da*
Since c > 0, dhj has the sign of the sum between square brackets in (14). Using the
expression for σ in (4) and simplifying we find that
∂σ(hj ,hk,hε) /
dhj 1 1
σ (hj,hk,hε^) у
(mk
2
mj)
σ2 (hj, hk, hε)
ʌ__L_
hj + hε
hk (2hj + hε) (hk + hε) 1 —
σ(mmk-m⅛y) - 2hj [hj (hj + hε) + hk (hk + hε)]
2hj (hj + hε) [hj (hj + hε) + hk (hk + hε)]
(16)
da* 1 1 . 1 /- „ч m1 . η η „ ,
dhj has the same sign as the numerator in (16). The partial derivative of the numerator
in (16) with respect to hj is
— 2hk (hk + hε)
(mk
2
mj)
σ2 (hj, hk, hε)
(mk — mj)2 ∂σ2
+ hk (2hj + hε) (hk + hε) 2 z, T rʒ TTT-
σ2 (hj, hk, hε) ∂hj
<0
— 2hj (hj + hε) — 2hj (2hj + hε),
which is clearly negative. Since limhj→0 dhj = ∞ and limhj→∞ dhj = 0, it follows from
da*
this that given the other parameters there exists a unique hj > 0 such that dj < 0 if and
only if hj = hj.
Finally,
da*
dhk
exp
( (mk-mj )
∖ 2σ2(hj ,hk,,
)2
,hε)
√2πσ (hj,hk,he)
'----------------------v------
>0
Wjh + /
c'' ⅛ ) V
-------zC_
∂σ(hj ,hk ,hε)
∂hk
σ (hj, hk, hε)
SZ
>0
)(■
S
(mk
2
mj)
σ2 (hj, hk, hε)
. (17)
From (17) it is easy to see that daj > 0 if and only if (mk — mj)2 < σ2 (hj, hk, hε). Since
limhk→0 σ2 = ∞ and σ2 (hj , hk, hε) is strictly decreasing in hk (see (4)), there exists a
unique threshold hk such that dhj ≥ 0 if and only if hk = hk. Observe that hk > hj. ■
Lemma 1
(Lemma A1) limhj→ da* = ∞
Proof.
daj*
limh, →o -г;—,
hj→0 dhj
First note that both limhj→0 σ = ∞ and limhj→0 dh- = ∞∙ To determine
we first simplify the limit of (14) as follows:
daj
lim -ɪ
hj→0 dhj
limhj→0 exp -
(mk-mj )2 ∖ "I W hε
2σ2(hj ,hk,hε) J W Wj hj +hε
lim
hj→0
2π
∂σ(hj ,hk,hε)
dhj
c' (⅛')
Wj
σ2 (hj, hk, hε)
lim 1
hj →0
(mk
2
mj)
σ2 (hj, hk, hε)
ʌ 1- (
- lim
hj→0
(hj +hε)σ(hj,hk,hε
2∏C^c" (α*)
lim
hj →0
∂σ(hj ,hk,hε)
dhj
σ2 (hj, hk, hε)
41
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