Bargaining and the provision of health services

Abstract

We model and compare the bargaining process between a purchaser of health services, such as a health authority, and a provider (the hospital) in three plausible scenarios: (a) activity bargaining: the purchaser sets the price and activity (number of patients treated) is bargained between the purchaser and the provider; (b) price bargaining: the price is bargained between the purchaser and the provider, but activity is chosen unilaterally by the provider; (c) efficient bargaining: price and activity are simultaneously bargained between the purchaser and the provider. We show that: (1) if the bargaining power of the purchaser is high (low), efficient bargaining leads to higher (lower) activity and purchaser’s utility, and lower (higher) prices and provider’s utility compared to price bargaining. (2) In activity bargaining, prices are lowest, the purchaser’s utility is highest and the provider’s utility is lowest; activity is generally lowest, but higher than in price bargaining for high bargaining power of the purchaser. (3) If the purchaser has higher bargaining power, this reduces prices and activity in price bargaining, it reduces prices but increases activity in activity bargaining, and it reduces prices but has no effect on activity in efficient bargaining.

Appendix

The model

Proof of Eq. 2. Activity bargaining

The result is obtained by differentiating $\gamma \log [B(y)-py-\overline{V}] +(1-\gamma) \log [py-C(y)-\overline{U}]$ with respect to y. The second order condition (SOC) is $\Upgamma=\gamma {\frac{B_{yy}V-( B_{y}-p) ^{2}}{\widetilde{V}^{2}}}-(1-\gamma) {\frac{C_{yy}\widetilde{U}+(p-C_{y})^{2}}{\widetilde{U}^{2}}}<0$, which is always satisfied. $\square$

Proof of Eq. 8. Price bargaining

By taking the log and differentiating with respect to p we obtain $\gamma {\frac{B_{y}y_{p}-y(p)-py_{p}}{\widetilde{V}}} +(1-\gamma) {\frac{y(p)+py_{p}-C_{y}y_{p}}{\widetilde{U}}}=0$. From the FOC of the provider we know that p = C _y. By simplifying, we obtain: $\gamma {\frac{B_{y}y_{p}-y(p)-py_{p}}{\widetilde{V}}}+(1-\gamma) {\frac{y(p)}{\widetilde{U}}}=0$. The SOC is $-{\frac{\gamma \widetilde{U}^{2}((B_{y}-p) y_{p}-y)^{2}+(1-\gamma) \widetilde{V}^{2}y^{2}}{\widetilde{V}^{2}\widetilde{U}^{2}}} -{\frac{\gamma(2-{\frac{B_{yy}}{C_{yy}}}) \widetilde{U}-(1-\gamma) \widetilde{V}}{\widetilde{V}\widetilde{U}}}y_{p}$. $\square$

Proof of Eq. 11. Efficient bargaining

Define

$$ \Upomega=\left[B(y)-py-\overline{V}\right]^{\gamma} \left[py-C(y)-\overline {U}\right]^{1-\gamma} $$

