One Draw Is One Possible Completion
A stochastic imputation can look more realistic than a deterministic one because it adds random variation to the missing values. But randomness in the imputation procedure does not make one completed dataset a record of all the values that could plausibly have been imputed. It is still just one draw. To understand what it leaves hidden, distinguish two kinds of variation: the spread of values within one completed dataset and the differences across possible completed datasets.
The first kind describes how dispersed the observed and imputed values are in a particular completed dataset. The second describes how the imputed values, or a statistic calculated from them, could change if the stochastic imputation were repeated while the observed data stayed fixed. These are different questions and can have very different answers.
The observed entries ordinarily stay fixed across repeated completions; it is the imputation draw that changes. For repeated draws to describe meaningful uncertainty, the imputation procedure must represent the relevant predictive uncertainty. In a regression-based procedure, this can include variation in outcomes around their predictions and, if the procedure is designed to include it, uncertainty in estimated model parameters. A draw that adds residual variation but holds estimated parameters fixed represents only the uncertainty included by that particular procedure.
A Small Repeated-Draw Example
Suppose a hypothetical dataset has four observed measurements and two missing measurements. The observed values are 2, 4, 6, and 8 units. Consider four possible random completions, each generated by the same stipulated imputation procedure. These draws are for illustration; they are not claims about a real dataset or a fitted model.
Worked Example: Comparing Within- and Across-Dataset Variation
For each completion below, calculate the completed-data mean and sample variance. Then compare the variation of the means across the four completions with the sample variance within any one completion.
| Completion | Imputed values | Completed-data mean | Completed-data sample variance |
|---|---|---|---|
| 1 | 4, 6 | 5.000 | 4.400 |
| 2 | 2, 8 | 5.000 | 7.600 |
| 3 | 5, 7 | 5.333 | 4.667 |
| 4 | 1, 3 | 4.000 | 6.800 |
For completion 1, the six values are 2, 4, 6, 8, 4, and 6. Their mean is \((2+4+6+8+4+6)/6=5\), and their sample variance is
For completion 2, the values are 2, 4, 6, 8, 2, and 8. Their mean is also 5, but their sample variance is \((9+1+1+9+9+9)/5=38/5=7.600\). The other table entries follow the same calculation. For completion 3, the mean is \(32/6=5.333\), and its squared deviations sum to \(23.333\), giving \(s^2=23.333/5=4.667\). For completion 4, the mean is \(24/6=4\), and the squared deviations sum to 34, giving \(s^2=34/5=6.800\).
The four completed-data means are 5, 5, 5.333, and 4. Their mean is \(19.333/4=4.833\), and their sample variance across these four draws is
The sample variance within completion 1, 4.400, describes spread among that completion’s six values. The across-draw variance, about 0.333, describes how much the completed-data mean varies among these four illustrative draws. They summarize different objects, so it is not meaningful to treat either as a substitute for the other. In particular, completion 1’s within-dataset variance cannot tell us that another completion would also have mean 5.
Why One Completion Cannot Show the Range of Possible Results
A single draw gives one value for each missing entry and one resulting value for any statistic calculated from the completed dataset. It does not reveal the distribution of those possible values across repeated draws. Even if the imputed values look plausible and the completed dataset has a reasonable spread, that appearance alone does not show whether the imputation procedure would also produce very different values on another run.
The distinction is especially clear when there is only one missing entry. The observed values can remain identical in every completion, while the imputed value—and therefore a calculated statistic—changes. Within-dataset spread is calculated from all the values in the selected completion; across-dataset variation asks what would change if the missing entry were redrawn.
Worked Example: A Plausible Single Draw Can Hide Predictive Spread
Suppose two hypothetical observed measurements are 4 and 6 units. For a transparent illustration, stipulate that the missing value \(X_{\text{mis}}\) is drawn with equal probability from 1, 5, or 9 units. A single run happens to draw 5. Calculate the completed-data sample variance for that run, and compare it with the variation in the imputed value across its stipulated predictive distribution.
