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Potential Outcomes and Experiments · Tutorial 615 of 952

Neyman Variance and Conservative Standard Errors

Learn how complete randomization determines the variance of the difference in means and why the standard Neyman variance estimator is conservative in expectation.

Intermediate 10 min read

What You'll Learn

  • State the finite-population assumptions behind Neyman variance.
  • Define the finite-population variances of both potential outcomes and of individual treatment effects.
  • Derive the randomization variance of the difference-in-means estimator.
  • Calculate its usual plug-in variance estimate from observed treatment and control outcomes.
  • Explain why the plug-in estimate is conservative in expectation, but not necessarily in a particular assignment.
  • Interpret the standard error as measuring assignment uncertainty about a finite-population effect.

Why the Variance Has an Unobservable Piece

In a completely randomized experiment, the treatment and control groups are formed by assigning a fixed number of units to treatment. The difference between their observed outcome means estimates the average treatment effect for the experimental units. Its uncertainty depends not only on how much outcomes vary, but also on how individual treatment effects vary. The difficulty is that each unit reveals only one of its two potential outcomes, so the variation in individual effects cannot generally be calculated from the observed data.

Neyman variance makes this limitation explicit. It gives the randomization variance of the difference-in-means estimator for a fixed finite population, then shows that the familiar variance estimate leaves out a nonnegative term. Thus the estimate is conservative in expectation over the random assignment. That qualification matters: it does not guarantee that the estimate is larger than the true randomization variance in every realized experiment.

Finite-Population Setup

Consider \(N\) experimental units. Unit \(i\) has two fixed potential outcomes: \(Y_i(1)\) if assigned to treatment and \(Y_i(0)\) if assigned to control. Its individual treatment effect is \(\tau_i=Y_i(1)-Y_i(0)\). A complete randomization assigns exactly \(n_1\) units to treatment and \(n_0=N-n_1\) to control, with every assignment of that size equally likely.

Let \(W_i=1\) when unit \(i\) is assigned to treatment and \(W_i=0\) otherwise. The observed outcome is \(Y_i=W_iY_i(1)+(1-W_i)Y_i(0)\). The difference-in-means estimator is \(\hat{\tau}=\bar{Y}_1-\bar{Y}_0\), where \(\bar{Y}_1\) and \(\bar{Y}_0\) are the observed treatment and control means. The target is the finite-population average treatment effect, \(\tau=N^{-1}\sum_{i=1}^N\tau_i\).

Definition: For any fixed finite-population values \(A_1,\ldots,A_N\), define \(S_A^2=(N-1)^{-1}\sum_{i=1}^N(A_i-\bar{A})^2\), where \(\bar{A}=N^{-1}\sum_i A_i\). Write \(S_1^2\), \(S_0^2\), and \(S_\tau^2\) for this variance calculated over \(Y_i(1)\), \(Y_i(0)\), and \(\tau_i\), respectively.

The randomization variance treats all these potential outcomes as fixed and considers variation across assignments. Its usual formula assumes complete randomization with fixed group sizes, a well-defined treatment and control condition, and no interference between units: one unit's observed outcome is determined by its own assignment and potential outcomes, not by other units' assignments. These are the conditions under which the observed outcomes correspond to the potential outcomes just defined.

The Neyman Randomization Variance

Theorem: Under complete randomization with \(n_1\) treated and \(n_0\) controls, the difference-in-means estimator is unbiased for \(\tau\), and its randomization variance is
$$ \operatorname{Var}(\hat{\tau}) = \frac{S_1^2}{n_1} + \frac{S_0^2}{n_0} - \frac{S_\tau^2}{N}. $$

Unbiasedness follows because each unit has treatment-assignment probability \(n_1/N\), and the treatment and control means therefore average the corresponding potential outcomes over the finite population. For the variance, randomization gives each arm a simple random sample without replacement. The variance of the treatment mean is \((1/n_1-1/N)S_1^2\); the control-mean variance is \((1/n_0-1/N)S_0^2\). The two arm means are dependent because a unit assigned to one arm cannot also enter the other. Their covariance is \(-S_{10}/N\), where \(S_{10}=(N-1)^{-1}\sum_i(Y_i(1)-\bar{Y}(1))(Y_i(0)-\bar{Y}(0))\).

