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Individual Health Insurance Market with an Entrant – The ACA Health Insurance Exchange Observations

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Published/Copyright: April 21, 2018

Abstract

In this paper, a general framework is built up to model the dynamic of consumer health plan choice and individual health insurance market competition. A primary goal is to identify driving forces to individual health insurance equilibrium market coverage and premium. In the baseline model, we introduce plan quality information search cost as an additional determinant to consumer’s plan choice. Health insurers compete under the Hotelling’s game theory framework. Equilibrium solutions of the baseline model highlight the importance of budget limit and information search cost to health plan enrollment. The more important objective is to examine the impact of market entrants on equilibrium insurance market coverage and plan prices. In the model with market entry, we add an additional dimension to the baseline model. Equilibrium solutions and numerical studies show positive impact of higher insurance market coverage and lower health plan prices. The Affordable Care Act (ACA) brought multiple unprecedented changes to the health insurance market and provided opportunities to study market dynamics and driving forces. The ACA health insurance exchange market experience shows consistency with our model findings even at the early stage of implementation. More importantly, market observations suggest that entry barriers of claim costs and information search cost are high for entrants.

Appendix

A Proofs

Proof of Theorem 1

From eq. (1), we have

(5)I+p1+p22\ltv.

From eq. (2), we have

(6)I+p1+p22\ltv2p1=αu1+p2+I2p2=αu2+p1+Iv\lt3I+αu1+u22p1=3I+αu1+2u23p2=3I+α2u1+u23

Since both insurance companies can only either cover the entire market or the partial market, the following three cases are all possible situation.

Case (i)s1,s2=p2p1+I2I,p1p2+I2I=αu1u2+3I6I,αu2u1+3I6I

From eq. (5), we have s1+s2=1 hence

I+p1+p22=v.

Under such prices, we can get vv+αu12I+vv+αu22I1vαu1+p2+I2I+vαu2+p1+I2I1αu1+p2+I2p1v+αu12αu2+p1+I2p2v+αu22p1+p2=2vIv[2I+αu1+αu22,3I+αu1+αu22p1=xp2=2vIx

Case (ii), xmax4v3Iαu23,v+αu12,min3v2Iαu22,2v+αu13. and s1,s2=p2p1+I2I,p1p2+I2I=vxI,xv+II From eq. (6), we have

s1+s2

Under such prices, we get

vδ1.

Case (iii), vδ1 and δ1

q1δ1,q2δ1,s1δ1,s2δ1.

Where vδ12I+αu1+u22,

Under such prices, we get

s1δ1+s1δ1=1s1+s2.

After checking the solution at the boundary of each case, we can conclude the proof.■

Proof of Proposition 1

To check whether vδ1\lt2I+αu1+u22. increases in v, we consider two systems v and v\ltvδ1\lt2I+αu1+u22 Assume that s1+s2=vαu12I+vαu22I\ltvδ1αu12I+vδ1αu22I=s1δ1+s1δ1 > v and the subscript s1+s2 indicates the corresponding equilibrium, i. e. s1+s2

First, if u1, then u1

Second, we consider the case when u1δ2.

From u1δ2 and Theorem 1, we have

δ2

From the above discussion, we can conclude that q1δ2,q2δ2,s1δ2,s2δ2. increases when v increases.

To check whether v2I+αu1+u22, decreases in s1+s2=1s1δ2+s1δ2. we consider two systems v\lt2I+αu1+u22. and v\lt2I+αu1+u22\lt2I+αu1δ1+u22 Assume that s1+s2=vαu12I+vαu22I\ltvαu1δ22I+vαu22I=s1δ2+s1δ2 > v and the subscript s1+s2 indicates the corresponding equilibrium, i. e. u1

First, if s1+s2 then u2,

Second, we consider the case when p3αu3,

From p3+dcI>v. and Theorem 1, we have

p1ε,p2ε,p3ε,s1ε,s2ε,s3ε=p1,p2,αu3,s1,s2,0

From the above discussion, we can conclude that 2I+αu1+u22y3I+αu1+u22 increases when p1ε,p2ε,p3ε,s1ε,s2ε,s3ε=x,2yIx,αu3,yxI,xy+II,0 increases.

Following similar argument as the above, we obtain that x=max4y3Iαu23,y+αu12,min3y2Iαu22,2y+αu13 decreases in p1=xα and I. We conclude the proof of Proposition 1. ■

Proof of Theorem 2

The proof of Theorem 2 follows these cases:

Case (i): y > v. For every p2=2yIx, we have minp1+dI,p2+1dIp1+p2+I2=y=αu3+dcIp3+dcI,p3αu3 It means that no customer chooses the CO-OP for any feasible p3.Combining with Theorem 1, we have the equilibrium:

(7)d=yxI

Case (ii):p3=αu3. and y ≤ v. We can show that

(8)p3=αu3

Where s3=0.

