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What should patients do if they miss a dose of medication? A theoretical approach

Abstract

Medication adherence is a major problem for patients with chronic diseases that require long term pharmacotherapy. Many unanswered questions surround adherence, including how adherence rates translate into treatment efficacy and how missed doses of medication should be handled. To address these questions, we formulate and analyze a mathematical model of the drug concentration in a patient with imperfect adherence. We find exact formulas for drug concentration statistics, including the mean, the coefficient of variation, and the deviation from perfect adherence. We determine how adherence rates translate into drug concentrations, and how this depends on the drug half-life, the dosing interval, and how missed doses are handled. While clinical recommendations require extensive validation and should depend on drug and patient specifics, as a general principle our theory suggests that nonadherence is best mitigated by taking double doses following missed doses if the drug has a long half-life. This conclusion contradicts some existing recommendations that cite long drug half-lives as the reason to avoid a double dose after a missed dose. Furthermore, we show that a patient who takes double doses after missed doses can have at most only slightly more drug in their body than a perfectly adherent patient if the drug half-life is long. We also investigate other ways of handling missed doses, including taking an extra fractional dose following a missed dose. We discuss our results in the context of hypothyroid patients taking levothyroxine.

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Acknowledgements

SDL was supported by the National Science Foundation (Grant Nos. DMS-1944574 and DMS-1814832). SDL thanks Jennifer Babin and Colt Schisler for helpful discussions.

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Correspondence to Sean D. Lawley.

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A Appendix

A Appendix

In this appendix, we analyze the mathematical model formulated in the main text.

A.1 General theory

Let \(\{\xi _{n}\}_{n\in {\mathbb {Z}}}\) be a bi-infinite sequence of iid Bernoulli random variables as in (10) (it is convenient to allow the index n to vary over positive and negative integers). The dose taken at dosing time n may depend on the patient’s behavior at time n and the prior m dosing times for some given memory parameter \(m\ge 0\). Toward this end, let \(\{X_{n}\}_{n\in {\mathbb {Z}}}\) be the history process,

$$\begin{aligned} X_{n}=(\xi _{n-m},\xi _{n-m+1},\dots ,\xi _{n-1},\xi _{n})\in \{0,1\}^{m+1}, \end{aligned}$$
(34)

which records whether or not the patient remembered at dosing time n and the prior m dosing times. It is immediate that \(\{X_{n}\}_{n\in {\mathbb {Z}}}\) is an irreducible discrete-time Markov chain on the state space \(\{0,1\}^{m+1}\) [72]. In particular, let

$$\begin{aligned} P =\{P(x,y)\}_{x,y\in \{0,1\}^{m+1}} \in {\mathbb {R}}^{2^{m+1}\times 2^{m+1}} \end{aligned}$$
(35)

denote the transition probability matrix of the Markov chain \(\{X_{n}\}_{n\in {\mathbb {Z}}}\) with entries defined by

$$\begin{aligned} P(x,y) ={\mathbb {P}}(X_{1}=y\,|\,X_{0}=x),\quad x,y\in \{0,1\}^{m+1}, \end{aligned}$$

where \(x\in \{0,1\}^{m+1}\) denotes the vector,

$$\begin{aligned} x =(x_{-m},x_{m+1},\dots ,x_{-1},x_{0})\in \{0,1\}^{m+1}, \end{aligned}$$

and \(y\in \{0,1\}^{m+1}\) is denoted analogously. The definition of \(\{\xi _{n}\}_{n\in {\mathbb {Z}}}\) then implies that the entries of P are

$$\begin{aligned} P(x,y) ={\left\{ \begin{array}{ll} p &{} \text {if }y_{0}=1,\,(x_{-m+1},\dots ,x_{0})=(y_{-m},\dots ,y_{-1}),\\ 1-p &{} \text {if }y_{0}=0,\,(x_{-m+1},\dots ,x_{0})=(y_{-m},\dots ,y_{-1}),\\ 0 &{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

Furthermore, the definition of \(\{\xi _{n}\}_{n\in {\mathbb {Z}}}\) implies that the distribution of \(X_{n}\) is

$$\begin{aligned} \pi (x)&:={\mathbb {P}}(X_{n}=x) \nonumber \\&=p^{s(x)}(1-p)^{m+1-s(x)}>0,\quad n\in {\mathbb {Z}},\,x\in \{0,1\}^{m+1}, \end{aligned}$$
(36)

where \(s(x):=\sum _{k=0}^{m}x_{k}\in \{0,1,\dots ,m+1\}\) is the number of 1’s in x.

