Information in Repeated Measurements

Why Repeated Measurements Do Not Multiply Information Automatically

Repeated measurements provide several observations on each participant, but those observations are often correlated. The precision of a mean contrast therefore depends not just on how many measurements were recorded, but on how the measurements covary and on which contrast is being estimated. Averaging two highly correlated measurements, for example, generally provides less additional precision than averaging two independent measurements. In contrast, a within-participant difference may be estimated precisely when the two measurements move together.

The calculation below treats each participant as the independent sampling unit and uses the covariance among that participant’s repeated measurements to find the variance of the target contrast. This is the key distinction: the measurement count describes the data collected within participants; the information calculation reflects the variability of the contrast across independent participants.

Model and Definition

Suppose each of $n$ independent participants has $m$ measurements. Write the measurement vector for participant $i$ as $Y_i=(Y_{i1},\ldots,Y_{im})^\top$, with mean vector $\mu$ and covariance matrix $\Sigma$. For the main calculation, assume the participants are identically distributed, have the same measurement schedule, and have a common covariance matrix. Treat $\Sigma$ as specified and known. A multivariate normal model, $Y_i\sim N_m(\mu,\Sigma)$, gives the usual normal-theory interpretation; the variance identity itself needs only finite second moments. The vectors $Y_i$ must be independent across participants, though measurements within a participant may be correlated.

For a fixed vector of contrast coefficients $c=(c_1,\ldots,c_m)^\top$, the target is the mean contrast $\theta=c^\top\mu$. Its participant-level value is $D_i=c^\top Y_i$. The contrast coefficients determine the target: for example, coefficients that sum to zero often describe a difference, while coefficients that sum to one can describe a weighted mean. The variance of $D_i$ is determined by the full covariance structure, not by the measurement count alone.

Definition: Let $v_c=c^\top\Sigma c$ be the variance of one participant’s contrast $D_i=c^\top Y_i$. If $v_c>0$, the sample contrast $c^\top\bar{Y}$, where $\bar{Y}=n^{-1}\sum_{i=1}^nY_i$, has variance $v_c/n$. Its variance-based information is the reciprocal of that variance: $$ I_c=\frac{n}{v_c}. $$ This information is for estimating $\theta=c^\top\mu$ from $n$ independent participants under the stated covariance structure. It is not the number of measurements.

Information here has units equal to the inverse square of the outcome unit. If each measurement is in millimeters, for example, $v_c$ is in squared millimeters and $I_c$ is in inverse squared millimeters. Sample size $n$ is a count, but it is the number of independent participants—not $n$ multiplied by the number of repeated measurements.

Deriving the Contrast Variance

The contrast can be calculated participant by participant and then averaged. Since the participants are independent and each has the same covariance matrix,

\operatorname{Var}(c^\top\bar{Y}) =\operatorname{Var}\left(\frac{1}{n}\sum_{i=1}^n c^\top Y_i\right) =\frac{1}{n^2}\sum_{i=1}^n\operatorname{Var}(c^\top Y_i) =\frac{c^\top\Sigma c}{n}.

The quadratic form expands to $c^\top\Sigma c=\sum_{j=1}^m\sum_{k=1}^m c_jc_k\operatorname{Cov}(Y_{ij},Y_{ik})$. Thus, each measurement variance contributes through a squared coefficient, and each within-participant covariance contributes through a cross-product of coefficients. Positive covariance can increase the variance of an average, but it can reduce the variance of a difference whose coefficients have opposite signs.

When $\Sigma$ is positive definite and the normal model holds, the Fisher information matrix for $\mu$ from $n$ participants is $n\Sigma^{-1}$. The Cramér–Rao bound for an unbiased estimator of $c^\top\mu$ is then

\operatorname{Var}(\widehat{\theta})\geq c^\top(n\Sigma^{-1})^{-1}c =\frac{c^\top\Sigma c}{n}.

