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Information Fraction, Sample Size and Events · Tutorial 83 of 699

Information in Paired Comparisons

Learn to calculate paired-comparison information from the variance of differences and assess how within-pair correlation affects precision and sample size.

Intermediate 10 min read

What You'll Learn

  • Define the paired difference and the assumptions needed to use it for inference.
  • Derive its variance from the marginal variances and within-pair correlation.
  • Calculate information and standard errors from the number of complete pairs.
  • Estimate how correlation changes a paired design’s planned sample size.
  • Distinguish information from the number of individual measurements.

Why Pairing Changes Information

In a paired comparison, each unit contributes two related measurements, such as outcomes under two conditions or measurements before and after an intervention. The analysis does not treat these measurements as independent observations. Instead, it forms one difference for each pair and estimates the mean of those differences. The precision of that estimate depends on the within-pair variance, which in turn depends on the correlation between the two measurements.

This matters for information-based planning: two designs with the same number of pairs can have different information if their within-pair correlations differ. Positive correlation often reduces the variability of the differences and increases information. But pairing does not automatically improve precision; the correlation and the comparability of pairs matter.

Definition: For pair \(i\), let \(X_i\) and \(Y_i\) denote the two outcomes, and define the paired difference \(D_i=Y_i-X_i\). The paired mean effect is \(\delta=E(D_i)\), estimated by \(\bar{D}=n^{-1}\sum_{i=1}^n D_i\), where \(n\) is the number of complete pairs. If the differences are independent and have variance \(\sigma_D^2\), the information for estimating \(\delta\), in the variance-of-estimator sense, is \(I=n/\sigma_D^2\). Under a normal model with known \(\sigma_D^2\), this is also the Fisher information for \(\delta\).

From Within-Pair Correlation to Information

Assume the pairs are independent of one another and identically distributed. Within a pair, the outcomes may be correlated. Let \(X_i\) and \(Y_i\) have variances \(\sigma_X^2\) and \(\sigma_Y^2\), and within-pair correlation \(\rho\). The variance of one difference follows from the variance of a subtraction:

$$ \sigma_D^2=\operatorname{Var}(Y_i-X_i) =\sigma_Y^2+\sigma_X^2-2\operatorname{Cov}(X_i,Y_i) =\sigma_Y^2+\sigma_X^2-2\rho\sigma_X\sigma_Y. $$

Because the \(n\) differences are independent, \(\operatorname{Var}(\bar{D})=\sigma_D^2/n\). Its standard error is therefore \(\sigma_D/\sqrt{n}\), and the reciprocal of its variance is \(n/\sigma_D^2\). This gives the information expression above. If both marginal standard deviations are \(\sigma\), then \(\sigma_D^2=2\sigma^2(1-\rho)\). Holding the marginal variability fixed, a larger positive \(\rho\) makes differences less variable, so each pair contributes more information.

The normal-model justification is direct. If \(D_i\sim N(\delta,\sigma_D^2)\) with known variance, the log likelihood for \(\delta\), apart from terms not involving \(\delta\), is \(-\sum_i(D_i-\delta)^2/(2\sigma_D^2)\). Its score is \(\sum_i(D_i-\delta)/\sigma_D^2\), whose variance is \(n/\sigma_D^2\). Thus the Fisher information is \(I=n/\sigma_D^2\). Without normality, the same formula describes the reciprocal variance of the sample mean under the stated finite-variance, independent-pairs model; it is the usual large-sample information measure for planning.

In a group sequential design with a fixed planned difference variance, \(n_k\) complete pairs at look \(k\) provide \(I_k=n_k/\sigma_D^2\). If the final analysis has \(N\) pairs, the planned information fraction is \(I_k/I_{\mathrm{final}}=n_k/N\). A pair supplies one difference, not two independent outcome contributions. The number of enrolled individuals alone is not the relevant count if some do not provide both measurements.

Worked Example: Calculating Information From the Paired Variance

Suppose a hypothetical study measures an outcome under two conditions in each of 40 complete pairs. For planning, assume each condition has standard deviation 10 outcome units and the within-pair correlation is 0.60. Assume pairs are independent and the planned variance is known.

The difference variance is \(\sigma_D^2=10^2+10^2-2(0.60)(10)(10)=200-120=80\) outcome-units squared. The difference standard deviation is \(\sqrt{80}=8.944\) outcome units.

The information is \(I=n/\sigma_D^2=40/80=0.5000\) inverse outcome-units squared. The standard error of the paired mean is \(\sqrt{80/40}=\sqrt{2}=1.414\) outcome units. These quantities agree because \(I=1/SE^2=1/(1.414^2)=0.5000\), up to rounding.

This calculation uses the correlation within each pair, not a correlation between different pairs. If the pairs were dependent, the variance of \(\bar{D}\) would not in general be \(\sigma_D^2/n\), and this information calculation would need adjustment.

Using Information for Sample-Size Planning

For a fixed-sample, two-sided test of \(H_0:\delta=0\) at significance level \(\alpha\), a normal-approximation benchmark for the number of pairs needed to detect a target difference \(\delta_1\) with power \(1-\beta\) is

$$ n \approx \frac{(z_{1-\alpha/2}+z_{1-\beta})^2\sigma_D^2}{\delta_1^2}, $$

where \(z_q\) is the \(q\)-quantile of the standard normal distribution. For a prespecified one-sided test, replace \(z_{1-\alpha/2}\) with \(z_{1-\alpha}\). This is a fixed-sample, known-variance normal approximation. A group sequential design’s final target information must also reflect its sequential testing plan; the formula here is a benchmark, not a substitute for that calibration.

