Calculating Relative Standard Errors

From a Covariance Matrix to Parameter Precision

A parameter estimate is more informative when accompanied by an estimate of its uncertainty. In population pharmacokinetic analysis, software may report an estimated covariance matrix for the fitted parameters. Its diagonal provides the estimated variances needed to calculate standard errors; dividing a standard error by the corresponding parameter estimate gives a relative standard error (RSE). These calculations turn a matrix output into parameter-specific summaries that can be compared across parameters with different units.

The calculations are meaningful only when the covariance matrix and parameter estimates refer to the same parameters, ordering, and scale. They also rely on the covariance matrix being a reasonable approximation to the sampling uncertainty of the estimates. A small RSE is not, by itself, proof that a model is correct or that a parameter is well identified under every plausible model.

Definition: Let $\hat{\boldsymbol{\theta}}$ be a vector of parameter estimates and let $\widehat{\mathrm{Cov}}(\hat{\boldsymbol{\theta}})$ be its estimated covariance matrix. The estimated standard error of parameter estimate $\hat{\theta}_i$ is the square root of the $i$th diagonal element of that matrix: $$ SE(\hat{\theta}_i)=\sqrt{\widehat{\mathrm{Cov}}(\hat{\boldsymbol{\theta}})_{ii}}. $$ For a positive estimate, its relative standard error is $$ RSE_i=100\frac{SE(\hat{\theta}_i)}{\hat{\theta}_i}\%. $$ RSE is a dimensionless percentage; the standard error has the same units as the parameter.

A covariance matrix is organized in the same order as its parameter vector. The diagonal entry for a parameter is its estimated variance, so its units are the square of the parameter’s units. For instance, if clearance is measured in L/h, its variance is in $(\mathrm{L/h})^2$, and its standard error is in L/h. Off-diagonal entries are covariances between pairs of estimates and have product units. For an individual parameter’s standard error, use its diagonal entry.

The RSE compares the estimated standard error with the estimate’s magnitude. For a positive PK parameter, the usual calculation divides by the estimate itself. If an estimate can be negative, a percentage intended to describe relative magnitude is commonly calculated using its absolute value in the denominator. An RSE is not an interval, and it does not give the probability that the parameter lies within a specified percentage of the estimate.

Assumptions and a Short Justification

A covariance matrix from an estimation procedure is typically a local, approximate summary of sampling uncertainty. For example, it may be obtained from the inverse of an estimated information matrix based on the curvature of the objective function near the optimum. Interpreting its diagonal square roots as standard errors presumes that the fitted solution is a suitable interior optimum, the local curvature is informative, and the covariance calculation is numerically reliable. The approximation is generally more defensible when estimates are not dominated by boundaries, severe nonlinearity, or weak identification.

The square-root step follows directly from the definition of variance. If the estimated variance of $\hat{\theta}_i$ is $v_i$, then the estimated standard deviation is the nonnegative square root of that variance, $\sqrt{v_i}$. This quantity has the parameter’s units because taking the square root also takes the square root of squared units. Dividing it by an estimate in the same units cancels the units, which is why RSE is dimensionless.

This is an approximate uncertainty calculation for an estimate under the fitted model and covariance method. It does not account automatically for model misspecification, uncertainty about the chosen structural model, or limitations in the data design. The earlier material on point estimates versus precision distinguishes the numerical value of an estimate from the uncertainty attached to it; the calculations here provide one common way to summarize that uncertainty.

Worked Example: Standard Errors for Clearance and Volume

Suppose a hypothetical population PK fit estimates clearance $CL=5.0$ L/h and volume $V=50$ L. In the parameter order $(CL,V)$, the estimated covariance matrix is

\widehat{\mathrm{Cov}}(\hat{\boldsymbol{\theta}}) = \begin{pmatrix} 0.64 & -4.0\\ -4.0 & 100 \end{pmatrix}.

The diagonal entries are the estimated variances. For clearance, the variance is $0.64\;(\mathrm{L/h})^2$, so

SE(\widehat{CL})=\sqrt{0.64}=0.80\ \mathrm{L/h}.