Then: ${\frac{\partial \log \Upomega}{\partial p}}=-{\frac{\gamma y}{B(y)-py-\overline{V}}}+{\frac{(1-\gamma) y}{py-C(y)-\overline{U}}}=0$ and ${\frac{\partial \log \Upomega}{\partial y}}={\frac{\gamma (B_{y}-p) }{B(y)-py-\overline{V}}}+{\frac{(1-\gamma) (p-C_{y})}{py-C(y)-\overline{U}}}=0$. From the first equation we obtain $p={\frac{\gamma [C(y)+\overline{U}]+(1-\gamma) [B(y)-\overline{V}]}{y}}$, which, substituted into the second one, yields: B _y = C _y. The SOCs are: ${\frac{\partial^{2}\log \Upomega}{\partial p^{2}}}=-y^{2}( {\frac{\gamma}{\tilde{V}^{2}}}+{\frac{1-\gamma }{\tilde{U}^{2}}})<0$, ${\frac{\partial^{2}\log \Upomega}{\partial y^{2}}}=\gamma {\frac{B_{yy}\tilde{V}-(B_{y}-p) ^{2}}{\tilde{V}^{2}}}-(1-\gamma) {\frac{C_{yy}\tilde{U} +(p-C_{y})^{2}}{\tilde{U}^{2}}}<0$, and ${\frac{\partial^{2}\log \Upomega}{\partial p^{2}}}{\frac{\partial^{2}\log \Upomega}{\partial y^{2}}}>({\frac{\partial^{2}\log \Upomega}{\partial p\partial y}})^{2}$. ${\frac{\partial^{2}\log \Upomega}{\partial p\partial y}}=-{\frac{\gamma}{\tilde{V}}}+{\frac{1-\gamma}{\tilde{U}}} +y(B_{y}-p)({\frac{\gamma}{\tilde{V}^{2}}} +{\frac{1-\gamma}{\tilde{U}^{2}}})={\frac{(1-\gamma) B+\gamma C-py}{\tilde{V}\tilde{U}}}+y(B_{y}-p) ({\frac{\gamma}{\tilde{V}^{2}}}+{\frac{1-\gamma}{\tilde{U}^{2}}}) =y(B_{y}-p)({\frac{\gamma}{\tilde{V}^{2}}}+{\frac{1-\gamma}{\tilde {U}^{2}}})$, where the last simplification follows from the FOC for price. ${\frac{\partial^{2}\log \Upomega}{\partial p^{2}}}{\frac{\partial^{2}\log \Upomega}{\partial y^{2}}}> ({\frac{\partial^{2}\log \Upomega}{\partial p\partial y}}) ^{2}=-y^{2}({\frac{\gamma}{\tilde{V}^{2}}} +{\frac{1-\gamma}{\tilde{U}^{2}}}) [\gamma {\frac{B_{yy}\tilde{V}}{\tilde{V}^{2}}}-(1-\gamma) {\frac{C_{yy}\tilde{U}}{\tilde{U}^{2}}}-(B_{y}-p)^{2} ({\frac{\gamma}{\tilde{V}^{2}}}+{\frac{1-\gamma}{\tilde{U}^{2}}})] -y^{2}(B_{y}-p)^{2}({\frac{\gamma}{\tilde{V}^{2}}} +{\frac{1-\gamma}{\tilde{U}^{2}}}) ^{2}=-\gamma {\frac{B_{yy}\tilde{V}}{\tilde{V}^{2}}}+(1-\gamma) {\frac{C_{yy}\tilde{U}}{\tilde{U}^{2}}}>0$. All three SOCs are always satisfied, since B _yy ≤ 0. $\square$

Constant marginal benefit

Activity bargaining. $p^{a}={\frac{a}{2}}$, $y^{a}={\frac{a}{c( 2-\gamma)}}$, $V^{a}={\frac{a^{2}}{2c( 2-\gamma)}}$, $U^{a}={\frac{a^{2}(1-\gamma)}{2c(2-\gamma)^{2}}}$.

Proof

The rule determining activity is, for a given price: $\gamma {\frac{a-p}{(a-p) y}}+(1-\gamma){\frac{p-cy}{( p-{\frac{c}{2}}y) y}}=0$, from which $y={\frac{2p}{c( 2-\gamma)}}$. The FOC for price is: ${\frac{2a}{c(2-\gamma)}}-{\frac{4p}{c( 2-\gamma)}}=0$, from which: $p^{a}={\frac{a}{2}}$ (the SOC is $-{\frac{4p}{c(2-\gamma)}}<0$). The bargained activity is therefore: $y^{a}={\frac{a}{c(2-\gamma)}}$. The utility of the purchaser and the provider are: $V^{a}=( a-p) y={\frac{a^{2}}{2c( 2-\gamma )}}$ and $U^{a}=(p-{\frac{c}{2}}y) y={ \frac{a^{2}(1-\gamma)}{2c(2-\gamma)^{2}}}$. $\square$

Price bargaining. $p^{p}={\frac{a(2-\gamma)}{2}}$, $y^{p}={\frac{a(2-\gamma)}{2c}}$, $V^{p}={\frac{\gamma a^{2}(2-\gamma)}{4c}}$, $U^{p}={\frac{a^{2}(2-\gamma)^{2}}{8c}}$.