For the one completed dataset, the values are 4, 6, and 5. Their mean is 5, so the sample variance is
Across the three equally likely imputed values, the mean of \(X_{\text{mis}}\) is \((1+5+9)/3=5\). Its variance under this stipulated distribution is
The within-dataset sample variance of 1.000 units squared is not an estimate of this predictive variance. It describes the spread of the three values in the one selected completion, where the imputed value happened to sit between the observed values. The across-draw calculation describes the stipulated uncertainty about the missing value. In an actual analysis, that predictive distribution would need to come from a justified imputation procedure, not be chosen merely to illustrate the arithmetic.
How a Missing Draw Changes a Statistic
The same distinction applies to statistics calculated from completed data. Consider an observed-data mean with one missing value. A different draw can produce a different completed-data mean even though the observed measurements do not change. The following calculation makes that link explicit.
Worked Example: Across-Draw Variation in a Completed Mean
Suppose a hypothetical dataset contains observed measurements of 10, 12, and 14 units, plus one missing measurement. For illustration, let the missing value be equally likely to be 8, 12, or 16 units. The completed-data mean is \((36+X_{\text{mis}})/4\), since the observed values sum to 36 and the completed dataset has four entries.
The three possible completed-data means are \((36+8)/4=11\), \((36+12)/4=12\), and \((36+16)/4=13\) units. Their mean is 12 units, and their variance across the stipulated distribution of draws is
This result can also be derived directly. Each completed mean is \((36+X_{\text{mis}})/4\). Adding a constant does not change variance, and multiplying a random variable by \(1/4\) multiplies its variance by \(1/16\). Here \(\operatorname{Var}(X_{\text{mis}})=32/3\), so
If one draw gives 12, the completed-data mean is 12 units. That single value does not show that the other possible completed means are 11 or 13 units. The across-draw variance here is calculated from the full stipulated distribution, not estimated from the one selected completion. With only a small number of random draws, a variance calculated across those draws would itself be imprecise.
Assumptions, Diagnostics, and Alternatives
These examples assume that the specified random draws represent the uncertainty of interest, that the imputation model is suitable for the data, and that the observed entries are held fixed while completions are repeated. The finite examples stipulate the distribution explicitly, so no model has to be estimated. In applied work, the distribution is not known in advance: its adequacy depends on the model and assumptions used to generate imputations. Repeating a flawed procedure can reveal its variability without correcting its bias or misspecification.
A useful diagnostic is to make repeated completions and inspect how key imputed values and completed-data statistics change. Large changes indicate that one completion is particularly unrepresentative of the range generated by the procedure. Little change does not prove the procedure is sound: a narrow or incorrectly specified predictive model can produce stable but misleading draws. Also check that the random part of the procedure is actually active. If repeated runs produce exactly the same imputed values despite an intended stochastic step, a fixed seed reset on every run or an implementation error may be responsible.
| Approach | What one result contains | What it does not show by itself |
|---|---|---|
| Deterministic single imputation | One fixed completed dataset | Random variation in possible imputed values |
| Stochastic single imputation | One random completed dataset | How results vary across other possible draws |
| Repeated stochastic completions | Several draws that can display variation across completions | Whether the imputation model is correctly specified |
Repeated stochastic completions are useful for exposing variation that a single completion conceals, but repetition alone does not guarantee valid inference. The model must be appropriate to the variables and uncertainty being represented, and results can still be limited by poor predictions or unsupported assumptions. How estimates and uncertainty are formally combined across multiple completed datasets is a separate procedure; the key point here is simply that one random completion does not display the variation across other completions.
Common Mistake
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- In the first example, what does the within-completion sample variance of 4.400 describe, and what does the across-draw variance of 0.333 describe?
- For observed values 4 and 6 with a missing value drawn from 1, 5, or 9 with equal probability, what is the mean of the imputed value’s predictive distribution?
- In the completed-mean example, why does one draw of 12 units not establish that the completed-data mean is always 12 units?
- Name one way repeated completions can be misleading even when they produce very similar results.
Solutions. (1) The value 4.400 units squared summarizes spread among the six values in completion 1. The value 0.333 units squared summarizes variation in the completed-data mean across the four illustrative completions. They do not measure the same variation. (2) The mean is \((1+5+9)/3=5\) units. (3) It is only one of three possible draws under the stipulated distribution; the other draws give completed means of 11 and 13 units. (4) Repeated results can be misleading if the imputation model is misspecified or too narrow; stability across draws does not establish that the model represents uncertainty adequately.