The market of wearables are growing explosively for the past few years. The majority of the devices are related to health care and fitness. It is embarrassing that users easily lose interest in these devices, and thus fail to improve health condition. Recently, the “be healthy and be rewarded” programs are gaining popularity in health insurance market. The insurance companies give financial rewards to its policyholders who take the initiative to keep healthy. It provides the policyholders with incentives to lead a healthier lifestyle and the insurer can also benefit from less medical claims. Unfortunately, there are hardly any studies discussing how to design the incentive mechanism in this new emerging health promotion program. Improper design would not change policyholders’ unhealthy behavior and the insurer cannot benefit from it. In this paper, we propose a mechanism for this health promotion program. We model it as a monopoly market using contract theory, in which there is one insurer and many policyholders. We theoretically analyze how all parties would behave in this program. We propose a design that can guarantee that policyholders would faithfully participate in the program and the insurer can maximize its profit. Simulation results show that the insurer can improve its profit by\(40\%\) using the optimal contract.

This research is supported in part by the Key-Area Research and Development Program of Guangdong Province (No. 2020B0101390001) and in part by the National Natural Science Foundation of China (No. 62002150).

Authors and Affiliations

1. Sun Yat-sen University, Guangzhou, Guangdong, China

   Qianyi Huang

2. Huazhong University of Science and Technology, Wuhan, Hubei, China

   Wei Wang

3. Hong Kong University of Science and Technology, Hong Kong, China

   Qian Zhang

Corresponding author

Correspondence toQianyi Huang.

Editors and Affiliations

1. Shanghai Jiao Tong University, Shanghai, China

   Chao Li

2. Tsinghua University, Beijing, Beijing, China

   Zhenhua Li

3. National University of Defense Technology, Nanjing, China

   Li Shen

4. Shanghai Jiao Tong University, Shanghai, China

   Fan Wu

5. Nankai University, Tianjin, China

   Xiaoli Gong

A Appendices

1.1A.1 Supplementary Proof for Lemma5

Here we provide supplementary proof for Lemma5. As\(i\le k-1\),\(t_{k-1}\ge t_i\).

To prove that

we distinguish five cases:

Case 1:\(t_{i}\ge S^e_{k}>S_{k-1}^e\).

The inequality follows from Lemma3.

Case 2:\(S_{k-1}^e \le t_{i}< S^e_{k}\)and\(t_{k-1}\ge S^e_{k}\).

The first inequality holds because\(u_0(\theta _{k-1},t)\) is decreasing when\(t>S^e_{k-1}\).

Case 3:\(S_{k-1}^e \le t_{i}< S^e_{k}\)and\(t_{k-1}< S^e_{k}\).

The inequality holds because\(u_0(\theta _{k-1},t)\) is decreasing when\(t>S^e_{k-1}\), and then\(u_0\left( \theta _{k-1},t_i\right) >u_0(\theta _{k-1},t_{k-1})\).

Case 4:\(t_{i}<S^e_{k-1}\)and\(t_{k-1}\ge S^e_{k}\).

The first inequality holds because\(u_0(\theta _{k-1},t)\) is decreasing when\(t>S^e_{k-1}\).

Case 5:\(t_{i}<S^e_{k-1}\)and\(t_{k-1}< S^e_{k}\).

The inequality holds because\(u_0(\theta _{k-1},t)\) is increasing when\(t<S^e_{k-1}\).

In summary, type-\(\theta _k\) PHs always prefer\(\pi _k\) over\(\pi _i\).

Next, we show that\(A\le B\).

If\(t_{j-1}>S_j^e\),

The inequality follows from Lemma3.

If\(t_{j-1}\le S_j^e\),

1.2A.2 Prove the Concavity of \(f_i(t_i)\)

Prove that

is a concave function.

Proof

For\({f_i^1}''<0\) and\({f_i^2}''<0\),\({f_i^1}'(t_i)\) and\({f_i^2}'(t_i)\) is decreasing. Furthermore,\(f_i^1(S^e_{i+1})=f_i^2(S^e_{i+1})\) and\({f_i^1}'(S^e_{i+1})={f_i^2}'(S^e_{i+1})\). But\({f_i^1}''(S^e_{i+1})\ne {f_i^2}''(S^e_{i+1})\). Thus,\(f_i'(t_i)\) is defined but\(f_i''(t_i)\) is undefined when\(t_i=S^e_{i+1}\).

To show that\(f_i(t)\) is concave, we need to show that\(\forall x_1,x_2\) and\(\forall \lambda \in [0,1]\),

Without loss of generality, we assume that\(x_1\le x_2\). We distinguish four cases:

Case 1:\(x_1<S^e_{i+1}\)and\(x_2<S^e_{i+1}\).

It is intuitive because\(f_i^1\) is concave.

Case 2:\(x_1\ge S^e_{i+1}\)and\(x_2\ge S^e_{i+1}\).

It is also intuitive because\(f_i^2\) is concave.

Case 3:\(x_1< S^e_{i+1}\),\(x_2\ge S^e_{i+1}\) and\( \lambda x_1+(1-\lambda )x_2<S^e_{i+1}\).

We write\(x_0=\lambda x_1+(1-\lambda )x_2\). Then,

According to Mean Value Theorem,\(a\in \left[ x_1,x_0\right] \),\(b\in \left[ x_0,S^e_{i+1}\right] \), and\(c\in \left[ S^e_{i+1},x_2\right] \). Because\({f_i^1}'(t)\) and\({f_i^2}'(t)\) is decreasing,\({f_i^1}'(a)\ge {f_i^1}'(b)\ge {f_i^1}'(S^e_{i+1})={f_i^2}'(S^e_{i+1})\ge {f_i^2}'(c)\). Furthermore, we have\(x_2\ge S^e_{i+1}>x_0\) and\(\lambda \le 1\). Thus,

meaning that\(f_i(t)\)s.t. the condition for concavity in this case.

Case 4:\(x_1< S^e_{i+1}\),\(x_2\ge S^e_{i+1}\) and\( \lambda x_1+(1-\lambda )x_2\ge S^e_{i+1}\).

We note that\(f_i^2(t)\ge f_i^1(t)\) always holds. Thus,

The last line follows because\(f_i^2\) is concave.

Therefore, we prove that\(f_i(t_i)\) is a concave function.

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