RECURSIVE KERNEL ESTIMATION OF THE INTENSITY FUNCTION
OF AN INHOMOGENEOUS POISSON PROCESS
Anna Kitayeva1, and Mihail Kolupaev2
1
Tomsk Polytechnic University, Tomsk, Russia, kit1157@yandex.ru
2
Tomsk State University, Tomsk, Russia, al13n@sibmail.com
In this paper we give an extension of the results considered in [1] to the recursive
algorithms. Recursive estimation is particularly useful in large sample size since the result can
be easily updated with each additional observation. The structure of the estimate is similar to the
recursive kernel estimate of density function introduced in [2, 3], and thoroughly examined in
[4]. The estimate is constructed on a single realization of a Poisson process on fixed interval
[0, T ] . Specificity of the statistic (1) is due to a random sample size.

Let t i , i  1, N ,

0  t i  T be a realization of a Poisson point process having
unknown intensity function  () on [0, T ] , where N is the number of points falling into [0, T ] ,
b

denote (a, b)  (t )dt . Consider the following expression as an estimate of the function
a
 () /  (0, T ) at a point t
 t  tN 
1 N 1  t  ti  
1
1
  1  S N 1 

(1)
K 
K 

N i 1 hi  hi   N 
NhN  hN 
where hn   0 is a sequence of real numbers, nhn   , K () is a kernel function; let S N  0
if N  0 .
SN 
Consider an asymptotic behavior of statistic (1) under following scheme of series: let series
of observations are done on [0, T ] with the intensity of the process in n-th trial equals to
 n ()  n() . Denote the value of the statistic (1) in n-th trial
Sn 
1 N n 1  t  tin 
,
 K
N n i 1 hin  hin 
(2)
where N n and ( t in ) − respectively the number of observations and the realization of the process
in n-th trial.
Theorem 1 (asymptotic unbiasedness). Let the kernel K () is a compact bounded real
T
valued Borel function on [T , T ] and
 K (u )u
m
du  M m  1,2,...; the intensity function
T
 () is continuous at the point t  (0, T ) , (t , T )  0; (hin ) is monotonically no increasing
with i sequence, (hin )  0 as i   , and
m
 (1) C
i
i 0
i
m
m
hmi  N n  ChNn
m with some constant
C. Then the estimate (1) is asymptotically unbiased: lim E ( S n )  (t ) / (0, T )  0.
n 
Proof. Let
pi ( x / n) be the conditional density function ti given N  n  1,
pn  P N  n   (0, T ) n e   ( 0,T ) / n!, n  0 . Then the expected value
1

E ( S N )   pn
n 1
T
t x
1 n 1
 pi ( x / n)dx 
K 


n i 1 hi 0  hi 
T
 t  x  (0, x)i 1 ( x, T )n i ( x)
1 n 1
e   ( 0 ,T )

(3)
    K 
dx
.
  ( 0 ,T )
n 1 n i 1 hi 0
 hi  (i  1)! (n  i)! (0, T ) 1  e
 1 n  (0, x)i 1  ( x, T ) n  i
  (0, T ) n 1
e (0,T )  1
Taking into account  
and


(n  i )!
n!
(0, T )
n 1 n i 1 (i  1)!
n 1

boundedness of the kernel we know that series (3) convergences absolutely. Let
hi  min t / T , 1  t / T  i, and since K () is a compact function on [T , T ] we have
T
(0, t  hi x) i 1 (t  hi x, T ) ni
1 T  t  x  (0, x) i 1 ( x, T ) ni

K 
 ( x)dx   K ( x)
 (t  hi x)dx.

hi 0  hi  (i  1)!
(n  i)!
(i  1)!
(n  i)!
T
~
Note that (t  hin x)  (t )  Chin x for sufficiently small hin , then
E S n  

e  n ( 0,T )
1  e  n ( 0,T )
e  n ( 0,T )
1  e  n ( 0,T )
T
 (0, t  hin x) i 1  (t  hin x, T ) k i
1 k k
n
K
(
x
)
(t  hin x)dx 

 
(i  1)!
(k  i )!
k 1 k i 1
T



T
 (0, t  hin x) i 1  (t  hin x, T ) k i
1 k k
~
n
K
(
x
)
(t )  Chin x dx  (t ) Ln  Lon .



