Why Check Innovation Sequences?
At each observation time, an SDE model gives a predictive distribution conditional on the observations and dosing information available so far. A residual is informative only in relation to that prediction: an error of one concentration unit may be large when predictive uncertainty is small and unremarkable when uncertainty is large. Serial-correlation checks therefore begin with standardized one-step innovations, not with raw differences between observed and predicted concentrations.
This distinction matters especially when observation times are irregular. Consecutive observations might be separated by minutes in one part of a study and days in another. The innovation index records the order of observations; elapsed time records how far apart they are. Both can be useful, but they answer different diagnostic questions.
Definition and a Basic Property
Let \(t_1<\cdots<t_n\) be observation times, and let \(\mathcal F_{i-1}\) contain the information available just before observing \(Y_i\), including earlier observations and the relevant dosing history. Suppose the one-step predictive distribution has conditional mean \(\mu_i\) and conditional variance \(s_i^2\), where \(0<s_i^2<\infty\). Define
The predictive moments must be conditional on \(\mathcal F_{i-1}\); in particular, \(s_i\) can depend on the elapsed time since the previous observation. For a non-Gaussian predictive distribution, this standardized residual is still defined using its mean and variance, when those moments exist. It need not have a normal distribution.
For the conditional statements, substitute the definition of \(Z_i\) and use that \(\mu_i\) and \(s_i\) are known given \(\mathcal F_{i-1}\). For the serial-covariance statement, \(Z_i\) is measurable with respect to \(\mathcal F_{j-1}\) when \(i<j\). Therefore
Thus correct one-step conditional means and variances imply uncorrelated innovations, even with irregular time intervals. They do not by themselves imply that the \(Z_i\) are independent or normally distributed. A serial-correlation check targets linear dependence; it is not a complete test of the predictive distributions.
Why Elapsed Time Changes the Diagnostic
An index-lag-one correlation pairs \(Z_i\) with \(Z_{i+1}\), regardless of whether their observation times are close together. That can be a useful check for dependence between consecutive measurements, but it combines transitions with potentially very different durations. If residual dependence decays with time, a pair separated by a short interval may behave differently from one separated by a long interval.
A time-aware correlogram instead groups pairs using their actual separation \(\Delta_{ij}=t_j-t_i\). Choose non-overlapping elapsed-time bins \(B_1,\ldots,B_K\). For bin \(b\), let \(\mathcal P_b=\{(i,j):i<j,\ \Delta_{ij}\in B_b\}\) and \(m_b\) be the number of pairs in that bin. A simple product estimate is
Because the innovations are standardized to have conditional variance one, these average products have a natural scale for describing residual linear dependence. They are not automatically unbiased estimates of a stationary correlation function: the predictive process, covariates, or sampling design may change over time. Report the bin boundaries and pair counts alongside the estimates.
Worked Example: Standardizing an Irregular OU Transition
Consider the mean-reverting process \(dX_t=-k(X_t-\theta)\,dt+\sigma\,dW_t\), observed without measurement error. Conditional on \(X_t=x\), its distribution after an interval \(\Delta\) is normal with mean \(\theta+(x-\theta)e^{-k\Delta}\) and variance \(\frac{\sigma^2}{2k}(1-e^{-2k\Delta})\). Take \(\theta=10\), \(k=0.5\), and \(\sigma=2\).
First suppose \(x=12\) and \(\Delta=2\). Since \(e^{-k\Delta}=e^{-1}\approx0.3679\), the predictive mean is \(10+2(0.3679)=10.7358\). The predictive variance is \(\frac{4}{1}(1-e^{-2})\approx3.4587\), so the predictive standard deviation is approximately \(1.8598\). If the next observation is \(12.5956\), then
Now use the same starting value but a shorter interval, \(\Delta=0.5\). The mean is \(10+2e^{-0.25}\approx11.5576\). The variance is \(4(1-e^{-0.5})\approx1.5739\), giving standard deviation \(1.2545\). An observation of \(10.3031\) gives \(Z=(10.3031-11.5576)/1.2545\approx-1.00\). The same raw concentration difference would not represent the same standardized departure under these two transition distributions.
A Portmanteau-Style Summary
A correlogram displays dependence estimates across several lags or elapsed-time bins. A portmanteau-style statistic compresses several estimates into one number. For the elapsed-time bins above, one useful descriptive form is
Larger values indicate that one or more bin-average products are large relative to the number of available pairs. This construction alone does not give \(Q\) a universal chi-square distribution. In particular, pairs in different products may share an observation, bins may contain very different numbers of pairs, and parameter estimation affects the distribution of fitted-model innovations.
