Step and Save: A Wearable Technology Based Incentive Mechanism for Health Insurance

Abstract

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.

A Appendices

1.1 A.1 Supplementary Proof for Lemma 5

Here we provide supplementary proof for Lemma 5. As $i\le k-1$, $t_{k-1}\ge t_i$.

To prove that

$$\begin{aligned} {} & {} \quad u_0\left[ \theta _{k},\max (t_{k-1},S^e_k)\right] +u_0\left[ \theta _{k-1},\max \left( t_{i},S_{k-1}^e\right) \right] \nonumber \\ {} & {} \quad -u_0(\theta _{k-1},t_{k-1})\nonumber \\ {} & {} \ge u_0\left[ \theta _{k},\max \left( t_{i},S_{k}^e\right) \right] , \nonumber \end{aligned}$$

we distinguish five cases:

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

$$\begin{aligned} &u_0\left( \theta _{k},t_{k-1}\right) +u_0\left( \theta _{k-1},t_{i}\right) -u_0(\theta _{k-1},t_{k-1}) &\ge u_0\left( \theta _k,t_i\right) .\nonumber \end{aligned}$$

The inequality follows from Lemma 3.

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

$$\begin{aligned} {} & {} \quad u_0\left( \theta _{k},t_{k-1}\right) +u_0\left( \theta _{k-1},t_{i}\right) -u_0(\theta _{k-1},t_{k-1})\nonumber \\ {} & {} \ge u_0\left( \theta _{k},t_{k-1}\right) +u_0\left( \theta _{k-1},S_{k}^e\right) -u_0(\theta _{k-1},t_{k-1})\nonumber \\ {} & {} \ge u_0\left( \theta _k,S_{k}^e\right) .\nonumber \end{aligned}$$

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 $t_{k-1}< S^e_{k}$.

$$\begin{aligned} u_0\left( \theta _{k},S^e_k\right) +u_0\left( \theta _{k-1},t_i\right) -u_0(\theta _{k-1},t_{k-1}) \ge u_0\left( \theta _k,S^e_k\right) .\nonumber \end{aligned}$$

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 $t_{k-1}\ge S^e_{k}$.

$$\begin{aligned} {} & {} \quad u_0\left( \theta _{k},t_{k-1}\right) +u_0\left( \theta _{k-1},S^e_{k-1}\right) -u_0(\theta _{k-1},t_{k-1})\nonumber \\ {} & {} \ge u_0\left( \theta _{k},t_{k-1}\right) +u_0\left( \theta _{k-1},S^e_{k}\right) -u_0(\theta _{k-1},t_{k-1})\nonumber \\ {} & {} \ge u_0\left( \theta _k,S^e_{k}\right) .\nonumber \end{aligned}$$

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 $t_{k-1}< S^e_{k}$.

$$\begin{aligned} u_0\left( \theta _{k},S^e_{k}\right) +u_0\left( \theta _{k-1},S^e_{k-1}\right) -u_0(\theta _{k-1},t_{k-1}) \ge u_0\left( \theta _k,S^e_{k}\right) .\nonumber \end{aligned}$$

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 $t_{j-1}>S_j^e$,

$$\begin{aligned} {} & {} A = \frac{1}{I}\left[ u_0(\theta _{j},t_{j-1})-u_0(\theta _{j},t_{j})\right] \nonumber \\ {} & {} \le \frac{1}{I}\left[ u_0(\theta _{j-1},t_{j-1})-u_0(\theta _{j-1},t_{j})\right] =B. \nonumber \end{aligned}$$

The inequality follows from Lemma 3.

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

$$\begin{aligned} {} & {} B = \frac{1}{I}\left[ u_0(\theta _{j-1},t_{j-1})-u_0(\theta _{j-1},t_{j})\right] \nonumber \\ {} & {} \quad \ge \frac{1}{I}\left[ u_0(\theta _{j-1},S_j^e)-u_0(\theta _{j-1},t_{j})\right] \ge A. \nonumber \end{aligned}$$

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

Prove that

$$\begin{aligned} f_i(t_i) = \left\{ \begin{array}{ll} f_i^1(t_i) &{} \text {if }t_i<S^e_{i+1},\\ f_i^2(t_i) &{} \text {otherwise,} \end{array} \right. \end{aligned}$$

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 $\forall \lambda \in [0,1]$,

$$\lambda f_i\left( x_1\right) +(1-\lambda ) f_i\left( x_2\right) \le f_i\left( \lambda x_1+(1-\lambda )x_2 \right) .$$

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,

$$x_1-x_0=(\lambda -1)(x_2-x_1), x_2-x_0=\lambda (x_2-x_1).$$

$$\begin{aligned} {} & {} \lambda f_i\left( x_1\right) +(1-\lambda ) f_i\left( x_2\right) -f_i\left( x_0 \right) \nonumber \\ = & {} \lambda \left[ f_i^1\left( x_1\right) - f_i^1\left( x_0\right) \right] +(\lambda -1) \left[ f_i^1\left( x_0\right) -f_i^1(S^e_{i+1})\right] \nonumber \\ {} & {} +(\lambda -1) \left[ f_i^2(S^e_{i+1})-f_i^2\left( x_2\right) \right] \nonumber \\ = & {} \lambda {f_i^1}'(a)(x_1-x_0) +(\lambda -1) {f_i^1}'(b)(x_0-S^e_{i+1}) \nonumber \\ {} & {} +(\lambda -1){f_i^2}'(c)(S^e_{i+1}-x_2)\nonumber \\ = & {} (\lambda -1)\left[ {f_i^1}'(a)-{f_i^2}'(c) \right] (x_2-S_{i+1}^e)\nonumber \\ {} & {} +(\lambda -1)\left[ {f_i^1}'(a)-{f_i^1}'(b) \right] (S_{i+1}^e-x_0).\nonumber \end{aligned}$$

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,

$$\lambda f_i\left( x_1\right) +(1-\lambda ) f_i\left( x_2\right) -f_i\left( x_0 \right) \le 0,$$

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,

$$\begin{aligned} {} & {} \quad \lambda f_i\left( x_1\right) +(1-\lambda ) f_i\left( x_2\right) -f_i\left( x_0 \right) \nonumber \\ {} & {} =\lambda f_i^1\left( x_1\right) +(1-\lambda ) f_i^2\left( x_2\right) -f_i^2\left( x_0 \right) \nonumber \\ {} & {} \le \lambda f_i^2\left( x_1\right) +(1-\lambda ) f_i^2\left( x_2\right) -f_i^2\left( x_0 \right) \nonumber \\ {} & {} \le 0.\nonumber \end{aligned}$$

The last line follows because $f_i^2$ is concave.

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