Law of Large Numbers
Etienne Pardoux
Aix–Marseille Université
Etienne Pardoux (AMU)
CIMPA, Ziguinchor
1/9
The Model
Suppose there is no latency period (T1 = 0 a.s.) and that
∆T ∼ Exp(α). Consider a SIR model with constant population size
equal to N. Let S(t) denote the number of susceptibles at time t, I (t)
the number of infectious, R(t) the number of “removed” (i.e. “healed
and immune”).
Hence the following equations, with P1 (t) and P2 (t) two standard
mutually independent Poisson processes :
Z t
β
S(t) = S(0) − P1
S(s)I (s)ds ,
N 0
Z t
Z t
β
S(s)I (s)ds − P2 α
I (s)ds .
I (t) = I (0) + P1
N 0
0
Etienne Pardoux (AMU)
CIMPA, Ziguinchor
2/9
The Model
Suppose there is no latency period (T1 = 0 a.s.) and that
∆T ∼ Exp(α). Consider a SIR model with constant population size
equal to N. Let S(t) denote the number of susceptibles at time t, I (t)
the number of infectious, R(t) the number of “removed” (i.e. “healed
and immune”).
Hence the following equations, with P1 (t) and P2 (t) two standard
mutually independent Poisson processes :
Z t
β
S(t) = S(0) − P1
S(s)I (s)ds ,
N 0
Z t
Z t
β
S(s)I (s)ds − P2 α
I (s)ds .
I (t) = I (0) + P1
N 0
0
Etienne Pardoux (AMU)
CIMPA, Ziguinchor
2/9
The proportions
Let us define sN (t) = S(t)/N, iN (t) = I (t)/N. The equations for the
proportions of susceptibles and infectious are written
Z t
1
sN (r )iN (r )dr ,
sN (t) = sN (0) − P1 βN
N
0
Z t
Z t
1
1
iN (t) = iN (0) + P1 βN
sN (r )iN (r )dr − P2 αN
iN (r )dr
N
N
0
0
Define the two martingales M1 (t) = P1 (t) − t, M2 (t) = P2 (t) − t.
We have
Z t
Z t
1
sN (t) = sN (0) − β
sN (r )iN (r )dr − M1 βN
sN (r )iN (r )dr ,
N
0
0
Z t
Z t
iN (t) = iN (0) + β
sN (r )iN (r )dr − α
iN (r )dr
0
0 Z t
Z t
1
1
sN (r )iN (r )dr − M2 αN
iN (r )dr .
+ M1 βN
N
N
0
0
Etienne Pardoux (AMU)
CIMPA, Ziguinchor
3/9
The proportions
Let us define sN (t) = S(t)/N, iN (t) = I (t)/N. The equations for the
proportions of susceptibles and infectious are written
Z t
1
sN (r )iN (r )dr ,
sN (t) = sN (0) − P1 βN
N
0
Z t
Z t
1
1
iN (t) = iN (0) + P1 βN
sN (r )iN (r )dr − P2 αN
iN (r )dr
N
N
0
0
Define the two martingales M1 (t) = P1 (t) − t, M2 (t) = P2 (t) − t.
We have
Z t
Z t
1
sN (t) = sN (0) − β
sN (r )iN (r )dr − M1 βN
sN (r )iN (r )dr ,
N
0
0
Z t
Z t
iN (t) = iN (0) + β
sN (r )iN (r )dr − α
iN (r )dr
0
0 Z t
Z t
1
1
sN (r )iN (r )dr − M2 αN
iN (r )dr .
+ M1 βN
N
N
0
0
Etienne Pardoux (AMU)
CIMPA, Ziguinchor
3/9
Consider the process
Z t
1
sN (r )iN (r )dr .
MN (t) := M1 βN
N
0
Let Ft = σ{sN (r ), iN (r ), 0 ≤ r ≤ t}.
We have
Lemma
{MN (t), t ≥ 0} is a Ft –martingale which satisfies
Z t
h
i
β
2
sN (r )iN (r )dr .
