Mean and Variance of Integrals with Deterministic Kernels

Why the Kernel Determines Variability

An Ito integral accumulates random Brownian increments after weighting them over time. When the weight is deterministic, the mean and variance can be calculated directly from the kernel. This is useful in pharmacokinetics because linear elimination weights earlier process noise differently from later noise: a perturbation introduced earlier has more time to be removed.

The key distinction is between accumulating the kernel itself and accumulating its squared size. The mean of the integral is zero, while its variance depends on the integral of the kernel squared. The result gives an exact moment calculation, not an approximation based on a particular sampling schedule.

For a deterministic step function, the integral is a finite sum. If \(k(s)=c_j\) on each interval \([t_j,t_{j+1})\), then \(I_T=\sum_j c_j(W_{t_{j+1}}-W_{t_j})\). The increments have mean zero, are independent, and have variance \(t_{j+1}-t_j\). These facts give the moment formulas; approximation extends them to continuous kernels.

Mean and Variance: Derivation

Proof. First take a step kernel with values \(c_j\) on intervals of lengths \(\Delta t_j=t_{j+1}-t_j\). Linearity of expectation gives \(E[I_T]=\sum_j c_j E[W_{t_{j+1}}-W_{t_j}]=0\). Independence of distinct Brownian increments makes their cross-covariances zero, so

For two step kernels on a common refinement of their partitions, the same increment calculation gives covariance \(\sum_j c_jd_j\Delta t_j=\int_0^T k(s)g(s)\,ds\), where \(d_j\) are the values of \(g\).

Now approximate continuous \(k\) and \(g\) uniformly by step functions \(k_m\) and \(g_m\). Uniform convergence on \([0,T]\) implies convergence in squared-integral distance. For step functions, the calculation above gives \(E[(\int_0^T(k_m-k_n)\,dW_s)^2]=\int_0^T(k_m-k_n)^2\,ds\). Thus the integrals form Cauchy sequences in mean square, and their limits are the Ito integrals for \(k\) and \(g\). Mean square convergence implies convergence of expectations by the Cauchy–Schwarz inequality. It also implies convergence of second moments: if \(X_m\to X\) in mean square, then \(|E[X_m^2]-E[X^2]|\leq (E[(X_m-X)^2])^{1/2}((E[X_m^2])^{1/2}+(E[X^2])^{1/2})\), which tends to zero. The integrals of \(k_m^2\) and \(k_mg_m\) converge to those of \(k^2\) and \(kg\), respectively, by uniform convergence. Passing to the limit proves the mean, variance, and covariance formulas. \(\square\)

Exponential Kernels in Linear Drug Elimination

Consider a concentration state satisfying the linear stochastic model \(dC_t=-k_e C_t\,dt+\sigma\,dW_t\), where \(k_e>0\) is a deterministic elimination rate, \(\sigma\) is a constant diffusion scale, and \(C_0\) is fixed. This model isolates continuous process noise around deterministic first-order elimination. Multiplication by the integrating factor \(e^{k_et}\), using the product rule with a deterministic differentiable function, gives

The first term is the deterministic concentration trajectory from the initial state. The second is the random contribution. A noise increment at time \(s\) is multiplied by \(e^{-k_e(t-s)}\) before it contributes to concentration at time \(t\). Applying the theorem to this deterministic kernel gives the moments.

To obtain the variance, the deterministic first term contributes no variance, and \(\operatorname{Var}(C_t)=\sigma^2\int_0^t e^{-2k_e(t-s)}\,ds\). Substitute \(u=t-s\); then the integral is \(\int_0^t e^{-2k_eu}\,du=(1-e^{-2k_et})/(2k_e)\). The mean follows because the Ito-integral term has mean zero.

Covariance Across Times

Concentrations at two times share some of the same Brownian increments. Consequently, they are generally correlated even though Brownian increments over disjoint intervals are independent. For \(0\leq s\leq t\), the random parts of \(C_s\) and \(C_t\) have kernels \(\sigma e^{-k_e(s-u)}\) and \(\sigma e^{-k_e(t-u)}\) over their shared interval \(0\leq u\leq s\). The covariance theorem therefore gives

The last equality follows by factoring out \(e^{-k_e(t-s)}\) and integrating \(e^{-2k_e(s-u)}\) from \(0\) to \(s\). This covariance is nonnegative for the stated parameters. It is not generally equal to the variance at the earlier time: later elimination attenuates the earlier random contribution.

Checks, Alternatives, and Limitations

A numerical left-endpoint sum can approximate an Ito integral, but it is not needed to obtain these moments. For a partition, the variance of the sum is the sum of squared kernel values times interval lengths; refining the partition approaches \(\int k^2\). Direct integration is usually simpler for a smooth deterministic kernel. An ODE without process noise gives a deterministic trajectory and zero process variance, while the stochastic model adds variance through the diffusion term. Neither calculation by itself specifies assay measurement error.

• Check the square. The variance uses \(k(s)^2\), not \(k(s)\). A negative kernel still produces a nonnegative variance.
• Check units. Since \(dW_s\) has units proportional to the square root of time, the kernel and diffusion scale must make the resulting integral have the intended concentration units.
• Check limiting cases. At \(t=0\), the process-noise variance is zero. As \(k_e\) approaches zero, \((1-e^{-2k_et})/(2k_e)\) approaches \(t\), consistent with a constant kernel.
• Check the covariance. The inequality \(|\operatorname{Cov}(C_s,C_t)|\leq\sqrt{\operatorname{Var}(C_s)\operatorname{Var}(C_t)}\) must hold. A violation indicates an algebra or numerical error.

These formulas rely on deterministic kernels and the stated Brownian-driven model. A random or state-dependent kernel cannot simply be treated as a deterministic weight in this calculation. Also, the variance result alone does not determine every feature of a more complicated pharmacokinetic model; it describes the contribution from this linear diffusion term under fixed parameters and initial concentration.