Preventives Versus Treatments Redux: Tighter Bounds on Distortions in Innovation Incentives with an Application to the Global Demand for HIV Pharmaceuticals

Abstract

Kremer and Snyder (Q J Econ 130:1167–1239, 2015) show that demand curves for a preventive and treatment may have different shapes though they target the same disease, biasing the pharmaceutical manufacturer toward developing the lucrative rather than the socially desirable product. This paper tightens the theoretical bounds on the potential deadweight loss from such biases. Using a calibration of the global demand for HIV pharmaceuticals, we demonstrate the dramatically sharper analysis achievable with the new bounds, allowing us to pinpoint potential deadweight loss at 62% of the global gain from curing HIV. We use the calibration to perform policy counterfactuals, assessing welfare effects of government policies such as a subsidy, reference pricing, and price-discrimination ban. The fit of our calibration is good: we find that a hypothetical drug monopolist would price an HIV drug so high that only 4% of the infected population worldwide would purchase, matching actual drug prices and quantities in the early 2000s before subsidies in low-income countries ramped up.

Appendix: Proofs of Propositions

Proof of Proposition 1

Assume benchmark values of the parameters. Suppose that the firm’s only choice is to develop product j or nothing. We will show that the supremum on relative deadweight loss is bounded by $1 - \rho ^0_j$ above and below, proving the two are equal and thus (6) holds.

A series of steps can be used to bound the supremum from below:

$$\begin{aligned} \sup _{ \{k_j \ge 0 \}} \left( \frac{{ DWL}^0}{B} \right)\ge & {} \lim _{ k_j \downarrow { PS}^0_j} \left( \frac{{ DWL}^0}{B} \right) \end{aligned}$$

(13)

$$\begin{aligned}&= \lim _{k_j \downarrow { PS}^0_j} \left( \frac{W^{00}}{B} \right) \end{aligned}$$

(14)

$$\begin{aligned}& = \lim _{k_j \downarrow { PS}^0_j} \left[ \frac{\max ({ TS}^{00}_j - k_j,0)}{B} \right] \end{aligned}$$

(15)

$$\begin{aligned}= & {} \frac{{ TS}^{00}_j - { PS}^0_j}{B} \end{aligned}$$

(16)

$$\begin{aligned}= & {} \frac{B - { PS}^0_j}{B} \end{aligned}$$

(17)

$$\begin{aligned}= & {} 1 - \rho ^0_j. \end{aligned}$$

(18)

Equation (13) follows since the limit point on the right-hand side is just one element of the closure of the larger set over which the supremum on the left-hand side is being taken. Equation (14) holds because nothing is developed in equilibrium in the limit, which implies $W^0 = 0$ and thus ${ DWL}^0 = W^{00} - W^0 = W^{00}$ in the limit. To see (15), note that either product j is developed, which yields the social first-best surplus ${ TS}^{00}_j - k_j$, or there is no product, which yields 0. The social optimum generates the maximum of these two social surpluses. Equation (16) follows from evaluating the limit, (17) follows from (2), and (18) follows from the definition of $\rho ^0_j$.

We next show that the supremum is bounded from above by $1 - \rho ^0_j$. If $W^{00}_j = 0$, then ${ DWL}^0_j = 0 \le 1 - \rho ^0_j$, and we are done. So suppose instead that $W^{00}_j > 0$. We have the following series of steps:

$$\begin{aligned} { DWL}^0= & {} W^{00} - W^0 \end{aligned}$$

(19)

$$\begin{aligned}= & {} W^{00}_j - W^0 \end{aligned}$$

(20)

$$\begin{aligned}\le & {} W^{00}_j - \varPi ^0_j \end{aligned}$$

(21)

$$\begin{aligned}= & {} TS^{00}_j - k_j - ({ PS}^0_j - k_j) \end{aligned}$$

(22)

$$\begin{aligned}= & {} B - { PS}^0_j. \end{aligned}$$

(23)

Equation (19) holds by definition. Equation (20) holds since $W^{00} = \max (0, W^{00}_j) = W^{00}_j$ by the maintained assumption that $W^{00}_j > 0$. Equation (21) holds since $W^0_j \ge \varPi ^0_j$. Equation (22) follows from substituting relevant definitions and (23) substituting from (2) and canceling terms. Dividing (19)–(23) by B yields

$$\begin{aligned} \frac{{ DWL}^0}{B} \le 1 - \rho ^0_j. \end{aligned}$$

(24)

Conditions (18) and (24) sandwich the supremum between $1 - \rho ^0_j$ above and below, which yields (6) as an exact equality. $\square$

Proof of Proposition 2

Similar to the previous proof, we will show that the supremum on relative deadweight loss is bounded by $1 - \rho _\mathrm {min}^0$ from above and below, which proves that the two are equal and thus (7) holds. For concreteness, suppose throughout the proof that

$$\begin{aligned} { PS}^0_v \le { PS}^0_d. \end{aligned}$$

(25)

Arguments establishing the bound for the reverse inequality are similar and omitted for brevity.

