which combined with 3.14 implies lim
→0
lim inf
→0
log
;
; ;V
{; ¡ } = lim
→0
lim sup
→0
log
;
; ;V
{; 6} = −J
V
: 3.15
Since the family {
;
; ;V
} is also exponentially tight, using Theorem 2.1 again, we get the result.
4. LDP for FV process with nite types
We now turn to the study of LDP at the path-level. In this section, we focus on FV process with nite types.
Let E
n
= {1; 2; : : : ; n}, and dene
S
n
= x = x
1
; : : : ; x
n −1
: x
i
¿ 0; i = 1; : : : ; n
− 1;
n −1
X
i=1
x
i
6 1
; with S
◦
n
denoting its interior. Then the FV process with n types is a FV process with neutral mutation and type space E
n
. It is a nite-dimensional diusion process statisfying the following system of stochastic dierential equations:
d x
k
t = b
k
x tdt +
√
n −1
X
l=1 kl
x tdB
l
t; 16k6n
− 1; 4.1
where x
t = x
1
t; : : : ; x
n −1
t, b
k
x t = =2p
k
− x
k
t, and
x t =
kl
x t
16k; l6n −1
is given by x
t
′
x t = Dx
t = x
k
t
kl
− x
l
t
16k; l6n −1
; where p
k
= k, and B
l
t; 16l6n − 1 are independent Brownian motions.
Let P
x
denote the law of x · starting at x. The LDP for P
x
on space C[0; T ]; S
n
as
goes to zero has been studied in Dawson and Feng 1998 under the assumptions that p
k
is strictly positive for all k and x is in the interior of S
n
. In this section we will consider the LDP for P
x
when these assumptions are not satised, i.e., some of p
k
are zero or x is on the boundary. This creates serious diculties because of the degeneracy
and the non-Lipschitz behaviour of the square root of the diusion coecient on the boundary. Let p
n
= 1 −
P
n −1
i=1
p
i
, x
n
= 1 −
P
n −1
i=1
x
i
, and b
n
x = =2p
n
− x
n
. If for a particular k, p
k
= x
k
∈ {0; 1}, then x
k
t will be zero or one for all positive t. Thus, without the loss of generality, we assume that p
k
+ x
k
is not zero or two for all k. For any given p = p
1
; : : : ; p
n −1
, dene Z
p
= {x = x
1
; : : : ; x
n −1
∈ S
n
: 0 ¡ x
k
+ p
k
¡ 2; k = 1; : : : ; n }:
For any x in Z
p
, let H
; x
is the set of all absolutely continuous element in C[; ]; S
n
starting at x, i.e., H
; x
= ’
∈ C[; ]; S
n
: ’t = x + Z
t
˙ ’s ds
:
Dene I
; x
’ =
1 2
Z
n
X
i=1 ˙
’
i
t −b
i
’t
2
’
i
t
dt; ’
∈ H
; x
; ∞;
’ 6∈ H
; x
; 4.2
where ’
n
t = 1 −
P
n −1
i=1
’
i
t, 0=0 = 0, c=0 = ∞ for c ¿ 0, and the integrations are the
Lebesgue integrals. We denote I
0; T x
H
0; T x
by I
x
resp. H
x
.
Lemma 4.1. For n = 2 and any ’
· in C[0; T ]; S
2
; we have lim sup
→0
lim sup
→0
log P
x
sup
t ∈ [0;T ]
|xt − ’t|6 6
− I
x
’: 4.3
Proof. Since n = 2, we have x = x
1
; p = p
1
. The result has been proved in the case of 0 ¡ x ¡ 1; 0 ¡ p ¡ 1 in Theorem 3:3 of Dawson and Feng 1998. Next we consider
the remaining cases: A x = 0; 0 ¡ p ¡ 1; B 06x ¡ 1; p = 1; C x = 1; 0 ¡ p ¡ 1; D 0 ¡ x61; p = 0. By dealing with 1
− x and 1 − p, we can derive C and D from A and B, respectively.
