Impact of a financial risk-sharing scheme on budget-impact estimations: a game-theoretic approach

Abstract

Background

As part of the process of updating the National List of Health Services in Israel, health plans (the ‘payers’) and manufacturers each provide estimates on the expected number of patients that will utilize a new drug. Currently, payers face major financial consequences when actual utilization is higher than the allocated budget. We suggest a risk-sharing model between the two stakeholders; if the actual number of patients exceeds the manufacturer’s prediction, the manufacturer will reimburse the payers by a rebate rate of α from the deficit. In case of under-utilization, payers will refund the government at a rate of γ from the surplus budget. Our study objective was to identify the optimal early estimations of both ‘players’ prior to and after implementation of the risk-sharing scheme.

Methods

Using a game-theoretic approach, in which both players’ statements are considered simultaneously, we examined the impact of risk-sharing within a given range of rebate proportions, on players’ early budget estimations.

Results

When increasing manufacturer’s rebate α to be over 50 %, then manufacturers will announce a larger number, and health plans will announce a lower number of patients than they would without risk sharing, thus substantially decreasing the gap between their estimates. Increasing γ changes players’ estimates only slightly.

Conclusion

In reaction to applying a substantial risk-sharing rebate α on the manufacturer, both players are expected to adjust their budget estimates toward an optimal equilibrium. Increasing α is a better vehicle for reaching the desired equilibrium rather than increasing γ, as the manufacturer’s rebate α substantially influences both players, whereas γ has little effect on the players behavior.

Appendix

We first show that if α is not too small then in equilibrium the manufacturer announces a number of patients that is strictly between the extreme possible values of the distribution.

Proposition 3

Assume that $\alpha \ge \left( {1 - \frac{c}{p}} \right)\frac{E\left( n \right)}{{E\left( n \right) - k_{1} }}$ then k ₁ < x _eq < k ₂.

Proof

Observe that under the assumption

$$\frac{\text{d}}{{{\text{d}}x}}U_{M} \left( {x,y} \right)|_{{x = k_{1} }} = \delta G^{{\prime }} \left( z \right)\left( {\left( {p\left( {1 - \alpha } \right) - c} \right)E\left( n \right) + \alpha pk_{1} } \right) + \alpha pG\left( z \right) > 0$$

and

$$\frac{\text{d}}{{{\text{d}}x}}U_{M} \left( {x,y} \right)|_{{x = k_{2} }} = \delta G'\left( z \right)\left( {p - c} \right)E\left( n \right) < 0.$$

We actually proved a stronger result.

We have shown here that x = k ₁ and x = k ₂ are dominated strategies for M.

Moreover, when k ₁ = 0, the assumption becomes $\alpha \ge \left( {1 - \frac{c}{p}} \right)$.

Under the same assumption on α, we can prove a stronger result by which not only the extreme value k ₁ will not be used in equilibrium but rather an interval of values [k ₁ , x ₀] for some $x_{0} \in \left[ {k_{1} ,k_{2} } \right]$ can be ruled out since each x in this interval is dominated by each $x \in (x_{0} ,k_{2} ]$.□

Thus, in equilibrium, given that α is not too small, the manufacturer announces a larger number of patients than he would with no such mechanism, i.e., $x_{\text{eq}} > k_{1}.$

Proposition 4

Assume that $\alpha \ge \left( {1 - \frac{c}{p}} \right)\frac{E\left( n \right)}{{E\left( n \right) - k_{1} }}$ and let $x_{0}$ be the unique solution in the interval $\left[ {k_{1} ,k_{2} } \right]$ of the equation

$$\left( {p - c} \right)E\left( n \right) - \alpha p\int\limits_{x}^{{k_{2} }} {nf\left( n \right){\text{d}}n} + \alpha px\left( {1 - F\left( x \right)} \right) = 0$$

(1)

Then any x<$x_{0}$ is dominated by any x > $x_{0}$. Moreover, $x_{0}$ is increasing with α.

