The name is absent



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 h
j 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 limhj0 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 h
j = hj.

Finally,

da*
dhk


exp


(   (mk-mj )

2(hj ,hk,,


)2
,h
ε)


√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
lim
hk→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;—,
hj0 dhj


First note that both limhj0 σ = and limhj0 dh- = ∙ To determine
we first simplify the limit of (14) as follows:

daj
lim
h
j0 dhj


limhj0 exp -


(mk-mj )2 "I W  hε

2σ2(hj ,hk,hε) J W Wj hj +hε


lim
hj0


∂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
hj0


(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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