E [MN (t)] = 0, E |MN (t)| = E
N
0
Etienne Pardoux (AMU)
CIMPA, Ziguinchor
4/9
Consider the process
Z t
1
sN (r )iN (r )dr .
MN (t) := M1 βN
N
0
Let Ft = σ{sN (r ), iN (r ), 0 ≤ r ≤ t}.
We have
Lemma
{MN (t), t ≥ 0} is a Ft –martingale which satisfies
Z t
h
i
β
2
sN (r )iN (r )dr .
E [MN (t)] = 0, E |MN (t)| = E
N
0
Etienne Pardoux (AMU)
CIMPA, Ziguinchor
4/9
Proof of the Lemma
For the martingale property, see the Notes. From this, E[MN (t)] = 0.
Let 0 = t0 < t1 < · · · < tn = t.
n
X
|MN (t)|2 =
|MN (ti ) − MN (ti−1 )|2
i=1
+2
X
[MN (ti ) − MN (ti−1 )][MN (tj ) − MN (tj−1 )]
1≤i<j≤n
E |MN (t)|2
2
=
n
X
E |MN (ti ) − MN (ti−1 )|2
i=1
When |ti − ti−1 | → 0,
n
X
|MN (ti ) − MN (ti−1 )|2 →
i=1
X
|∆MN (r )|2
0≤r ≤t
=N
−2
Z t
P1 βN
sN (r )iN (r )dr
0
Etienne Pardoux (AMU)
CIMPA, Ziguinchor
5/9
Proof of the Lemma
For the martingale property, see the Notes. From this, E[MN (t)] = 0.
Let 0 = t0 < t1 < · · · < tn = t.
n
X
|MN (t)|2 =
|MN (ti ) − MN (ti−1 )|2
i=1
+2
X
[MN (ti ) − MN (ti−1 )][MN (tj ) − MN (tj−1 )]
1≤i<j≤n
E |MN (t)|2
2
=
n
X
E |MN (ti ) − MN (ti−1 )|2
i=1
When |ti − ti−1 | → 0,
n
X
|MN (ti ) − MN (ti−1 )|2 →
i=1
X
|∆MN (r )|2
0≤r ≤t
=N
−2
Z t
P1 βN
sN (r )iN (r )dr
0
Etienne Pardoux (AMU)
CIMPA, Ziguinchor
5/9
Proof of the Lemma
For the martingale property, see the Notes. From this, E[MN (t)] = 0.
Let 0 = t0 < t1 < · · · < tn = t.
n
X
|MN (t)|2 =
|MN (ti ) − MN (ti−1 )|2
i=1
+2
X
[MN (ti ) − MN (ti−1 )][MN (tj ) − MN (tj−1 )]
1≤i<j≤n
E |MN (t)|2
2
=
n
X
E |MN (ti ) − MN (ti−1 )|2
i=1
When |ti − ti−1 | → 0,
n
X
|MN (ti ) − MN (ti−1 )|2 →
i=1
X
|∆MN (r )|2
0≤r ≤t
=N
−2
Z t
P1 βN
sN (r )iN (r )dr
0
Etienne Pardoux (AMU)
CIMPA, Ziguinchor
5/9
The above cv is a.s., by uniform integrability we can interchange E
and the limit.
Hence
Z t
h
i
2
−2
E |MN (t)| = N EP1 βN
sN (r )iN (r )dr
0
Z t
β
sN (r )iN (r )dr .
= E
N
0
This implies that sup0≤t≤T |MN (t)| → 0 in probability. In fcat this is
true a.s.
This follows from
P1 (Nt)
sup − t → 0 a.s. as N → ∞.
N
0≤t≤T
Etienne Pardoux (AMU)
CIMPA, Ziguinchor
6/9
The above cv is a.s., by uniform integrability we can interchange E
and the limit.
Hence
Z t
h
i
2
−2
E |MN (t)| = N EP1 βN
sN (r )iN (r )dr
0
Z t
β
sN (r )iN (r )dr .
= E
N
0
This implies that sup0≤t≤T |MN (t)| → 0 in probability. In fcat this is
true a.s.
