Erratum / Correction to : “Minimization of the Kullback information of diffusion processes”

A NNALES DE L ’I. H. P., SECTION B

P. C ATTIAUX

C. L ^ÉONARD

Erratum / Correction to : “Minimization of the Kullback information of diffusion processes”

Annales de l’I. H. P., section B, tome 31, n

^o

4 (1995), p. 705-707

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705

ERRATUM Correction to:

Minimization of the Kullback information of diffusion _processes (*)

P. CATTIAUX and C.

LÉONARD

December 1994 Ann. Inst. Henri Poincaré,

Vol. 31, ^n° 4, 1995, p. -707 Probabilités et Statistiques

The

supermartingale

^{Z defined}

by ( 1.11 )

^{does not} vanish for t &#x3E;

Too,

^if

00

w

belongs

^{to the set A =}

U ^{Tm

⁼

^{Too T},}

^{contrary to what is written}

m=l

at the bottom of page 90. Of ^course the associated Follmer-measure

Q

^and

the initial P are

equivalent

in restriction to

A,

^{so that if}

T)

⁼

0,

then

P(A)

^{= 0.} This holds if we know that

Q

^{is a}

probability

^measure

on 0 such

that Q(Too T)

^{= 0.}

As a consequence, ^{we cannot} conclude in

Proposition ^(2.3)

^that

Too

⁼

T, Q-a.s. since

^on

^A, T~0 03B2s. ^a(XS,

^{ds +(0)’ the rest of}

the

proposition being

^{true. So the trick}

(2.6)

^is

incomplete,

^{and to both the}

energy condition and the domination

condition,

^{one has to} add the third condition:

Q(Too T) =

^0.

This additional condition is checked in the

proof

^{of Theorem}

(3.1 ),

^thanks

to

(3.47),

^{so that the}

proof

of the theorem is

complete.

(*) Vol. 30, ^n° 1, 1994, pp. 83-132.

Annales de l ’lnstitut Henri Poincaré - Probabilités et Statistiques - ^0246-0203

Vol. 31 /95/04/$ 4.00/@ Gauthier-Villars

706 P. CATTIAUX AND C. LEONARD

In order to

complete

^the

proofs

in sections 4 and

5,

^{we have to check} that this condition is satisfied. We shall

present

^this

below,

^{in the context}

of Theorem

(4.18),

^{but the}

proof

still works for _any other case considered in the _paper.

~

Denote

where

and

One

clearly

^has

D Am

^for

all j.

Now,

^denote

P ( Am )

^{= a}

&#x3E;_

^{0. We are}

going

^{to build a subset}

Bm

of with a

positive

^{mass and which is an} intersection of To this

end,

^choose

k1

such that

P(Am

ⁿ &#x3E;

a(1 - 1/2).

^This

is

possible

^{because is an}

increasing

sequence of ^sets such that

Am.

^We put

Bm

⁼

^{Am n}

^C

^Am,

^{and we can choose}

^k2

such that

P(B

ⁿ

Am~’~2 )

&#x3E;

cx ( 1 - 1 / 2 ) ( 1 - 1 / 4 ) , by

^{the same} argument

as above. Now,

by

^{induction we build a}

nonincreasing

sequence

Bin

^of

subsets such that

00 B

Finally

⁼

Q ~

^{satisfies G}

^A~.,2 n Q ^~~

^and

j&#x3E;i 00 v=i

/

ca &#x3E; 0,

^{with c}

= JY(1 - ^2’’)

&#x3E; 0.

~=i

But for each n

&#x3E; 1,

^{P and}

~n

^are

equivalent

measures so that

since

Bm

^is

-measurable,

^which

yields

^{for all n} &#x3E; 1

Annales de l’lnstitut Henri Poincaré - Probabilités et Statistiques

707

ERRATUM

since

Now,

^{we can}

evaluate

by

^{means of}

(4.26). Indeed,

^for &#x3E; 0

It follows that for a

fixed i,

hence

inf

⁼

^0,

^{which is in} contradiction with

(*)

^{unless a = 0.}

We

thus have proved

^that

P(Am) = ^0, henceP ( A)

^{= 0.}

Consequently,

all the results of the _paper are true, except ^for

Propo-

sition

(2.3).

^{All our}

apologies.

Vol. 31, n° ^4-1995.