A quantitative analysis of risk-sharing agreements with patient support programs for improving medication adherence

Abstract

Medication adherence is a challenge for patients, drugmakers, and payers. To promote adherence, some payers adopt a form of pay-for-success value-based risk-sharing agreements. Drugmakers reduce prices for meaningful improvement in adherence and share patient information and resources with payers; as a return, payers run patient support programs and put drugs on a tier with lower copays. We use a game-theoretic approach to investigate the optimal program effect and the optimal prices with and without improvement in adherence, measured by Proportion of Days Covered, under such an agreement. Since negotiation power impacts how prices are determined, we consider several pricing settings: the payer or the drugmaker sets both prices or sets one price simultaneously or sequentially. Although a discount for improved adherence tends to promote adherence, it may not always be achievable nor guarantees better adherence. The drugmaker with strong negotiation power can align its interest with social welfare but the payer may not. The payer with strong negotiation power can improve more adherence than the drugmaker. Balanced negotiation power contributes either the most or the least to adherence depending on contract form and decision sequence. Although cost-sharing by the drugmaker expects to increase program efforts, it may not be true. We find that the policymaker prefers different levels of cost-sharing under different pricing settings. The payer may have a first-mover advantage when setting the price without improved adherence; the drugmaker, however, does not have such an advantage.

Appendices

Appendix 1

Fig. 1

A classification scheme for patients receiving treatment

Table 1 Summary of notation

Fig. 2

The optimal program effect e^∗ for combinations of p₁ and p₂

Fig. 3

The change of h_D(p₁, p₂) and h_P(p₁, p₂) w.r.t prices

Fig. 4

The optimal program effect under different pricing settings. a With respect to the unit health benefit b. b With respect to the expected refill rate r. c With respect to the marginal program cost k. d With respect to the baseline L. e With respect to cost-sharing t.

Fig. 5

The impact of cost-sharing t on the optimal prices

Fig. 6

The impact of cost-sharing t on both parties’ utilities under different pricing settings

Fig. 7

The optimal social welfare under different pricing settings. a With respect to the unit health benefit b. b With respect to the expected refill rate r. c With respect to the marginal program cost k. d With respect to the baseline L. e With respect to cost-sharing t.

Appendix 2

1.1 Proof for Lemma 1

The FOC ofV_P w.r.t. e is:

$$\frac{dV_P}{de}=\left(b-{p}_2\right)\left(1-r\right)F\left(1-e\right)- tke-\left({p}_1-{p}_2\right)\left(1-r\right)F\left(L-e\right)+\left({p}_1-{p}_2\right)\left[L\left(1-r\right)+r\right]f\left(L-e\right)=0.$$

The second derivative of V_P w.r.t. e is:

$$\frac{d^2{V}_P}{de^2}=-\left(b-{p}_2\right)\left(1-r\right)f^{\prime}\left(1-e\right)- tk+\left({p}_1-{p}_2\right)\left(1-r\right)f^{\prime}\left(L-e\right)-\left({p}_1-{p}_2\right)\left[L\left(1-r\right)+r\right]{f}^{\prime}\left(L-e\right).$$

Let e^P be the solution to$\frac{dV_P}{de}=$0. When a₀ is from a uniform distribution U[0, 1], we have ${e}^P=\frac{b\left(1-r\right)+{p}_1r-{p}_2}{tk+\left(b-{p}_1\right)\left(1-r\right)}$ and $\frac{d^2{V}_P}{de^2}=$ −tk − (b − p₁)(1 − r).

If p₁ < b + tk/(1 − r), then V_P is concave in e; and the optimal effect is max(0,min(e^P, L)). 0<e^P<L when l₁(p₁) < p₂ < l₂(p₁). If p₁ > b + tk/(1 − r), then V_P^P is convex in e. e^∗=0 if e^P > L/2, e^∗=L if e^P < L/2, e^∗is either 0 or L otherwise. If p₁ = b + tk/(1 − r), then e^∗= L if p₂ < l₂(p₁), e^∗= 0 if p₂ > l₂(p₁), and e^∗is any value ifp₂ = l₂(p₁).

