Deriving the Variance After Mean Imputation

Why the Variance Changes

Replacing missing values by the observed mean creates a completed data set, but it does not add new variation. Each replacement lies exactly at the center of the observed values. When sample variance is calculated on the completed data, those replacements contribute zero squared deviation, while the denominator reflects the larger completed sample size. The result is a smaller variance whenever at least one value is imputed and the observed values have nonzero spread.

This is an exact algebraic consequence of mean imputation, not a claim that the true variance in the full population must be smaller. The calculation compares two different summaries: the sample variance among the observed values and the sample variance in the completed data set. Keeping those summaries distinct is essential.

Definition: Suppose $n_o$ values of a quantitative variable are observed and $n_m$ are missing, with total sample size $n=n_o+n_m$. Let $\bar{x}_{obs}$ be the mean of the observed values and $s_{obs}^2$ their sample variance, calculated with denominator $n_o-1$. Replace every missing value by $\bar{x}_{obs}$, and let $s_{comp}^2$ be the sample variance of all $n$ completed values, calculated with denominator $n-1$. When $n_o\geq 2$, $$ s_{comp}^2=\frac{n_o-1}{n-1}s_{obs}^2. $$

Deriving the Completed-Data Variance

Write the observed values as $x_1,\ldots,x_{n_o}$. Their squared deviations from their own mean sum to $(n_o-1)s_{obs}^2$. Each imputed value is $\bar{x}_{obs}$, so its deviation from that mean is zero. Also, the mean of the completed data remains $\bar{x}_{obs}$: the observed values sum to $n_o\bar{x}_{obs}$, and the $n_m$ replacements sum to $n_m\bar{x}_{obs}$. Thus the total is $n\bar{x}_{obs}$, and dividing by $n$ gives the same mean.

The completed-data sample variance is the completed sum of squared deviations divided by $n-1$. Substituting the observed sum of squares gives the result directly:

\begin{aligned} \bar{x}_{comp} &=\frac{\sum_{i=1}^{n_o}x_i+n_m\bar{x}_{obs}}{n}\\ &=\frac{n_o\bar{x}_{obs}+n_m\bar{x}_{obs}}{n} =\bar{x}_{obs},\\[4pt] \sum_{i=1}^{n}(z_i-\bar{x}_{comp})^2 &=\sum_{i=1}^{n_o}(x_i-\bar{x}_{obs})^2 +\sum_{j=1}^{n_m}(\bar{x}_{obs}-\bar{x}_{obs})^2\\ &=(n_o-1)s_{obs}^2,\\[4pt] s_{comp}^2 &=\frac{(n_o-1)s_{obs}^2}{n-1} =\frac{n_o-1}{n-1}s_{obs}^2. \end{aligned}

The derivation requires only that the observed values are finite, there is at least one observed value to define $\bar{x}_{obs}$, and the missing values are replaced exactly by that mean. To define $s_{obs}^2$ in the usual way, at least two values must be observed. No assumption about MCAR, MAR, or MNAR is needed for this arithmetic identity. Those mechanisms matter when interpreting observed data or making inferences about the unobserved values; they do not change the sum-of-squares calculation.

When $n_o\geq2$, the shrinkage factor can also be written as $\frac{n_o-1}{n-1}=1-\frac{n_m}{n-1}$. If $n_m>0$, it is less than one. Therefore $s_{comp}^2\leq s_{obs}^2$, with strict inequality if $s_{obs}^2>0$. The absolute reduction is

s_{obs}^2-s_{comp}^2 =\left(1-\frac{n_o-1}{n-1}\right)s_{obs}^2 =\frac{n_m}{n-1}s_{obs}^2.

The units remain squared units throughout. If the variable is measured in milligrams, both variances are in squared milligrams. The factor is dimensionless; it changes the size of the variance, not its units.