For any ¿ 0; N ¿1, and 06a6b6T , set B’; ; a; b =
∈ C[0; T ]; S
n
: sup
a6t6b
| t − ’t|6 ;
B
◦
’; ; a; b = ∈ C[0; T ]; S
n
: sup
a6t6b
| t − ’t| ¡ :
If ’ is not in H
x
, either ’0 6= x or ’ is not absolutely continuous. Clearly the
result holds in the case of ’0 6= x. If ’ is not absolutely continuous, then there exist
c ¿ 0 and disjoint subintervals [a
m 1
; b
m 1
]; : : : ; [a
m k
m
; b
m k
m
] such that P
k
m
l=1
b
m l
− a
m l
→ 0; while
P
k
m
l=1
|’b
m l
− ’a
m l
|¿c: By Chebyshev’s inequality and martingale property, we get
lim sup
→0
lim sup
→0
log P
x
{B’; ; 0; T }
6 lim sup
→0
lim sup
→0
log E
P
x
exp
k
m
X
l=1
1
l
xb
m l
− xa
m l
− Z
b
m l
a
m l
l
bxs −
2 l
2 xs1
− xs ds
× inf
∈ B’;;0;T
exp −
k
m
X
l=1
1
l
b
m l
− a
m l
− Z
b
m l
a
m l
l
b s
−
2 l
2 s1
− s ds
= −
k
m
X
l=1 l
’b
m l
− ’a
m l
− Z
b
m l
a
m l
l
b’s −
2 l
2 ’s1
− ’s ds
6 − c + C
k
m
X
l=1
b
m l
− a
m l
; where
l
= sign’b
m l
− ’a
m l
; l = 0; : : : ; k
m
− 1, and C is a positive constant depending on . Here
signc =
1; c ¿ 0;
−1; c ¡ 0;
0; c = 0:
Now let m go to innity, and then let go to innity, one ends up with lim sup
→0
lim sup
→0
log P
x
{B’; ; 0; T }6 − ∞ = −I
x
’: 4.4
Next we assume that ’ ∈ H
x
. A x=0; 0 ¡ p ¡ 1. Four cases need to be treated separately based on the behaviour
of ’. Case
I: There is a t ¿ 0 such that ’t = 0 for all t in [0; t
]. In this case I
x
’ = ∞. On the other hand, choose small enough such that
¡ min p
2 ;
pt 8
: Then for any x
· in C[0; T ]; S
n
satisfying sup
t ∈ [0; t
]
xt6, we have sup
t ∈ [0; t
]
Z
t
bxs ds − xt
¿ pt
8 :
Next by choosing c = p=8
, we get sup
t ∈ [0; t
]
−cxt − Z
t
bxs ds −
c
2
2 Z
t
xs1 − xs ds
¿
2
p
2
t 128
; which combined with Doob’s inequality implies
P
x
sup
06t6T
|xt − ’t|6 6
P
x
sup
06t6t
xt6 6
P
x
sup
06t6t
− xt
− Z
t
bxs d s −
2
2 Z
t
xs1 − xs ds
¿
2
p
2
t 128
6 exp
−
2
p
2
t 128
: Letting
go to zero, then go to zero, we get 4.3.
Case II: For all t in 0; T ], ’t stays away from 0 and 1.
For any N ¿1, choose small enough such that no functions in the set B’; 2; 1=N; T hit zero or one in the time interval [1=N; T ]. Let be the law of
x
1=N
under P
x
. Then one gets log P
x
{B’; ; 0; T }6
log P
x
{B’; ; 1=N; T } =
log
Z
1
P
y
{B’; ; 1=N; T } dy 6
log sup
|y−’1=N |6
P
y
{B’; ; 1=N; T } =
log P
y
{B’; ; 1=N; T } for some |y − ’1=N |6;
where in the last equality we used the property that the supremum of an upper semi- continuous function over a closed set can be reached at certain point inside the set.
