Health Care Provider Networks: Are Insurers Better Off?

Taking Proposition 1 into account, providers’ profits and policyholders’ net utility are respectively given by

From the insurer’s point of view, the cost incurred and the total net utility are respectively given by

To prove that ΔV = V ^s − V^B > 0 when k ≤ k ̂ , we have to show that ΔV is strictly concave in k. The second derivative of ΔV has the same sign as a sum

where Y < 0 if k < k ̂ and 4 − 4γ − 2γ² + γ³ < 0 if γ > 0.806. So, when k < k ̂ and γ > 0.806, the second derivative is negative. Replacing X with ( 4 ( α − c ) ( 1 − γ ) γ + k ( 2 − γ 2 ) ) 2 > X , we obtain a sum of negative terms for any γ and k < k ̂ . Consequently, ΔV is strictly concave in k when k < k ̂ . As ΔV > 0 when k = 0 and ΔV > 0 when k = k ̂ , ΔV > 0 for any k < k ̂ .

The right-hand member of (12) is defined if c/(βρ) ≥ p₂ > c/β and is positive when c > k. The left-hand member of (12) is positive if p 2 ∈ A = p 2 − = c ( 2 − β ρ ( 1 + β ) − ( 1 − β ) ( 4 − 8 β ρ + 5 β 2 ρ 2 − β 3 ρ 2 ) 2 β ( 1 + ρ ( 1 − 2 β ) ) ) , p 2 + = c ( 2 − β ρ ( 1 + β ) + ( 1 − β ) ( 4 − 8 β ρ + 5 β 2 ρ 2 − β 3 ρ 2 ) 2 β ( 1 + ρ ( 1 − 2 β ) ) ) . Thus, p₂ solution of (12) belongs to [ c / β , p 2 + . The left-hand member of the equality is decreasing and concave when p₂ ∈ A. It is equal to c²(1 − β)(1 − ρ)/β when p₂ = c/β and to 0 when p 2 = p 2 + . The right-hand member of (12) is decreasing and convex in this interval. It tends to + ∞ when p₂ tends to c/β and is positive when p 2 = p 2 + , as c / ( β ρ ) > p 2 + > c / β . It tends to 0 when p₂ tends to c/βρ. When c > k, the right-hand member of (12) is positive and (12) has at most two solutions belonging to A. Figure 3 shows the two members of (12) for c = 20, k = 10, β = 1/2, and ρ = 1/2.

Figure 3:

Right and left-hand members of (12).

Moreover, the equilibrium solution must satisfy (10). When k ^s = k, this imply

Taking (12) and (13) into account, we must have p 2 ( β − c − 2 ) ( 1 + β ρ ) c ( 2 − β ( 1 + ρ ) ) − p 2 β ( 1 + ρ ( 1 − 2 β ) ) > 0 , which implies p 2 < p ̄ 2 = c 2 − β ( 1 + ρ ) β ( 1 + ρ ( 1 − 2 β ) ) . As the left-hand member of (12) is higher (resp. lower) than the right-hand member of (12) when p 2 = p ̄ 2 (resp. p₂ = c/β), there is only one solution belonging to] c / β , p ̄ 2 .

The equilibrium price is solution of (12). (12) is an implicit equation F(p_2,c, k, β, ρ) = 0. For each value of ρ, ρ ∈ {1/2, 3/5, 7/10, 4/5, 9/10}, and β, β ∈ {1/2, 3/5, 7/10, 4/5, 9/10}, using the Mathematica computation program, we can obtain m expressions p₂ = Root[F, m] representing the exact root of F(p_2,c, k) = 0. Each root is a function of c and k and can be computed for any value of (c, k). Among the m roots, we select the root corresponding to a real value satisfying the constraints c / β < p 2 < p ̄ 2 and c > k” > k and ensuring the highest profit. Using this root and replacing in the implicit function (9), we obtain p₁(p_2,c, k) for any (β, ρ). Reporting these values in q 1 s , we can calculate q 1 s − 2 q B for any c and k < k” and obtain numerical values showing that q 1 s > 2 q B for any c, k, β, and ρ. As an example, let us consider ρ = 1/2 and β = 1/2. (12) becomes

whose Mathematica solution is given by

where #1 represents the first argument supplied to a pure function. For each couple (c, k), we obtain a value of p₂. Replacing (14) in p₁, we obtain p₁(c, k) and we can calculate q 1 s ( c , k ) and q 1 s ( c , k ) − 2 q B . Table 6 summarizes these results when k < k”.

For any c and k < k^′′, q 1 s ( c , k ) > 2 q B : the insurer’s cost increases when a selective contracting mechanism is implemented. Moreover, this cost increases with the reimbursement k and decreases with the marginal cost c. Similar results are obtained with other values of β and ρ.