The value of mHealth for managing chronic conditions

Abstract

Chronic conditions place a high cost burden on the healthcare system and deplete the quality of life for millions of Americans. Digital innovations such as mobile health (mHealth) technology can be used to provide efficient and effective healthcare. In this research we explore the use of mobile technology to manage chronic conditions such as diabetes and hypertension. There is ample empirical evidence in the healthcare literature showing that patients who use mHealth observe improvement in their health. However, an analytical study that quantifies the benefit of using mHealth is lacking. The benefit of using mHealth depends on many factors such as the current health condition of the patient, pattern of disease progression, frequency of measurement and intervention, the effectiveness of intervention, and the cost of measuring. Stochastic modeling is a suitable approach to take these factors into consideration to evaluate the benefit of mHealth. In this paper, we model the disease progression with the help of a Markov chain and quantify the benefits of measuring and intervention taking into consideration the above-mentioned factors. We compare two different modes for measuring and intervention, mHealth mode and conventional office visit mode, and evaluate the impact of these factors on health outcome.

Appendices

Appendix A: Markov model for gaps in intervention

In Section 3 we assume that there is an intervention every time a patient measures. This assumption can be easily relaxed by expanding our state space. Assume that there is an intervention only when the patient is in state 2. We consider one more dimension to our current state space. The second dimension denotes whether there is an intervention. The state at period n is given by {X_n, Y_n}, where X_n ∈ {1, 2, 3}, as before, Y_n = 0 if there is no intervention, and Y_n = 1 if there is an intervention. Note that there is no intervention when in states 1 and 3. Therefore, Y_n = 0. That is, P(X_n+ 1, 1|X_n) = 0 for X_n+ 1 = {1, 3} and X_n = {1, 2, 3}. Hence, we have 4 possible states: (1, 0), (2, 0), (2, 1), and (3,0).

Then the patient’s immediate transition over the next base period follows the transition probability matrix P_m, given by

$$\begin{array}{@{}rcl@{}} P_{m} = \left[\begin{array}{cccccccccc} 1- a & 0 & a & 0\\ b & 1- b - a & 0 & a \\ \beta_{m} b & \frac{1 - \beta_{m} b - e_{m} a}{2} & \frac{1 - \beta_{m} b - e_{m} a}{2} & e_{m} a \\ 0 & 0 & 0 & 1 \end{array}\right]. \end{array} $$

(14)

In Eq. 14, note that we accommodate the possibility that the intervention (for a patient who just got sick and transitioned to state 2) happens over multiple base periods, especially if the intervention is not effective. If the patient stays in (2,1), they continue to receive intervention and therefore increase their chance of getting back to state 1 (from which they originally transitions from). Of course, in reality, not everyone at state 2 might be continuously eligible for such interventions and hence we also let them transition to state (2,0) which means no more interventions as long as they are in state 2. We have arbitrarily chosen the 50-50 split between (2, 0) and (2, 1). This can of course be varied.

In case measurements are taken only every k periods, since we already consider a second dimension in the state space, let the second dimension denote either there is an intervention or there is going to be one during the next measurement period. Hence, we can still do with just 4 possible states: (1, 0), (2, 0), (2, 1), and (3, 0). State (2, 1) denotes that the patient received an intervention if he measures (in the case of P_m matrix) and will receive an intervention when he measures next at the k th period (in the case of P_w matrix). P_r in this case will be simply P_m for one period and the following modified P_w for the remaining k − 1 periods,

$$\begin{array}{@{}rcl@{}} P_{w} = \left[\begin{array}{cccccccccc} 1- a & 0 & a & 0\\ b & 1- b - a & 0 & a \\ b & \gamma (1- b - a) & (1-\gamma) (1- b - a) & a \\ 0 & 0 & 0 & 1 \end{array}\right]. \end{array} $$

(15)

In Eq. 15, states (2, 0) and (2, 1) are treated the same when it pertains to transitions to states 1 and 3. The same state transition for state (2, 0) is also unchanged, implying that these patients do not receive any interventions as long as they are in state 2. For patients moving to state (2,1) they will get intervention if they measure in the next period. But, with probability γ(1 − b − a) they go to state (2, 0). If we make γ to be 0, then patients in regular mode will all get an intervention in the next period if they are in state 2.