This follows from
P1 (Nt)
sup − t → 0 a.s. as N → ∞.
N
0≤t≤T
Etienne Pardoux (AMU)
CIMPA, Ziguinchor
6/9
The above cv is a.s., by uniform integrability we can interchange E
and the limit.
Hence
Z t
h
i
2
−2
E |MN (t)| = N EP1 βN
sN (r )iN (r )dr
0
Z t
β
sN (r )iN (r )dr .
= E
N
0
This implies that sup0≤t≤T |MN (t)| → 0 in probability. In fcat this is
true a.s.
This follows from
P1 (Nt)
sup − t → 0 a.s. as N → ∞.
N
0≤t≤T
Etienne Pardoux (AMU)
CIMPA, Ziguinchor
6/9
The above cv is a.s., by uniform integrability we can interchange E
and the limit.
Hence
Z t
h
i
2
−2
E |MN (t)| = N EP1 βN
sN (r )iN (r )dr
0
Z t
β
sN (r )iN (r )dr .
= E
N
0
This implies that sup0≤t≤T |MN (t)| → 0 in probability. In fcat this is
true a.s.
This follows from
P1 (Nt)
sup − t → 0 a.s. as N → ∞.
N
0≤t≤T
Etienne Pardoux (AMU)
CIMPA, Ziguinchor
6/9
We prove the above results in three steps. Let P(t) be a rate λ
Poisson process
n
−1
P(n) = n
−1
n
X
[P(i) − P(i − 1)]
i=1
→ λ a.s. as n → ∞.
Second step
t −1 P(t) = t −1 P([t]) + t −1 (P(t) − P([t]))
−1
t P(t) − λ ≤ t −1 P([t]) − λ + t −1 P([t] + 1) − t −1 P([t]).
We have just proved that N −1 P(Nt) → λt a.s. for all t > 0. We have
a sequence of increasing functions which converges to a continuous
function, hence by the second Dini theorem, the convergence is
uniform.
Etienne Pardoux (AMU)
CIMPA, Ziguinchor
7/9
We prove the above results in three steps. Let P(t) be a rate λ
Poisson process
n
−1
P(n) = n
−1
n
X
[P(i) − P(i − 1)]
i=1
→ λ a.s. as n → ∞.
Second step
t −1 P(t) = t −1 P([t]) + t −1 (P(t) − P([t]))
−1
t P(t) − λ ≤ t −1 P([t]) − λ + t −1 P([t] + 1) − t −1 P([t]).
We have just proved that N −1 P(Nt) → λt a.s. for all t > 0. We have
a sequence of increasing functions which converges to a continuous
function, hence by the second Dini theorem, the convergence is
uniform.
Etienne Pardoux (AMU)
CIMPA, Ziguinchor
7/9
We prove the above results in three steps. Let P(t) be a rate λ
Poisson process
n
−1
P(n) = n
−1
n
X
[P(i) − P(i − 1)]
i=1
→ λ a.s. as n → ∞.
Second step
t −1 P(t) = t −1 P([t]) + t −1 (P(t) − P([t]))
−1
t P(t) − λ ≤ t −1 P([t]) − λ + t −1 P([t] + 1) − t −1 P([t]).
We have just proved that N −1 P(Nt) → λt a.s. for all t > 0. We have
a sequence of increasing functions which converges to a continuous
function, hence by the second Dini theorem, the convergence is
uniform.
Etienne Pardoux (AMU)
CIMPA, Ziguinchor
7/9
We have the Law of Large Numbers :
sup {|sN (t) − s(t)| + |iN (t) − i(t)|} → 0 a.s., where
0≤t≤T
(s(t), i(t)) solves

ds

 (t) = −βs(t)i(t), t > 0,
dt

 di (t) = βs(t)i(t) − αi(t), t > 0.
dt
s(t)
sN (t)
Define X (t) =
, XN (t) =
, X N (t) = X (t) − XN (t),
i(t)
iN (t)
MN (t)
x
−βxy
YN (t) =
, and finally F
=
. For
NN (t) − MN (t)
y
βxy − αy
0 ≤ x, y , x 0 , y 0 ≤ 1,
0 0 F x − F x ≤ C (α, β) x − x .