$$\frac{\partial {e}^P}{\partial {p}_1}=\frac{rtk+\left(1-r\right)b-\left(1-r\right){p}_2}{{\left[ tk+\left(1-r\right)\left(b-{p}_1\right)\right]}^2}>0,\frac{\partial {e}^P}{\partial {p}_2}=-\frac{1}{tk+\left(1-r\right)\left(b-{p}_1\right)}<0$$

1.2 Proof for Lemma 3a)

$\frac{\partial {V}_P}{\partial {p}_1}=-{\int}_0^{L-e}\left[\left({a}_0+e\right)\left(1-r\right)+r\right]f\left({a}_0\right){da}_0\le 0$, (and)

$\frac{\partial {V}_P}{\partial {p}_2}=-{\int}_{L-e}^{1-e}\left[\left({a}_0+e\right)\left(1-r\right)+r\right]f\left({a}_0\right){da}_0-{\int}_{1-e}^1f\left({a}_0\right){da}_0<0$.

$\frac{\partial {h}_P\left({p}_1,{p}_2\right)}{\partial {p}_1}=\frac{\partial {V}_P}{\partial {p}_1}+\frac{\partial {V}_P}{\partial {e}^{\ast }}\frac{\partial {e}^{\ast }}{\partial {p}_1}=\frac{\partial {V}_P}{\partial {p}_1}\le 0$. Similarly, $\frac{\partial {h}_P\left({p}_1,{p}_2\right)}{\partial {p}_2}\le 0$.

1.3 Proof for Lemma 3b)

Given p₂, the drugmaker’s profit as a function of p₁ has three segments, corresponding to e^∗=0, e^P and L, respectively. Since $\frac{\partial {e}^P}{\partial {p}_1}>0$, h_D(p₁, p₂)increases with p₁ for relatively small p₁ (e^∗=0). When e^∗=L, $\frac{\partial {V}_D}{\partial {p}_1}$=0 and h_D(p₁, p₂) is a constant. When e^∗=e^P, we have $\frac{\partial {h}_D\left({p}_1,{p}_2\right)}{\partial {p}_1}=\frac{\partial {V}_D}{\partial {p}_1}+\frac{\partial {V}_D}{\partial {e}^P}\frac{\partial {e}^P}{\partial {p}_1}=\frac{\partial {V}_D}{\partial {p}_1}+\frac{rtk+\left(1-r\right)\left(b-{p}_2\right)}{{\left[\left(b-{p}_1\right)\left(1-r\right)+ tk\right]}^2}\left(b\left(1-r\right)-\left(b\left(1-r\right)+k\right){e}^P\right)$.

Since $\frac{\partial {V}_D}{\partial {p}_1}$>0 and $\frac{\partial {e}^P}{\partial {p}_1}>0$, h_D(p₁, p₂) always increases with p₁ if b(1 − r) − (b(1 − r) + k)L > 0; it first increases then declines with p₁ otherwise.

1.4 Proof for Lemma 3c)

Given p₁, the first derivative of h_D(p₁, p₂) w.r.t p₂ is:

$\frac{\partial {h}_D\left({p}_1,{p}_2\right)}{\partial {p}_2}={\int}_{L-e}^{1-e}\left[\left({a}_0+e\right)\left(1-r\right)+r\right]f\left({a}_0\right){da}_0+{\int}_{1-e}^1f\left({a}_0\right){da}_0+\frac{\partial e}{\partial {p}_2}\left[b\left(1-r\right)\left(1-e\right)- ke\right]$, where ${\int}_{L-e}^{1-e}\left[\left({a}_0+e\right)\left(1-r\right)+r\right]f\left({a}_0\right){da}_0+{\int}_{1-e}^1f\left({a}_0\right){da}_0=\left(1-L\right)\left(r+\frac{\left(1-r\right)\left(1+L\right)}{2}\right)+e$>0. In particular, when e = e^P, $\frac{\partial e}{\partial {p}_2}=-\frac{1}{\left(b-{p}_1\right)\left(1-r\right)+ tk}$<0 if (b − p₁)(1 − r) + k>0. When e^∗ = 0ore^∗ = L, $\frac{\partial e}{\partial {p}_2}=0$ and $\frac{\partial {h}_D\left({p}_1,{p}_2\right)}{\partial {p}_2}>0$. When e^∗ = e^P, since b(1 − r)(1 − e^P) − ke^P decreases with e^P while e^P decreases with p₂, b(1 − r)(1 − e^P) − ke^P increases with p₂. ${e}^P=\frac{b\left(1-r\right)+{rp}_1}{\left(b-{p}_1\right)\left(1-r\right)+ tk}$ at p₂ = 0 so that b(1 − r)(1 − e^P) − ke^P<0 at p₂ = 0, and increases with p₂ thereafter. That is, $\frac{\partial {h}_D\left({p}_1,{p}_2\right)}{\partial {p}_2}>0$ at p₂ = 0 and the drugmaker’s profit increases with p₂ from p₂ = 0. As p₂ rises, e^P decreases and b(1 − r)(1 − e^P) − ke^P increases so that $\frac{\partial {h}_D\left({p}_1,{p}_2\right)}{\partial {p}_2}$ may be reduced to 0 with e^P<e^S. When e^P declines to 0, $\frac{\partial {h}_D\left({p}_1,{p}_2\right)}{\partial {p}_2}=\frac{1}{2}\left(1-r\right)\left(1-{L}^2\right)+r\left(1-L\right)-\frac{b\left(1-r\right)}{\left(b-{p}_1\right)\left(1-r\right)+ tk}$. If this value is negative, i.e., ${p}_1>b-\frac{2b}{\left(1-r\right)\left(1-{L}^2\right)+2r\left(1-L\right)}+\frac{tk}{1-r}$, then h_D(p₁, p₂) is concave in p₂; otherwise, it increases with p₂ until e^P=0.

1.5 Proof for Lemma 4a)

From h_P(p₁, p₂)=0, we get a relationship between p₁ and p₂, denoted by p₁ = g(p₂), where g^′(p₂) satisfies:

$$\frac{\partial {V}_P}{\partial {p}_2}+\frac{\partial {V}_P}{\partial {p}_1}{g}^{\prime}\left({p}_2\right)+\frac{\partial {V}_P}{\partial e}\left(\frac{\partial {e}^P}{\partial {p}_2}+\frac{\partial {e}^P}{\partial {p}_1}{g}^{\prime}\left({p}_2\right)\right)=0.$$

Since $\frac{\partial {V}_P}{\partial e}=0$, we have ${g}^{\prime}\left({p}_2\right)=-\frac{\partial {V}_P}{\partial {p}_2}{\left(\frac{\partial {V}_P}{\partial {p}_1}\right)}^{-1}$<0.

By deriving t on both sides of h_P(p₁, p₂)=0, we have $\frac{dg\left({p}_2\right)}{dt}=-\frac{\frac{{\partial V}_P}{\partial t}+\frac{{\partial V}_P}{\partial e}\frac{{\partial e}^P}{\partial t}}{\frac{{\partial V}_P}{{\partial p}_1}+\frac{{\partial V}_P}{\partial e}\frac{{\partial e}^P}{{\partial p}_1}}$ , where $\frac{\partial {V}_P}{\partial {p}_1}+\frac{\partial {V}_P}{\partial e}\frac{\partial {e}^P}{\partial {p}_1}=\frac{\partial {h}_P\left({p}_1,{p}_2\right)}{\partial {p}_1}$<0. Thus $\frac{\partial {V}_P}{\partial t}+\frac{\partial {V}_P}{\partial e}\frac{\partial {e}^P}{\partial t}<\frac{\partial {V}_P}{\partial t}-\frac{\partial {e}^P}{\partial t}\frac{\partial {V}_P/\partial {p}_1}{\partial {e}^P/\partial {p}_1}$<0 because $\frac{\partial {V}_P}{\partial t}<0,\frac{\partial {e}^P}{\partial t}<0,\frac{\partial {V}_P}{\partial {p}_1}<0,\frac{\partial {e}^P}{\partial {p}_1}>0$. Thus, $\frac{dg\left({p}_2\right)}{dt}$<0 when e^∗ = e^P and $\frac{dg\left({p}_2\right)}{dt}$=0 when e^∗ = 0. The same logic applies to g⁻¹(p₁).