The sample contrast $c^\top\bar{Y}$ attains this bound in that model. This matrix-information argument requires positive-definite $\Sigma$; it does not apply when $\Sigma$ is singular. The variance identity above does not require an inverse, so it remains valid for positive semidefinite $\Sigma$. For a nondegenerate contrast with $c^\top\Sigma c>0$, $n/(c^\top\Sigma c)$ remains a well-defined variance-based information measure. If $c^\top\Sigma c=0$, the contrast is deterministic under the model: its reciprocal variance is undefined, and the ordinary information formula must not be used. Such a target is exactly determined under the assumed model, rather than having a finite positive sampling variance.

Worked Example: Information for the Average of Two Measurements

Suppose each participant has two measurements of an outcome in millimeters. Assume each has variance $16$ square millimeters and their covariance is $9.6$ square millimeters, corresponding to correlation $9.6/16=0.60$. The target is the average of the two population means, so $c=(0.5,0.5)^\top$. There are $n=50$ independent participants. These values specify a valid covariance matrix: its variances are positive and its determinant is $16(16)-9.6^2=163.84>0$.

The covariance matrix and contrast variance are

\Sigma= \begin{pmatrix} 16&9.6\\ 9.6&16 \end{pmatrix}, \qquad v_c =(0.5)^2(16)+(0.5)^2(16)+2(0.5)(0.5)(9.6) =12.8.

Thus, the variance of the estimated average of the two means is $12.8/50=0.256$ square millimeters. Its standard error is $\sqrt{0.256}=0.5060$ millimeters, and its information is $50/12.8=3.906$ inverse squared millimeters. If the two measurements within a participant were incorrectly treated as independent, the calculated participant-level variance would be $8$, not $12.8$. That calculation would overstate the precision of this average because it discards the positive within-participant covariance.

Worked Example: A Within-Participant Change

Consider two measurements, in centimeters, from each of $n=40$ independent participants. Assume the first measurement has variance $25$, the second has variance $16$, and their covariance is $10$. The correlation is $10/\sqrt{25\cdot16}=0.50$, so this is a valid two-measurement covariance structure. The target is the mean at the second occasion minus the mean at the first, with $c=(-1,1)^\top$.

For one participant, the change variance is

v_c=(-1)^2(25)+(1)^2(16)+2(-1)(1)(10) =25+16-20=21\text{ cm}^2.

The estimated mean change has variance $21/40=0.525\text{ cm}^2$, standard error $\sqrt{0.525}=0.7246$ cm, and information $40/21=1.905\text{ cm}^{-2}$. The positive covariance reduces the variance of the change: if both measurements tend to be high or low together, subtracting one from the other cancels some of that shared variation. The result describes precision under the stated model; it does not by itself establish that the mean change is different from zero.

Worked Example: A Late Measurement Compared With an Earlier Average

Suppose each of $n=36$ independent participants has three measurements, each with variance $9$ units squared. Assume every pair has correlation $0.50$, so each covariance is $0.50\sqrt{9\cdot9}=4.5$ units squared. This equicorrelation structure is valid: its eigenvalues are $18$, $4.5$, and $4.5$, all positive. Compare the third measurement with the average of the first two using $c=(-0.5,-0.5,1)^\top$.

The contrast variance includes three variance terms and three covariance terms:

\begin{aligned} v_c &=9\left[(-0.5)^2+(-0.5)^2+1^2\right]\\ &\quad+2(4.5)\left[(-0.5)(-0.5)+(-0.5)(1)+(-0.5)(1)\right]\\ &=9(1.5)+9(0.25-0.5-0.5)\\ &=13.5-6.75=6.75\text{ units}^2. \end{aligned}

The estimated contrast has variance $6.75/36=0.1875$ units squared and standard error $\sqrt{0.1875}=0.4330$ units. Its information is $36/6.75=5.333$ inverse squared units. If the within-participant measurements were incorrectly treated as independent, the calculated contrast variance per participant would be $9(1.5)=13.5$, twice the correct value. Here the positive correlation helps because the contrast subtracts earlier measurements from the later one.