A dosing protocol \(f_{n}=f(X_{n})\) is any function

$$\begin{aligned} f:\{0,1\}^{m+1}\mapsto [0,\infty ). \end{aligned}$$
(37)

While we are most interested in the single dose and double dose protocols in (13) and (14), we also investigate a few other protocols. First, consider the “boost” dosing protocol,

$$\begin{aligned} \begin{aligned} f_{n}^{boost } :={\left\{ \begin{array}{ll} 0 &{} \text {if }\xi _{n}=0,\\ 1 &{} \text {if }\xi _{n}=1,\,\xi _{n-1}=1,\\ 1+b &{} \text {if }\xi _{n}=1,\,\xi _{n-1}=0, \end{array}\right. } \end{aligned} \end{aligned}$$
(38)

in which the patient takes a standard single dose of size D plus a “boost” dose of size bD if they missed the prior dose, for some \(b\ge 0\). Notice that the boost protocol reduces to the single dose protocol if \(b=0\) and the double dose protocol if \(b=1\). Another protocol is the “triple dose” protocol,

$$\begin{aligned} \begin{aligned} f_{n}^{triple } :={\left\{ \begin{array}{ll} 0 &{} \text {if }\xi _{n}=0,\\ 1 &{} \text {if }\xi _{n}=1,\,\xi _{n-1}=1,\\ 2 &{} \text {if }\xi _{n}=1,\,\xi _{n-1}=0,\,\xi _{n-2}=1,\\ 3 &{} \text {if }\xi _{n}=1,\,\xi _{n-1}=0,\,\xi _{n-2}=0, \end{array}\right. } \end{aligned} \end{aligned}$$
(39)

in which the patient takes a double dose to make up for a single missed dose and a triple dose to make up for two or more consecutive missed doses. Finally, consider the “all dose” protocol in which the patient takes all of their missed doses,

$$\begin{aligned} \begin{aligned} f_{n}^{all } :={\left\{ \begin{array}{ll} 0 &{} \text {if }\xi _{n}=0,\\ k+1 &{} \text {if }\xi _{n}=1,\,\xi _{n-1}=0,\dots ,\xi _{n-k}=0,\,\xi _{n-k-1}=1. \end{array}\right. } \end{aligned} \end{aligned}$$
(40)

The all dose protocol does not fit into the framework of (34), and thus an alternative analysis is developed in the section below.

For a dosing protocol f, a real number \(a\ge 0\), integers \(M\le N\), and time \(t\in [0,\tau )\), define the random variable

$$\begin{aligned} C_{M,N}(a,t) :=\alpha ^{t/\tau }\frac{DF}{V}\Big (\alpha ^{N-M+1}a+\sum _{n=M}^{N}\alpha ^{N-n}f(X_{n})\Big ), \end{aligned}$$
(41)

which is the drug concentration if time t has elapsed since dosing time \(n=N\), where \(a\ge 0\) describes the concentration at dosing time \(n=M-1\). We are interested in the drug concentration after a long time, which corresponds to taking \(N\rightarrow \infty\) in (41). We will see that this limiting distribution is independent of a and M.

Since \(\{X_{n}\}_{n\in {\mathbb {Z}}}\) is a stationary sequence, we have that

$$\begin{aligned} C_{M,N}(a,t) =_{\text {d}}C_{-(N-M),0}(a,t),\quad \text {for integers }N\ge M, \end{aligned}$$
(42)

where \(=_{\text {d}}\) denotes equality in distribution. Define

$$\begin{aligned} \begin{aligned} {C}(t) :=\lim _{N\rightarrow \infty }C_{-(N-M),0}(a,t) =\alpha ^{t/\tau }\frac{DF}{V}A,\quad \text {for }t\in [0,\tau ), \end{aligned} \end{aligned}$$
(43)

where

$$\begin{aligned} A :=\sum _{n=0}^{\infty }\alpha ^{n}f(X_{-n}). \end{aligned}$$
(44)

The function f must be bounded since the state space \(\{0,1\}^{m+1}\) is finite, and thus the Weierstrass M-test ensures that C(t) exists almost surely, and it is immediate that C(t) does not depend on \(M\in {\mathbb {Z}}\) or \(a\ge 0\). Random variables of the form in (43)-(44) are sometimes called random pullback attractors because they take an initial condition (in this case, \(a\ge 0\)) and pull it back to the infinite past [23,24,25,26,27].

Therefore, (42) and (43) imply that for any \(M\in {\mathbb {Z}}\) and \(a\ge 0\), the random variable \(C_{M,N}(a,t)\) converges in distribution to C(t) as \(N\rightarrow \infty\) [73], which we denote by

$$\begin{aligned} C_{M,N}(a,t)\rightarrow _{\text {d}}{C}(t),\quad \text {as }N\rightarrow \infty . \end{aligned}$$
(45)

Since f is bounded, \(C_{M,N}(a,t)\) can be bounded by a nonrandom constant independent of N, and thus (42), (43), and the Lebesgue dominated convergence theorem ensure the convergence of every moment of \(C_{M,N}(a,t)\),

$$\begin{aligned} {\mathbb {E}}[(C_{M,N}(a,t))^{j}] \rightarrow {\mathbb {E}}[({C(t)})^{j}],\quad \text {as }N\rightarrow \infty \text { for all }j>0. \end{aligned}$$
(46)

Summarizing, the large N distribution and statistics of \(C_{M,N}(a,t)\) are independent of \(a\ge 0\) and \(M\in {\mathbb {Z}}\), and we can study them by studying the distribution and statistics of C(t).

