Demand for Insurance

Compare two versions of life in one picture: sick and healthy. Insurance is a promise that fattens your share in the bad case.

At a fair premium, covering the bad case completely is best. Both cases end up holding the same šŸ’° money. Add a loading to the premium and the math changes. Covering only part of the loss becomes the smarter deal.

For example, 30 šŸ§‘ classmates each chip in 100 won every month. When someone gets hurt, the pot helps pay the hospital bill. If nobody gets hurt, the 100 won is simply gone. Still, the hurt friend is spared a scary bill.

W1=Wāˆ’pĪ²,W2=Wāˆ’L+(1āˆ’p)Ī²W_1 = W - p\beta, \quad W_2 = W - L + (1-p)\beta
maxā”Ī²Ā (1āˆ’Ļ€)W1+Ļ€W2\max_\beta\ (1-\pi)\sqrt{W_1} + \pi\sqrt{W_2}

Part of these learning paths

Examples & Exercises

A consumer with wealth W=100W = 100, loss L=60L = 60 (probability Ļ€=0.3\pi = 0.3), and u(x)=xu(x) = \sqrt{x} buys Ī²\beta dollars of coverage at premium pp per dollar.

Show via the first-order condition that at the fair premium p=Ļ€p = \pi the optimum is full insurance Ī²āˆ—=60\beta^* = 60.

maxā”Ī²āˆˆ[0,60](1āˆ’Ļ€)Wāˆ’pĪ²+Ļ€Wāˆ’L+(1āˆ’p)Ī²\max_{\beta \in [0,60]} (1-\pi)\sqrt{W - p\beta} + \pi\sqrt{W - L + (1-p)\beta}
dā€‰EUdĪ²=āˆ’(1āˆ’Ļ€)p2W1+Ļ€(1āˆ’p)2W2\frac{d\,EU}{d\beta} = -\frac{(1-\pi)p}{2\sqrt{W_1}} + \frac{\pi(1-p)}{2\sqrt{W_2}}
Zero loading ā€” the fair premium p=0.3p = 0.3.
Show proof

  1. Contingent wealth at coverage Ī²\beta is W1=100āˆ’pĪ²W_1 = 100 - p\beta, W2=40+(1āˆ’p)Ī²W_2 = 40 + (1-p)\beta. Expected utility EU(Ī²)=0.7W1+0.3W2EU(\beta) = 0.7\sqrt{W_1} + 0.3\sqrt{W_2} is concave in Ī²\beta (a concave function of affine functions), so the first-order condition is sufficient.

  2. The first-order condition is 0.7ā€‰p2W1=0.3ā€‰(1āˆ’p)2W2\frac{0.7\,p}{2\sqrt{W_1}} = \frac{0.3\,(1-p)}{2\sqrt{W_2}}. At the fair premium p=Ļ€=0.3p = \pi = 0.3 this becomes 0.21W1=0.21W2\frac{0.21}{\sqrt{W_1}} = \frac{0.21}{\sqrt{W_2}} ā€” i.e. W1=W2W_1 = W_2.

  3. W1=W2W_1 = W_2 means 100āˆ’0.3Ī²=40+0.7Ī²100 - 0.3\beta = 40 + 0.7\beta, i.e. Ī²=60=L\beta = 60 = L ā€” full insurance covering the entire loss. Then W1=W2=100āˆ’18=82W_1 = W_2 = 100 - 18 = 82, on the 45-degree line.

    In the figure the budget line touches the highest indifference curve on the 45-degree line.

In the same setting, suppose a 20% loading is added so p=0.3Ɨ1.2=0.36p = 0.3 \times 1.2 = 0.36.

Find the optimal coverage and verify it is not full insurance (partial insurance).

p=0.36,0.7Ɨ0.36W1=0.3Ɨ0.64W2p = 0.36, \qquad \frac{0.7 \times 0.36}{\sqrt{W_1}} = \frac{0.3 \times 0.64}{\sqrt{W_2}}
W2=40+1āˆ’pp(100āˆ’W1)=40+1.778ā€‰(100āˆ’W1)W_2 = 40 + \frac{1-p}{p}(100 - W_1) = 40 + 1.778\,(100 - W_1)
Loading 0.20.2 ā€” premium rate p=0.36p = 0.36.
Show proof

  1. The first-order condition is 0.252W1=0.192W2\frac{0.252}{\sqrt{W_1}} = \frac{0.192}{\sqrt{W_2}}. Rearranging, W2W1=(0.1920.252)2ā‰ˆ0.5805\frac{W_2}{W_1} = \left(\frac{0.192}{0.252}\right)^2 \approx 0.5805 ā€” the optimum lies below the 45-degree line (less wealth in the loss state).

  2. Intersecting with the budget line W2=40+1.778ā€‰(100āˆ’W1)W_2 = 40 + 1.778\,(100 - W_1) gives W1ā‰ˆ92.35W_1 \approx 92.35, W2ā‰ˆ53.61W_2 \approx 53.61. Optimal coverage is Ī²āˆ—=(100āˆ’92.35)/0.36ā‰ˆ21.26<60\beta^* = (100 - 92.35)/0.36 \approx 21.26 < 60 ā€” partial insurance covering only part of the loss.

  3. With loading, insurance is priced above the fair premium, so even a risk-averse consumer does not fully insure. In the figure the optimum sits on the budget line below the 45-degree line, where an indifference curve is tangent to the budget.

7. Choice under Uncertainty