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Monotonicity of the stochastic discount factor and expected option ...

Monotonicity of the stochastic discount factor and expected option ...

Using integration by

Using integration by parts:Cov (Y, g (Y )| A) =∫ b= −=a−∫ ba−∫ ba−(y − E (Y | A)) 1 {y∈A} g (y)g (y) dh (y; A)h (y−; A) dg (y) .dF (y)P (A)If g is strictly decreasing then the left hand side must be strictly negative.Part b) The sufficiency of g strictly decreasing follows from Part a.To prove necessity,suppose (A7) holds, but g is not strictly decreasing.Then there must exist points c < esuch that g (c) ≤ g (e), and therefore a point d ∈ (c, e) such that g ′ (d) > 0 (we can rule outg ′ (x) = 0 for all x ∈ (c, e), because this contradicts Cov ( )Y, g (Y ) 1 {Y ∈[c,e]} < 0). Continuityof g ′ implies g ′ (x) > 0 for x in a neighborhood of d, implying Cov ( )Y, g (Y ) 1 {Y ∈[d−ε,d+ε]} > 0for some ε > 0, contradicting (A7).Proof of Proposition 8The proof of (i) ⇐⇒ (ii) follows from Lemma A1 above.The proof of the equivalenceof (ii), (iii) and (iv) is the same as in Proposition 3 (the weak monotonicity case). Thesufficiency part of (ii) follows from Proposition 7 above. The necessity proof of (ii) follows.Suppose m () is not strictly decreasing. Then there must exist points c < e such thatm (c) ≤ m (e).If m () is constant in the interval [c, e], then fix some ¯K and define some Gsatisfying Condition 6 so that G ( s − ¯K ) is positive only for s within some interval withinthe interior of [c, e].Then it is easy to show that R (G (S T − K)) will be constant for Kwithin some neighborhood of ¯K.If m () is not constant in the interval [c, e], there mustexist a point d ∈ (c, e) such that g ′ (d) > 0.Continuity of g ′ implies g ′ (x) > 0 for x in aninterval [d 1 , d 2 ] containing d.Fix some ¯K and construct a G that satisfies Condition 6 sothat G ( s − ¯K ) is positive only for s ∈ [d 1 , d 2 ]. It is easy to show from the sufficiency part ofthe proof of Proposition 7 that R (G (S T − K)) is decreasing in K within some neighborhood34

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