Furthermore, it is immediate from the definitions in (8) and (43) that

$$\begin{aligned} Z :=\frac{AUC }{AUC ^{perf }} =\frac{C(t)}{C^{perf }(t)} =\frac{A}{A^{perf }},\quad \text {for all }t\in [0,\tau ). \end{aligned}$$
(47)

Therefore, studying how drug concentrations are affected by imperfect adherence amounts to studying Z. Since \(A^{perf }=1/(1-\alpha )\), note that

$$\begin{aligned} {\mathbb {E}}[Z]=(1-\alpha ){\mathbb {E}}[A],\quad {\mathbb {E}}[Z^{2}]=(1-\alpha )^{2}{\mathbb {E}}[A^{2}]. \end{aligned}$$

For the single dose protocol in (13), the analysis of Z is straightforward since elements of the sequence \(\{f(X_{n})\}_{n\in {\mathbb {Z}}}\) are independent in this special case. The following theorem computes statistics of Z for a general dosing protocol.

Theorem 2

The first and second moments of Z are

$$\begin{aligned} {\mathbb {E}}[Z]&=\sum _{x}f(x)\pi (x), \end{aligned}$$
(48)
$$\begin{aligned} {\mathbb {E}}[Z^{2}]&=\frac{1-\alpha }{1+\alpha }\Big (\sum _{x}f(x)2{{u}}(x)-\sum _{x}(f(x))^{2}\pi (x)\Big ), \end{aligned}$$
(49)

where \(\sum _{x}\) denotes the sum over all \(x\in \{0,1\}^{m+1}\), \(\pi\) is in (36), and

$$\begin{aligned} {{u}} :=(I-\alpha P^{\top })^{-1}v, \end{aligned}$$
(50)

where \(I\in {\mathbb {R}}^{2^{m+1}\times 2^{m+1}}\) is the identity matrix, \(P^{\top }\) is the transpose of P in (35), and \(v\in {\mathbb {R}}^{2^{m+1}}\) is the vector with entries \(v(x)=f(x)\pi (x)\) for \(x\in \{0,1\}^{m+1}\).

Remark 1

The random variable C(t) in (43) generalizes an infinite Bernoulli convolution [13]. If we let \(C^{single }(t)\) denote C(t) in the case that f is the single dose protocol in (13), then an infinite Bernoulli convolution is merely a shift and rescaling of \(C^{single }(t)\),

$$\begin{aligned} \varTheta =\sum _{n=0}^{\infty }\alpha ^{n}(2\xi _{n}-1) =\frac{2C^{single }(t)-C^{perf }(t)}{\alpha ^{t/\tau }\frac{DF}{V}}. \end{aligned}$$

Dating back to Erdős and others in the 1930s [18,19,20] and continuing in more recent years [13,14,15,16,17], mathematicians have studied the distribution of \(\varTheta\). Though the definition of \(\varTheta\) is quite simple, its distribution is often quite irregular and depends very delicately on the parameters \(\alpha\) and p.

Proof of Theorem 2

Define

$$\begin{aligned} A_{1} :=\sum _{n=0}^{\infty }\alpha ^{n}f(X_{-n+1}). \end{aligned}$$

The definition of A in (44) and the stationarity of \(\{X_{n}\}_{n\in {\mathbb {Z}}}\) imply that

$$\begin{aligned} {A} =_{\text {d}}A_{1} =\alpha {A}+f(X_{1}), \end{aligned}$$
(51)

where \(=_{\text {d}}\) denotes equality in distribution. The invariance relation in (51) plays a key role in our analysis.

Taking the expectation of (51) and rearranging implies that

$$\begin{aligned} {\mathbb {E}}[{A}] =\frac{{\mathbb {E}}[f(X_{1})]}{1-\alpha }, \end{aligned}$$
(52)

where we have used that \({A}=_{\text {d}}A_{1}\). We note that (52) can also be obtained by taking the expectation of (44). Combining (36), (47), and (52) gives (48) in Theorem 2.