Appendix B: Distribution of first passage time

In this section we present the distribution of the first passage time for a healthy (State 1) patient to the absorbing state. We first restate a result pertaining to the distribution of the absorption time for general birth and death processes, from Theorem 1.2 of [57]:

Theorem 4

Consider a discrete birth and death chain with transition kernelP^∗on the state space {0, … , d} started at 0, whered is an absorbing state, and suppose that the other birthprobabilities$p^{*}_{i}, 0\leq i \leq d-1$, and death probabilities$q_{i}^{*}, 1\leq i \leq d-1$arepositive. If$p^{*}_{i-1}+q^{*}_{i} < 1$for1 ≤ i ≤ d, then the absorption time starting from state 0 has the probability generating function

$$\begin{array}{@{}rcl@{}} f(u) = \prod\limits_{j = 0}^{d-1}\frac{(1-\theta_{j})u}{1-\theta_{j} u}, \end{array} $$

where − 1 ≤ θ_j < 1 are the d nonuniteigenvalues ofP^∗.

Note that Theorem 4 implies that the absorption time starting from state 0 can be expressed as the sum of d independent geometric random variables each with parameter 1 − θ_j, since this sum has a probability generating function of the form of f.

We begin by finding the distribution of the absorption time for a Markov chain following transition matrix P_w, by calculating the eigenvalues of P_w and applying Theorem 4. If a + b < 1, then the random time to absorption starting from state 1, denoted as $\tilde {T}(1)$, has a distribution of the form

$$\begin{array}{@{}rcl@{}} P(\tilde{T}(1)=n) &=& \frac{(1-p_{1})(1-p_{2})(p_{2}^{n-1}-p_{1}^{n-1})}{p_{2}-p_{1}},\\ &&{}\text{ for} n\geq 2 \text{and if} p_{1} \neq p_{2}, \end{array} $$

(16)

$$\begin{array}{@{}rcl@{}} &&{}=\frac{(n-1)(1-p_{1})(1-p_{2})p_{2}^{n-1}}{p_{1}},\\ &&{}\text{ for} n\geq 2 \text{if} p_{1} = p_{2} \end{array} $$

(17)

where

$$\begin{array}{@{}rcl@{}} p_{1} = \frac{1}{2}(2a+b+\sqrt{b(4a+b)}), \end{array} $$

$$\begin{array}{@{}rcl@{}} p_{2} = \frac{1}{2}(2a+b-\sqrt{b(4a+b)}). \end{array} $$

With this result, we can easily derive the distribution of the absorption time starting from state 1, for patients using mHealth mode or office visits, since these processes are Markov chains with transition probabilities of a similar form as P_w. The distribution of the absorption time for a patient using mHealth mode following transition matrix P_m is given by replacing a by e_ma and b with β_mb in Eqs. 16 and 17. The distribution of the absorption time for a patient using office visits following transition matrix $\tilde {P}$ defined in Eq. 3 is given by replacing a by a(e_r(k − 1)/k + e_m/k) and b with b(β_r(k − 1)/k + β_m/k) in Eqs. 16 and 17.

Appendix C: Proofs of results in Section 3

Proof of Lemma 1

We construct the coupling. We generate a sequence of uniform random variables, denoted by {U_n}. Given X_n and U_n, we define X_n+ 1 by

$$\begin{array}{@{}rcl@{}} X_{n + 1}&=&2 \text{ if} X_{n}= 1 \text{and} 0\leq U_{n} \leq A,\\ &=&1 \text{ if} X_{n}= 1 \text{and} A\leq U_{n} \leq 1,\\ X_{n + 1}&=&3 \text{if} X_{n}= 2 \text{and} 0\leq U_{n} \leq A,\\ &=&2 \text{ if} X_{n}= 2 \text{and} A\leq U_{n} \leq 1-B,\\ &=&1 \text{ if} X_{n}= 2 \text{and} 1-B\leq U_{n} \leq 1. \end{array} $$

Given Y_n, we generate Y_n+ 1 in a similar fashion:

$$\begin{array}{@{}rcl@{}} Y_{n + 1}&=&2 \text{ if} Y_{n}= 1 \text{and} 0\leq U_{n} \leq A^{\prime},\\ &=&1 \text{ if} Y_{n}= 1 \text{and} A^{\prime}\leq U_{n} \leq 1,\\ Y_{n + 1}&=&3 \text{ if} Y_{n}= 2 \text{and} 0\leq U_{n} \leq A^{\prime},\\ &=&2 \text{ if} Y_{n}= 2 \text{and} A^{\prime}\leq U_{n} \leq 1-B^{\prime},\\ &=&1 \text{ if} Y_{n}= 2 \text{and} 1-B^{\prime}\leq U_{n} \leq 1. \end{array} $$