Noting that P
y
coincides with a non-degenerate diusion over the interval [1=N; T ] on any set that does not hit the boundary of [0; 1]. By the uniform large deviation
principle for non-degenerate diusions cf. Dembo and Zeitouni, 1993, we get lim sup
→0
log P
x
{B’; ; 0; T }6 − inf
|y−’1=N |6
inf
∈ B’;;1=N;T
I
1=N; T y
: 4.5
Assume that inf
|y−’1=N |6
inf
∈ B’; ;1=N; T
I
1=N; T y
is nite for small . Otherwise I
x
’¿I
1=N; T ’1=N
’ = ∞, and the upper bound is trivially true. For any y satisfying
|y − ’1=N |6, in B’; ; 1=N; T satisfying 1=N = y, I
1=N; T y
¡ ∞, we dene
for t in [1=N; T ] t = t + ’1=N
− y: Then it is clear that
is in B’; 2; 1=N; T and thus does not hit zero or one. By direct calculation, we get that
I
1=N; T ’1=N
6I
1=N; T y
+
N
; where
N
goes to zero as goes to zero for any xed N . This combined with 4.5 implies that
lim sup
→0
lim sup
→0
log P
x
sup
06t6T
|xt − ’t|6 6
− lim
→0
inf
∈ B’; 2;1=N;T
I
1=N; T ’1=N
= −I
1=N; T ’1=N
’; 4.6
where the equality follows from the lower semicontinuity of I
1=N; T ’1=N
· at non-degenerate paths. Finally by letting N go to innity we end up with 4.3.
More generally, if ’t is in 0; 1 over [; ] ⊂ [0; 1], then the above argument
leads to lim sup
→0
lim sup
→0
log P
x
sup
06t6T
|xt − ’t|6 6
− I
; ’
’: 4.7
Case III: ’t is in 0; 1] for all t in 0; T ], i.e., ’ may hit boundary point 1.
Let
1
= inf {t ∈ [0; T ]: ’t = 1};
= inf {t ∈ [0; T ]: I
0; t x
’ = ∞}:
If ¡
1
, 4.3 is proved by applying the arguments in Case II to the time interval [0; +
1
=2]. Now assume that ¿
1
. Since p ¡ 1, one can nd 0 ¡ t
3
¡ t
2
¡
1
satisfying inf
s ∈ [t
3
;
1
]
’s ¿ p. By using result in Case II, we have
lim sup
→0
log P
x
sup
t ∈ [0;T ]
|xt − ’t|6 6
− I
0; t
2
x
’: Since I
0; t
2
x
’ is nite for all t
2
, we get I
x
’ ¿ I
0; t
2
x
’ = 1
2 Z
t
2
˙ ’t
− =2p − ’t
2
’t1 − ’t
dt ¿
2 Z
t
2
t
3
’t − p
˙ ’t
1 − ’t
dt → ∞ as t
2
ր
1
; 4.8
which implies 4.3. Case
IV: A second visit to zero by ’t occurs at a strictly positive time. Let
= inf {t ¿ 0: ’t = 0} ¿ 0:
Choosing 0 ¡ t
1
¡ t
2
¡ such that inf
t ∈ [t
1
; ]
p − ’t ¿ 0. Then we have
I
x
’ ¿ lim
t
2
ր
1 2
Z
t
2
t
1
˙ ’t
− =2p − ’t
2
’t dt
¿ − lim
t
2
ր
2 Z
t
2
t
1
p − ’t
˙ ’t
’t dt =
∞: This combined with 4.7 implies the result.
B 06x ¡ 1; p = 1. First assume that I
x
’ is nite i.e. = ∞. For small , dene
’ = {t ∈ [0; T ]: ’t ¡ 1 − } =
∞
[
i=1
a
i
; b
i
: Since the set
{t ∈ [0; T ]: ˙ ’t
− =21 − ’t
2
=’t1 − ’t = 0} has no con-
tribution to the value of I
x
’ and the niteness of I
x
’ implies that the set N = {t ∈ [0; T ]: ˙
’t − =21 − ’t
2
=’t1 − ’t = ∞} has zero Lebesgue measure,
we may redene the value of ˙ ’t
−=21−’t
2
=’t1 −’t to be zero on N
without changing the value of the rate function. After this modication we can apply the monotone convergence theorem and get that I
x
’ converges to I
x
’ as goes to zero.
By the Markov property and an argument similar to that used in deriving 4.5 and 4.6, we have for any m¿1 and ¿ 0,
lim sup
→0
lim sup
→0
log P
x
sup
t ∈ [0;T ]
|xt − ’t|6
6
m
X
i=1
lim sup
→0
lim sup
→0
sup
|y−’a
i
|6
log P
x
sup
t ∈ [a
i
;b
i
]
|xt − ’t|6
6 −
m
X
i=1
lim inf
→0
inf
|y−’a
i
|6
inf
∈ B’;;a
i
;b
i
I
a
i
; b
i
y
= −
m
X
i=1
I
a
i
; b
i
’a
i
’: 4.9
By letting m go to innity, and then go to zero, we get 4.3. Next we assume the rate function is innity. By an argument similar to that used
in Case IV, we get the result for all paths that hit zero at a positive time. We now assume that ’ does not hit zero at any positive time. If ¡
1
the result is true by using the argument in Case III. Let us now assume that
1
6 6T .