Interpreting and Checking the Calculation

The covariance structure affects different contrasts in different ways. There is no general rule that correlation always reduces information. For a weighted average with coefficients of the same sign, positive covariance tends to increase the contrast variance. For a difference with coefficients of opposite signs, positive covariance can reduce it. The coefficients and the covariance matrix must therefore be considered together.

Useful checks before using the result include the following:

Check the target. Confirm that the entries of $c$ represent the intended comparison and that the measurement occasions are in the correct order. A reversed difference changes the sign of the estimate, though not its variance.
Check the covariance matrix. It must be symmetric and positive semidefinite. For the ordinary reciprocal-variance information calculation, also verify $c^\top\Sigma c>0$. A negative computed variance signals an invalid covariance specification or an arithmetic error.
Check the independent unit. The formula uses $n$ independent participants. Repeated observations on one participant are not additional independent participants. Dependence between participants would require a different variance calculation.
Check that the design matches the model. The formula assumes a common measurement schedule and covariance structure. If participants have different observed occasions or covariance structures, applying the same $v_c$ to all of them may not describe the actual estimator.
Check the units. The variance must be in squared outcome units, the standard error in outcome units, and the information in inverse squared outcome units.

A common mistake is to count $n m$ measurements and treat them as $n m$ independent observations. This ignores within-participant covariance and can seriously misstate precision. The correction is to form the participant-level contrast first, calculate its variance using the full covariance matrix, and then use the number of independent participants to divide that variance. Another mistake is to apply $1/v_c$ when $v_c=0$. The correction is to recognize that the contrast is degenerate under the stated model and not use a reciprocal-variance formula for it.

How This Information Measure Fits With Other Approaches

Approach	What it uses	Key distinction
Participant-level contrast	The specified contrast and the within-participant covariance matrix	Directly gives the variance and information for the target contrast.
Counting all measurements as independent	The total number of recorded values	Generally incorrect when measurements on the same participant are correlated.
Using only one occasion	One measurement per participant	Can be valid for a one-occasion target, but discards information relevant to a repeated-measurement contrast.
Paired comparison	A two-occasion within-participant difference	A special case of the contrast calculation, with coefficients $(-1,1)$.

This calculation is intentionally tied to a specified contrast and covariance structure. It does not determine whether a particular contrast is scientifically important, and its precision can be wrong if the covariance assumptions are wrong. It also does not address uncertainty from estimating the covariance matrix; that issue is distinct from treating the specified covariance structure as known.

Key takeaway: For a nondegenerate mean contrast from $n$ independent participants with common within-participant covariance $\Sigma$, calculate $v_c=c^\top\Sigma c$ first. The contrast variance is $v_c/n$ and its variance-based information is $n/v_c$. Repeated-measurement count alone does not determine information.

Check Your Understanding

Answer each question, then compare with the short solutions below.

Two measurements each have variance $4$ and covariance $1$. For the average contrast $c=(0.5,0.5)^\top$, calculate the participant-level contrast variance.
For the same covariance structure, calculate the variance of the difference with $c=(-1,1)^\top$. Why is it smaller than the sum of the two variances?
There are 25 independent participants, and a specified contrast has variance $v_c=10$. What are the variance and information for its sample mean contrast?
What should be concluded about the reciprocal-variance information formula if $c^\top\Sigma c=0$?
Why is it generally wrong to use the number of participants multiplied by the number of repeated measurements as the independent sample size?

Solutions. (1) $v_c=0.25(4)+0.25(4)+2(0.5)(0.5)(1)=2.5$ squared outcome units. (2) $v_c=4+4-2(1)=6$ squared outcome units; the positive covariance cancels part of the shared variation in the difference. (3) The variance is $10/25=0.400$ squared outcome units, and information is $25/10=2.5$ inverse squared units. (4) The contrast is deterministic under the specified covariance model, so the reciprocal is undefined and the ordinary formula does not apply. (5) Measurements within a participant may be correlated, so they are not generally independent sampling units; the calculation must use their covariance and the number of independent participants.