1.6 Proof for Lemma 4b)

First, we consider h_D(p₁, p₂) given p₂. When e^∗ = 0, h_D(p₁, p₂)=0 only if p₁ ≤ 0; when e^∗ = L, h_D(p₁, p₂) is a constant. Thus, the solution to h_D(p₁, p₂)=0 must happen when e^∗ = e^P. When e^∗ = e^P, given p₁, h_D(p₁, p₂) first increases with and then decreases with p₂. Thus, when h_D(p₁, p₂)increases with p₂, l(p₂) has to be reduced for a higher p₂; when h_D(p₁, p₂)decreases with p₂, l(p₂)has to be increased for a higher p₂.

By deriving t on both sides of h_D(p₁, p₂)=0 given e^∗ = e^P, we have $\frac{dl\left({p}_2\right)}{dt}=-\frac{\frac{{\partial V}_D}{\partial t}+\frac{{\partial V}_D}{\partial e}\frac{{\partial e}^P}{\partial t}}{\frac{{\partial V}_D}{{\partial p}_1}+\frac{{\partial V}_D}{\partial e}\frac{{\partial e}^P}{{\partial p}_1}}$ , where $\frac{\partial {V}_D}{\partial {p}_1}+\frac{\partial {V}_D}{\partial e}\frac{\partial {e}^P}{\partial {p}_1}>0$ because h_D(p₁, p₂) increases with p₁ at p₁ = l(p₂). Thus, $\frac{\partial {V}_D}{\partial t}+\frac{\partial {V}_D}{\partial e}\frac{\partial {e}^P}{\partial t}>\frac{\partial {V}_D}{\partial t}-\frac{\partial {e}^P}{\partial t}\frac{\partial {V}_D/\partial {p}_1}{\partial {e}^P/\partial {p}_1}$>0 because $\frac{\partial {V}_D}{\partial t}>0$, $\frac{\partial {e}^P}{\partial t}<0$, $\frac{\partial {V}_D}{\partial {p}_1}>0$, and $\frac{\partial {e}^P}{\partial {p}_1}>0$. Therefore, $\frac{dl\left({p}_2\right)}{dt}$<0. The same logic applies to l⁻¹(p₁).

1.7 Proof for Lemma 4c)

${p}_1^D\left({p}_2\right)$ is obtained from the following FOC:

$$m=\frac{\left(1-r\right){\left(L-{e}^P\right)}^2}{2}+\left({e}^P+\left(1-{e}^P\right)r\right)\left(L-{e}^P\right)+\frac{rtk+\left(1-r\right)\left(b-{p}_2\right)}{{\left[\left(b-{p}_1\right)\left(1-r\right)+ tk\right]}^2}\left(b\left(1-r\right)-\left(b\left(1-r\right)+k\right){e}^P\right)=0.$$

By deriving p₂ on both sides, we have.

$\frac{dp_1^D\left({p}_2\right)}{dp_2}=-\frac{\frac{\partial m}{\partial {p}_2}+\frac{\partial m}{\partial e}\frac{\partial {e}^P}{\partial {p}_2}}{\frac{\partial m}{\partial {p}_1}+\frac{\partial m}{\partial e}\frac{\partial {e}^P}{\partial {p}_1}}$, where $\frac{\partial m}{\partial e}=-{e}^P-\left(1-{e}^P\right)r<0$, b(1 − r) − (b(1 − r) + k)e^P<0, $\frac{\partial m}{\partial {p}_2}>0,\frac{\partial m}{\partial {p}_1}<0$, $\frac{\partial {e}^P}{\partial {p}_1}>0$ and $\frac{\partial {e}^P}{\partial {p}_2}<0$. Thus, $\frac{dp_1^D\left({p}_2\right)}{dp_2}$>0.

${p}_2^D\left({p}_1\right)$ is obtained from the following FOC:

$$\left(1-L\right)\left(\frac{1}{2}\left(1-r\right)\left(1+L\right)+r\right)+{e}^P-\frac{b\left(1-r\right)-\left(b\left(1-r\right)+k\right){e}^P}{\left(b-{p}_1\right)\left(1-r\right)+ tk}=0.$$

By deriving p₁for both sides, we have.