If ’t = 1 over [; + ] for some ¿ 0, then we have lim
t ր
I
0; t x
’ = ∞ and the
result follows by approaching from below. If ’ is in 0; 1, then the result is obtained by using 4.7 in a small two-sided
neighborhood of since the rate function over the neighborhood is innity. The only possibility left is that ’ = 1, and 0 ¡ ’t ¡ 1 over ; ] for some
¡ 6T . By applying 4.7 over [; ] with ∈ ; and letting approach from
above, we get the result.
Lemma 4.2. For any n¿2 and any ’
· in C[0; T ]; S
n
; we have that for any x
∈ Z
p
; lim sup
→0
lim sup
→0
log P
x
sup
t ∈ [0;T ]
|xt − ’t|6 6
− I
x
’; 4.10
for any x ∈ S
◦
n
∩ Z
p
; lim inf
→0
lim inf
→0
log P
x
sup
t ∈ [0;T ]
|xt − ’t| ¡ ¿
− I
x
’: 4.11
Proof. If ’t
∈ C[0; T ]; S
n
and I
x
i
’
i
= ∞ for some i = 1; : : : ; n, where I
x
i
’
i
represents the rate function for the two-type process x
i
·; P
j 6=i
x
j
·, then we have lim sup
→0
lim sup
→0
log P
x
sup
t ∈ [0;T ]
|xt − ’t|6
6 lim sup
→0
lim sup
→0
log P
x
sup
t ∈ [0;T ]
|x
i
t − ’
i
t |6
6 − I
x
i
’
i
= −∞:
4.12 Since for any k; l
˙ ’
k
t + ˙ ’
l
t − b
k
’t − b
l
’t
2
’
k
t + ’
l
t
6 ˙
’
k
t − b
k
’t
2
’
k
t +
˙ ’
l
t − b
l
’t
2
’
l
t ;
we conclude that I
x
’ = ∞ implies that I
x
k
’
k
= ∞ for some 16k6n. Without the
loss of generality, let us assume that k = 1. Thus by 4.12 and by applying Lemma 4.1 to the two-type process x
1
·; P
n j=2
x
j
· we get both 4.10 and 4.11 when I
x
’ = ∞.
Assume that I
x
’ ¡ ∞ in the sequel, and thus ’ is in H
x
. For any ¿ 0, let
T ’ =
t ∈ [0; T ]: inf
16i6n
{’
i
t; 1 − ’
i
t } ¿
=
∞
[
j=1 j
;
j
: By an argument similar to that used in the proof of B of Lemma 4.1, one can
apply the monotone convergence theorem and get I
x
’ = lim
→0
1 2
Z
t ∈ T
’ n
X
i=1
˙ ’t
− b
i
’t
2
’
i
t dt;
and so 4.10 follows. Next assume that x is in S
◦
n
∩ Z
p
. For convenience we assume that p
1
¿ 0. Since I
x
’ is nite, by applying the results in the proof of Lemma 4.1, ’
1
t will not hit zero at a later time. Thus d = inf
06t6T
’
1
t ¿ 0. For any ¿ 0, choose small such that n
− 1 ¡ min{=2; d=2; x
i
; i = 1; : : : ; x
n
}. For any 06t6T and i = 2; : : : ; n, set h
i
t = 0 for ’
i
t ¿ ; h
i
t = − ’
i
t for ’
i
t6. Let h
1
t =
n
X
i=2
h
i
t; ’
t = ’t + −h
1
t; h
2
t; : : : ; h
n −1
t: Then it is clear that ’
0 = ’0 = x, and ’ t
∈ S
◦
n
for t in [0; T ]. It is also not hard to see that ’
i
t = if ’t6. Note that for any i = 2; : : : ; n, if p
i
¿ 0, then h
i
t ≡ 0
for small enough . Let K
p
= {i ∈ {2; : : : ; n}: p
i
= 0 }.