For completeness, the simple-random-sample variance identity follows from assignment indicators. For a sample of size \(n\), each indicator \(I_i\) has variance \((n/N)(1-n/N)\), while distinct indicators have covariance \(-n(N-n)/(N^2(N-1))\). Expanding the variance of \(n^{-1}\sum_i I_i(A_i-\bar{A})\), and using \(\sum_i(A_i-\bar{A})=0\), gives \((1/n)(1-n/N)S_A^2=(1/n-1/N)S_A^2\). The cross-arm covariance follows from the same assignment probabilities: the two groups are disjoint, so their centered sample means have covariance \(-S_{10}/N\).

Subtracting the covariance when taking the variance of a difference gives

$$ \begin{aligned} \operatorname{Var}(\hat{\tau}) &=\left(\frac{1}{n_1}-\frac{1}{N}\right)S_1^2 +\left(\frac{1}{n_0}-\frac{1}{N}\right)S_0^2 +\frac{2S_{10}}{N}\\ &=\frac{S_1^2}{n_1}+\frac{S_0^2}{n_0} -\frac{S_1^2+S_0^2-2S_{10}}{N}\\ &=\frac{S_1^2}{n_1}+\frac{S_0^2}{n_0}-\frac{S_\tau^2}{N}. \end{aligned} $$

The last equality uses the variance identity \(S_\tau^2=S_1^2+S_0^2-2S_{10}\), obtained by expanding the squared deviations of \(Y_i(1)-Y_i(0)\). Since \(S_\tau^2\geq 0\), variation in treatment effects reduces the randomization variance relative to the sum of the two outcome-variance terms. The reduction is not directly available from observed data: for each unit, one of \(Y_i(1)\) and \(Y_i(0)\) is missing.

Worked Example: Deriving the Variance for a Small Population

Suppose a hypothetical population has four units with potential outcomes as follows. Assume two units are assigned to treatment and two to control.

Unit\(Y_i(1)\)\(Y_i(0)\)\(\tau_i\)
1514
2734
3954
41174

The treatment-outcome mean is 8 and the control-outcome mean is 4. Their sums of squared deviations are each 20, so \(S_1^2=S_0^2=20/(4-1)=6.667\). Every individual effect equals 4, so \(S_\tau^2=0\). Therefore,

$$ \operatorname{Var}(\hat{\tau}) = \frac{6.667}{2}+\frac{6.667}{2}-\frac{0}{4} =6.667, \qquad SE(\hat{\tau})=\sqrt{6.667}=2.582. $$

The standard error is in the same units as the outcome. Although treatment effects are constant in this example, the estimator varies because different units have different outcome levels and different pairs are assigned to the two arms.

The Usual Plug-In Estimate

Within each observed arm, calculate the sample variance using the usual denominator \(n_z-1\): call these \(s_1^2\) and \(s_0^2\). Under complete randomization, each arm is a simple random sample from its corresponding fixed potential-outcome values. Consequently, \(s_1^2\) is unbiased for \(S_1^2\), and \(s_0^2\) is unbiased for \(S_0^2\). This motivates the usual Neyman variance estimate and its standard error:

$$ \widehat{V}_{N}=\frac{s_1^2}{n_1}+\frac{s_0^2}{n_0}, \qquad \widehat{SE}(\hat{\tau})=\sqrt{\widehat{V}_{N}}. $$

Taking expectation over assignments and comparing with the true variance gives the conservativeness result:

$$ \mathbb{E}[\widehat{V}_{N}] = \frac{S_1^2}{n_1}+\frac{S_0^2}{n_0} = \operatorname{Var}(\hat{\tau})+\frac{S_\tau^2}{N} \geq \operatorname{Var}(\hat{\tau}). $$

The excess in expectation is exactly \(S_\tau^2/N\). If effects are constant across units, \(S_\tau^2=0\) and the variance estimate is unbiased in expectation. If effects vary, the estimate overstates the randomization variance on average. But \(s_1^2\) and \(s_0^2\) fluctuate with the assignment, so a particular \(\widehat{V}_{N}\) can fall below the true variance. “Conservative” here describes the expectation of the variance estimate, not a guarantee for every dataset or every confidence interval.