For volume, the variance is $100\;\mathrm{L}^2$, giving

SE(\widehat{V})=\sqrt{100}=10\ \mathrm{L}.

The negative off-diagonal entry is not used in either of these individual standard-error calculations. It does not change the square root of either diagonal variance. Both results have units consistent with their respective parameters.

Calculating and Comparing RSEs

Once the standard error is calculated, the RSE is obtained by dividing by the corresponding estimate and multiplying by 100. The pairing must be exact: use the standard error for clearance with the clearance estimate, not with volume or another parameter. RSE is useful for comparing relative uncertainty when parameters have different units or scales, but it should not be treated as a universal pass/fail threshold.

Worked Example: Relative Standard Errors for Clearance and Volume

In a second hypothetical fit, the estimates are $CL=6.0$ L/h and $V=40$ L. The corresponding diagonal variances in the estimated covariance matrix are $0.81\;(\mathrm{L/h})^2$ for clearance and $64\;\mathrm{L}^2$ for volume. First calculate the standard errors:

SE(\widehat{CL})=\sqrt{0.81}=0.90\ \mathrm{L/h}, \qquad SE(\widehat{V})=\sqrt{64}=8.0\ \mathrm{L}.

Then calculate each RSE using its matching estimate:

RSE_{CL}=100\frac{0.90\ \mathrm{L/h}}{6.0\ \mathrm{L/h}}=15.0\%, \qquad RSE_V=100\frac{8.0\ \mathrm{L}}{40\ \mathrm{L}}=20.0\%.

Under the covariance approximation used for this fit, the estimated standard error is 15.0% of the clearance estimate and 20.0% of the volume estimate. These are relative precision summaries, not probabilities or guarantees about the error in either estimate.

The RSE for volume is unchanged if volume is reported in millilitres instead of litres, provided the estimate and standard error are both converted. Here, $40$ L is $40{,}000$ mL and $8.0$ L is $8{,}000$ mL, so

100\frac{8{,}000\ \mathrm{mL}}{40{,}000\ \mathrm{mL}}=20.0\%.

This illustrates why RSE can support comparisons across parameters or unit conventions. The absolute standard error remains essential for understanding the uncertainty in the parameter’s actual units.

Residual-Error Parameters

RSEs can also be calculated for residual-error parameters, provided the estimate and covariance matrix refer to the same parameterization. Consider a hypothetical combined additive and proportional residual model:

Y_{ij}=f_{ij}+\varepsilon_{ij}, \qquad \varepsilon_{ij}\sim N\left(0,\sigma_{\mathrm{add}}^2+ (\sigma_{\mathrm{prop}}f_{ij})^2\right).

Here, $Y_{ij}$ is an observed concentration, $f_{ij}$ is its model prediction, $\sigma_{\mathrm{add}}$ is an additive residual standard deviation in concentration units, and $\sigma_{\mathrm{prop}}$ is a dimensionless proportional residual standard deviation. The variance expression specifies the residual model; it does not mean that the estimated covariance matrix is a matrix of residual variances. The latter is the covariance matrix of the parameter estimates.

Worked Example: RSEs for Additive and Proportional Error

Suppose a hypothetical fit gives $\hat{\sigma}_{\mathrm{add}}=0.12$ mg/L and $\hat{\sigma}_{\mathrm{prop}}=0.18$. In that parameter order, the estimated covariance matrix is

\widehat{\mathrm{Cov}}(\hat{\boldsymbol{\theta}}) = \begin{pmatrix} 0.0009 & -0.00012\\ -0.00012 & 0.0004 \end{pmatrix}.

The first diagonal entry has units $(\mathrm{mg/L})^2$, and the second is dimensionless squared. Therefore,

SE(\hat{\sigma}_{\mathrm{add}})=\sqrt{0.0009}=0.030\ \mathrm{mg/L}, \qquad SE(\hat{\sigma}_{\mathrm{prop}})=\sqrt{0.0004}=0.020.