By direct calculation, one gets lim inf
→0
log P
x
{B
◦
’; ; 0; T }¿ lim inf
→0
log P
x
{B
◦
’ ; =2; 0; T
} ¿
lim
˜ →0
lim inf
→0
log P
x
{B
◦
’ ; ˜
=2; 0; T }¿ − I
x
’ :
4.13 Observe that
I
x
’ =
1 2
n
X
i=1
Z
T
˙ ’
i
t − b
i
’ t
2
’
i
t dt
= 1
2
X
i 6∈K
p
∪{1}
Z
T
˙ ’
i
t − b
i
’ t
2
’
i
t dt +
Z
T
˙ ’
1
t − b
1
’ t
2
’
1
t dt
+ X
i ∈ K
p
Z
T
˙ ’
i
t − b
i
’ t
2
’
i
t dt
6 1
2
X
i 6=1; i6∈K
p
Z
T
˙ ’
i
t − b
i
’ t
2
’
i
t dt +
Z
T
˙ ’
1
t − b
1
’ t
2
’
1
t dt
+ X
i ∈ K
p
Z
T
˙ ’
i
t − b
i
’t
2
’
i
t dt +
nT
2
4
;
4.14 where in the last inequality we used the fact that ’
i
t = if ’t6. For any i 6∈ K
p
and any t, we have ’
i
t = ’t for small enough . Hence by letting go to zero, we have
X
i 6=1; i6∈K
p
Z
T
˙ ’
i
t − b
i
’ t
2
’
i
t dt
→ X
i 6=1; i6∈K
p
Z
T
˙ ’
i
t − b
i
’t
2
’
i
t dt: 4.15
Let N
p
= {t ∈ [0; T ] : ’
i
t = 0 for some i ∈ K
p
}. Noting that ˙
’
1
t
2
6 n
X
i ∈ K
p
∪{1}
˙ ’
i
t
2
: 4.16
The niteness of I
x
’ · implies that I
N
p
x
’ · = 0. By the dominated convergence
theorem, we get that lim
→0
Z
T
˙ ’
1
t − b
1
’ t
2
’
1
t dt = lim
→0
Z
N
c p
+ Z
N
p
˙ ’
1
t − b
1
’ t
2
’
1
t dt
= Z
T
˙ ’
1
t − b
1
’t
2
’
1
t dt:
4.17 This, combined with 4.14 and 4.15, implies
lim sup
→0
I
x
’ 6I
x
’: 4.18
Finally, by letting go to zero, then go to zero in 4.13, we get 4.11.
Remark. More detailed information about the way that a path leaves or approaches
the boundary can be obtained by an argument similar to that used in Feng 2000, where the one-dimensional continuous branching processes are studied.
Let Y
p
= {x ∈ S
n
: x
i
¿ 0 whenever p
i
¿ 0 }: Then we get the following:
Theorem 4.3. For any x
∈ Y
n
; the family {P
x
}
¿0
satises a LDP on space C[0; T ]; S
n
with speed
and good rate function I
x; p
· given by I
x; p
’ =
1 2
Z
T n
X
i=1
˙ ’
i
t − b
i
’t
2
’
i
t dt;
’ ∈ H
x; p
; ∞;
’ 6∈ H
x; p
; 4.19
where H
x; p
= ’
∈ C[0; T ]; S
n
: ’t = x + Z
t
˙ ’s ds; t
∈ [0; T ] and ’
i
t ≡ 0 if x
i
= p
i
= 0 :
Proof. If x
i
= p
i
= 0 for some i = 1; : : : ; n − 1, then x
i
t ≡ 0 for all t in [0; T ]. This
explains the denition of the set H
x; p
. By projection to lower dimension, the result is reduced to the case where x
i
+ p
i
¿ 0 for all i. Then by applying Lemma 4.2 and Theorem 2.1, we get the result.
Remark. This result has been proved in Theorem 3:3 in Dawson and Feng 1998
under the assumption that x
i
¿ 0; p
i
¿ 0 for all i = 1; : : : ; n. Here we removed all restrictions on x and p in the upper bound, and extend the lower bound to cases
where p
i
can be zero for some i = 1; : : : ; n.
5. LDP for FV processes
