Interpreting Pooled Variance Under Heteroscedasticity

Why the pooled variance needs careful interpretation

When treatment labels are masked, the observed outcomes can still provide information about variability. But with unequal variances in the two arms, “the pooled variance” does not mean there is one common variance shared by both arms. Instead, a variance computed across all participants combines the arm variances according to the allocation and, if the arm means differ, also includes variation between those means. These components matter when the estimate is used in a sample-size update.

The key distinction is between (1) the variance of the masked mixture of observations and (2) the variance combination that governs the treatment-contrast estimator. They agree under equal allocation in some familiar settings, but need not agree under unequal allocation and heteroscedasticity. The earlier material on pooled variance estimation with treatment labels masked and correcting pooled variance for allocation and assumed effect motivates the mixture and treatment-effect components; here the focus is the variance target when the arm variances themselves differ.

Definition: Let arm 1 contain $n_1$ independent outcomes with mean $\mu_1$ and variance $\sigma_1^2$, and arm 0 contain $n_0$ independent outcomes with mean $\mu_0$ and variance $\sigma_0^2$. Write $N=n_1+n_0$, $\pi_1=n_1/N$, and $\pi_0=n_0/N$. Heteroscedasticity here means $\sigma_1^2\ne\sigma_0^2$. The masked overall sample variance is $S^2_{\mathrm{all}}=\sum_{i,j}(Y_{ij}-\bar{Y})^2/(N-1)$, where $\bar{Y}$ is the mean of all $N$ outcomes.

Assume for the derivations that allocation counts are fixed, outcomes are independent between and within arms, and each arm has a finite variance. For the exact expectation formula, assume the arm-specific means and variances are well-defined; normality is not required. Independence and fixed arm sizes are important: with clustering, repeated measurements, or data-dependent allocation, the calculation may not apply as written.

What the masked overall variance estimates

The total sum of squares around the overall mean separates into variation around each arm mean and variation of the arm means around the overall mean. The within-arm part reflects the two arm variances; the between-arm part reflects the difference in arm means. Taking expectations gives the exact result below.

Theorem: Under the fixed-allocation, independence assumptions, $$ E(S^2_{\mathrm{all}}) = \pi_1\sigma_1^2+\pi_0\sigma_0^2 + \frac{n_1n_0}{N(N-1)}(\mu_1-\mu_0)^2. $$ The first term is the allocation-weighted combination of the arm variances. The second term is the contribution from the difference in arm means.

To see why, let $\bar{Y}_1$ and $\bar{Y}_0$ be the arm means. The sum-of-squares identity is

\sum_{i,j}(Y_{ij}-\bar{Y})^2 = \sum_{i=1}^{n_1}(Y_{1i}-\bar{Y}_1)^2 + \sum_{i=1}^{n_0}(Y_{0i}-\bar{Y}_0)^2 + \frac{n_1n_0}{N}(\bar{Y}_1-\bar{Y}_0)^2.

Each within-arm sum of squares has expectation $(n_a-1)\sigma_a^2$. Also, independence gives

E\{(\bar{Y}_1-\bar{Y}_0)^2\} = (\mu_1-\mu_0)^2+\frac{\sigma_1^2}{n_1}+\frac{\sigma_0^2}{n_0}.

Substituting this expectation into the identity, the between-arm contribution to the variance terms in the numerator is $(n_0\sigma_1^2+n_1\sigma_0^2)/N$. Adding this to the within-arm terms gives coefficients $(N-1)n_1/N$ and $(N-1)n_0/N$ for $\sigma_1^2$ and $\sigma_0^2$, respectively. Dividing by $N-1$ gives $\pi_1\sigma_1^2+\pi_0\sigma_0^2$, along with the stated mean-separation term.

When the means are equal, the expectation is exactly the allocation-weighted variance combination in the theorem. For large arm sizes with allocation proportions near $\pi_1$ and $\pi_0$, the variance component is

\pi_1\sigma_1^2+\pi_0\sigma_0^2.

Thus the arm with more participants contributes more to the large-sample masked mixture variance. If the treatment means differ, the overall sample variance has an additional positive component. A correction based on an assumed treatment effect addresses that mean-separation contribution; it does not make the two arm variances equal.

Worked Example: The Exact Finite-Sample Target

Suppose a hypothetical trial allocates $n_1=40$ participants to arm 1 and $n_0=20$ to arm 0. For illustration, suppose $\sigma_1^2=100$ and $\sigma_0^2=25$, in squared outcome units, and the arm means are equal. Find the expectation of the masked overall variance.

