7E/ (y — vr(w))ƒ(y | w)dy | ▲2 Specimenaj Klausuloj

7E/ (y — vr(w))ƒ(y | w)dy | ▲2 k r E/ l, , +7E/ = I — Pr [w c ▲ ] E yƒ(y | w)dy + vƒ(y | w)dy | ▲ / f— Prw c ▲2k, r E/ E/yƒ(y | w)dy +, +7E/ lvr(w)ƒ(y | w)dy | ▲2(10)k f r E/ l , +7E/ — Pr c ▲ E s(y | vr(w), o(w, v, ))ƒ(y | w)dy | ▲ w / k f l , +7E/ — Pr c ▲e E s(y | vr(w), (w, v, ))ƒ(y | w)dy | ▲e .w / fGiven our assumptions about the probability distribution for w, both sides of (10) are continuous in . When = 0, the left-hand side is zero. When = 0, the right-hand side is zero if v = v¯ and strictly positive if v < v¯ (by the definition of v¯ and using Assumption 3). When = 1, o(w, v, ) = (w, v, ) = 1, so the left-hand side of(10) is Pr [w c ▲ ] Ek, l+7E/ (y — v)ƒ(y | w)dy | ▲ /+ Pr w c ▲2 Ek, l+7E/ (y — vr(w))ƒ(y | w)dy | ▲2 .(11) and the right-hand side is/ r E/