Squaring (51), taking expectation, and rearranging implies that

$$\begin{aligned} {\mathbb {E}}[A^{2}] =\frac{1}{1-\alpha ^{2}}\Big (2\alpha {\mathbb {E}}\big [{A}f(X_{1})\big ]+{\mathbb {E}}\big [(f(X_{1}))^{2}\big ]\Big ), \end{aligned}$$
(53)

where we have again used \({A}=_{\text {d}}A_{1}\). By definition of expectation, we have that

$$\begin{aligned} {\mathbb {E}}\big [(f(X_{1}))^{2}\big ] =\sum _{x}(f(x))^{2}\pi (x). \end{aligned}$$
(54)

Computing \({\mathbb {E}}[{A}f(X_{1})]\) is more challenging since \({A}\) and \(X_{1}\) are in general correlated if \(m\ge 1\).

Let \(1_{E}\in \{0,1\}\) denote the indicator function on an event E, meaning

$$\begin{aligned} 1_{E} :={\left\{ \begin{array}{ll} 1 &{} \text {if } E \text { occurs},\\ 0 &{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

Decomposing \({\mathbb {E}}[{A}f(X_{1})]\) based on the value of \(X_{1}\) gives

$$\begin{aligned} {\mathbb {E}}[{A}f(X_{1})] =\sum _{x}{\mathbb {E}}[{A}f(X_{1})1_{X_{1}=x}] =\sum _{x}f(x){\mathbb {E}}[{A}1_{X_{1}=x}]. \end{aligned}$$
(55)

Multiplying (51) by the indicator function on the event \(X_{1}=x\), taking expectation, and using that \(({A},X_{0})=_{\text {d}}(A_{1},X_{1})\) yields

$$\begin{aligned} {\mathbb {E}}[{A}1_{X_{0}=x}]&={\mathbb {E}}[A_{1}1_{X_{1}=x}] =\alpha {\mathbb {E}}[{A}1_{X_{1}=x}]\nonumber \\&\quad +f(x)\pi (x),\quad x\in \{0,1\}^{m+1}. \end{aligned}$$
(56)

Using the tower property of conditional expectation [74], it follows that

$$\begin{aligned} {\mathbb {E}}[{A}1_{X_{1}=x}] =\sum _{y}{\mathbb {E}}[{A}1_{X_{1}=x}1_{X_{0}=y}] =\sum _{y}{\mathbb {E}}[{A}1_{X_{0}=y}]P(y,x), \end{aligned}$$
(57)

where P is the transition matrix in (35). Combining (56) and (57) yields the following system of linear algebraic equations for \({\mathbb {E}}[{A}1_{X_{0}=x}]\),

$$\begin{aligned} {\mathbb {E}}[{A}1_{X_{0}=x}]&=\alpha \sum _{y}{\mathbb {E}}[{A}1_{X_{0}=y}]P(y,x)\nonumber \\&\quad +f(x)\pi (x),\quad x\in \{0,1\}^{m+1}. \end{aligned}$$
(58)

If we define the vectors \({{u}},v\in {\mathbb {R}}^{2^{m+1}}\) by

$$\begin{aligned} {{u}}(x)&:={\mathbb {E}}[{A}1_{X_{0}=x}],\quad v(x) :=f(x)\pi (x), \end{aligned}$$

then (50) solves (58). Note that the Perron-Frobenius theorem guarantees that \(I-\alpha P^{\top }\) in (50) is invertible since \(I-\alpha P^{\top }=\alpha (\alpha ^{-1}I-P^{\top })\) and \(\alpha \in (0,1)\). Putting this together by combining (47), (50) and (53)-(56) yields (49) in Theorem 2 and completes the proof. \(\square\)

A.2 An alternative history process

The history process in (34) assumes that the patient remembers whether or not they took their medication at the previous \(m\ge 0\) dosing times. Here, we assume instead that when the patient remembers to take the drug, they know how many consecutive doses they have missed. This modification will allow us to consider the “all dose” protocol in (40).

Define a new history process \(\{X_{n}\}_{n\in {\mathbb {Z}}}\) to encode how much time has passed since the patient last took their medication. Specifically, for integers \(n\in {\mathbb {Z}}\) and \(k\ge 1\), define

$$\begin{aligned} X_{n}&={\left\{ \begin{array}{ll} 0 &{} \text { if }\xi _{n}=\xi _{n-1}=1,\\ k &{} \text { if }\xi _{n}=1,\,\xi _{n-1}=\cdots =\xi _{n-k}=0,\,\xi _{n-(k+1)}=1,\\ -k &{} \text { if }\xi _{n}=\cdots =\xi _{n-(k-1)}=0,\,\xi _{n-k}=1. \end{array}\right. } \end{aligned}$$
(59)

In words, \(X_{n}=0\) if the patient takes the drug at time n and \(n-1\), \(X_{n}=k\ge 1\) if the patient takes the drug at time n after missing the last k doses, and \(X_{n}=-k\le -1\) if the patient misses their kth consecutive dose at time n.