(i  1)!
(k  i )!
k 1 k i 1
T

For the proof of the theorem we establish the convergence lim Ln  (0, T ) 1 .
n
Using binomial expansion and the linearization (t  hin x, t ) 
t
  (t )dt  rhin x,
where
t  hin x
r > 0 – some constant, we obtain
(0, t  hin x)i 1 (t  hin x, T ) k i
1 k k
n
K
(
x
)
dx 

 
(i  1)!
(k  i)!
k 1 k i 1
T

T
l
i 1
k i
1 k k
 (t , T ) m rhin x 
i 1 l  (0, t ) rhin x 
n
K
(
x
)
(

1
)
  
0 m! k  i  m !dx 
l
!
(
i

1

l
)!
k 1 k i 1
l

0
m
T


i 1 l
T
rh x 
n k  (0, t )l k k  l
n  (1)i  K ( x) in

l!
i!
k 0 k  1 l 0
i 0
T

n k  (0, t )l nl

 l!
k 0 k  1 l 0

i k i l
T


m0
k i  m
 (t , T ) m rhin x 
dx 
m! k  i  l  m  !
k i l  m
rhin x 
 (t , T ) m n m k  l  m k  l  m
n
(1)i  K ( x)
dx 


m!
i!k  i  l  m  !
m0
i 0
T
k l
T
k l  m
(4)
n k  (0, t )l nl  (t , T ) k  l n k  l
 n ,
 l!
(k  l )!
k 0 k  1 l 0


where
n k 1 (0, t ) l n l
 l!
k 1 k  1 l 0

n  
k l 1

m 0
T
rhin x 
(t , T ) m n m k l m k l m
i
n
(

1
)
K ( x)
dx 




m!
i
!
k

i

l

m
!
i 0
T
k l  m
(0, t ) k n k 1
n
.
k  1!
k 0

In the next step we prove the convergence lim e  n ( 0,T )  n  0 :
n 
2
n k 1 (0, t ) l n l  (t , T ) k l n k l
 l!
(k  l )!
k 1 k  1 l  0

lim e  n ( 0,T ) 
n 
 MC lim e
 n ( 0,T )
n 
 MC lim e
n ( 0 ,T )
n
k l
T
Ckml r m m
(1) i Cmi hmi l 1n  K ( x)x m dx 

m
m 1  (t , T ) i  0
T
n k 1 (0, t )l nl (t , T ) k  l n k l

 l!
(k  l )!
k 1 k  1 l  0

n k 1 (0, t )l n l (t , T ) k l n k l

 l!
(k  l )!
k 1 k  1 l 0

k l
Ckm l r m hml 1n


m
m 1  (t , T )
(5)
k l


 1  rhl 1n   1  0.
  (t , T ) 




It follows from

n k 1 (0, t )l n l (t , T ) k l n k l
n k (0, t )l n l (t , T ) k l n k l
 lim e n ( 0,T ) 


 l!
n
l!
(k  l )!
(k  l )!
k 1 k  1 l 0
k 1 k  1 l 0

lim e n ( 0,T ) 
n


n k 1 (0, T ) k
e n ( 0,T )  1  n(0, T )
1
 lim e n ( 0,T )

,
n

k!
(0, T )
(0, T )
k 1 k  1

 lim e n ( 0,T ) 
n

1
n k 1 (0, t )l nl (t , T ) k  l n k  l
 lim e  n ( 0,T ) 

 l!
(0, T ) n 
(k  l )!
k 1 k  1 l  0
 lim e
 n ( 0 ,T )
n 
rhl 1n 
n k 1 (0, t )l nl (t , T ) k  l n k  l 

