Estimating Individualized Treatment Regimes to Optimize Incremental Cost-Effectiveness Ratio

Abstract

Medical decision making can be challenging due to the trade-off between improving clinical efficacy and the associated medical costs. Evaluation of the incremental cost-effectiveness ratio (ICER) of different treatment programs is important for cost-effectiveness analysis. Individualized treatment regimes (ITRs) that consider patient heterogeneity can lead to varying health benefits and costs. To identify a promising ITR that balances efficacy and cost, the ICER criterion can be used to evaluate the quality of the ITR. We propose a method that considers both health benefits and costs to derive the the optimal ITR. We utilize Dinkelbach’s algorithm to transform a fractional program into a parametric program, which is easier to handle. We compare our method to ITRs that only optimize a single outcome (benefits or costs) through extensive simulation studies and show that our approach performs satisfactorily. To demonstrate the practical application of our method, we apply it to the Multicenter Automatic Defibrillator Implantation Trial with Cardiac Resynchronization Therapy (MADIT-CRT) study, a randomized trial. Our approach can help identify an optimal ITR that balances the trade-off between clinical efficacy and medical costs. Overall, our method provides a valuable tool for medical decision making that takes into account patient heterogeneity and cost-effectiveness analysis.

Appendices

Appendix A Proof

1.1 Proof of Proposition 2.1

Proof

We first note that (3) is equivalent to $\underset{d \in D }{\max }\ \quad \{E^d(T) - T_0\}/\{E^d(max(R, R_0)) - R_0\}$, given the maximizer of (3) must be in the set D. Given any fixed $d \in D$, we have

$$\begin{aligned} \underset{\theta \in \Theta }{\max }\ E^d(\max (R, \theta ))&=E^d(max(R, R_0)) \\ \underset{\theta \in \Theta }{\text {max}} \{E^d(\max (R, \theta )) - R_0\}/\{E^d(T) - T_0\}&= \{E^d(\max (R, R_0)) - R_0\}/\{E^d(T) - T_0\}\\ E^d(T) - T_0&>0 \\ E^d(\max (R, R_0)) - R_0&>0.\\ \end{aligned}$$

Thus, $\underset{d\in D}{\max }\ \{E^d(T) - T_0\}/\{E^d(max(R, R_0)) - R_0\}$ is equivalent to minimizing its inverse, which is $\underset{d\in D}{\min }\{E^d(max(R, R_0)) - R_0\} /\{E^d(T) - T_0\}$. The proof of the equivalence between (4) and (??) is similar, after observing $\underset{\theta \in \Theta }{\min } E^d(\max (R, \theta ))=E^d(max(R, -\infty ))=E^d(R), \forall d\in D$. $\square$

1.2 Proof of Theorem 2.2

The proof of Theorem 2.2 directly follows from that of [19]. We define $N(d)= E^{d}(T)-T_0$, and $D(d)= E^{d}(\textbf{R})-R_0$. Our proposed objective can be formulated as

$$\begin{aligned} \underset{d \in {S}}{\max } \quad \frac{N(d)}{D(d)}. \end{aligned}$$

(A1)

We consider another related problem

$$\begin{aligned} \underset{d \in {S}}{\max } \quad \{N(d)- qD(d)\} , \end{aligned}$$

(A2)

where $q\in \mathbb {R}$. We would want to prove $q^*=\frac{N(d^*)}{D(d^*)}=\underset{d \in {S}}{\max } \quad \frac{N(d)}{D(d)}$ iff $F(q^*)=\underset{d \in {S}}{\max } \quad \{N(d)- q^*D(d)\}=0$.

Proof

We first establish lemmas regarding $F(q)=\underset{d \in {S}}{\max } \quad \{N(d)- qD(d)\}$.

• 1. F(q) is convex over $\mathbb {R}$.

• 2. F(q) is continuous for $q \in \mathbb {R}$.

• 3. F(q) is strictly monotonic decreasing function of q.

The proofs of lemmas are detailed in [19]. We can then prove $F(q) = 0$ has an unique solution, say $q^*$. This assertion results from (2), (3), and the following fact $\lim _{q\rightarrow -\infty } F(q)=+\infty$, and $\lim _{q\rightarrow \infty } F(q)=-\infty$.

$\boxed {\hbox {Part 1}}$ In part 1, we prove if $q^*=\frac{N(d^*)}{D(d^*)}=\underset{d \in {S}}{\max } \quad \frac{N(d)}{D(d)}$, then $F(q^*)=\underset{d \in {S}}{\max } \quad \{N(d)- q^*D(d)\}=0$. We let $d^*$ to be solution to (1). We then have $N(d^*)/D(d^*)=q^*\ge {N(d)}/{D(d)}, \forall d\in S$. Given $D(d)>0, \forall d\in S$, we have

$$\begin{aligned} N(d)-q^* D(d)&\le 0, \forall d\in S \nonumber \\ N(d^*)-q^* D(d^*)&=0. \end{aligned}$$

(A3)

From (3) we see that $F(q^*)=\underset{d \in {S}}{\max } \quad \{N(d)- q^*D(d)\}=0$, and $d^*$ is an maximizer of $\underset{d \in {S}}{\max } \quad \{N(d)- q^*D(d)\}$.

$\boxed {\hbox {Part 2}}$ In part 2, we prove if $F(q^*)=\underset{d \in {S}}{\max } \quad \{N(d)- q^*D(d)\}=0$, then $q^*=\frac{N(d^*)}{D(d^*)}=\underset{d \in {S}}{\max } \quad \frac{N(d)}{D(d)}$. We let $d^*$ to be a solution to (2) such that $N(d^*)- q^*D(d^*)=0$. The definition of (2) implies that $N(d)- q^*D(d)\le N(d^*)- q^*D(d^*)=0, \forall d\in S$. Therefore, we have

$$\begin{aligned} N(d)-q^* D(d)&\le 0, \forall d\in S \nonumber \\ N(d^*)-q^* D(d^*)&=0. \end{aligned}$$

(A4)

From (4) we have $q^*\ge N(d)/D(d),\forall d\in S$, that is $q^*$ is the maximum of problem (3). Given that $N(d^*)/D(d^*) =q^*$, we can conclude $d^*$ is a solution for (1). Therefore, $d^*$ that solves (2), with $q=q^*$, also solves (1). $\square$

The above theorem does not guarantee the uniqueness of $d^*$. While the strict conditions of N(d) and D(d) being continuous and S being compact may not always hold, we find practical success in solving a sequence of convex problems $\underset{d \in {S}}{\max } \quad {N(d)- qD(d)}$ indexed by q, leading to convergence.

Appendix 2

1.1 Additional Simulation Results Under Small Sample Size (See Table 6)

Table 6 Estimated average iwICER(d) (sd) in Scenario 3