How Hot-Deck Imputation Reuses Observed Values
When a value is missing, a hot-deck procedure fills it with a value taken from an observed case, called a donor. Donors may be selected from the full set of observed cases or from a more restricted group, such as cases with the same recorded characteristics as the recipient. A common version samples donors with replacement: the same donor can be selected more than once. Consequently, several missing cases may receive identical values, even when their unobserved values would have differed.
This tutorial focuses on what that reuse can do to a single completed dataset. It does not imply that repeated values always cause a large distortion, or that every repeat is an error. The effect depends on the donor pool, the number of missing cases, which donors are selected, and how the imputed variable relates to other variables.
For the calculations below, suppose \(r\) observed values \(x_1,\ldots,x_r\) are eligible donors for one variable, and \(m\) cases have missing values. Unless stated otherwise, each donor is equally likely to be selected at each draw, selections are made with replacement, and the draws are independent conditional on the donor pool. Let \(N=r+m\) be the number of cases in the completed data. These are explicit simplifying assumptions, not requirements for every hot-deck implementation.
Assumptions Behind a Donor Assignment
A donor value can stand in for a recipient's missing value only if the donor-selection scheme makes the donor plausible for that recipient. In a simple unstratified hot deck, that requires the missing cases and observed donors to be sufficiently comparable for the variable being imputed. More commonly, donors are restricted or sampled according to observed characteristics. The key assumption is then that, conditional on those characteristics, the observed donors represent the missing values well enough for the intended analysis.
This assumption is not guaranteed by using observed values. If cases with missing values differ systematically from eligible donors even after conditioning on the variables used for donor selection, the donor distribution may be a poor representation of the missing values. There must also be eligible donors for the recipients. If donor groups are too narrow, some recipients may have no plausible donor; broadening groups solves that availability problem only by potentially making donors less comparable.
Sampling with replacement avoids exhausting a donor group, but allows reuse. Sampling without replacement prevents reuse of a donor record within a set of assignments, but requires enough eligible donors and can impose artificial dependence between assignments. Neither choice, by itself, establishes that the donor values are appropriate.
Worked Example: How Likely Is Donor Reuse?
Suppose four equally likely donor records are available, and three missing cases each receive an independently selected donor, with replacement. What is the probability that at least one donor record is selected more than once?
There are \(4^3=64\) equally likely ordered sequences of three donor selections. To have no reuse, the first selection has 4 possibilities, the second has 3 remaining possibilities, and the third has 2 remaining possibilities. Thus there are \(4\cdot3\cdot2=24\) sequences with no reuse, and \[ P(\text{at least one repeated donor})=1-\frac{24}{64}=0.625. \]
So even with three recipients and four donors, there is a 62.5% chance of selecting at least one donor record more than once under these assumptions. The expected number of distinct donor records used is \[ 4\left[1-\left(\frac{3}{4}\right)^3\right] =4\left(\frac{37}{64}\right)=2.3125. \] For each of the four donors, the probability of being used at least once is \(1-(3/4)^3\); summing those probabilities gives the expected number used. This is a probability statement about the selection procedure, not a claim that every dataset will use exactly 2.3125 donors.
How Repeated Values Affect Variance
A single completed dataset treats every imputed value as if it were observed. The assigned values therefore contribute to the completed-data variance as realized numbers, not as uncertain possibilities. Depending on which donors are selected, repeats can concentrate the completed data around some values or overrepresent values near the extremes.
The following derivation gives the expected sample variance in a simple setting. It averages over the random donor draws while holding the observed donor pool fixed. Define the donor mean and donor variance using divisor \(r\):
Let \(Y_1,\ldots,Y_m\) be the imputed values. Under equal-probability independent sampling with replacement, each \(Y_k\) has mean \(\bar{x}_D\) and variance \(v_D\). Write \(Z_1,\ldots,Z_N\) for the completed values: the \(r\) observed donor values and the \(m\) draws. The sample variance of the completed data is \(s_{\mathrm{comp}}^2=\sum_{i=1}^{N}(Z_i-\bar{Z})^2/(N-1)\).
Use the identity \(\sum_i(Z_i-\bar{Z})^2=\sum_i Z_i^2-(\sum_i Z_i)^2/N\). The expected sum of squares about zero is \(N(v_D+\bar{x}_D^2)\), since each of the \(N\) values has expected square \(v_D+\bar{x}_D^2\). The total sum has mean \(N\bar{x}_D\) and variance \(m v_D\): the observed values are fixed, and only the \(m\) independent draws vary. Therefore its expected square is \(N^2\bar{x}_D^2+m v_D\). Substitution gives
This is an expectation over the random donor selections, not a formula for the variance in every realized completed dataset. It also depends on the specified equal-probability, with-replacement scheme. For comparison, the observed donors' sample variance is \(s_D^2=\frac{r}{r-1}v_D\) when \(r>1\). The two quantities use different data and denominators; neither comparison tells us the unknown variance that would have been obtained if all missing values had been observed.
Worked Example: A Repeated Extreme Donor Can Raise the Realized Variance
Suppose the observed donor values are \(0,1,1,2\), and one missing value is imputed. The donor mean is \(1\), and the sum of squared deviations from the mean is \(2\). The observed donors' sample variance is therefore \(2/(4-1)=0.6667\).