Given that \(\{\xi _{n}\}_{n\in {\mathbb {Z}}}\) are iid as in (10), it follows that \(\{X_{n}\}_{n\in {\mathbb {Z}}}\) is a discrete-time Markov chain on \({\mathbb {Z}}\) that evolves according to the following transition matrix \(P=\{P(x,y)\}_{x,y\in {\mathbb {Z}}}\) with \(P(x,y):={\mathbb {P}}(X_{1}=y\,|\,X_{0}=x)\). For \(x\in {\mathbb {Z}}\) and \(x\ge 0\),

$$\begin{aligned} \begin{aligned} P(x,y)&={\left\{ \begin{array}{ll} p &{} \text {if }y=0,\\ 1-p &{} \text {if }y=-1,\\ 0 &{} \text {otherwise}, \end{array}\right. } \qquad P(-x,y) ={\left\{ \begin{array}{ll} p &{} \text {if }y=x,\\ 1-p &{} \text {if }y=-(x+1),\\ 0 &{} \text {otherwise}. \end{array}\right. } \end{aligned} \end{aligned}$$
(60)

It is straightforward to check that the distribution of \(X_{n}\) is

$$\begin{aligned} \pi (k) :={\mathbb {P}}(X_{n}=k) ={\left\{ \begin{array}{ll} p^{2}(1-p)^{k} &{} k\ge 0,\\ p(1-p)^{|k|} &{} k\le -1, \end{array}\right. } \qquad n,k\in {\mathbb {Z}}. \end{aligned}$$
(61)

For this alternative history process \(\{X_{n}\}_{n\in {\mathbb {Z}}}\) in (59), a dosing protocol is a function \(f:{\mathbb {Z}}\rightarrow [0,\infty )\). Since the state space of \(\{X_{n}\}_{n\in {\mathbb {Z}}}\) (namely \({\mathbb {Z}}\)) is infinite, we assume for technical reasons that dosing protocols cannot grow faster than linearly,

$$\begin{aligned} 0\le f(k)\le B_{0}|k|+B_{1},\quad k\in {\mathbb {Z}}, \end{aligned}$$
(62)

for some constants \(B_{0},B_{1}>0\).

Note that (62) and the value of \(\pi\) in (61) ensure that all the moments of \(X_{n}\) are finite. Furthermore, the bound in (62) ensures that the definition of \(A\) in (43) exists almost surely. To see this, note that (61) implies

$$\begin{aligned} \sum _{n\ge 0}{\mathbb {P}}(|X_{-n}|\ge n)&=\sum _{n\ge 0}\sum _{k\ge 0}{\mathbb {P}}(|X_{0}|= n+k) \\&\le 2\sum _{n\ge 0}\sum _{k\ge 0}p(1-p)^{n+k} =\frac{2}{p}<\infty . \end{aligned}$$

Therefore, the Borel-Cantelli lemma [74] implies that there is an almost surely finite random integer \(N_{0}\ge 1\) so that \(|X_{-n}|<n\) for all \(n\ge N_{0}\). Hence, (62) implies that \(f(X_{-n})\le B_{0}n+B_{1}\) for all \(n\ge N_{0}\), and the almost sure existence of \(A\) in (43) follows, as well as the convergence in distribution in (45). Furthermore, the moment convergence in (46) follows from the Lebesgue dominated convergence theorem upon noting that we have almost sure convergence of moments and using some simple bounds on \((C_{M,N}(a,t))^{j}\).

The definitions of \(C_{M,N}(a,t)\), \(A\), and \(A_{1}\) and the analysis in the section above carry over directly to this definition of \(\{X_{n}\}_{n\in {\mathbb {Z}}}\) if we use the definition of P and \(\pi\) in (60) and (61). In particular, the formula for the mean in (48) and the formula for the second moment in (49) hold. The benefit of the structure of P in (60) is that we can solve for u in (50) in closed form.

To simplify the formulas for u, we take \(f(-k)=0\) for all \(k\ge 1\), which means the patient cannot take medication when they forget. Equation (58) then implies

$$\begin{aligned} {{u}}(-k)&=\alpha (1-p){{u}}(-(k-1)) ,\quad k\ge 2,\\ {{u}}(k)&=\alpha p {{u}}(-k)+f(k)\pi (k) ,\quad k\ge 1,\\ {{u}}(0)&=\alpha p\sum _{k\ge 0}{{u}}(0)+f(0)\pi (0),\\ {{u}}(-1)&=\alpha (1-p)\sum _{k\ge 0}{{u}}(k). \end{aligned}$$

It is straightforward to solve these equations and obtain that

$$\begin{aligned} {{u}}(-1)&=\frac{\alpha (1-p) (1-\alpha (1-p)) \sum _{k\ge 0}f(k)\pi (k)}{1-\alpha },\\ {{u}}(-k)&=\alpha ^{k-1}(1-p)^{k-1}{{u}}(-1) ,\quad k\ge 1,\\ {{u}}(k)&=\alpha ^{k}p(1-p)^{k-1}{{u}}(-1)+f(k)\pi (k),\quad k\ge 0. \end{aligned}$$