Here $N=60$, so the theorem gives

E(S^2_{\mathrm{all}}) = \frac{(40-1)(100)+(20-1)(25)+50}{59} = \frac{4425}{59} = 75.00.

The allocation-weighted combination is $(40/60)(100)+(20/60)(25)=75.00$ squared units, equal to the exact expectation when the arm means are equal. This is not the unweighted average of the arm variances, $62.50$. The larger-variance arm has more participants, so it has more weight in the masked variance.

The variance combination for a treatment contrast

The masked mixture variance is not automatically the right single variance to insert into every sample-size formula. For the difference in arm means, independence gives the exact variance

\operatorname{Var}(\bar{Y}_1-\bar{Y}_0) = \frac{\sigma_1^2}{n_1}+\frac{\sigma_0^2}{n_0}.

If a calculation is written in the common-variance form $\sigma^2(1/n_1+1/n_0)$, the value of $\sigma^2$ that makes that expression equal to the true contrast variance is

\sigma^2_{\mathrm{eff}} = \frac{\sigma_1^2/n_1+\sigma_0^2/n_0}{1/n_1+1/n_0} = \pi_0\sigma_1^2+\pi_1\sigma_0^2.

The weights are opposite to those in the large-sample masked mixture variance. A variance in the smaller arm is divided by a smaller sample size, so it has more influence on the contrast variance. By contrast, the overall masked variance gives more weight to the arm with more observations. This difference is central when interpreting a blinded update for a treatment contrast.

Worked Example: Mixture Variance Versus Contrast-Effective Variance

Use the hypothetical variances $\sigma_1^2=100$ and $\sigma_0^2=25$, but allocate $n_1=40$ and $n_0=20$. Compare the large-sample masked mixture variance with the contrast-effective variance.

The allocation proportions are $\pi_1=40/60=2/3$ and $\pi_0=20/60=1/3$. The masked mixture target is

\pi_1\sigma_1^2+\pi_0\sigma_0^2 = \frac{2}{3}(100)+\frac{1}{3}(25) = 75.00

squared outcome units. The contrast-effective variance instead uses the reversed weights:

\sigma^2_{\mathrm{eff}} = \frac{1}{3}(100)+\frac{2}{3}(25) = 50.00

squared outcome units. Directly, the difference in means has variance $100/40+25/20=3.75$ squared outcome units. The common-variance representation gives $50(1/40+1/20)=50(0.075)=3.75$, as required. Substituting the mixture target of 75 instead would give $75(0.075)=5.625$, which is not the true contrast variance for these arm variances and sample sizes.

This does not mean that 75 is a bad estimate of the masked mixture variance: it targets a different quantity. It means that a sample-size calculation must connect its variance input to its own variance formula and allocation convention.

Worked Example: Why Allocation Can Change the Direction of a Plug-In Error

Consider a second hypothetical setting with $n_1=20$, $n_0=40$, $\sigma_1^2=100$, and $\sigma_0^2=25$. The arm with the larger variance now has fewer participants. Calculate both variance combinations and the contrast variance.

Now $\pi_1=1/3$ and $\pi_0=2/3$. The large-sample masked mixture target is

\frac{1}{3}(100)+\frac{2}{3}(25)=50.00

squared units. The contrast-effective variance is

\frac{2}{3}(100)+\frac{1}{3}(25)=75.00

squared units. The direct contrast variance is $100/20+25/40=5.625$ squared units, and $75(1/20+1/40)=75(0.075)=5.625$, so the common-form identity holds. The larger variance in the smaller arm increases the contrast variance relative to the previous allocation, and the exact calculation remains internally consistent.

The variance combination matching $\sigma_1^2/n_1+\sigma_0^2/n_0$ is

\frac{\sigma_1^2/n_1+\sigma_0^2/n_0}{1/n_1+1/n_0} = \frac{n_0\sigma_1^2+n_1\sigma_0^2}{N} = 75.00,

Multiplying by $1/n_1+1/n_0=0.075$ gives $75(0.075)=5.625$, which equals the direct contrast variance $100/20+25/40=5.625$. The common-form identity therefore holds.

Assumptions, diagnostics, and alternatives

The identities above rely on independent outcomes and the stated arm variances. In a practical blinded re-estimation, also verify that the estimator’s definition, denominator, and any treatment-effect correction match the intended target. When arm sizes are small, the $n_a-1$ weights in the exact expectation may matter. When arm means differ, a raw overall sample variance includes a between-arm component; under unequal variances, removing an assumed effect does not remove uncertainty about whether that assumed effect is correct.