$\left(1+\frac{b\left(1-r\right)+k}{\left(b-{p}_1\right)\left(1-r\right)+ tk}\right)\frac{\partial {e}^P}{\partial {p}_1}-\left(1-r\right)\frac{b\left(1-r\right)-\left(b\left(1-r\right)+k\right){e}^P}{{\left[\left(b-{p}_1\right)\left(1-r\right)+ tk\right]}^2}+\left(1+\frac{b\left(1-r\right)+k}{\left(b-{p}_1\right)\left(1-r\right)+ tk}\right)\frac{\partial {e}^P}{\partial {p}_2}\frac{dp_2^D}{dp_1}=0.$Since $\frac{\partial {e}^P}{\partial {p}_1}>0$ and $\frac{\partial {e}^P}{\partial {p}_2}<0$, and $\left(1+\frac{b\left(1-r\right)+k}{\left(b-{p}_1\right)\left(1-r\right)+ tk}\right)\frac{\partial {e}^P}{\partial {p}_1}-\left(1-r\right)\frac{b\left(1-r\right)-\left(b\left(1-r\right)+k\right){e}^P}{{\left[\left(b-{p}_1\right)\left(1-r\right)+ tk\right]}^2}=\frac{r\beta k+r\left(1-r\right)b-\left(1-r\right){p}_2}{{\left[\beta k+\left(1-r\right)\left(b-{p}_1\right)\right]}^2}+\left(k+b\left(1-r\right)\right)\frac{r\beta k+\left(1-r\right)\left(2-r\right)b-2\left(1-r\right){p}_2+\left(1-r\right){p}_1r}{{\left[\beta k+\left(1-r\right)\left(b-{p}_1\right)\right]}^3}$. Thus, $\frac{dp_2^D}{dp_1}$ p₂ is uncertain.

1.8 Proof for case 1 (Proposition 1)

The payer’s problem is:

$$\underset{p_1,{p}_2}{\max }{h}_P\left({p}_1,{p}_2\right)$$

s.t. h_P≥ 0, h_D≥ 0, and p₁, p₂ ≥ 0.

Since h_P(p₁, p₂) decreases with p₁and p₂, the optimal prices should be the lowest possible feasible solution. When both prices are 0, h_D(p₁, p₂)<0 if t < 1, making it infeasible. Thus, the prices have to be reduced until h_D(p₁, p₂)=0. Otherwise, if h_D(p₁, p₂)>0, then p₁and p₂ can be lowered without changing the program effect so that h_P(p₁, p₂) is increased. Thus, h_P(p₁, p₂) = V_S.

