Examining terms on the right hand side and using the definitions of V(X) and Vt(X), we get
Xc —cc ∞ Xc —cc ∞
I V,(c + v + u)fι(u)du g(v)dv = / [ [—Bk — 2[c + v + u]]fι(u)du g(v)dv
∞ √—∞ √—∞ √—∞
Xc —cc
∞
'∞
—Bk — 2c — 2v — 2
Zufι(u)du g(v)dv = /
∞ - —
4.
Xc —cc
-∞
[-Bk — 2c — 2v] g(v)dv
— [Bk + 2c]Φ Xc—--\+2^συφ(Xc—^-
σ σv J σ σv J
as well as
∞∞
X Xc—c — -∞
∞∞
V,(c + r + v + u)f2(u)du g(v)dv = / I [—Bk — 2[c + r + v + u]]f2(u)du g(v)dv
X Xc—c — -∞
∞
Xc
—cc
'∞
—Bk — 2c — 2v — 2r — 2
Zuf2(u)du g(v)dv = /
∞ ∙ΛX
∙∞
4.
X.
[—Bk — 2c — 2r — 2v] g(v)dv
c —cc
[Bk + 2c+2r] 1
—*i ⅛ )l
— 2σv φ(JXσV(1 )
and finally
V V (Xc + r + u)f2(u)du — f
— -∞ — —∞
V (Xc + u)fι(u)du
∖ ∖α — Bk[Xc + r + u] — [Xc + r + u]2] f2(u)du — f ∖α — Bk[Xc + u] — [Xc + u]2] fι(u)du
-∞ -∞
[α — BkXc
z∞ fi∞) ∞∞
f2(u)du —[Bk + 2Xc] / [r + u]f2(u)du — / [r + u]2f2(u)du
∞ J—∞ J—∞
4.
4.
v
σ22 +r2
— [a — BkXc
4.
∞∞
fι(u)du +[Bk + 2Xc] / ufι(u)du + /
∞ —-∞ ——
u2fι(u)du
4.
4.
σ≡ι
—[Bk + 2Xc ]r —
r2
— σ;
.2
U2
+ σ'.
.2
U1
Putting things together, the first-order condition becomes
k—
δ—φ f —c ^ ∖Bkr + 2Xcr + r2 + σ;
συ ∖ συ J
U2
— σ(
U1
] — δ[Bk + 2c]Φ χx- c
+2σ.δφ(XcσvC )
δ[Bk + 2c + 2r] 1
— 2avδφ(Xσ^ )
k—
δ—φ f —c---ʌ ∖Bkr + 2Xcr + r2 + σ,
σ21 ] — δ[Bk + 2c] — 2δr 1
συ \ συ y
30
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