Useful checks include comparing the planned allocation proportions with those actually observed, tracking how sensitive the update is to plausible arm-variance patterns, and checking whether the design’s sample-size formula assumes a common variance or permits arm-specific variances. A suspiciously stable pooled estimate does not establish homoscedasticity: masking prevents direct inspection of each arm’s variance. Conversely, a change in the pooled estimate can reflect changes in allocation or mean separation as well as changes in within-arm variability.

There are two main planning alternatives. If arm-specific variance estimates are available and their use is permitted by the design, calculate the contrast variance directly as $\sigma_1^2/n_1+\sigma_0^2/n_0$. If the design requires a single common variance input, define explicitly which common-form expression it multiplies and choose a variance combination algebraically consistent with that expression. A masked overall variance can be useful for updating a mixture-scale nuisance quantity, but it is not by itself a diagnostic for equal arm variances or a universal substitute for the contrast variance.

Quantity	Large-sample or exact target	Role
Masked overall variance, equal arm means	Exact: $\pi_1\sigma_1^2+\pi_0\sigma_0^2$; asymptotically: $\pi_1\sigma_1^2+\pi_0\sigma_0^2$	Variance of the pooled masked mixture
Contrast variance	$\sigma_1^2/n_1+\sigma_0^2/n_0$	Sampling variance of $\bar{Y}_1-\bar{Y}_0$
Single variance in a common-form contrast expression	$(\sigma_1^2/n_1+\sigma_0^2/n_0)/(1/n_1+1/n_0)$	Value that reproduces the contrast variance when multiplied by $1/n_1+1/n_0$

Common pitfall and limitations

Wrong reasoning: “The masked pooled variance is the variance for each arm, so I can insert it into any two-sample sample-size formula.” This treats a weighted mixture as if it were a shared arm variance and ignores how each arm’s variance is scaled by its own sample size. Correction: state the target first. For the masked mixture, account for allocation weights and possible mean-separation contamination. For the treatment contrast, use the arm-specific variance formula or derive the single variance that makes the chosen common-form expression equivalent.

A blinded pooled estimate cannot identify both arm variances separately from one pooled number. Many pairs $(\sigma_1^2,\sigma_0^2)$ can produce the same allocation-weighted mixture but different contrast variances, especially when allocation is unequal. The estimate’s usefulness therefore depends on the planning assumptions and the degree of heteroscedasticity. It should not be interpreted as evidence that the arms have equal variances, nor should a pooled update be treated as a guarantee that the updated sample size attains its nominal precision under every arm-variance configuration.

Check Your Understanding

Answer each question, then compare with the solutions below.

With $\pi_1=0.7$, $\pi_0=0.3$, $\sigma_1^2=16$, and $\sigma_0^2=4$, what is the large-sample masked mixture variance when the arm means are equal?
For the same allocation and arm variances, calculate the single variance that reproduces $\sigma_1^2/n_1+\sigma_0^2/n_0$ when multiplied by $1/n_1+1/n_0$.
Why can a raw masked overall variance exceed its within-arm variance component?
Does observing a pooled variance identify whether the arm variances are equal? Explain.

Solutions. (1) The mixture target is $0.7(16)+0.3(4)=12.4$ squared outcome units. (2) The contrast-effective combination uses reversed allocation weights: $0.3(16)+0.7(4)=7.6$ squared units. (3) If $\mu_1\ne\mu_0$, the overall variance includes the nonnegative between-arm term $\frac{n_1n_0}{N(N-1)}(\mu_1-\mu_0)^2$ in its expectation. (4) No. One pooled quantity combines the two variances with allocation weights, so distinct arm-variance pairs can yield the same pooled target.

Quantity	Large-sample or exact target	Role
Masked overall variance, equal arm means	Exact: \(\pi_1\sigma_1^2+\pi_0\sigma_0^2\); asymptotically: \(\pi_1\sigma_1^2+\pi_0\sigma_0^2\)	Variance of the pooled masked mixture
Contrast variance	\(\sigma_1^2/n_1+\sigma_0^2/n_0\)	Sampling variance of \(\bar{Y}_1-\bar{Y}_0\)
Single variance in a common-form contrast expression	\((\sigma_1^2/n_1+\sigma_0^2/n_0)/(1/n_1+1/n_0)\)	Value that reproduces the contrast variance when multiplied by \(1/n_1+1/n_0\)