Worked Example: Two Replacements Among Five Values

Consider hypothetical observed measurements of 2, 4, and 6 units, with two values missing. The mean of the observed values is $\bar{x}_{obs}=(2+4+6)/3=4$ units. The observed squared-deviation sum is $(2-4)^2+(4-4)^2+(6-4)^2=4+0+4=8$ squared units. There are $n_o=3$ observed and $n_m=2$ missing values, so $n=5$.

The observed-data sample variance is $s_{obs}^2=8/(3-1)=4$ squared units. After replacing each missing value by 4, the completed values are 2, 4, 6, 4, 4. Their mean is still 4, and their squared-deviation sum is still 8. The completed-data sample variance is $s_{comp}^2=8/(5-1)=2$ squared units.

The formula gives the same answer: $s_{comp}^2=[(3-1)/(5-1)]4=(2/4)4=2$ squared units. Here the completed-data variance is half the observed-data variance. This describes the two calculated variances; it does not establish that the full-data variance is 2 squared units.

Worked Comparisons and the Role of the Denominator

The amount of shrinkage depends on the number of observed and missing values through the sample-variance denominators. The observed squared-deviation sum itself does not change when values are replaced by the observed mean. More replacements increase the completed sample size and therefore increase the denominator used for the completed-data sample variance.

Worked Example: One Replacement Among Five Values

Suppose the hypothetical observed values are 10, 12, 14, and 16 units, with one missing value. Their mean is $\bar{x}_{obs}=(10+12+14+16)/4=13$ units. The squared-deviation sum is $(10-13)^2+(12-13)^2+(14-13)^2+(16-13)^2=9+1+1+9=20$ squared units.

For the four observed values, $s_{obs}^2=20/(4-1)=20/3=6.667$ squared units. Replacing the missing value by 13 gives five completed values with the same squared-deviation sum, 20. Thus $s_{comp}^2=20/(5-1)=5.000$ squared units. Equivalently, $s_{comp}^2=[(4-1)/(5-1)](20/3)=(3/4)(20/3)=5.000$ squared units.

The completed variance is lower by $6.667-5.000=1.667$ squared units. The mean remains 13 units; the change occurs because the completed-data variance uses a denominator of 4 rather than 3.

Worked Example: More Replacements, More Shrinkage

Take the hypothetical observed values 3, 5, 7, and 9 units. Their mean is 6 units, and the squared-deviation sum is $(3-6)^2+(5-6)^2+(7-6)^2+(9-6)^2=9+1+1+9=20$ squared units. The observed-data variance is $s_{obs}^2=20/3=6.667$ squared units.

If there are two missing values, replacing both by 6 gives $n=6$ and $s_{comp}^2=20/(6-1)=4.000$ squared units. The factor is $(4-1)/(6-1)=3/5$, and $(3/5)(6.667)=4.000$ squared units.

If there are six missing values instead, the same observed values and same squared-deviation sum produce $n=10$ and $s_{comp}^2=20/(10-1)=2.222$ squared units. The factor is $3/9=1/3$, giving $(1/3)(6.667)=2.222$ squared units. This comparison isolates the mechanical effect of adding more mean replacements; it does not supply information about what the missing measurements would have been.

The denominator convention matters. The result above concerns the usual sample variance, with denominators one less than the relevant sample sizes. If instead a calculation uses the uncorrected empirical variance, dividing the sum of squared deviations by the number of values, then its observed and completed versions have ratio $n_o/n$, not $(n_o-1)/(n-1)$. These are different summaries. Always identify the denominator before comparing variances.

What the Calculation Means—and What It Cannot Establish

The observed-data variance $s_{obs}^2$ describes spread among the cases whose values were recorded. The completed-data variance $s_{comp}^2$ describes spread in a constructed data set where every missing entry has been set to the observed mean. The latter is smaller by construction when values are missing and the observed values vary. It is not an independent estimate of the missing values' variability.