From e^∗ = e^P and h_D(p₁, p₂)=0, we havep₂ = l⁻¹(p₁), which is replaced into V_P so that V_P = H_P(p₁). Thus, ${H}_P^{\prime}\left({p}_1\right)=\frac{\partial {V}_P}{\partial {p}_1}+\frac{\partial {V}_P}{\partial {p}_2}\frac{dl^{-1}\left({p}_1\right)}{dp_1}+\frac{\partial {V}_P}{\partial e}\left[\frac{\partial {e}^P}{\partial {p}_1}+\frac{\partial {e}^P}{\partial {p}_2}\frac{dl^{-1}\left({p}_1\right)}{dp_1}\right]=\frac{\partial {V}_P}{\partial {p}_1}+\frac{\partial {V}_P}{\partial {p}_2}\frac{dl^{-1}\left({p}_1\right)}{dp_1}$, where $\frac{dl^{-1}\left({p}_1\right)}{dp_1}$satisfies$\frac{\partial {V}_D}{\partial {p}_1}+\frac{\partial {V}_D}{\partial {p}_2}\frac{dl^{-1}\left({p}_1\right)}{dp_1}+\frac{\partial {V}_D}{\partial e}\left[\frac{\partial {e}^P}{\partial {p}_1}+\frac{\partial {e}^P}{\partial {p}_2}\frac{dl^{-1}\left({p}_1\right)}{dp_1}\right]=0$. Thus, ${H}_P^{\prime}\left({p}_1\right)=\frac{\partial {V}_D}{\partial e}\left[\frac{\partial {e}^P}{\partial {p}_1}+\frac{\partial {e}^P}{\partial {p}_2}\frac{dl^{-1}\left({p}_1\right)}{dp_1}\right]$. If $\frac{\partial {V}_D}{\partial e}=0$, then e^P = e^S, i.e., ${p}_2=\frac{p_1\left[b\left(1-r\right)+ kr\right]+b\left(1-r\right)k\left(1-t\right)}{b\left(1-r\right)+k}$, which is replaced into V_D so that ${\displaystyle \begin{array}{c}{V}_D=\frac{p_1\left[b\left(1-r\right)+ kr\right]+b\left(1-r\right)k\left(1-t\right)}{b\left(1-r\right)+k}\left[{\int}_{L-{e}^S}^{1-{e}^S}\left[\left({a}_0+{e}^S\right)\left(1-r\right)+r\right]f\left({a}_0\right)d{a}_0+{e}^S\right]\\ {}+{p}_1{\int}_0^{L-{e}^S}\left[\left({a}_0+{e}^S\right)\left(1-r\right)+r\right]f\left({a}_0\right)d{a}_0-\left(1-t\right)\frac{k{\left({e}^S\right)}^2}{2}\\ {}=\frac{p_1\left[b\left(1-r\right)+ kr\right]}{b\left(1-r\right)+k}+{p}_1{\int}_0^{L-{e}^S}\left[\left({a}_0+{e}^S\right)\left(1-r\right)+r\right]f\left({a}_0\right)d{a}_0\\ {}+\frac{b\left(1-r\right)k\left(1-t\right)}{b\left(1-r\right)+k}\left[{\int}_{L-{e}^S}^{1-{e}^S}\left[\left({a}_0+{e}^S\right)\left(1-r\right)+r\right]f\left({a}_0\right)d{a}_0+{e}^S/2\right]\end{array}}$.

V _D=0 only when t = 1 and p₁=0. When t < 1 and p₁=0, V_D>0. Thus, to reduce V_D to 0, p₂ has to be lowed untilp₂ = l⁻¹(0)and the resulting program effect is larger than e^S. Any combination of p₁ and p₂ = l⁻¹(p₁)generates a program effect larger than e^S i.e., $\frac{\partial {V}_D}{\partial e}<0$. Thus, $\frac{\partial {e}^P}{\partial {p}_1}+\frac{\partial {e}^P}{\partial {p}_2}\frac{d{l}^{-1}\left({p}_1\right)}{d{p}_1}={\left(\frac{\partial {V}_D}{\partial {p}_2}+\frac{\partial {V}_D}{\partial e}\frac{\partial {e}^P}{\partial {p}_2}\right)}^{-1}\left[\frac{\partial {e}^P}{\partial {p}_1}\left(\frac{\partial {V}_D}{\partial {p}_2}+\frac{\partial {V}_D}{\partial e}\frac{\partial {e}^P}{\partial {p}_2}\right)-\frac{\partial {e}^P}{\partial {p}_2}\left(\frac{\partial {V}_D}{\partial {p}_1}+\frac{\partial {V}_D}{\partial e}\frac{\partial {e}^P}{\partial {p}_1}\right)\right]=\Big({}^{\frac{\partial {V}_D}{\partial {p}_2}}\left[\frac{\partial {e}^P}{\partial {p}_1}\frac{\partial {V}_D}{\partial {p}_2}-\frac{\partial {e}^P}{\partial {p}_2}\frac{\partial {V}_D}{\partial {p}_1}\right]>0$, and ${H}_P^{\prime}\left({p}_1\right)<0$. Thus, the optimal prices are p₁=0 and p₂ = l⁻¹(0).

When e^∗ = 0, h_D(p₁, p₂) > 0and the payer will not set prices to e^∗ = 0; when e^∗ = L, from h_D(p₁, p₂) = 0, we have${p}_2\left[\left(1-L\right)\left(\frac{1}{2}\left(1-r\right)\left(1+L\right)+r\right)+L\right]-\frac{\left(1-t\right){kL}^2}{2}=0$, so that ${p}_2=\frac{\left(1-t\right){kL}^2}{1+r+L\left(1-r\right)\left(2-L\right)}$. p₁can be any feasible value.