Proof

Observe first that for x = k ₁,

$$\left( {p\left( {1 - \alpha } \right) - c} \right)E\left( n \right) + \alpha pk_{1} \le 0$$

and for x = k ₂,

$$\left( {p - c} \right)E\left( n \right) > 0$$

Moreover, $\frac{\text{d}}{{{\text{d}}x}}\left( {\left( {p - c} \right)E\left( n \right) - \alpha p\int\limits_{x}^{{k_{2} }} {nf\left( n \right){\text{d}}n} + \alpha px\left( {1 - F\left( x \right)} \right)} \right) = \alpha p\left( {1 - F\left( x \right)} \right) > 0$, and therefore there exists a unique $x_{0} \in \left[ {k_{1} ,k_{2} } \right]$ that solves Eq. (1). Now, for each x < x ₀,

$$U_{M} \left( {x,y} \right) = G\left( z \right)\left( {\left( {p - c} \right)E\left( n \right) - \alpha p\int\limits_{x}^{{k_{2} }} {nf\left( n \right){\text{d}}n} + \alpha px\left( {1 - F\left( x \right)} \right)} \right) \le 0$$

for any y. Thus, M is better off by announcing a sufficiently large x such that $U_{M} \left( {x,y} \right) \ge 0$. Since any x > x ₀ guarantees a positive payoff the result obtains.

Finally, differentiating Eq. (1) with respect to α, we get

$$\frac{{{\text{d}}x_{0} }}{{{\text{d}}\alpha }} = \frac{{\int_{x}^{{k_{2} }} {nf\left( n \right){\text{d}}n - x\left( {1 - F\left( x \right)} \right)} }}{{\alpha \left( {1 - F\left( x \right)} \right)}}$$

and since at x = x ₀, we have $\int\nolimits_{x}^{{k_{2} }} {nf\left( n \right){\text{d}}n - x\left( {1 - F\left( x \right)} \right)} = \frac{{\left( {p - c} \right)E\left( n \right)}}{\alpha p} > 0$ we know that $\frac{{{\text{d}}x_{0} }}{{{\text{d}}\alpha }} > 0$.□

Moreover, a direct consequence of the last proposition is that:

Corollary 5

In equilibrium $U_{M} \ge 0$.

Thus, M can always ensure a positive payoff in equilibrium.

We now turn to analyze the HCP’s behavior in equilibrium given the risk-sharing mechanism: We first show that if δ is sufficiently small (i.e., the weight given to the manufacturer’s announcement) then in equilibrium the HCP announces y strictly larger than the extreme value $k_{1}$.

Proposition 6

If δ is sufficiently smaller than 1 such that $\delta k_{2} + \left( {1 - \delta } \right)k_{1} < E\left( n \right)$ then $k_{1} < y_{eq}$.

Proof

Substituting y = k ₁ in the derivative of $U_{\text{HCP}} \left( {x,y} \right)$ gives

$$\begin{aligned} \frac{\text{d}}{{{\text{d}}y}}U_{\text{HCP}} \left( {x,y} \right)|_{{y = k_{1} }} & = \left( {1 - \delta } \right)G^{{\prime }} \left( z \right)\left( {pz - pE\left( n \right) + \alpha p\int\limits_{x}^{{k_{2} }} {nf\left( n \right){\text{d}}n} - apx\left( {1 - F\left( x \right)} \right)} \right) \\ & \quad + \left( {1 - \delta } \right)pG\left( z \right). \\ \end{aligned}$$

Since the second term is positive. it is sufficient to show that the first term is also positive. Since $G^{{\prime }} \left( z \right) < 0$ it is sufficient to show that $p\left( {z - E\left( n \right) + \alpha \int\limits_{x}^{{k_{2} }} {nf\left( n \right){\text{d}}n} - ax\left( {1 - F\left( x \right)} \right)} \right)|_{{y = k_{1} }} < 0$.

Note that the only term affected by y is z. Therefore, we define the function

$$\begin{aligned} h\left( x \right) & = \left.\left( {z - E\left( n \right) + \alpha \int\limits_{x}^{{k_{2} }} {nf\left( n \right){\text{d}}n} - ax\left( {1 - F\left( x \right)} \right)} \right)\right|_{{y = k_{1} }} \\ & = \left( {\delta x + \left( {1 - \delta } \right)k_{1} } \right) - E\left( n \right) + \alpha \int\limits_{x}^{{k_{2} }} {nf\left( n \right){\text{d}}n} - ax\left( {1 - F\left( x \right)} \right) \\ \end{aligned}$$

And it is sufficient to show that $h\left( x \right) < 0$ for all $k_{1} \le x \le k_{2}$.

Differentiating h(x) with respect to x, we get

$$h^{\prime } \left( x \right) = \delta - \alpha xf\left( x \right) - \alpha \left( {1 - F\left( x \right)} \right) + \alpha xf\left( x \right) = \delta - \alpha \left( {1 - F\left( x \right)} \right)$$

and

$$h^{{\prime \prime }} \left( x \right) = \alpha xf\left( x \right) > 0$$

Therefore, h(x) is convex and attends its maximum at either x = k ₁ or x = k ₂.