To verify this, write \(S_b=\sum_{(i,j)\in\mathcal P_b}Z_iZ_j\), so that \(m_b\widehat r_b^{\,2}=S_b^2/m_b\). Each product has mean zero and variance one. For two distinct pairs with no shared index, their products are independent and have covariance zero. If they share one index, for example \(Z_iZ_j\) and \(Z_iZ_k\), their product has expectation \(E[Z_i^2Z_jZ_k]=E[Z_i^2]E[Z_j]E[Z_k]=0\). Hence all distinct pair products in a bin have covariance zero, and \(E[S_b^2]=m_b\). It follows that \(E[S_b^2/m_b]=1\) for each bin; summing gives \(E[Q]=K\). This expectation does not establish a chi-square reference distribution, nor does it account for fitted parameters. \(\square\)
Worked Example: Calculating an Elapsed-Time Statistic
Suppose observations occur at times \(0,1,3,4,7\), with standardized innovations \(0.5,-1,1.5,0,-0.5\), respectively. Use bins \(B_1=(0,1.5]\), \(B_2=(1.5,3.5]\), and \(B_3=(3.5,\infty)\).
In \(B_1\), the pairs are \((1,2)\) and \((3,4)\), with products \((0.5)(-1)=-0.5\) and \((1.5)(0)=0\). Thus \(m_1=2\), \(\widehat r_1=-0.5/2=-0.25\), and the contribution is \(2(-0.25)^2=0.125\).
In \(B_2\), the pairs are \((1,3),(2,3),(2,4),(4,5)\). Their products are \(0.75,-1.5,0,0\), giving \(m_2=4\), \(\widehat r_2=-0.75/4=-0.1875\), and contribution \(4(-0.1875)^2=0.140625\).
In \(B_3\), the pairs are \((1,4),(1,5),(2,5),(3,5)\). Their products are \(0,-0.25,0.5,-0.75\), so \(m_3=4\), \(\widehat r_3=-0.5/4=-0.125\), and contribution \(4(-0.125)^2=0.0625\). Therefore \(Q=0.125+0.140625+0.0625=0.328125\). This is a summary for these bins, not a p-value: its null distribution must be justified for the sampling design and fitting procedure in use.
Practical Interpretation and Failure Modes
For a fitted pharmacokinetic SDE, a defensible workflow is to calculate each innovation from its one-step predictive distribution, plot innovations in observation order and against time, then inspect both index-lag and elapsed-time summaries where scientifically relevant. Mark dosing events, changes in sampling intensity, and other known design features. A run of same-sign innovations may signal a missed mean pattern even if a pooled correlation is modest; dependence that changes with elapsed time may be hidden by a single index-lag summary.
When a formal reference distribution is needed, simulate datasets under the fitted model using the actual dosing and observation schedule, refit the model if parameter estimation is part of the intended procedure, and recompute the chosen statistic for each simulated dataset. Compare the observed value with that simulated reference distribution. This approach preserves the irregular schedule and can reflect parameter fitting. It is only as credible as the fitted model and simulation procedure; it does not turn a misspecified model into a valid one.
Do not substitute raw concentration residuals \(Y_i-\mu_i\) for \(Z_i\). Their scale changes with predictive uncertainty, which can vary with elapsed time, dose history, and the state. A raw-residual pattern may therefore reflect changing variability rather than serial dependence. Conversely, standardized residuals can have zero linear correlation and still show misspecified tails, skewness, or other distributional features. Those require diagnostics beyond a correlogram.
Common alternatives answer related but distinct questions. An ordinary correlogram by observation index assesses dependence across successive recorded observations. An elapsed-time correlogram compares pairs by duration and is preferable when the duration itself may shape dependence. A model-based simulation reference is generally safer than applying a standard equally spaced, large-sample portmanteau calibration unchanged. None of these checks alone establishes overall model adequacy.
Check Your Understanding
Answer each question, then check the solutions below.
- State the conditional mean and variance properties of a correctly standardized one-step innovation.
- Why does correct standardization imply zero covariance between \(Z_i\) and \(Z_j\) for \(i<j\), under the stated conditions?
- With times \(0,1,5\), which pairs belong to the index-lag-one set, and what elapsed-time separations do they have?
- Why is \(E[Q]=K\) under the independent standard normal assumptions insufficient to claim that \(Q\) has a chi-square distribution?
- Give one reason a raw concentration-residual correlogram may be misleading.
Solutions. (1) \(E[Z_i\mid\mathcal F_{i-1}]=0\) and \(E[Z_i^2\mid\mathcal F_{i-1}]=1\). (2) \(Z_i\) is measurable given \(\mathcal F_{j-1}\), so \(E[Z_iZ_j]=E[Z_iE(Z_j\mid\mathcal F_{j-1})]=0\). (3) The pairs are \((1,2)\) and \((2,3)\), separated by 1 and 4 time units, respectively. (4) A mean does not determine a distribution; shared indices among pair products and model fitting can affect the distribution of \(Q\). (5) Predictive uncertainty can vary across observations, so raw residuals may change scale with time or dose history even when standardized innovations are the appropriate objects to assess.