This distinction matters because a variance from the completed data may be used in later calculations as if it summarized all $n$ observed measurements. Mean replacement has not, however, supplied the omitted cases with their unknown deviations from the mean. In particular, the exact formula alone cannot tell us whether the observed cases represent the full sample well, nor how the full-data variance compares with either calculated variance. Such conclusions require information or assumptions beyond the arithmetic identity.

Complete-case analysis avoids inserting artificial mean values, but its variance summarizes only the observed cases and can be unrepresentative if observation is associated with the variable or relevant characteristics. Mean imputation retains all rows in the completed table but mechanically reduces its variance. Approaches that account for uncertainty about missing values aim to avoid treating a single fixed replacement as if it were observed; their details are separate from this derivation.

Diagnostics, Edge Cases, and a Common Mistake

A simple diagnostic is to calculate the predicted completed-data variance from $s_{obs}^2$ and the counts, then compare it with the variance computed directly from the completed values. They should agree, apart from rounding, if every missing value was replaced by the observed mean and both calculations use the usual sample-variance denominator. Disagreement can indicate a different imputation value, a calculation using a different subset, a denominator mismatch, or an arithmetic error.

No missing values: If $n_m=0$, then $n=n_o$, so the factor is 1 and the two variances agree.
One observed value: The observed sample variance $s_{obs}^2$ is undefined because its denominator would be zero. If there are additional missing values and all are replaced by that one observed value, the completed-data sample variance is 0, provided $n>1$. The ratio formula using $s_{obs}^2$ does not apply.
No observed values: The observed mean cannot be calculated, so this form of mean imputation is not defined without an additional rule.
Zero observed spread: If $n_o\geq2$ and all observed values are identical, then $s_{obs}^2=0$, and the completed variance is also zero. The inequality is not strict in this case.

Common mistake: “The completed variance is lower, so the true population variance must be lower than the observed-data variance.” This confuses a deterministic property of the constructed data with a conclusion about unobserved values or a population. The correction is to state only what the calculation establishes: mean replacement leaves the observed squared-deviation sum unchanged while increasing the sample-variance denominator. The full-data variance cannot be determined from that identity alone.

Key takeaway: With at least two observed values, exact replacement of every missing value by $\bar{x}_{obs}$ gives $s_{comp}^2=[(n_o-1)/(n-1)]s_{obs}^2$. This is a precise description of variance shrinkage in the completed data, not a verdict about the true variability of the unobserved values.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

There are 5 observed values and 3 missing values. If $s_{obs}^2=12$ squared units, what is $s_{comp}^2$ after mean imputation?
Why does replacing missing values by the observed mean leave the completed-data mean equal to the observed-data mean?
With 4 observed values and 2 missing values, the observed squared-deviation sum is 30 squared units. Find both sample variances.
Can the formula using $s_{obs}^2$ be applied when there is exactly one observed value? Explain.
A colleague says that a lower completed-data variance proves the missing values had less spread than the observed values. What is the error?

Solutions. (1) Here $n=5+3=8$, so $s_{comp}^2=[(5-1)/(8-1)]12=(4/7)12=6.857$ squared units. (2) The observed values sum to $n_o\bar{x}_{obs}$, and the $n_m$ replacements sum to $n_m\bar{x}_{obs}$. Their combined sum is $(n_o+n_m)\bar{x}_{obs}=n\bar{x}_{obs}$; dividing by $n$ gives $\bar{x}_{obs}$. (3) The observed variance is $30/(4-1)=10.00$ squared units. The completed sample size is 6, so the completed variance is $30/(6-1)=6.000$ squared units. (4) No. The usual observed sample variance is undefined with one observed value because its denominator is $1-1=0$. If additional values are imputed with that single observed value, the completed variance is zero when the completed sample size exceeds one, but the ratio formula based on $s_{obs}^2$ is unavailable. (5) The colleague mistakes a mechanical feature of the constructed data for evidence about unobserved values. The calculation shows that mean replacements add zero squared deviations while increasing the denominator; it does not reveal the missing values' actual spread.