Thus, it is sufficient to show that h(x) < 0 for x = k ₁ and x = k ₂. Now

$$h\left( {k_{1} } \right) = - \left( {1 - \alpha } \right)\left( {E\left( n \right) - k_{1} } \right) < 0$$

and

$$h\left( {k_{2} } \right) = \delta k_{2} + \left( {1 - \delta } \right)k_{1} - E\left( n \right)$$

If δ is not too large such that $\delta k_{2} + \left( {1 - \delta } \right)k_{1} < E\left( n \right)$ then $h\left( {k_{2} } \right) < 0$.□

Since the HCP’s payoff depends on M’s announcement x through z, a natural question of interest is whether the manufacturer can inflict a negative payoff on the HCP by announcing a smaller x (given that the drug is approved). We therefore rewrite the HCP’s expected payoff given that the drug is approved:

$$\begin{aligned} U_{\text{HCP}} \left( {x,y} \right) & = p\left( {\delta x - \delta E\left( n \right) + \alpha \int\limits_{x}^{{k_{2} }} {nf\left( n \right){\text{d}}n} - \alpha x\left( {1 - F\left( x \right)} \right)} \right) \\ & \quad + p\left( {\left( {1 - \delta } \right)y - \left( {1 - \delta } \right)E\left( n \right) - \gamma yF\left( y \right) + \gamma \int\limits_{{k_{1} }}^{y} {nf\left( n \right){\text{d}}n} } \right) \\ \end{aligned}$$

where the first component is the one affected by x. Let

$$T = \delta x - \delta E\left( n \right) + \alpha \int\limits_{x}^{{k_{2} }} {nf\left( n \right){\text{d}}n} - \alpha x\left( {1 - F\left( x \right)} \right)$$

Then we have the following:

Proposition 7

If

$$\alpha \ge \delta \frac{E\left( n \right)}{{\int_{{x_{\hbox{min} } }}^{{k_{2} }} {nf\left( n \right){\text{d}}n} }}$$

(2)

where $x_{\hbox{min} } = F^{ - 1} \left( {1 - \frac{\delta }{\alpha }} \right)$ then T(x) ≥ 0 for every $x \in \left[ {k_{1} ,k_{2} } \right]$.

Proof

Differentiating z w.r.t. x shows that the minimum of T is obtained for x _min [note that condition (2) implies that α > δ]. Substituting x _min in T gives

$$T\left( {x_{\hbox{min} } } \right) = \alpha \int\limits_{{x_{\hbox{min} } }}^{{k_{2} }} {nf\left( n \right){\text{d}}n} - \delta E\left( n \right),$$

which is positive if condition (2) is satisfied.□

Following Proposition 7 we have the following result.

Proposition 8

Assume that $\left( {1 - \delta } \right) < \gamma$ and

$$\gamma \int\limits_{{k_{1} }}^{{y_{\hbox{max} } }} {nf\left( n \right){\text{d}}n} + \alpha \int\limits_{{x_{\hbox{min} } }}^{{k_{2} }} {nf\left( n \right){\text{d}}n} \ge E\left( n \right)$$

(3)

where $y_{\hbox{max} } = F^{ - 1} \left( {\frac{1 - \delta }{\gamma }} \right)$ then in equilibrium the HCP’s expected payoff is always non-negative, i.e.: $U_{\text{HCP}} \left( {x,y} \right) \ge 0$.

Proof

The probability term G(z) in $U_{\text{HCP}}$ is always positive. We have

$$U_{\text{HCP}} \left( {x,y} \right) = pT + p\left( {\left( {1 - \delta } \right)y - \left( {1 - \delta } \right)E\left( n \right) - \gamma yF\left( y \right) + \gamma \int\limits_{{k_{1} }}^{y} {nf\left( n \right){\text{d}}n} } \right)$$

Note that the second term is maximized at y = y _max. Substituting x = x _min and y = y _max we have

$$\min_{x,y} U_{\text{HCP}} \left( {x,y} \right) = U_{\text{HCP}} \left( {x_{\hbox{min} } ,y_{\hbox{max} } } \right) = \gamma \int\limits_{{k_{1} }}^{{y_{\hbox{max} } }} {nf\left( n \right){\text{d}}n} + \alpha \int\limits_{{x_{\hbox{min} } }}^{{k_{2} }} {nf\left( n \right){\text{d}}n} - E\left( n \right) \ge 0.$$

□