Worked Example: Calculating the Plug-In Standard Error

In a hypothetical complete-randomization experiment, four units are assigned to each arm. The observed treatment outcomes are 12, 15, 13, and 16 units; the observed control outcomes are 10, 11, 12, and 13 units. The difference in means is \(\hat{\tau}=14-11.5=2.5\) outcome units.

For the treatment group, the mean is 14 and the sum of squared deviations is \(4+1+1+4=10\). Thus \(s_1^2=10/(4-1)=3.333\). For the control group, the mean is 11.5 and the sum of squared deviations is \(2.25+0.25+0.25+2.25=5\), so \(s_0^2=5/3=1.667\). The estimated variance and standard error are

$$ \widehat{V}_{N} = \frac{3.333}{4}+\frac{1.667}{4} =1.250, \qquad \widehat{SE}(\hat{\tau})=\sqrt{1.250}=1.118 $$

Thus the observed difference is 2.5 outcome units, with a plug-in estimated standard error of 1.118 outcome units. The calculation uses only observed outcomes; it does not recover the unobserved variance of individual effects.

Worked Example: Seeing Why Conservativeness Is Not Guaranteed in One Assignment

Return to the four-unit population in the first example. Suppose units 1 and 2 are assigned to treatment, leaving units 3 and 4 in control. The observed treatment outcomes are 5 and 7, and the observed control outcomes are also 5 and 7. Each arm has sample variance \(s^2=((5-6)^2+(7-6)^2)/(2-1)=2\). The estimated variance is therefore

$$ \widehat{V}_{N}=\frac{2}{2}+\frac{2}{2}=2, \qquad \operatorname{Var}(\hat{\tau})=6.667. $$

For this assignment, the estimated variance is below the true randomization variance. There is no contradiction: the theorem concerns the average of \(\widehat{V}_{N}\) over all assignments, not its value in a particular assignment. In this example the estimate happens to be 2 for the selected allocation, even though its expectation over the six possible two-versus-two allocations is 6.667.

Interpretation and Common Pitfall

Neyman variance is useful because it supplies an observable, design-based estimate of uncertainty without requiring a model for how outcomes were generated. The finite-population target, assignment mechanism, and source of the missing term are all explicit. In larger samples, the corresponding standard error is commonly used for approximate normal-based inference; the quality of that approximation depends on the design and the distribution of outcomes and effects. The variance identity alone does not establish a finite-sample normal distribution.

A common mistake is to say, “The estimated variance is conservative, so it must exceed the true variance in this experiment.” That conclusion confuses an expectation with a realized value. The correct statement is \(\mathbb{E}[\widehat{V}_{N}]\geq\operatorname{Var}(\hat{\tau})\) under complete randomization. The omitted \(S_\tau^2/N\) is nonnegative, but sample variances from a particular allocation can be high or low. Also, the standard error is uncertainty due to assignment for the stated finite population; it is not a measure of uncertainty about which units were included in that population.

Key takeaway: Under complete randomization, the difference-in-means variance subtracts the unobserved treatment-effect variance \(S_\tau^2/N\). The usual plug-in estimate omits that term, so it is conservative in expectation, not necessarily for every realized assignment.

Check Your Understanding

Use the definitions and results above to answer the following questions.

  1. What is held fixed when calculating the randomization variance in the finite-population framework?
  2. If \(S_\tau^2=0\), how does the Neyman variance formula simplify, and what does that imply for the expectation of the plug-in variance estimate?
  3. Why can \(S_\tau^2\) generally not be calculated from the observed outcomes of a randomized experiment?
  4. In what precise sense is \(\widehat{V}_{N}\) conservative, and why does that not imply it exceeds the true variance for every assignment?
  5. Explain why a design-based standard error is not, by itself, a claim that the estimator has an exact normal sampling distribution.
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