Plugging into (49) yields

$$\begin{aligned} \begin{aligned} {\mathbb {E}}[Z^{2}]&=\frac{2\alpha p(1-\alpha (1-p))}{1-\alpha }\Big (\sum _{k\ge 0}f(k)\pi (k)\Big )\sum _{k\ge 0}\alpha ^{k}(1-p)^{k}f(k)+\sum _{k\ge 0}(f(k))^{2}\pi (k). \end{aligned} \end{aligned}$$
(63)

A.3 First and second moments of Z

We now work out the first and second moments of Z for a few different choices of the dosing protocol f. We begin by considering the history process in (34) with some given memory parameter \(m\ge 0\). The simplest case is \(m=0\), which corresponds to the patient having no recollection of their behavior at prior dosing times. It is natural to suppose that the patient takes no medication when they forget (\(f(0)=0\)) and that they take their normal dose when they remember (\(f(1)=1\)). Using (48) and (49), we obtain in this case,

$$\begin{aligned} \begin{aligned} {\mathbb {E}}[Z^{single }]&=p,\quad {\mathbb {E}}[(Z^{single })^{2}] =\frac{p (1+\alpha (2 p-1))}{1+\alpha }. \end{aligned} \end{aligned}$$
(64)

A more interesting case is \(m=1\), which allows the patient to potentially take a higher dose if they missed their prior dose. In this case, we need to specify f(ij) for \(i,j\in \{0,1\}\), where f(ij) is the dose taken at the nth dosing time if \(\xi _{n}=j\) and \(\xi _{n-1}=i\). Let \(f(0,0)=f(1,0)=0\) to impose that the patient must miss their dose when they forget. Further, suppose \(f(1,1)=1\), which means the patient takes their normal dose if they remember and they did not miss their prior dose. If \(f(0,1)=1+b>0\), then we obtain the “boost” dosing protocol in (38), and (48) and (49) yield

$$\begin{aligned} \begin{aligned} {\mathbb {E}}[Z^{boost }]&=p (1+b (1-p)),\\ {\mathbb {E}}[(Z^{boost })^{2}]&=\frac{p}{1+\alpha } \Big [b^2 (p-1) (\alpha +2 \alpha ^2 (p-1) p-1)\\&\quad +2 b (1-p) (\alpha (\alpha p+p-1)+1)+\alpha (2 p-1)+1\Big ]. \end{aligned} \end{aligned}$$
(65)

Note that the cases \(b=0\), \(b=1\), \(b=\alpha\), and \(b=0.5\) correspond respectively to the single, double, fractional, and 1.5 dosing protocols in (13), (14), (32), and (33).

Computing statistics of Z gets more complicated for larger values of m. If \(m=2\), then we need to specify f(ijk) for \(i,j,k\in \{0,1\}\), where f(ijk) is the dose taken at the nth dosing time if \(\xi _{n}=k\), \(\xi _{n-1}=j\), and \(\xi _{n-2}=i\). We set \(f(i,j,0)=0\) for \(i,j\in \{0,1\}\), and \(f(1,1,1)=1\) by the same reasoning as above. It follows then from (48) that the mean amount is

$$\begin{aligned} {\mathbb {E}}[Z] =p \big [f_{001} (1-p)^2+p (-p (f_{011}+f_{101})+f_{011}+f_{101}+p)\big ], \end{aligned}$$
(66)

where we have set \(f_{ijk}:=f(i,j,k)\) to simplify notation. We can similarly use (49) to obtain a complicated, but explicit formula for \({\mathbb {E}}[Z^{2}]\), which we omit for simplicity. These formulas allow us to investigate dosing protocols in which the patient takes even higher doses following two missed doses compared to a single missed dose. For example, setting

$$\begin{aligned} f(0,1,1)=1,\quad f(1,0,1)=2,\quad f(0,0,1)=3, \end{aligned}$$
(67)

yields the triple dose protocol in (39).

We now consider the alternative history process in (59) in order to consider the “all dose” dosing protocol in (40). In this case, using standard results for summing infinite series, (48) and (63) yield

$$\begin{aligned} \begin{aligned} {\mathbb {E}}[Z^{all }]&=1,\\ {\mathbb {E}}[(Z^{all })^{2}]&=\frac{\alpha ^2 (2-p) (1-p)+\alpha (p (p+4)-4)-p+2}{p (1+\alpha ) (1-\alpha (1-p))}. \end{aligned} \end{aligned}$$
(68)

A.4 Drug concentration statistics

We now use the calculations above to compute pharmacologically relevant statistics. Recall that \(\mu ={\mathbb {E}}[Z]\) in (18) compares the mean drug concentration to the perfectly adherent patient.