1.9 Proof for case 2 (Proposition 2)

The drugmaker’s problem is:

$$\underset{p_1,{p}_2}{\max }{h}_D\left({p}_1,{p}_2\right)$$

s.t. h_P≥ 0, h_D≥ 0, and p₁, p₂ ≥ 0.

First, we prove that V_P = 0 is always satisfied. Otherwise, if V_P > 0, we should be able to find higher prices while keeping the program effect e^P unchanged and increasing the value of V_D. Thus, V_P = 0. When e^∗ = 0or e^∗ = L, V_S = V_D + V_P is a constant and any prices making V_P = 0 are the same to the drugmaker.

Now we replace p₁ = g(p₂) and e^∗ = e^P into the drugmaker’s objective function and denote it as G(p₂). Thus,

$${G}^{\prime}\left({p}_2\right)=\frac{\partial {V}_D}{\partial {p}_2}+\frac{\partial {V}_D}{\partial {p}_1}{g}^{\prime}\left({p}_2\right)+\frac{\partial {V}_D}{\partial e}\left(\frac{\partial {e}^P}{\partial {p}_2}+\frac{\partial {e}^P}{\partial {p}_1}{g}^{\prime}\left({p}_2\right)\right).$$

g ^′(p₂)is obtained from:

$$\frac{\partial {V}_P}{\partial {p}_2}+\frac{\partial {V}_P}{\partial {p}_1}{g}^{\prime}\left({p}_2\right)+\frac{\partial {V}_P}{\partial e}\left(\frac{\partial {e}^P}{\partial {p}_2}+\frac{\partial {e}^P}{\partial {p}_1}{g}^{\prime}\left({p}_2\right)\right)=0.$$

Since $\frac{\partial {V}_P}{\partial e}$=0, $\frac{\partial {V}_P}{\partial {p}_1}=-\frac{\partial {V}_D}{\partial {p}_1},\mathrm{and}\ \frac{\partial {V}_P}{\partial {p}_2}=-\frac{\partial {V}_D}{\partial {p}_2}$, ${G}^{\prime}\left({p}_2\right)=\frac{\partial {V}_D}{\partial e}\left(\frac{\partial {e}^P}{\partial {p}_2}+\frac{\partial {e}^P}{\partial {p}_1}{g}^{\prime}\left({p}_2\right)\right)$.

G ^′(p₂) = 0when $\frac{\partial {V}_D}{\partial e}$=0 as g^′(p₂)<0. Thus, the optimal p₂ should satisfy ${e}^P=\frac{b\left(1-r\right)}{b\left(1-r\right)+k}$=e^S.

Since e^S>0, e^∗=0 will not be optimal. If e^S≤L, the optimal decisions are solutions to e^P = e^S, and V_P = 0. The optimal solution is unique because the prices satisfying V_P = 0 change in the oppositie directions and the prices satisfying e^P = e^S change in the same direction. In this case, $\frac{dp_1^{2\ast }}{dt}=-{\left(\frac{\partial {V}_P}{\partial {p}_1}+\frac{\partial {V}_P}{\partial {p}_2}\frac{dp_2}{dp_1}\right)}^{-1}\frac{\partial {V}_P}{\partial t}>0$ because $\frac{\partial {V}_P}{\partial t}<0$ and $\frac{dp_2}{dp_1}>0$.

If e^S>L, then e = L and V_P = 0. Thus, ${V}_P=\left(b-{p}_2\right)\left[\left(1-L\right)\left(\frac{1}{2}\left(1-r\right)\left(1+L\right)+r\right)+L\right]-\frac{tkL^2}{2}=0$ so that ${p}_2=b-\frac{tkL^2}{1+r+L\left(1-r\right)\left(2-L\right)}$. p₁ can be any non-negative value.

