Pd/Pf\. Since the covariance operator is linear, equation (15) implies Cov[U,(∏*),S] =
Cov[U,(H*), S — Pd/Pf] = 0. Using the fact that (S — Pd/Pf ) = max(S — Pd∕Pf, 0) —
max(Pd∕Pf — S, 0) and the linearity again results in Cov[U,(∏*), max(S, — Pd/Pf, 0)] =
Cov[U,(∏*), max(Pd∕Pf — S, 0)]. Using the definition of the covariance operator again
yields
Cov [U,(∏:), max(Pd/Pf — S, 0)]
= /P‘/P’ " '(πj<s )) — e[u,(∏ :)]}
× {(Pd/Pf — S) — e[ max(Pd/P, — S, 0)]} dG(S)
+ /’/pι {u'(n:(S)) — e[u'(∏:)]}{ — e[max(Pd/Pf — S,0)]} dG(S)
= /Pd/P' {u '(n:(S )) — e[u'(∏ :)]}(Pd/P, — s ) dG(S )
= /Pd/P' {u '(n:(s )) — e[u'(∏ : )]}{(Pd/Pf — s ) — (Pd/Pf — s #)} <jg(s )
+/Pd/P' {u ' (n:(s)) — e[u'(∏ :)]}(Pd/Pf — s #) dG(s)
< (Pd/Pf — S#) /f {U'(∏J(S)) — e[u'(∏:)]} dG(S)
= (Pd/Pf — S#)1 y' U'(∏((S)) dG(S) [1 — G(Pd/Pf)]
∣pS U'(∏(S)) dG(S) G(Pd/Pf )!
ʧd/4 )
= (Pd/Pf — S#) G(PdJPl) [1 — G(PdJPl)]
×1e[u'(∏()∣ S ≤ Pd/Pf] — e[u'(∏:)| S ≥ Pd/Pf] 1,
where the inequality follows from the fact that {U,(∏i(S)) — E[U,(∏:)]} and (Pd/Pf —
S) — (Pd/Pf — S#) = (S# — S) have opposite signs for all S < Pd/Pf. Since the curly
bracketed term in the last line is negative by assumption and (Pd/Pf — S#) is positive,
Cov[U,(∏:), max(S — Pd/Pf, 0)] is negative for case (b) as well. □
22
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