0
0
y
y
y y Etienne Pardoux (AMU)
CIMPA, Ziguinchor
8/9
We have the Law of Large Numbers :
sup {|sN (t) − s(t)| + |iN (t) − i(t)|} → 0 a.s., where
0≤t≤T
(s(t), i(t)) solves

ds

 (t) = −βs(t)i(t), t > 0,
dt

 di (t) = βs(t)i(t) − αi(t), t > 0.
dt
s(t)
sN (t)
Define X (t) =
, XN (t) =
, X N (t) = X (t) − XN (t),
i(t)
iN (t)
MN (t)
x
−βxy
YN (t) =
, and finally F
=
. For
NN (t) − MN (t)
y
βxy − αy
0 ≤ x, y , x 0 , y 0 ≤ 1,
0 0 F x − F x ≤ C (α, β) x − x .
0
0
y
y
y y Etienne Pardoux (AMU)
CIMPA, Ziguinchor
8/9
We have
Z
t
[F (X (r )) − F (XN (r ))]dr + YN (t).
X N (t) = X N (0) +
0
We have proved that sup0≤t≤T kYN (t)k → 0 a.s. as N → ∞.
Let εN (t) = sup0≤r ≤t kYN (r )k. We have
Z
kX N (t)k ≤ kX N (0)k + C (α, β)
t
kX N (r )kdr + εN (t).
0
It then follows from Gronwall’s Lemma that
sup kX N (r )k ≤ kX N (0)k + εN (t) exp (C (α, β)t) ,
0≤r ≤t
hence the result, provided that .X N (0) → 0.
Etienne Pardoux (AMU)
CIMPA, Ziguinchor
9/9
We have
Z
t
[F (X (r )) − F (XN (r ))]dr + YN (t).
X N (t) = X N (0) +
0
We have proved that sup0≤t≤T kYN (t)k → 0 a.s. as N → ∞.
Let εN (t) = sup0≤r ≤t kYN (r )k. We have
Z
kX N (t)k ≤ kX N (0)k + C (α, β)
t
kX N (r )kdr + εN (t).
0
It then follows from Gronwall’s Lemma that
sup kX N (r )k ≤ kX N (0)k + εN (t) exp (C (α, β)t) ,
0≤r ≤t
hence the result, provided that .X N (0) → 0.
Etienne Pardoux (AMU)
CIMPA, Ziguinchor
9/9
We have
Z
t
[F (X (r )) − F (XN (r ))]dr + YN (t).
X N (t) = X N (0) +
0
We have proved that sup0≤t≤T kYN (t)k → 0 a.s. as N → ∞.
Let εN (t) = sup0≤r ≤t kYN (r )k. We have
Z
kX N (t)k ≤ kX N (0)k + C (α, β)
t
kX N (r )kdr + εN (t).
0
It then follows from Gronwall’s Lemma that
sup kX N (r )k ≤ kX N (0)k + εN (t) exp (C (α, β)t) ,
0≤r ≤t
hence the result, provided that .X N (0) → 0.
Etienne Pardoux (AMU)
CIMPA, Ziguinchor
9/9
We have
Z
t
[F (X (r )) − F (XN (r ))]dr + YN (t).
X N (t) = X N (0) +
0
We have proved that sup0≤t≤T kYN (t)k → 0 a.s. as N → ∞.
Let εN (t) = sup0≤r ≤t kYN (r )k. We have
Z
kX N (t)k ≤ kX N (0)k + C (α, β)
t
kX N (r )kdr + εN (t).
0
It then follows from Gronwall’s Lemma that
sup kX N (r )k ≤ kX N (0)k + εN (t) exp (C (α, β)t) ,
0≤r ≤t
hence the result, provided that .X N (0) → 0.
Etienne Pardoux (AMU)
CIMPA, Ziguinchor
9/9