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The Birth of Insurance Contracts - The Ataturk Institute for Modern ...

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Show that dc2(y)<br />

<br />

<br />

d(k2−k) = −<br />

ˆ H (k2 −k)<br />

<br />

<br />

ˆ <br />

<br />

H<br />

<br />

<br />

<br />

∈ (0, 1) evaluated at ˆ λ and (ĉ2(y), ĉ2(x)).<br />

<br />

<br />

ˆ <br />

<br />

H (k2−k) = (1 − py) U ′ 2[c2(x)] 2 py U ′ 1[w(y) − c2(y)] R1[w(x) − c2(x)] −<br />

<br />

(1 − py) U ′ 2[c2(x)] 2 py U ′ 1[w(y) − c2(y)] R1[w(y) − c2(y)] < 0 (19)<br />

because U ′ (.) > 0, w(y) − c2(y) < w(x) − c2(x) < w(x) − c2(x), 18 and utility functions exhibit<br />

DARA, so that R1[w(y) − c2(y)] > R1[w(x) − c2(x)].<br />

<br />

<br />

Comparing (19) with (18) , it follows that 0 < ˆ <br />

<br />

H<br />

+ ˆ <br />

<br />

H (k2−k) . QED<br />

Show that dc2(y)<br />

dy<br />

<br />

<br />

= − <br />

<br />

ˆ <br />

<br />

H(y) <br />

ˆ <br />

<br />

H<br />

∈ (0, 1) evaluated at ˆ λ and (ĉ2(y), ĉ2(x)).<br />

<br />

<br />

ˆ <br />

<br />

H(y) = (1 − py) U ′ 2[c2(x)] 2 py U ′′<br />

1 [w(y) − c2(y)] =<br />

= − (1 − py) U ′ 2[c2(x)] 2 py U ′ 1[w(y) − c2(y)] R1[w(y) − c2(y)] < 0 (20)<br />

<br />

<br />

Comparing (20) with (18), it follows that 0 < ˆ <br />

<br />

H<br />

+ ˆ <br />

<br />

H (k2−k) . QED<br />

Show that dc2(y)<br />

dpy<br />

<br />

<br />

= − <br />

<br />

ˆ <br />

<br />

Hpy <br />

ˆ <br />

<br />

H<br />

> 0 evaluated at ˆ λ and (ĉ2(y), ĉ2(x)).<br />

<br />

<br />

ˆ <br />

<br />

Hpy = [U2[c2(y)] − U2[c2(x)]] py U ′ 2 [c2(y)] U ′ 1 [w(x) − c2(x)] [R1[w(x) − c2(x)] + R2[c2(x)]]<br />

− (1 − py) U ′ 2 [c2(x)] py U ′ 2 [c2(y)] U ′ 1 [w(x)−c2(x)]<br />

1−py<br />

because Ri(.) > 0, Ui ′(.) > 0, and c2(y) ≤ c2(x), so that [U2[c2(y)] − U2[c2(x)]] < 0. QED<br />

Show that dc2(y)<br />

dx<br />

<br />

<br />

= − <br />

<br />

ˆ <br />

<br />

H(x) <br />

ˆ <br />

<br />

H<br />

< 0 and dc2(y)<br />

d¯x<br />

<br />

<br />

= − <br />

<br />

ˆ <br />

<br />

H(¯x) <br />

ˆ <br />

<br />

H<br />

< 0<br />

< 0 evaluated at ˆ λ and (ĉ2(y), ĉ2(x)).<br />

<br />

<br />

ˆ <br />

<br />

H(x) = − (1 − py) U ′ 2 [c2(x)] py U ′ 2 [c2(y)] px U ′′<br />

1 [w(x) − c2(x)] > 0.<br />

<br />

<br />

H(¯x) = − (1 − py) U ′ 2 [c2(x)] py U ′ 2 [c2(y)] p¯x U ′′<br />

1 [w(¯x) − c2(x)] > 0.<br />

<strong>The</strong>se follow from U ′ (.) > 0 and U ′′ (.) < 0. QED<br />

Show that dc2(y)<br />

d p¯x<br />

<br />

<br />

<br />

<br />

= −<br />

px+p¯x<br />

ˆ <br />

<br />

H p¯x <br />

px+p¯x <br />

<br />

<br />

ˆ <br />

< 0 evaluated at<br />

H<br />

ˆ λ and (ĉ2(y), ĉ2(x)).<br />

<br />

<br />

<br />

ˆ H p¯x<br />

<br />

<br />

<br />

px+p¯x<br />

= (1 − py) U ′ 2 [c2(x)] py U ′ 2 [c2(y)] [U ′ 1 [w(x) − c2(x)] − U ′ 1 [w(¯x) − c2(x)]] > 0<br />

18 <strong>The</strong> optimal sharing rule gives some insurance to both parties, <strong>for</strong> both are risk-averse. Appendix B proves this<br />

result under full in<strong>for</strong>mation. <strong>The</strong> pro<strong>of</strong> under hidden in<strong>for</strong>mation runs similarly.<br />

27

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