42
Armando F. Mendoza-P´erez
Observe that (32)
τ1 (x, a) = ψ1 (x, a) + C(x, a) − ρ∗ ,
and (33)
τ2 (x, a) = ψ2 (x, a) + Λ(x, a) − σ∗2
for all (x, a) ∈ K. To prove the theorem we have to verify that
! u " 1 lim Exf∗ χn √ = exp{− σ∗2 u2 }. n→∞ n 2
(34)
To this end, first notice that ψl (xm , am ) for l = 1, 2, is the conditional expectation of hl (xm+1 ) − hl (xm ) given xm , am , that is, ψl (xm , am ) = Exf∗ [hl (xm+1 ) − hl (xm )|xm , am ]. This yields for l = 1, 2, with χm := χm (u) and ψl := ψl (xm , am ), the equations (35)
0=
iuExf∗
# n−1 $
m=0
and (36)
χm ψ1 −
n−1 $
m=0
% "
!
χm h1 (xm+1 ) − h1 (xm )
#
%
! " n−1 $ $ u2 f∗ n−1 Ex 0= χm h2 (xm+1 ) − h2 (xm ) − χm ψ2 . 2 m=0 m=0 !
"
To simplify the notation, let C := C(xm , am ), e1 := e1 u(C − ρ∗ ) !
"
and e2 := e2 u(C − ρ∗ ) . Moreover, notice that (37)
&
From (30), (31) and (37) we have Exf∗ χn − 1 = Exf∗ = Exf∗
(38)
n−1 $
and n−1 $
m=0
!
(χm+1 − χm )
m=0 n−1 $& m=0
−iuExf∗
'
χm+1 − χm = exp{iu(C − ρ∗ )} − 1 χm .
' 1 iu(C − ρ∗ ) − u2 (C − ρ∗ )2 + e2 χm , 2 "
χm h1 (xm+1 ) − h1 (xm ) =