1.10 Proof for case 3 (Proposition 3): p ₁ is set by the drugmaker and p ₂ by the payer

Since h_P(p₁, p₂) decreases with p₂, the payer will reduce p₂ as much as possible and p₂=0 if the drugmaker’s participation constraint is met. When e^S≤L, h_D(p₁, p₂) is first concave in p₁ and then becomes a constant. Thus, the drugmaker’s best response is to choose the local optimum ${p}_1^D\left({p}_2\right)$. If the drugmaker’s local maximum ${h}_D\left({p}_1^D(0),0\right)\ge 0$, then l(0) ≥ 0 and the payer sets p₂ = 0. Otherwise, the payer has to increase p₂until the maximum of h_D(p₁, p₂) reaches 0. The corresponding h_P(p₁, p₂) ≥ 0because g(p₂) ≥ l(p₂). The optimal effect is higher than e^S.

When e^S>L, ${p}_1^D\left({p}_2\right)>{e}^S$. h_D(p₁, p₂) reaches its maximum at e^∗ = L, where it is negative if t < 1. Thus, when t < 1, the payer has to set p₂above 0 until h_D(p₁, p₂) =0 at e^∗ = L, i.e., ${p}_2=\frac{\left(1-t\right){kL}^2}{1+r+L\left(1-r\right)\left(2-L\right)}$. The optimal p₁ can be any value between l(p₂) and g(p₂).

1.11 Proof for case 4 (Proposition 4)

1.11.1 Proof for a)

(1). The prices are decided simultaneously.

When ${p}_1>\hat{p_1}$, h_D(p₁, p₂) is first concave in and then increases with p₂. When p₂ is increased untile^∗ = 0, h_D(p₁, p₂)>0 regardless how low p₁ is. The maximum value ${h}_D\left({p}_1,{p}_2^D\right)$ is larger than the value when e^P is reduced to 0. Thus, ${h}_D\left({p}_1,{p}_2^D\right)$>0 and the payer can reduce p₁ to 0. The drugmaker thus set p₂ at either g⁻¹(0)or ${p}_2^D(0)$ depending on which one brings a higher h_D(p₁, p₂) while satisfying h_P(p₁, p₂) ≥ 0. When ${g}^{-1}(0)<{p}_2^D(0)$, p₂ = g⁻¹(0), and the corresponding program effect may be higher or lower than e^S. When ${p}_1\le \hat{p_1}$, h_D(p₁, p₂) increases with p₂ so that the drugmaker always prefers the highest feasible p₂and the payer sets p₁ to 0.

(2). The drugmaker is the first-mover.

The payer will always sets p₁= 0. The drugmaker’s decision is the same as in (1).

1.11.2 Proof for b)

When ${p}_1>\hat{p_1}$, the drugmaker chooses ${p}_2^D\left({p}_1\right)$ if ${p}_2^D\left({p}_1\right)<{g}^{-1}\left({p}_1\right)<{p}_2^0\left({p}_1\right)$, where V_P>0; he chooses g⁻¹(p₁)otherwise, where V_P=0. When ${p}_1\le \hat{p_1}$, h_D(p₁, p₂) increases with p₂ and the drugmaker chooses g⁻¹(p₁). The payer, as the first-mover, prefers p₁that brings a higher V_P. Thus, she tries to induce the drugmaker to choose ${p}_2^D\left({p}_1\right)$ instead of g⁻¹(p₁) by setting the lowest p₁ satisfying ${p}_1>\hat{p_1}$. If $\hat{p_1}\ge \overline{p_1}$, ${p}_1>\hat{p_1}$ leads to ${g}^{-1}\left({p}_1\right)<{p}_2^D\left({p}_1\right)$ so that the payer cannot find a p₁ to induce the drugmaker choosing ${p}_2^D\left({p}_1\right)$. The drugmaker must choose g⁻¹(p₁) and the payer thus improves her utility by setting p₁ to 0. If $\hat{p_1}<\overline{p_1}$, the payer can set p₁ to the lowest value satisfying both${p}_1>\hat{p_1}$and ${p}_2^D\left({p}_1\right)<{g}^{-1}\left({p}_1\right)<{p}_2^0\left({p}_1\right)$ (i.e., $\underline {p_1}<\hat{p_1}<\overline{p_1}$) so that the drugmaker chooses ${p}_2^D\left({p}_1\right)$. In this case, the resulting program effect is less than e^S.