The RSEs are

RSE_{\mathrm{add}}=100\frac{0.030\ \mathrm{mg/L}}{0.12\ \mathrm{mg/L}}=25.0\%, \qquad RSE_{\mathrm{prop}}=100\frac{0.020}{0.18}=11.1\%.

For this fit, the additive-error estimate has a larger standard error relative to its estimate than the proportional-error estimate. The interpretation is conditional on these estimates, their parameterization, and the validity of the covariance approximation.

Checks, Alternatives, and Common Pitfalls

Before reporting results, check that the covariance matrix’s parameter order matches the estimates, that each diagonal entry is nonnegative, and that the units are consistent. A negative diagonal entry cannot represent an estimated variance and usually signals an invalid or failed covariance calculation. A matrix that is not positive semidefinite may also indicate numerical or estimation problems. Small numerical deviations can occur in computation, but they should not be silently converted into plausible-looking standard errors.

Check as well whether the reported matrix is on the same scale as the estimates used in the RSE calculation. If the matrix refers to a transformed parameter but the estimate is reported on its natural scale, taking a diagonal square root and dividing by the natural-scale estimate does not generally produce the natural-scale RSE. Do not combine entries and estimates from different parameterizations without a justified conversion.

RSE becomes unstable as a positive estimate approaches zero: even a moderate standard error can produce a very large percentage. If the estimate is zero, the usual ratio is undefined. In these situations, report the estimate and standard error in their units and explain the limitation rather than presenting a misleading percentage. Estimates at a boundary, poorly determined curvature, or a nonconverged fit also call for caution; a numerical RSE should not conceal those warning signs.

The absolute standard error and RSE answer different questions. The standard error expresses approximate uncertainty in the parameter’s own units; RSE expresses that uncertainty relative to the estimate’s magnitude. A resampling-based standard error, such as the empirical standard deviation of estimates across bootstrap replicates, is an alternative uncertainty summary when a local covariance approximation is questionable. It is not automatically superior: its reliability depends on the resampling procedure and the stability of the fitted models. Neither approach repairs a misspecified model or inadequate data.

Common mistake: Treating a diagonal covariance entry as the standard error, then dividing that number by the estimate. The diagonal entry is a variance, so it must first be square-rooted. For example, a variance of $0.81\;(\mathrm{L/h})^2$ gives an SE of $0.90$ L/h, not $0.81$ L/h. With an estimate of $6.0$ L/h, the RSE is $100(0.90/6.0)=15.0\%$, not $13.5\%$.

Practical Calculation Sequence

Match the parameter.
Confirm the order of the parameter estimates and the rows and columns of the estimated covariance matrix.

Extract its diagonal entry.
Read the estimated variance for the selected parameter and check that it is nonnegative and in squared units.

Take the square root.
Report the resulting standard error in the same units as the parameter estimate.

Calculate and qualify the RSE.
For a positive estimate, divide the standard error by that estimate and multiply by 100; check for scale mismatch, boundary estimates, and numerical instability.

Check Your Understanding

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

A clearance estimate is $4.0$ L/h and its estimated variance is $0.36\;(\mathrm{L/h})^2$. What are its SE and RSE?
A volume estimate is $25$ L and its estimated standard error is $5$ L. What is its RSE?
Why is a covariance-matrix diagonal entry not itself the standard error?
An estimated variance is negative. Should its square root be reported as a standard error? Explain.
What happens to an RSE if both a positive estimate and its SE are converted from litres to millilitres?

Solutions. (1) $SE=\sqrt{0.36}=0.60$ L/h, so $RSE=100(0.60/4.0)=15.0\%$. (2) $RSE=100(5/25)=20.0\%$. (3) The diagonal entry is a variance in squared units; its square root is the standard error in the parameter’s units. (4) No. An estimated variance cannot be negative; investigate the covariance calculation or fitting problem rather than reporting an imaginary or altered SE. (5) It is unchanged because the common unit-conversion factor cancels in the ratio, as long as both quantities are converted consistently.