Considering the optimal reaction of the CO-OP given that p3=αu3. and y2I+αu1+u22,3I+αu1+u22, we will have

p1+s1Iy

We also notice that the equality only holds when s1+s21. and p1,p2,s1,s2=x,2yIx,yxI,xy+II. In other words, the optimal reaction of the CO-OP is vy3I+αu1+u22 and p1ε,p2ε,p3ε,s1ε,s2ε,s3ε=3I+α2u1+u23,3I+αu1+2u23,αu3,3I+αu1u26I,3I+αu2u16I,0

Let’s consider the optimal choices of the two existing insurers given that p1=3I+α2u1+u23 Notice that p2=3I+αu1+2u23,minp1+dI,p2+1dIp1+p2+I2=3I+αu1+u22yαu3+dcI,p3αu3 and d=3I+αu1u26I Following similar discussion in Theorem 1, we have p3=αu3. We verified the equilibrium for Case (ii).

Case (iii):p3=αu3 we will show that

s3=0.

Consider the optimal reaction of the CO-OP given that p3=αu3. and y3I+αu1+u22, we have

p1+s1Iy

And the equality only holds when s1+s21. and p1,p2,s1,s2=3I+α2u1+u23,3I+αu1+2u23,3I+αu1u26I,3I+αu2u16I. In other words, the optimal choice of the CO-OP is y>minv,2I+αu1+u22. and yminv,2I+αu1+u22.

Let’s consider the optimal choices of the two existing insurers given that yminv,2I+αu1+u22 Notice that πipi=1I2pi+p3+dcI+αui,i=1,2;π3p3=1I4pi+12dc)I+2αu2+(p1+p2 and 12dcI+2αu3+p1+p24vdcI\hfillp1=v+αu12,p2=v+αu22\hfillp3=vdcI\hfillyv+12v2I+αu1+u22\hfillp1=v+αu12,p2=v+αu22\hfillp3=vdcI\hfill Following similar discussion in Theorem 1, we have s1,s2,s3=p3+dcIp1I,p3+dcIp2I,1s1s2=vαu12I,vαu22I,2I+αu1+u22v2I And we verified the equilibrium for Case (iii).

According to Cases (i), (ii), and (iii), we can see that the CO-OP will not enter the market if 12dcI+2αu3+p1+p24vdcIp3=12dcI+2αu3+p1+p242p1=p3+dcI+αu12p2=p3+dcI+αu2yv+12v2I+αu1+u22p1=I+2y+α7u1+u2/26p2=I+2y+αu1+7u2/26p3=1dcI+α2u3+u1+u223 Now, let’s consider when s1,s2,s3=p3+dcIp1I,p3+dcIp2I,1s1s2\hfill=I+2y+αu25u1/26I,I+2y+αu15u2/26I,2I+αu1+u22y3I

Case (iv)ymin{v,2I+α(u1+u2)2} From eqs (3) and (4), we have

πipi=1I[2pi+(p3+dcI+αui)],i=1,2;
π3p3=1I[4pi+(12dc)I+2αu2+(p1+p2)]

Hence, there are two sets of solutions. The first set of solutions:

(9){(12dc)I+2αu3+p1+p24vdcIp1=v+αu12,p2=v+αu22p3=vdcI{yv+12[v2I+α(u1+u2)2]p1=v+αu12,p2=v+αu22p3=vdcI

Under such prices, we have (s1,s2,s3)=(p3+dcIp1I,p3+dcIp2I,1s1s2)=(vαu12I,vαu22I,2I+α(u1+u2)2v2I)

The second set of solutions:

(10){(12dc)I+2αu3+p1+p24vdcIp3=(12dc)I+2αu3+p1+p242p1=p3+dcI+αu12p2=p3+dcI+αu2{yv+12[v2I+α(u1+u2)2]p1=I+2y+α(7u1+u2)/26p2=I+2y+α(u1+7u2)/26p3=(1dc)I+α[2u3+u1+u22]3

Under such prices, we have

(s1,s2,s3)=(p3+dcIp1I,p3+dcIp2I,1s1s2)=(I+2y+α(u25u1)/26I,I+2y+α(u15u2)/26I,2I+α(u1+u2)2y3I)

The results can be obtained by combining cases.

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Published Online: 2018-04-21

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