We proceed by induction. Assume that X_n ≥ Y_n, then we must show that X_n+ 1 ≥ Y_n+ 1. We divide our analysis into subcases depending on the values of X_n, Y_n, and U_n, and show that in each subcase, X_n+ 1 ≥ Y_n+ 1. Note that 0 ≤ A′ ≤ A ≤ 1 − B′ ≤ 1 − B ≤ 1. The inequality A ≤ 1 − B′ is true because 1 − B′ − A ≥ 1 − B′ − A′ ≥ 0, where the first inequality follows from the fact that A ≥ A′, and the second inequality follows from the fact that 1 − B′ − A′ is a probability in the P′ transition matrix. Therefore, there are five possibilities for the value of U_n:

• Case I: 0 ≤ U ≤ A′.

• Case II: A′ ≤ U ≤ A.

• Case III: A ≤ U ≤ 1 − B′.

• Case IV: 1 − B′ ≤ U ≤ 1 − B.

• Case V: 1 − B ≤ U ≤ 1.

If X_n = 1 and Y_n = 1, then X_n+ 1 = 2 and Y_n+ 1 = 2 in Case I, X_n+ 1 = 2 and Y_n+ 1 = 1 in Case II, X_n+ 1 = 1 and Y_n+ 1 = 1 in Case III, X_n+ 1 = 1 and Y_n+ 1 = 1 in Case IV, and X_n+ 1 = 1 and Y_n+ 1 = 1 in Case V.

If X_n = 2 and Y_n = 1, then X_n+ 1 = 3 and Y_n+ 1 = 2 in Case I, X_n+ 1 = 3 and Y_n+ 1 = 1 in Case II, X_n+ 1 = 2 and Y_n+ 1 = 1 in Case III, X_n+ 1 = 2 and Y_n+ 1 = 1 in Case IV, and X_n+ 1 = 1 and Y_n+ 1 = 1 in Case V.

If X_n = 2 and Y_n = 2, then X_n+ 1 = 3 and Y_n+ 1 = 3 in Case I, X_n+ 1 = 3 and Y_n+ 1 = 2 in Case II, X_n+ 1 = 2 and Y_n+ 1 = 2 in Case III, X_n+ 1 = 2 and Y_n+ 1 = 1 in Case IV, and X_n+ 1 = 1 and Y_n+ 1 = 1 in Case V.

In all cases, it is true that X_n+ 1 ≥ Y_n+ 1. □

Proof of Theorem 1

We first prove that T_m(s) is non-increasing in s. Let {X_n} be the sequence of states visited by the patient using mHealth mode, with X₀ = 2, and similarly let {Y_n} be the sequence of states visited by the patient using mHealth mode, with Y₀ = 1. The conditions of Lemma 1 are met by setting P = P′ = P_m and so there is a coupling of {X_n} and {Y_n} such that X_n ≥ Y_n for all n.

From this observation, we are able to conclude that:

• The X process must reach the absorbing state sooner than the Y process. Thus time to absorption is higher for any given path {U_n} in the Y process compared to the X process, and so the average time to absorption is higher as well. For the sake of completeness, we prove this statement formally below, however we omit it for the other claims since the arguments are almost identical. Let X be defined on probability space Ω with measure P_X and Y be defined on probability space Ω with measure P_Y. Then Lemma 1 gives a coupling of X and Y such that for all n, X_n(ω) ≥ Y_n(ω) and P_X(ω) = P_Y(ω). Define $\tilde {T}_{X}(\omega )$ to be the random time to absorption in X and $\tilde {T}_{Y}(\omega )$ to be the random time to absorption in Y . Then we need to show that

  $$\begin{array}{@{}rcl@{}} {\int}_{\Omega}\tilde{T}_{X}(\omega) d P_{X}(\omega) \geq {\int}_{\Omega}\tilde{T}_{Y}(\omega) d P_{Y}(\omega). \end{array} $$

  This inequality holds since for all ω ∈Ω, $\tilde {T}_{X}(\omega ) \geq \tilde {T}_{Y}(\omega )$ and P_X(ω) = P_Y(ω) by our coupling construction.

• Since R(s) is decreasing in s, for every period n, the Y process yields higher utility than the X process. Furthermore, the average time until absorption is higher for the Y process than the X process. Thus, the total utility over the patient’s lifetime is higher for the Y process compared to the X process as well.