1



l!
(k  l )!  (t , T ) 
k 1 k  1 l  0

k l

k l


1  rh2 n  
 lim e
 (t , T ) 
n 


exp n (0, t )  (t , T ) 1  rh2 n / (t , T )  n(0, T )
1
 lim

.
n 
(0, t )  (t , T ) 1  rh2 n / (t , T )
(0, T )
 n ( 0,T )

n k 1 (0, t )l nl (t , T ) k l n k l


l!
(k  l )!
k 1 k  1l  0






(6)
The theorem is proved.
Theorem 2 (the variance convergence). Let the assumptions of Theorem 1 hold,
T
K
T
2
~
( x) x m dx  M
m  1,2,...; ihi   as i   . Then lim Var ( S n )  0.
n
Proof. By PN n  2  1 as n   the variance Var ( S n ) 

 t  tin   t  t jn  
1
 1  N 1 2  t  tin 
 
  1 N 1  t  tin  







  
 E
K

2
K
K

   E   K 

2  2






h

 hin  i  j hinh jn  hin   h jn  
  N n i 1 hin  hin  

 N n   i 1 in
 
2
n
n
n k 1
(0, t  hin x)i (t  hin x, T )k i k  n ( 0,T )
2
K
(
x
)

(
t

h
x
)
n dxe

in
2 

(i  1)!
(k  i)!
k 1 k i 1 hin T

 C1 (n)

T
1 k 1 k
1
2  
k  2 k i 1 j  i 1 hin h jn
 2C2 (n)
i 1
j  i 1
t  y
t  x
 ( y, T ) k  j

( y ) (0, x) ( x, y )


K

(
x
)
K

0 x  hin 
 h 
(i  1)! ( j  i  1)! (k  j )!
 jn 
T T
  1 n 1 T  t  x   (0, x)i 1  ( x, T ) n  i


 n k dydxe n ( 0,T )  C3 (n)    K 
( x)dxe n ( 0,T ) ,
 n 1 n i 1 hin 0  hin  (i  1)! (n  i )!

 (t ) 

where Ci (n)  1 as n  , i  1,2,3. The second term in (7) convergences to 
 (0, T ) 
k 1 k
 (0, x) i 1  ( x, y ) j i 1  ( y, T ) k  j

since according to Newton’s polynom  
( j  i  1)! (k  j )!
i 1 j i 1 (i  1)!
3
(7)
2

(0, T ) k  2
(the main part is finding similarly (4)).
k  2 !
To
establish

n
2
k 1 k
Qn  

oe
k
1
the
h K
 n ( 0,T )
i 1

statement
T
2
( x)(t  hin x)
in T
of
the
theorem
(0, t  hin x) (t  hin x, T )
(i  1)!
(k  i)!
i
k i
we
must
show
n k dx 
k
n
(0, t )l nl (t , T ) k l n k l
as n   . Let us consider Qn  

2
l!
(k  l )!
k 0 k  1 l 0

n k 1 (0, t )l nl

2
l!
k 1 k  1 l  0

k  l 1

m0
(t , T ) m n m k l  m k l  m (1)i
n

m!
hin
i 0
T
2
 K ( x)
T
rhin x k l  m dx 
i!k  i  l  m !
1
(0, T ) n
1
Ei( n(0,T ))  ln n(0,T )  ,


(0, T ) k 1
kk!
(0, T )
where Ei( n ) – integral exponent,  – Euler constant, lim Ei( n)e n  0 [5]. Thus we obtain

 Qn(1)  Qn( 2 ) , Qn(1) 
k
k
n 
lim
n
Q
Qn(1)e  n
( 2)
n
 0. Similarly (5) we have
 (0, t ) l n l  (t , T ) k l n k l
2
l!
(k  l )! hl 1n
k 1 k  1 l  0
~ 
 MC 
n
k 1
n k 1  (0, t ) l n l  (t , T ) k l n k l
 l! (k  l )!(l  1)h
k 1 k  1 l  0
l 1n
~ 
 MC 
n k 1 (0, t ) l n l (t , T ) k l n k l
 l!
(k  l )!
k 1 k  1 l 0
~ 
 MC 
k l


rhl 1n 



 1   (t , T )   1  o(n) 



k l


rhl 1n 

  1  o(n) 
1

   (t , T ) 





k l


rhl 1n 



 1  (t , T )   1  o(n).



Taking (6) into account we have that lim Qn( 2) en  0. The theorem is proved.
n 
References
1. A. Kitayeva. Mean-square convergence of a kernel type estimate of the intensity function of
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This work was supported by FP “Scientific and scientific-pedagogical cadres of innovative
Russia”, project no. 402.740.11.5190
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