Corollary 1

Using superscripts to denote the dosing protocol, we have that

$$\begin{aligned} \mu ^{single }&=p,\quad \mu ^{double } =p+p(1-p),\\ \mu ^{boost }&=p+bp(1-p),\quad \mu ^{triple } =3p-3p^{2}+p^{3},\quad \mu ^{all } =1. \end{aligned}$$
Fig. 7
figure7

The plots compare the mean (a) and second moment (b) computed from stochastic simulations (square markers) to the exact analytical formulas as functions of the adherence p for various dosing protocols. We take \(\alpha :=e^{-k_{e }\tau }=0.85\)

In Fig. 7a, we plot \(\mu\) as a function of p for various dosing protocols. Notice that nonadherence causes a reduction in the average drug concentrations for the single, double, and triple dose protocols since \(\mu ^{single }\), \(\mu ^{double }\), and \(\mu ^{triple }\) are all strictly less than 1 for all \(p\in (0,1)\). Naturally, taking more doses following missed doses increases the average drug concentration, and thus

$$\begin{aligned} \mu ^{single }<\mu ^{double }<\mu ^{triple }<\mu ^{all },\quad \text {for all }p\in (0,1). \end{aligned}$$

However, notice that while \(\mu ^{double }\) is much larger than \(\mu ^{single }\), the additional increase is relatively small for the triple and all dose protocols, as \(\mu ^{double }\approx \mu ^{triple }\approx \mu ^{all }\) if the adherence p is not too small.

In Fig. 7b, we plot \(\sqrt{{\mathbb {E}}[Z^{2}]}\) as a function of p for various dosing protocols. The curves in Fig. 7 use the analytical formulas for \({\mathbb {E}}[Z]\) and \({\mathbb {E}}[Z^{2}]\) obtained from Theorem 2 and the squares markers are results from stochastic simulations with \(\alpha =0.85\). In particular, the square markers are obtained from \(10^{5}\) independent realizations of \(C_{M,N}(a,0)\) with \(a=M=0\) and \(N=100\). The simulation results agree with the exact analytical results.

To measure the variability in drug concentrations that stems from imperfect adherence, we introduce the coefficient of variation of Z,

$$\begin{aligned} c_{v }&:=\frac{\sqrt{{\mathbb {E}}[(Z-{\mathbb {E}}[Z])^{2}]}}{{\mathbb {E}}[Z]} =\frac{\sqrt{{\mathbb {E}}[(AUC -{\mathbb {E}}[AUC ])^{2}]}}{{\mathbb {E}}[AUC ]}\nonumber \\&=\frac{\sqrt{{\mathbb {E}}[(C(t)-{\mathbb {E}}[C(t)])^{2}]}}{{\mathbb {E}}[C(t)]}, \end{aligned}$$
(69)

which is defined as the ratio of the standard deviation to the mean. Notice that (17) implies that the coefficient of variation of Z is equal to the coefficient of variation of \(AUC\) or C(t) for any \(t\in [0,\tau )\). Applying Theorem 2 gives explicit formulas for the coefficient of variation for the single dose and double dose protocols, which we give in the following corollary (the formulas for the other dosing protocols are omitted for brevity).

Corollary 2

Using superscripts to denote the dosing protocol, we have that

$$\begin{aligned} c_{v }^{single }&=\sqrt{\frac{1-\alpha }{1+\alpha }}\sqrt{p(1-p)},\\ c_{v }^{double }&=c_{v }^{single }\sqrt{4-3 p+p^2-2 p (2-p) \alpha }. \end{aligned}$$

The coefficient of variation measures the variability induced by nonadherence by measuring how drug concentrations deviate from their average value. From a pharmacological standpoint, a small coefficient of variation is desirable. However, a small coefficient of variation does not necessarily imply that the effects of nonadherence are small. Indeed, the coefficient of variation vanishes if the patient never takes their medication (\(p=0\)).

Hence, a more useful statistic for measuring the effects of nonadherence is how drug concentrations deviate from the drug concentrations of a perfectly adherent patient, which is the deviation \(\varDelta\) defined in (19). Applying Theorem 2 gives explicit formulas for the deviation \(\varDelta\) for different dosing protocols, which we give in the following corollary.

