d\) denote dropout pattern \(d\), and let \(Y_t\) be a positive continuous outcome at visit \(t\). A pattern-specific marginal model specifies the distribution of \(Y_t\mid D=d\), possibly conditional on covariates, separately for each relevant \(d\) and \(t\). A lognormal model specifies a normal distribution for the logarithm of the outcome; a gamma model specifies a gamma distribution for the outcome itself. Both distributions have support on positive values.

Lognormal Models and Original-Scale Summaries

For a lognormal model, logs should be taken of a dimensionless quantity. Let \(c\) be a fixed reference value with the same units as \(Y_t\), such as \(1\) mg/L when the outcome is a concentration. A convenient pattern- and visit-specific model is

$$ \log(Y_t/c)\mid D=d \sim N(\eta_{dt},\tau_{dt}^{2}). $$

Here \(\eta_{dt}\) is the mean of the log outcome and \(\tau_{dt}>0\) is its standard deviation. If covariates are included, \(\eta_{dt}\) may be a regression predictor; the formulas below apply at a fixed covariate value. The normality assumption is on the log scale, not the original scale.

Result: Under the lognormal model, the original-scale median is \(c\exp(\eta_{dt})\), the arithmetic mean is \(c\exp(\eta_{dt}+\tau_{dt}^{2}/2)\), and the \(q\)th quantile is \(c\exp(\eta_{dt}+\tau_{dt}z_q)\), where \(z_q\) is the \(q\)th standard normal quantile. These summaries have the units of \(Y_t\).

The mean formula follows by writing \(Y_t=c\exp(Z)\), where \(Z\sim N(\eta_{dt},\tau_{dt}^{2})\). The moment-generating function of a normal variable gives \(E[\exp(Z)]=\exp(\eta_{dt}+\tau_{dt}^{2}/2)\). For a quantile, the event \(Y_t\le y\) is equivalent to \(\log(Y_t/c)\le\log(y/c)\); applying the normal quantile gives the stated expression. The median is the special case \(z_{0.5}=0\).

The squared scale parameter matters: \(c\exp(\eta_{dt})\) is the median, not the arithmetic mean, unless the log-scale variance is zero. Increasing \(\tau_{dt}\) while holding \(\eta_{dt}\) fixed leaves the median unchanged but raises the arithmetic mean and spreads out the quantiles.

Worked Example: Converting a Lognormal Fit to Concentration Units

Suppose a hypothetical study models a positive concentration within dropout pattern B at one visit. The fitted model is \(\log(Y_t/c)\mid D=B\sim N(2.00,0.600^2)\), where \(c=1\) mg/L. Find the arithmetic mean, median, and 90th percentile.

The arithmetic mean is

$$ 1\text{ mg/L}\times\exp(2.00+0.600^2/2) =1\text{ mg/L}\times\exp(2.180) \approx 8.846\text{ mg/L}. $$

The median is \(1\text{ mg/L}\times\exp(2.00)\approx 7.389\) mg/L. Since \(z_{0.90}\approx1.2816\), the 90th percentile is

$$ 1\text{ mg/L}\times\exp(2.00+0.600(1.2816)) =1\text{ mg/L}\times\exp(2.76896) \approx 15.94\text{ mg/L}. $$

These are model-implied summaries for pattern B at this visit. The 90th percentile is not a confidence limit; it describes the fitted outcome distribution.

Gamma Models and Their Parameters

A gamma distribution is another positive-support model that can represent right-skewed outcomes. We use the shape–scale parameterization: \(Y_t\mid D=d\) has shape \(\kappa_{dt}>0\) and scale \(\theta_{dt}>0\). The scale has the units of the outcome, while the shape is dimensionless. Its density for \(y>0\) is proportional to \(y^{\kappa_{dt}-1}\exp(-y/\theta_{dt})\), with the normalizing constant \(\{\Gamma(\kappa_{dt})\theta_{dt}^{\kappa_{dt}}\}^{-1}\).

Result: Under the shape–scale gamma parameterization, \(E(Y_t\mid D=d)=\kappa_{dt}\theta_{dt}\) and \(\operatorname{Var}(Y_t\mid D=d)=\kappa_{dt}\theta_{dt}^{2}\). Its \(q\)th quantile is the inverse gamma cumulative distribution function evaluated at \(q\), with the fitted shape and scale. Except in special cases, this quantile is obtained numerically.

An alternative parameterization uses the mean \(m_{dt}=\kappa_{dt}\theta_{dt}\) and a dispersion parameter \(\phi_{dt}=1/\kappa_{dt}\). Then \(\operatorname{Var}(Y_t\mid D=d)=\phi_{dt}m_{dt}^{2}\), the scale is \(\phi_{dt}m_{dt}\), and the shape is \(1/\phi_{dt}\). This is useful when a regression model uses a log link for the mean: \(\log(m_{dt})\) can depend on pattern, visit, and covariates, while dispersion controls relative variability. The specified parameterization must be stated because “scale” and “rate” conventions are not interchangeable.

Worked Example: A Gamma Model for Positive Durations

For a hypothetical pattern-specific analysis, suppose a duration in hours follows a gamma distribution with shape \(\kappa=1\) and scale \(\theta=3\) hours. This gamma special case is exponential. Calculate its mean, variance, median, and 90th percentile.

The mean is \(\kappa\theta=1(3)=3\) hours, and the variance is \(\kappa\theta^2=1(3^2)=9\) hours squared. For shape 1, the cumulative distribution function is \(F(y)=1-\exp(-y/3)\). Setting \(F(y)=0.5\) gives \(y=-3\log(0.5)\approx2.079\) hours; setting \(F(y)=0.90\) gives \(y=-3\log(0.10)\approx6.908\) hours.