A series of steps can be used to bound the supremum from below:

$$\begin{aligned} \sup _{ \{k_v, k_d \ge 0 \}} \left( \frac{{ DWL}^0}{B} \right)\ge & {} \lim _{ k_v \downarrow { PS}^0_v, k_d \uparrow \infty } \left( \frac{{ DWL}^0}{B} \right) \end{aligned}$$

(26)

$$\begin{aligned}= & {} \lim _{k_v \downarrow { PS}^0_v} \left( \frac{W^{00}}{B} \right) \end{aligned}$$

(27)

$$\begin{aligned}= & {} \lim _{k_v \downarrow { PS}^0_v} \left[ \frac{\max (W^{00}_v,0)}{B} \right] \end{aligned}$$

(28)

$$\begin{aligned}= & {} \lim _{k_v \downarrow { PS}^0_v} \left[ \frac{\max ({ TS}^{00}_v - k_v, 0)}{B} \right] \end{aligned}$$

(29)

$$\begin{aligned}= & {} 1 - \rho ^0_v \end{aligned}$$

(30)

$$\begin{aligned}= & {} 1 - \rho _\mathrm {min}^0. \end{aligned}$$

(31)

The arguments for (26) and (27) are similar to those for (13) and (14), respectively. Equation (28) holds because the socially efficient product strategy cannot involve development of a drug, alone or together with a vaccine, for sufficiently large $k_d$. Hence $\sigma ^{00} \in \{v,n\}$, which implies that $W^{00} = \max (W^{00}_v, 0)$. Equation (29) holds by definition. Equation (30) follows from steps that are similar to (15)–(18), and (31) holds by assumption (25).

We next bound the supremum from above. We start by establishing that the following inequality,

$$\begin{aligned} { DWL}^0 \le B - { PS}_\mathrm {min}^0, \end{aligned}$$

(32)

holds regardless of which value—$\sigma ^{00} \in \{ v, d, b, n \}$—the socially optimal product strategy takes on. First consider the trivial case in which $\sigma ^{00} = n$. Then ${ DWL}^0 = W^{00} - W^0 = 0$ since $W^{00} = W^0 = 0$. But then (32) trivially holds because $B - { PS}_\mathrm {min}^0 \ge 0 = { DWL}^0$.

Next, consider the non-trivial case in which $\sigma ^{00} \in \{ v, d, b \}$. We can establish the following series of steps:

$$\begin{aligned} { DWL}^0&= W^{00} - W^0 \end{aligned}$$

(33)

$$\begin{aligned}= & { TS}^{00} - k_{\sigma ^{00}} - (\varPi ^0 + { CS}^0) \end{aligned}$$

(34)

$$\begin{aligned} & \le { TS}^{00} - k_{\sigma ^{00}} - \varPi ^0 \end{aligned}$$

(35)

$$\begin{aligned}\le & {} { TS}^{00} - k_{\sigma ^{00}} - \varPi _{\sigma ^{00}}^0 \end{aligned}$$

(36)

$$\begin{aligned}= & {} { TS}^{00} - { PS}_{\sigma ^{00}}^0. \end{aligned}$$

(37)

$$\begin{aligned}= & {} B - { PS}_{\sigma ^{00}}^0. \end{aligned}$$

(38)

Equations (33) and (34) follow from the substitution of the definitions of the relevant variables, and (35) follows from ${ CS}^0 \ge 0$. Equation (36) holds because the equilibrium product strategy is the most profitable, implying $\varPi ^0 \ge \varPi _{\sigma ^{00}}^0$. Equation (37) follows from $\varPi _{\sigma ^{00}}^0 = { PS}_{\sigma ^{00}}^0 - k_{\sigma ^{00}}$ and (38) from (2).

We next show that

$$\begin{aligned} { PS}^0_{\sigma ^{00}} \ge { PS}_\mathrm {min}^0 \end{aligned}$$

(39)

for all $\sigma ^{00} \in \{v, d, b\}$. If $\sigma ^{00} = v$, then ${ PS}^0_{\sigma ^{00}} = { PS}^0_v \ge { PS}_\mathrm {min}^0$, which implies that (39) holds. Similar arguments show that (39) holds if $\sigma ^{00} = d$. Suppose $\sigma ^{00} = b$. The firm can replicate producer surplus from a drug if both products have been developed by setting $p_{bv} = \infty$ and $p_{bd} = p^0_d$. Hence ${ PS}^0_b \ge { PS}^0_d \ge { PS}_\mathrm {min}^0$, which implies that (39) holds, which completes the proof that (39) holds for all $\sigma ^{00} \in \{v, d, b\}$.

We can now complete the proof: combining (38) and (39), we have ${ DWL}^0 \le B - { PS}_\mathrm {min}^0$ for all $\sigma ^{00} \in \{v, d, b\}$. Combining this fact with the argument in the text following (32) implies that (32) holds for all $\sigma ^{00} \in \{v, d, b, n\}$. Dividing (32) by B,

$$\begin{aligned} \frac{{ DWL}^0}{B} \le 1 - \rho _\mathrm {min}^0 \end{aligned}$$

(40)

for all $k_v, k_d \ge 0$, which implies that

$$\begin{aligned} \sup _{ \{k_v, k_d \ge 0 \}} \left( \frac{{ DWL}^0}{B} \right) \le 1 - \rho _\mathrm {min}^0. \end{aligned}$$

(41)

Conditions (31) and (41) sandwich the supremum from above and from below at $1 - \rho _\mathrm {min}^0$, which yields (7) as an exact equality. $\square$