The name is absent



Examining terms on the right hand side and using the definitions of V(X) and Vt(X), we get

Xccc ∞                                  Xccc

I   V,(c + v + u)fι(u)du g(v)dv = /       [   [Bk 2[c + v + u]]fι(u)du g(v)dv

∞ √—∞                        √—∞ √—∞

Xccc


'∞


Bk 2c 2v 2


Zufι(u)du g(v)dv = /

∞                  - —


4.


Xccc


-∞


[-Bk 2c 2v] g(v)dv


[Bk + 2c]Φ Xc—--\+2^συφ(Xc—^-
σ σv J σ σv J

as well as

∞∞

X Xcc — -∞


∞∞

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 Xcc — -∞

Xc


cc


'∞


Bk 2c 2v 2r 2


Zuf2(u)du g(v)dv = /

∞                     ∙ΛX


∙∞


4.


X.


[Bk 2c 2r 2v] g(v)dv
ccc


[Bk + 2c+2r] 1

*i ⅛ )l

v 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
-∞                               -∞

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.


σι


[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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