The median is below the mean, as expected for this right-skewed distribution. This calculation also illustrates why the scale is not the mean: the scale is 3 hours, and the mean happens to equal 3 hours here only because the shape equals 1.

Pattern-Specific Fits and Population Summaries

The pattern-specific distribution answers a conditional question: what distribution is being modeled among individuals in pattern \(d\)? A population summary combines these distributions using the pattern weights \(w_d=P(D=d)\), where the weights are nonnegative and sum to one. By the law of total expectation, if the pattern-specific means exist, the population mean at visit \(t\) is the weighted average of those means. This is an average of means, not an average of log-scale parameters.

$$ E(Y_t)=\sum_d w_d E(Y_t\mid D=d). $$

Worked Example: Weighting Lognormal Means Across Two Patterns

Suppose a hypothetical analysis has pattern weights \(w_A=0.700\) and \(w_B=0.300\). At a visit, both patterns have lognormal models with \(c=1\) mg/L and \(\eta=2.00\). Pattern A has \(\tau_A=0.400\), and pattern B has \(\tau_B=0.600\). Find the population mean concentration.

The pattern-specific means are

$$ E(Y_t\mid D=A)=1\text{ mg/L}\times\exp(2.00+0.400^2/2) =\exp(2.080)\text{ mg/L} \approx8.004469\text{ mg/L}, $$
$$ E(Y_t\mid D=B)=1\text{ mg/L}\times\exp(2.00+0.600^2/2) =\exp(2.180)\text{ mg/L} \approx8.846306\text{ mg/L}. $$

Weighting these on the original scale gives

$$ E(Y_t)=0.700(8.004469)+0.300(8.846306) =5.603128+2.653892 \approx8.257020\text{ mg/L}. $$

The population mean is therefore about \(8.257\) mg/L. Although the two patterns have the same log-scale mean and thus the same conditional median, their different log-scale variances give different conditional arithmetic means. This is why simply averaging the log-scale means would not produce the population arithmetic mean.

Assumptions, Checks, and Alternatives

A model is appropriate only if its assumptions make sense for the outcome and the intended interpretation. Lognormal and gamma distributions both require strictly positive values; ordinary versions do not directly accommodate exact zeros, negative values, or a point mass at zero. If zeros arise, their scientific meaning and recording process need to be addressed rather than hidden by an arbitrary small constant. A lognormal model assumes approximate normality of log outcomes within each pattern and visit; a gamma model assumes the gamma shape of the distribution on the original scale. Neither is automatically correct merely because the data are positive.

For a proposed fit, inspect the observed ranges and compare the fitted distribution with the data within patterns and visits where observations are available. A log-scale histogram or quantile plot can reveal substantial departures from lognormality; comparisons of empirical and fitted quantiles can expose systematic mismatch for a gamma fit. Very sparse patterns provide little information about their own distributional shape, so apparently precise pattern-specific summaries may be unstable. These checks flag possible problems; formal distributional diagnostics and uncertainty calculations require additional methods.

A Gaussian model on the original scale is a useful alternative when outcomes are reasonably symmetric and negative fitted values are scientifically plausible. A lognormal model is natural when multiplicative variation is plausible and the log outcome is reasonably normal. A gamma model is often convenient for positive data when a mean–variance relationship proportional to the squared mean is appropriate. Their means and upper quantiles can differ, so choice should be informed by the data and scientific context, not only by which model is easiest to fit.

A crucial limitation concerns dropout. Outcomes observed before dropout can inform pattern-specific distributions at those visits. A distribution for an outcome after members of a pattern have dropped out cannot be learned from that pattern’s observed outcomes at that visit alone. Such predictions require additional assumptions or information; fitting a lognormal or gamma distribution does not itself resolve that identification problem. Likewise, visit-specific distributions do not specify correlations among repeated outcomes. Those are separate modeling questions.

Common mistake: “The fitted lognormal location parameter is 2, so the mean is \(\exp(2)\).” This confuses the median with the arithmetic mean. Under \(\log(Y/c)\sim N(\eta,\tau^2)\), the median is \(c\exp(\eta)\), whereas the mean is \(c\exp(\eta+\tau^2/2)\). For a pattern-weighted population mean, first calculate each original-scale conditional mean, then weight and sum them.

Check Your Understanding

Try each question, then compare your reasoning with the solutions.

  1. A hypothetical lognormal outcome has \(c=1\) unit, \(\eta=1.50\), and \(\tau=0.400\). Find its median and arithmetic mean.
  2. A hypothetical gamma outcome has shape 1 and scale 2 hours. Find its mean and 90th percentile.
  3. Two patterns have weights 0.250 and 0.750 and conditional means 4 and 10 units. Find the population mean.
  4. Why does a pattern-specific distribution at observed visits not, by itself, determine the distribution after all members of that pattern have dropped out?

Solutions. (1) The median is \(\exp(1.50)\approx4.482\) units. The mean is \(\exp(1.50+0.400^2/2)=\exp(1.58)\approx4.855\) units. (2) The mean is \(1(2)=2\) hours. For shape 1, \(F(y)=1-\exp(-y/2)\), so the 90th percentile solves \(0.90=1-\exp(-y/2)\), giving \(y=-2\log(0.10)\approx4.605\) hours. (3) The weighted mean is \(0.250(4)+0.750(10)=1+7.5=8.5\) units. (4) No outcomes at that visit are observed for those individuals, so the observed data within the pattern do not determine their post-dropout distribution without additional assumptions or information.

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