Figure 2: Expected utility as a function of the severity of the illness
Δ = H⅜)>'(y — t) У δ(u)dF(y, €)
— [1+ t'(p)]u'(У — t — P) 2" δ(υ)dF(y,€) (15)
Moreover, (13) can in this case be written as
ʃ δ(υ)dF(y,€) = ʃ δ(υ)dF(y,€) < 0
(16)
since δ(v) > 0 for € > L. The first integral in (15) is in other words negative.
The second integral in (15) is positive, since δ(v~) > 0 for € > L. Since the
terms [—t'(p)] and [1 + t'(p)] are both positive, it follows that Δ < 0 in the
present case. From our previous discussion we thus have the the following
proposition:
10
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