Proof

Since $y={\frac{p}{c}}$ with $y_{p}={\frac{1}{c}}$, the FOC for the bargained price is: $\gamma {\frac{(a-p) {\frac{1}{c}}-{\frac{p}{c}}} {{\frac{ap}{c}}-{\frac{p^{2}}{c}}}}+(1-\gamma) {\frac{{\frac{p}{c}}}{{\frac{p^{2}}{c}}-{\frac{p^{2}}{2c}}}}=0$, which gives: $p^{p}={\frac{a(2-\gamma )}{2}}$ (the SOC is $-{\frac{1}{(a-p)^{2}p^{2}}}((a-p) ^{2}+p^{2})-{\frac{2}{p^{2}}}(1-\gamma)<0$). Hence $y^{p}={\frac{a(2-\gamma)}{2c}}$, $V^{p}=(a-p) y=\gamma {\frac{a^{2}(2-\gamma)}{4c}}$ and $U^{p}=(p-{\frac{c}{2}}y) y={\frac{a^{2}(2-\gamma)^{2}}{8c}}$. $\square$

Efficient bargaining. $p^{e}={\frac{a(2-\gamma) }{2}}$, $y^{e}={\frac{a}{c}}$, $V^{e}={\frac{\gamma a^{2}}{2c}}$, $U^{e}={ \frac{a^{2}(1-\gamma)}{2c}}$.

Proof

The FOC with respect to price implies: $p=(1-\gamma) a+\gamma {\frac{c}{2}}y$. The FOC with respect to activity implies: $y^{e}={\frac{a}{c}}$. Therefore $p^{e}={\frac{a(2-\gamma)}{2}}$ and $V^{e}=(a-p) y=\gamma {\frac{a^{2}}{2c}}$ and $U^{e}=(p-{\frac{c}{2}}y) y=(1-\gamma) {\frac{a^{2}}{2c}}$. $\square$

Proof of Proposition 1

(a) $p^{p}={\frac{a(2-\gamma)}{2}}\geq {\frac{a}{2}}=p^{a}$ if γ ≤ 1. (b) $y^{a}={\frac{a}{c (2-\gamma)}}\leq y^{e}={\frac{a}{c}}$ if ${\frac{a}{c(2-\gamma)}}\leq {\frac{a}{c}}$ or γ ≤ 1; $y^{p}={\frac{a(2-\gamma)}{2c}}\leq y^{e}={\frac{a}{c}}$ if γ ≥ 0; $y^{p}={\frac{a(2-\gamma)}{2c}}\geq y^{a}={\frac{a}{c(2-\gamma)}}$ if (2 − γ)² ≥ 2 or γ ≤ 0.59. (c) $V^{a}={\frac{a^{2}}{2c(2-\gamma)}}\geq V^{e}={\frac{\gamma a^{2}}{2c}}$ if 2γ − γ² − 1 ≤ 0 or −(γ − 1)² ≤ 0; $V^{e}={\frac{\gamma a^{2}}{2c}}\geq V^{p}={\frac{\gamma a^{2}(2-\gamma)}{4c}}$ if γ ≥ 0. (d) $U^{p}={\frac{a^{2}(2-\gamma)^{2}}{8c}}\geq U^{e}={\frac{a^{2}}{2c}}(1-\gamma)$ if ${\frac{(2-\gamma)^{2}}{4}}\geq (1-\gamma)$ or 4 + γ² − 4γ ≥ 4 − 4γ, or if γ² > 0; $U^{e}={\frac{a^{2}}{2c}}(1-\gamma) \geq U^{a}={\frac{a^{2}}{2c}}{\frac{1-\gamma}{(2-\gamma)^{2}}}$ if (2 − γ)² ≥ 1, which is always the case, since 0 ≤ γ ≤ 1. $\square$

Decreasing marginal benefit

Price bargaining. $p^{p}={\frac{ac(2-\gamma)}{b+2c}}$, $y^{p}={\frac{a(2-\gamma)}{b+2c}}$, $V^{p}={\frac{\gamma a^{2}(2-\gamma)}{2(b+2c)}}$, $U^{p}={\frac{a^{2}c(2-\gamma)^{2}} {2(b+2c)^{2}}}$.