To prove the remaining claims in the theorem, we repeat the same argument. For instance, to show that (T_m(s), v_m(s)) are decreasing in e_m, consider two values of e_m, denoted as $e_{m}^{(1)}$ and $e_{m}^{(2)}$, such that $e_{m}^{(1)}> e_{m}^{(2)}$. Create two corresponding processes {X_n} and {Y_n} where X₀ = Y₀ = s, and where the transition probabilities for the process X follow the P_m matrix with $e_{m} = e_{m}^{(1)}$ (denote this matrix as $P_{m}^{(1)}$), and the transition probabilities for the process Y follow the P_m matrix with $e_{m} = e_{m}^{(2)}$ (denote this matrix as $P_{m}^{(2)}$). Then we can couple {X_n} and {Y_n} according to Lemma 1 by setting $P=P_{m}^{(1)}$ and $P^{\prime }=P_{m}^{(2)}$, ensuring that for all n, X_n ≥ Y_n. The argument proceeds in exactly the same way as above to conclude that since the state is always higher in the X process than in the corresponding Y process, the average time to absorption and the average lifetime utility is higher for the Y process compared to the X process.

We omit the proofs for the remaining claims since they are identical to the proofs we have outlined. □

Proof of Theorem 2

We proceed with a similar proof as that of Theorem 1. Let X_n represent the nth period health state of a patient using regular mode, let Y_n represent the nth period health state of a patient using mHealth mode. Set X₀ = Y₀ = s. If n is a multiple of k, then the transition matrix used in the next step of the X process is P_r, and we can apply Lemma 1 by setting P = P_r and P′ = P_m. In the other case, if n is not a multiple of k, then the transition matrix used in the next step of the X process is P_w and we can apply Lemma 1 by setting P = P_w and P′ = P_m. In either case, there exists a coupling such that X_n ≥ Y_n for all n.

Thus, we conclude that T_m(s) ≥ T_r(s) since the time to absorption is always greater in process Y than process X, and must be greater in expectation as well. Denote $\tilde {T}_{X}$ to be the random absorption time in the X process, $\tilde {T}_{Y}$ to be the random absorption time in the Y process, $\tilde {v}(\{X_{n}\},s)$ to be the lifetime utility of a patient traversing the X process, starting in state s, and $\tilde {v}(\{Y_{n}\},s)$ to be the lifetime utility of a patient traversing the Y process, starting in state s. We append “∼” to the utility function v to emphasize that these quantities are random and dependent on the path of X and Y, and are different from the average utility functions (v_m(s), v_r(s)). The expressions for $\tilde {v}(\{X_{n}\},s)$ and $\tilde {v}(\{Y_{n}\},s)$ are

$$\begin{array}{@{}rcl@{}} \tilde{v}(\{X_{n}\},s) = \sum\limits_{n = 0}^{\tilde{T}_{X}}{R(X_{n})} - \lfloor T_{X} / k \rfloor C_{r}, \end{array} $$

$$\begin{array}{@{}rcl@{}} \tilde{v}(\{Y_{n}\},s) = \sum\limits_{n = 0}^{\tilde{T}_{Y}}{R(Y_{n})} - T_{Y} C_{m}. \end{array} $$

Note that ${\sum }_{i=n}^{\tilde {T}_{X}}{R(X_{n})}\leq {\sum }_{n = 1}^{\tilde {T}_{Y}}{R(Y_{n})}$ since $\tilde {T}_{X}\leq \tilde {T}_{Y}$, X_i ≥ Y_i for all i, and R(s) is decreasing in s. From this observation and our assumption that kC_m ≤ C_r, we conclude that $\tilde {v}(\{X_{n}\},s)\leq \tilde {v}(\{Y_{n}\},s)$. For each possible path, the lifetime utility is higher in the Y process than the X process, and must be higher on average as well. □

Proof of Theorem 3

From Eq. 13,

$$\begin{array}{@{}rcl@{}} \dfrac {\mathrm{d}g}{\mathrm{d}a} = - \frac{b (1 - 2 a + a^{2} e) (k + e -1)}{(1 - a)^{2} a^{2} (1 - e) e (k -1)}. \end{array} $$

The equation dg/da = 0 has two solutions:

$$\begin{array}{@{}rcl@{}} \frac{k+e-1 - \sqrt{(1 - e) (k+e-1)^{2}}}{e (k+e-1)} \end{array} $$

and

$$\begin{array}{@{}rcl@{}} \frac{k+e-1 + \sqrt{(1 - e) (k+e-1)^{2}}}{e (k+e-1)}. \end{array} $$