Corollary 3

The deviation in (19) for the boost protocol in (38) for any \(b\ge 0\) is

$$\begin{aligned} \varDelta ^{boost }&=\sqrt{\frac{1-p}{1+\alpha }}\sqrt{\alpha +p \left( b^2+2 \alpha ^2 b p (1+b- bp)-\alpha (b (b-2 p+4)+2)\right) +1}. \end{aligned}$$

Note that setting \(b=0\), \(b=1\), \(b=\alpha\), and \(b=0.5\) yield the respective deviations for the single, double, fractional, and 1.5 dosing protocols in (13), (14), (32), and (33). The deviation for the triple dose protocol in (39) is

$$\begin{aligned} \varDelta ^{triple }&=\sqrt{\frac{1-p}{1+\alpha }}\Big [1-3 p^2+4 p+2 \alpha ^3 (1-p) p^2 ((p-3) p+3) \end{aligned}$$
(70)
$$\begin{aligned}&\quad +2 \alpha ^2 p^2 ((p-3) p+3)+\alpha \left( 1-2 p^3+11 p^2-14 p\right) \Big ]^{1/2}. \end{aligned}$$
(71)

The deviation for the all dose protocol in (40) is

$$\begin{aligned} \varDelta ^{all }&=\sqrt{\frac{1-p}{1+\alpha }}\sqrt{\frac{2(1-\alpha )^{2}}{1-p (\alpha (1-p))}}. \end{aligned}$$

For certain drugs, it is important to ensure that the dosing protocol cannot cause the drug concentration in the patient to rise too high. We thus consider \(\lambda\) in (20), which is the largest possible drug concentration compared to the perfectly adherent patient. The following theorem calculates \(\lambda\) for the dosing protocols above.

Theorem 3

Using superscripts to denote the dosing protocol, we have that

$$\begin{aligned} \lambda ^{boost }&=\max \Big \{1,\frac{1+b}{1+\alpha }\Big \},\quad \lambda ^{triple } =\frac{3}{1+\alpha +\alpha ^{2}},\quad \lambda ^{all } =\infty . \end{aligned}$$

Note that setting \(b=0\), \(b=1\), \(b=\alpha\), and \(b=0.5\) corresponds respectively to the single, double, fractional, and 1.5 dosing protocols in (13), (14), (32), and (33).

Notice that if we set \(b=\alpha\) in the boost protocol in (38), then \(\lambda ^{boost }=1\) and thus Theorem 3 ensures that a patient following the boost protocol with \(b=\alpha\) will never have more drug in their body than the perfectly adherent patient.

We note that Theorem 1 in the main text follows immediately from Corollaries 1 and 3 and Theorem 3.

Proof of Theorem 3

For the single dose protocol, it is immediate that \(\lambda ^{single }=1\), and this corresponds to a patient who never misses a dose.

For the double dose protocol, observe that if \(\xi _{2n}=1\) and \(\xi _{2n+1}=0\) for all \(n\in {\mathbb {Z}}\), then \(A=\frac{2}{1-\alpha ^{2}}\), and thus

$$\begin{aligned} \lambda ^{double } \ge \frac{2/(1-\alpha ^{2})}{{A^{perf }}} =\frac{2}{1+\alpha }. \end{aligned}$$
(72)

This describes a patient who misses a dose at every odd dosing time, and thus always takes a double dose at even dosing times.

To see that \(\lambda ^{double }\le \frac{2}{1+\alpha }\), suppose that the patient has concentration \(\frac{2}{1-\alpha ^{2}}\frac{DF}{V}\) just after dosing time \(n=1\). If they take the drug at dosing time \(n=2\), then the concentration in their body will be lower than \(\frac{2}{1-\alpha ^{2}}\frac{DF}{V}\) since

$$\begin{aligned} \alpha \frac{2}{1-\alpha ^{2}}+1<\frac{2}{1-\alpha ^{2}}. \end{aligned}$$

Hence, suppose they miss taking the drug at dosing time \(n=2\). If they take the drug at dosing time \(n=3\), then they will take a double dose and the concentration in their body will return to \(\frac{2}{1-\alpha ^{2}}\frac{DF}{V}\). If they miss taking the drug at dosing time \(n=3\), then the concentration will be even lower. Therefore, \(\lambda ^{double }\le \frac{2}{1+\alpha }\), which upon combining with (72) yields \(\lambda ^{double }=\frac{2}{1+\alpha }\).

The proof that \(\lambda ^{boost }=\max \{1,\frac{1+b}{1+\alpha }\}\) is almost identical to the proof that \(\lambda ^{double }=\frac{2}{1+\alpha }\). The proof that \(\lambda ^{triple }=\frac{3}{1+\alpha +\alpha ^{2}}\) is also almost identical, upon noting that this value of \(\lambda ^{triple }\) is attained by a patient who takes medication at every third dosing time.

The proof for the “all dose” protocol follows from noting that if the patient misses k consecutive doses and then takes the next dose, then the drug concentration in their body just after that dose is at least \((k+1)\frac{DF}{V}\). Since this is true for every positive integer k, the result \(\lambda ^{all }=\infty\) follows. \(\square\)

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Counterman, E.D., Lawley, S.D. What should patients do if they miss a dose of medication? A theoretical approach. J Pharmacokinet Pharmacodyn (2021). https://doi.org/10.1007/s10928-021-09777-6

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Keywords

  • Medication adherence
  • Missed doses
  • Levothyroxine
  • Stochastics