Proof

Since $y={\frac{p}{c}}$ with $y_{p}={\frac{1}{c}}$, the FOC for the bargained price is: $\gamma {\frac{(a-b{\frac{p}{c}}-p) {\frac{1}{c}}-{ \frac{p}{c}}}{(a{\frac{p}{c}}-{\frac{b}{2}} {\frac{p^{2}}{c^{2}}}-{\frac{p^{2}}{c}}) }}+(1-\gamma) {\frac{{\frac{p}{c}}}{{\frac{p^{2}}{c}}-{\frac{p^{2}}{2c}}}}=0$, which simplifies to $\gamma {\frac{(a-b{ \frac{p}{c}}-p) -p}{(a-{\frac{b}{2}}{\frac{p}{c}}-p)}}+2(1-\gamma) =0$ or $\gamma (a-b{\frac{p}{c}}-p)-\gamma p+2(1-\gamma) (a-{\frac{b}{2c}}p-p)=0$, giving: $p^{p}={\frac{ac(2-\gamma)}{b+2c}}$. Hence $y^{p}={\frac{a(2-\gamma)}{b+2c}}$, $V^{p}=(a-{\frac{b}{2}}y-p) y={\frac{\gamma a^{2}(2-\gamma )}{2(b+2c)}}$ and $U^{p}=(p-{\frac{c}{2}}y) y={\frac{a^{2}c(2-\gamma)^{2}}{2(b+2c)^{2}}}$. $\square$

Efficient bargaining. $p^{e}={\frac{a((1-\gamma)b +(2-\gamma) c)}{2(b+c)}}$, $y^{e}={\frac{a}{b+c}}$, $V^{e}={\frac{\gamma a^{2}}{2(b+c)}}$, $U^{e}={\frac{a^{2}(1-\gamma)}{2(b+c)}}$.

Proof

The FOC with respect to price implies: $p^{e}=(1-\gamma) (a-{\frac{b}{2}}y)+\gamma {\frac{cy}{2}}$. The FOC with respect to activity implies: $y^{e}={\frac{a}{b+c}}$. Therefore $p^{e}={\frac{a((1-\gamma)b+(2-\gamma) c)}{2(b+c)}}$ and $V^{e}=(a-{\frac{b}{2}}y-p) y={\frac{\gamma a^{2}}{2(b+c)}}$ and $U^{e}=(p-{\frac{c}{2}}y) y={\frac{a^{2}(1-\gamma)}{2(b+c)}}$. $\square$

Proof of Proposition 2

(a) p ^p > p ^e if ${ \frac{ac(2-\gamma)}{b+2c}}>{\frac{a((1-\gamma) b+( 2-\gamma) c)}{2(b+c)}}$ or b(cγ + bγ − b) > 0 or $\gamma>{\frac{b}{b+c}}$. (b) y ^e > y ^p if ${\frac{a}{b+c}}>{\frac{a(2-\gamma)}{b+2c}}$ or b + 2c − (2 − γ) (b + c) > 0 or $\gamma>{\frac{b}{b+c}}$. (c) U ^p > U ^e if ${\frac{a^{2}c(2-\gamma)^{2}}{2(b+2c)^{2}}}> {\frac{(1-\gamma)a^{2}}{2(b+c)}}$ or b ²γ + bcγ² + c ²γ² − b ² > 0 or $\gamma= \{-{\frac{b}{c}},{\frac{b}{b+c}}\}$. (d) V ^e > V ^p if ${\frac{\gamma a^{2}}{2(b+c)}}>{\frac{\gamma a^{2}(2-\gamma)}{2(b+2c)}}$ or (b + 2c) − (b + c) (2 − γ) > 0 or $\gamma>{\frac{b}{b+c}}$. $\square$

Decreasing marginal benefit and activity bargaining

Proof

From FOC with respect to y we have $\gamma {\frac{(a-by)-p}{ay-{\frac{b}{2}}y^{2}-py}}=-(1-\gamma) {\frac{p-cy}{py-{\frac{c}{2}}y^{2}}}$ or $\gamma (a-by-p)(p-{\frac{c}{2}}y)+(1-\gamma) (p-cy)(a-{\frac{b}{2}}y-p)=0$. Upon expanding, we obtain ${\frac{bc}{2}}y^{2}-y(c{\frac{2-\gamma}{2}}(a-p) +bp{\frac{1+\gamma}{2}})+p(a-p)=0$, with solution $y={\frac{(c{\frac{2-\gamma}{2}}(a-p) +bp{\frac{1+\gamma}{2}})-\sqrt{(c{\frac{2-\gamma}{2}} (a-p)+bp{\frac{1+\gamma}{2}})^{2}-4{\frac{bc}{2}}p (a-p)}}{bc}}$. $\square$