A simple insurance model: optimal coverage and deductible

Abstract

An insurance model, with realistic assumptions about coverage, deductible and premium, is studied. Insurance is shown to decrease the variance of the cost to the insured, but increase the expected cost, a tradeoff that places our model in the Markowitz mean-variance model.

Appendix:  Proof of Theorem 1

We prove (42)

$$\frac{\partial}{\partial C} \operatorname{Var}L(\mathbf{X}|C,D)<0 $$

given that (22) holds,

$$\int_D^C f(x) \,dx < 1, $$

which implies that either

$$\begin{aligned} \int_C^\infty f(x) \,dx > 0, \end{aligned}$$

(50)

or

$$\begin{aligned} \int_0^D f(x) \,dx > 0, \end{aligned}$$

(51)

hold.

From (39),

$$ \operatorname{E}\phi(\mathbf{X}|C,D)=\int_D^C (D-x)f(x) \,dx+(D-C) \int_C^\infty f(x) \,dx, $$

(52)

and

$$\begin{aligned} \operatorname{Var}\phi(\mathbf{X}|C,D) &=\operatorname{E}\phi ^2( \mathbf{X}|C,D)-\bigl(\operatorname{E}\phi(\mathbf{X}|C,D)\bigr)^2 \\ &=\int_D^C (D-x)^2f(x) \,dx+(D-C)^2 \int_C^\infty f(x) \,dx \\ &\quad - \biggl(\int_D^C (D-x)f(x) \,dx+(D-C) \int _C^\infty f(x) \,dx \biggr)^2. \end{aligned}$$

(53)

By (41) the partial derivative of $\operatorname {Var}L(\mathbf{X})$ w.r.t. C is

$$ \frac{\partial}{\partial C} \operatorname{Var}L(\mathbf{X})=\frac {\partial}{\partial C} \operatorname{Var}\mathbf{X}+ \frac{\partial }{\partial C}\operatorname{Var}\phi( \mathbf{X})+2 \frac{\partial }{\partial C} \bigl(\operatorname{E}\bigl[\mathbf{X}\phi( \mathbf {X})\bigr]-\operatorname{E}\mathbf{X}\operatorname{E}\phi(\mathbf {X}) \bigr). $$

(54)

We calculate next the three derivatives on the right side.

• (a) Clearly,

  $$ \frac{\partial}{\partial C}\operatorname{Var}\mathbf{X}=0. $$

  (55)
• (b)
  $$\begin{aligned} \frac{\partial}{\partial C}\operatorname{Var}\phi(\mathbf {X})&=(2C-2D) \int _C^\infty f(x) \,dx+2 \int_D^C (D-x)f(x) \,dx \int_C^\infty f(x) \,dx \\ &\quad -2(C-D) \biggl(\int_C^\infty f(x) \,dx \biggr)^2 \\ &=2 \int_C^\infty f(x) \,dx \biggl[C \int _0^C f(x) \,dx \\ &\quad -D \int_0^D f(x) \,dx-\int_D^C xf(x) \,dx \biggr] \end{aligned}$$

  (56)

• (c) From (39),

  $$\begin{aligned} \operatorname{E} \bigl(\mathbf{X}\phi(\mathbf{X})\bigr) - \operatorname {E} \mathbf{X}\operatorname{E} \phi(\mathbf{X}) &= D \biggl[\int_D^\infty x f(x) \,dx - (\operatorname{E} \mathbf{X}) \int_D^\infty f(x) \,dx \biggr] \\ &\quad -C \biggl[\int_C^\infty x f(x) \,dx - ( \operatorname{E} \mathbf {X}) \int_C^\infty f(x) \,dx \biggr] \\ &\quad - \biggl[\int_D^C x^2 f(x)\,dx-( \operatorname{E} \mathbf{X}) \int_D^Cx f(x) \,dx \biggr]. \end{aligned}$$

Therefore

$$ 2 \frac{\partial}{\partial C} \bigl(\operatorname{E}\bigl[\mathbf{X}\phi (\mathbf{X}) \bigr]-\operatorname{E}\mathbf{X}\operatorname{E}\phi (\mathbf{X}) \bigr)=2 \biggl[\operatorname{E}\mathbf{X}\int_C^\infty f(x) \,dx-\int_C^\infty xf(x) \,dx \biggr]. $$

(57)

Substituting (55), (56), and (57) in (54) we get

$$\begin{aligned} \frac{1}{2} \frac{\partial}{\partial C}\operatorname{Var} L(\mathbf {X}) &=\int _C^\infty f(x) \,dx \biggl[C \int _0^C f(x) \,dx-D \int_0^D f(x) \,dx-\int_D^C xf(x) \, dx+\operatorname{E} \mathbf{X} \biggr] \\ &\quad -\int_C^\infty xf(x) \,dx \\ &=\int_C^\infty f(x) \,dx \biggl[C \int _0^C f(x) \,dx-D \int_0^D f(x) \,dx \\ &\quad +\int_0^D xf(x) \,dx +\int _C^\infty xf(x) \,dx \biggr] -\int _C^\infty xf(x) \,dx \end{aligned}$$

(58)

Rearrange (58) to get

$$\begin{aligned} \frac{1}{2} \frac{\partial}{\partial C}\operatorname{Var L(\mathbf {X})} &=C \int _C^\infty f(x) \,dx \int_0^C f(x) \,dx+\int_C^\infty f(x) \,dx \int _C^\infty xf(x) \,dx \end{aligned}$$

(59)

$$\begin{aligned} &\quad -\int_C^\infty xf(x) \,dx +\int_C^\infty f(x) \,dx \biggl(\int _0^D xf(x) \,dx-D \int_0^D f(x) \,dx \biggr). \end{aligned}$$

(60)

The right side of (59) can be written as

$$\begin{aligned} &=C \int_C^\infty f(x) \,dx \int _0^C f(x) \,dx+\int_C^\infty xf(x) \, dx \biggl(\int_C^\infty f(x) \,dx -1 \biggr) \\ &=C \int_C^\infty f(x) \,dx \int _0^C f(x) \,dx-\int_C^\infty xf(x) \, dx \int_0^C f(x) \,dx \\ &=\int_0^C f(x) \,dx \biggl(C \int _C^\infty f(x) \,dx-\int_C^\infty xf(x) \,dx \biggr) \\ &< 0\quad \text{if (50) holds, because then}\ \int _C^\infty xf(x) \,dx > C \int_C^\infty f(x) \,dx. \end{aligned}$$

Similarly, (60) is negative if (51) holds, because then

$$D \int_0^D f(x) \,dx > \int _0^D xf(x) \,dx. $$

This completes the proof of (42) if any of the conditions (50)–(51) holds.

The inequality (43) is similarly proved.