When the marginal standard deviations are equal, substituting \(\sigma_D^2=2\sigma^2(1-\rho)\) makes the role of pairing explicit. Relative to an independent two-arm design with the same marginal standard deviation, a positive within-pair correlation can substantially reduce the required number of pairs. The benefit comes from comparing measurements that share pair-specific variation, not simply from collecting two outcomes per pair.

Worked Example: How Correlation Affects the Required Number of Pairs

Consider a hypothetical two-sided test with \(\alpha=0.05\), target power \(0.80\), target mean difference \(\delta_1=5\) outcome units, and marginal standard deviation 12 outcome units in each condition. Assume independent pairs, known planning variances, and within-pair correlation \(\rho=0.50\).

The within-pair variance is \(\sigma_D^2=12^2+12^2-2(0.50)(12)(12)=288-144=144\) outcome-units squared. Using \(z_{0.975}=1.960\) and \(z_{0.80}=0.8416\),

$$ n \approx \frac{(1.960+0.8416)^2(144)}{5^2} =\frac{(2.8016)^2(144)}{25} =\frac{7.849(144)}{25} =45.21. $$

Rounding up gives a planning target of 46 complete pairs, or 92 individuals if each pair contains two individuals. The calculation is an approximation; it does not account for estimating the variance or for any sequential-design adjustment.

For comparison, under the same marginal standard deviations, an independent two-arm design requires approximately \(2(12^2)(2.8016^2)/5^2=90.42\) individuals per arm, rounded up to 91 per arm. Under these assumptions, the paired design needs fewer individuals because the correlation of 0.50 reduces the difference variance. This comparison is meaningful only if the pairing is appropriate and the two designs target the same effect.

Information Fractions, Diagnostics, and Alternatives

With a fixed \(\sigma_D^2\), information accumulates linearly with the number of complete pairs. For example, if the planned final analysis has \(N\) pairs, the information fraction after \(n_k\) pairs is \(n_k/N\). This simple relation depends on the difference variance remaining as planned. If the variance of differences is higher than anticipated, the same number of pairs yields less information than planned.

Worked Example: Translating a Pair Count Into an Information Fraction

Suppose a hypothetical trial plans 80 complete pairs and uses a planned difference variance of 64 outcome-units squared. After 32 complete pairs, the information is \(I_k=32/64=0.5000\) inverse outcome-units squared. The planned final information is \(I_{\mathrm{final}}=80/64=1.250\) inverse outcome-units squared.

The information fraction is \(I_k/I_{\mathrm{final}}=0.5000/1.250=0.4000\), which equals \(32/80\). Although 64 individual measurements have been observed at this point, they represent 32 paired differences, not 64 independent contributions to the mean-difference information.

Useful checks and failure modes include the following:

  • Check the pairing. Each \(X_i\) must be matched with the corresponding \(Y_i\) for the intended comparison. Incorrect matching changes the differences and their variance.
  • Check the variance input. The relevant quantity is \(\sigma_D^2\), not either marginal variance alone. Ignoring the covariance term can greatly misstate precision.
  • Check independence across pairs. Shared sites, households, or other dependencies between pairs can make \(n/\sigma_D^2\) too large as an information estimate.
  • Check that correlation is plausible for the target setting. A planning value borrowed from a different population or measurement process may not predict the variance of future differences. A negative correlation increases \(\sigma_D^2\) when the marginal variances are fixed.
  • Check the estimand and study conditions. Pairing supports inference about the mean difference for the defined pairs; it does not, by itself, justify a causal interpretation. In an intervention study, assignment and design assumptions still matter.

The main alternative is an unpaired comparison, which estimates a difference between two group means without using within-pair covariance. It can be appropriate when observations cannot be meaningfully matched or when pairing is infeasible. A paired analysis can be more precise when the pairs are well formed and the correlation is positive, but may be less efficient if correlation is negative or pairing adds substantial variability. The choice should follow the design and target comparison, not a presumption that paired data are always superior.

Common mistake: “There are two measurements in every pair, so each pair contributes two independent observations of the treatment difference.” This is incorrect: the comparison is based on one difference per pair, whose variance includes the within-pair covariance. Use \(I=n/\sigma_D^2\), not a count of \(2n\) independent differences.
Key takeaway: Paired-comparison information is governed by the variability of the within-pair differences. Under independent pairs, \(I=n/\sigma_D^2\); calculating \(\sigma_D^2\) correctly requires both marginal variances and their within-pair correlation.

Check Your Understanding

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

  1. Each measurement has standard deviation 8 units, and the within-pair correlation is 0.25. What is the variance of the paired difference?
  2. If the difference variance is 36 units squared, how many independent pairs provide information \(I=2\) inverse units squared?
  3. With equal marginal variances, what happens to paired-difference variance as within-pair correlation increases?
  4. A planned study has 100 pairs, but only 40 complete pairs are available at an interim analysis. Under a fixed planned difference variance, what is the information fraction?

Solutions. (1) \(\sigma_D^2=8^2+8^2-2(0.25)(8)(8)=128-32=96\) units squared. (2) Since \(I=n/\sigma_D^2\), \(n=I\sigma_D^2=2(36)=72\) pairs. (3) From \(\sigma_D^2=2\sigma^2(1-\rho)\), increasing \(\rho\) decreases the difference variance and increases information per pair. (4) The fraction is \(40/100=0.4000\), provided the planned variance applies at both analyses.

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