Clearly, $\frac {k+e-1 + \sqrt {(1 - e) (k+e-1)^{2}}}{e (k+e-1)} \ge 1$ since e ≤ 1. Since $0 \le \frac {k+e-1 - \sqrt {(1 - e) (k+e-1)^{2}}}{e (k+e-1)} \le 1$, this is the a^∗ we want. □

Appendix D: Justifying $\tilde P$ approximation

In this section we run numerical experiments to show that the office visit process for a patient is well approximated by a Markov chain with transition matrix $\tilde {P}$ defined in Eq. 3. To do so, for a given set of parameters (a, b, e_r, k) we calculate the average absorption time using T_r(1) defined in Eq. 5, and compare it to the average absorption time calculated using simulation. For the latter, in every run we simulate a Markov process starting in state 1 where the transition probability follows P_r in each kth period, and follows P_w otherwise, and track the number of transitions needed until the Markov chain reaches the absorbing state 3. This process is repeated 500,000 times and we report the average number of transitions until absorption. We remark that a similar analysis can be performed to check the accuracy of N_r in Eq. 4, but the results are similar and omitted for brevity.

We run five different experiments by choosing different parameters in each setting:

• Experiment I: a = 0.2, b = 0.1, e_r = 0.75.

• Experiment II: a = 0.2, b = 0.1, e_r = 0.5.

• Experiment III: a = 0.2, b = 0.1, e_r = 0.25.

• Experiment IV: a = 0.8, b = 0.1, e_r = 0.5.

• Experiment V: a = 0.1, b = 0.8, e_r = 0.5.

For all the five experiments, we vary k from 1 to 10 and report the average absorption time using both simulation and based on Eq. 5. The results of these experiments are shown in Figs. 7, 8, 9, 10 and 11.

Fig. 7

Experiment I, a = 0.2, b = 0.1, e_r = 0.75

Fig. 8

Experiment II, a = 0.2, b = 0.1, e_r = 0.5

Fig. 9

Experiment III, a = 0.2, b = 0.1, e_r = 0.25

Fig. 10

Experiment IV, a = 0.8, b = 0.1, e_r = 0.5

Fig. 11

Experiment V, a = 0.1, b = 0.8, e_r = 0.5

We are mainly interested in Experiments I, II, III. For these experiments we set a and b, respectively the probability of a patient’s health condition worsening or improving in the absence of physician intervention, to a realistic level. We vary e_r and k since different values for these two parameters will affect the frequency with which we make transitions according to P_r instead of P_w, and the magnitude of the discrepancy in the transition probabilities between P_r and P_w. Overall, our approximation works well in these experiments. By assuming that the office visit process follows a Markov chain with transition matrix $\tilde {P}$, we do not lose much in overall accuracy. At maximum, the error between the simulated absorption time and the approximated absorption time is 3.4%. For all the settings that we tested in experiments I, II, and III, the average error was 1.6%.

Experiments IV and V represent extreme cases where in the absence of treatment, either (i) it is likely that the patient condition worsens quickly, or (ii) it is likely that the patient’s condition improves quickly. Although the parameters may be unrealistic, we nevertheless report them to test the limits of our approximation. For Experiment IV, the maximum error between the simulated absorption time and the approximated absorption time is 13.5%, while the average error amongst all the settings tested in experiment IV was 8.5%. One explanation for the higher error is that in this setting when k is large, it is likely that in the simulation the patient reaches the absorbing state before even the first intervention at period k is reached. However in the approximation, it is assumed that the patient can get an intervention in every period with probability 1/k, and so the approximation is likely to overestimate the absorption time. In any case, the error is unlikely to have any practical impact since the magnitudes of the absorption time are small to begin with. For instance, the maximum error between simulated absorption time and approximated absorption time occurs when the simulated absorption time is 2.97 periods and the approximated absorption time is 3.375 periods.

For Experiment V, the maximum error between the simulated absorption time and the approximated absorption time is 8%, while the average error amongst all the settings tested in experiment V was 1.5%. Overall the approximation works well, except for one instance when k = 2. When k is small, it is possible that having an intervention once every k periods is more beneficial than having a 1/k chance of having an intervention every period. In the latter case, the patient has a risk of not receiving intervention within k periods. However, we remark this setting is largely irrelevant to the analysis of mHealth mode; there is little point of continuous monitoring of the patient when a is small and b is large since the patient is usually healthy even without intervention.