If the donor value \(0\) is selected, the completed values are \(0,1,1,2,0\), so the donor value at the lower end is repeated. The completed mean is \(4/5=0.8\). Using the computational formula for the sample variance, \[ s_{\mathrm{comp}}^2 =\frac{\sum_i Z_i^2-(\sum_i Z_i)^2/5}{4} =\frac{6-16/5}{4} =0.7000. \] Here the realized completed-data variance is slightly larger than the observed donors' sample variance. If instead the donor value \(1\) is selected, the completed values are \(0,1,1,2,1\), with mean \(1\), and \(s_{\mathrm{comp}}^2=2/4=0.5000\). Thus the realized variance depends on which value is repeated; donor reuse does not force variance to move in only one direction.
In this example, \(r=4\), \(m=1\), \(N=5\), and \(v_D=2/4=0.5\). The derived expectation is \[ E[s_{\mathrm{comp}}^2] =\frac{5-1/5}{4}(0.5)=0.6000. \] That agrees with the three possible donor values and their probabilities: \(0.25(0.7000)+0.50(0.5000)+0.25(0.7000)=0.6000\). The average over repeated applications of the procedure is not the same as any one realized completed-data variance.
Why Correlations Can Change Too
A donor carries the donor's value for the imputed variable, but the recipient keeps the recipient's other recorded values. Thus a donor's value is paired with the recipient's covariates, not with the donor's covariates. If a donor is selected repeatedly, its imputed value contributes repeatedly to those recipient-specific pairs. This can change the covariance and correlation in the completed data. The direction of change is not fixed: it depends on the recipient covariates and the donor assignments.
Worked Example: Reusing a Donor Changes a Correlation
Consider four complete donor cases with paired values \((X,Y)=(1,1),(2,2),(3,3),(4,4)\). Two additional cases have known \(X\) values \(1\) and \(4\), but missing \(Y\). Suppose hot-deck draws assign donor value \(Y=1\) to both recipients, reusing the first donor. The completed pairs are \((1,1),(2,2),(3,3),(4,4),(1,1),(4,1)\).
For these six pairs, \(\sum X=15\), \(\sum Y=12\), \(\sum X^2=47\), \(\sum Y^2=32\), and \(\sum XY=35\). The centered sums are \[ S_{XX}=47-\frac{15^2}{6}=9.5,\qquad S_{YY}=32-\frac{12^2}{6}=8,\qquad S_{XY}=35-\frac{15\cdot12}{6}=5. \] Therefore the completed-data correlation is \[ r_{XY}=\frac{S_{XY}}{\sqrt{S_{XX}S_{YY}}} =\frac{5}{\sqrt{9.5\cdot8}} \mathrel{\approx}0.5735. \] The four donor pairs alone have correlation \(1\). The completed-data correlation is lower because the repeated \(Y=1\) value is paired with a recipient having \(X=4\), weakening the linear pairing in this particular assignment. This calculation does not establish that hot-deck imputation always attenuates correlation; different assignments can produce different results.
Checks, Alternatives, and Limitations
Useful checks focus on both donor selection and its consequences. Review which donor records were used and how often, especially when many recipients draw from a small pool. Compare the observed and imputed value distributions, including their ranges and the frequencies of common values. For analyses involving more than one variable, compare completed-data means, variances, and correlations with summaries from the observed cases, while recognizing that observed cases need not represent the full sample. These checks can reveal suspicious concentrations or mismatches; they cannot confirm that unobserved values have been recovered correctly.
Repeated values are not automatically evidence of a faulty procedure. A variable may genuinely have many tied values, and legitimate sampling with replacement can select the same donor again. Conversely, a completed distribution that looks plausible does not show that the donor-selection assumptions hold. In particular, a single completed dataset does not display the range of plausible missing values or carry forward all uncertainty about which donors would have been selected.
| Approach | How it differs | Important limitation |
|---|---|---|
| Hot deck with replacement | Copies observed values; a donor can serve multiple recipients. | Repeated assignments can alter variance and pairings; one completed dataset does not represent imputation uncertainty. |
| Hot deck without replacement | Prevents selecting the same donor record more than once in the assignment set. | Requires enough eligible donors and can make assignments dependent; it does not guarantee suitable donor-recipient matches. |
| Mean imputation | Uses a fixed mean rather than observed donor values. | Can create a point mass at the mean and alter distributional shape and variance. |
| Regression imputation | Uses a prediction based on other variables, with or without added residual variation. | Predicted values can have an inappropriate distribution or relationship to other variables if the model or residual treatment is unsuitable. |
The distinction between one random assignment and representing uncertainty across multiple imputations matters here. Random donor selection makes the completed dataset vary from run to run, but choosing and analyzing just one run still leaves the analysis conditional on that particular set of donor assignments. A random draw is not, by itself, a summary of the uncertainty caused by missing values.
Check Your Understanding
Answer each question, then compare with the solutions below.
- With three equally likely donors and two independent draws with replacement, what is the probability that both draws use the same donor?
- For donor values \(2,4,6\), calculate the donor mean and \(v_D\), using divisor \(r\).
- Using those three donors, suppose one missing value is imputed by repeating \(2\). Calculate the completed sample variance.
- Why can a repeated donor change the correlation between the imputed variable and a recipient's observed covariate?
Solutions. (1) After the first draw, the probability the second draw matches it is \(1/3\), so the probability is \(1/3\). (2) The mean is \((2+4+6)/3=4\), and \(v_D=[(2-4)^2+(4-4)^2+(6-4)^2]/3=8/3\). (3) The completed values are \(2,4,6,2\), with mean \(3.5\). Their squared deviations sum to \(2.25+0.25+6.25+2.25=11\), so the sample variance is \(11/(4-1)=3.667\). (4) The imputed donor value is paired with each recipient's own covariate values. Reusing a donor therefore repeats the imputed value across potentially different covariate values, changing the covariance and potentially the correlation; the direction depends on the assignments.