Time Spent below a Random Threshold by a Poisson Driven Sequence of Observations
Author(s): S. N. U. A. Kirmani and Jacek Wesołowski
Source: Journal of Applied Probability, Vol. 40, No. 3 (Sep., 2003), pp. 807-814
Published by: Applied Probability Trust
Stable URL: http://www.jstor.org/stable/3215955
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J. Appl. Prob. 40, 807-814 (2003)
Printed in Israel
? AppliedProbabilityTrust2003
TIME SPENT BELOW A RANDOM THRESHOLD BY
A POISSON DRIVEN SEQUENCE OF OBSERVATIONS
S. N. U. A. KIRMANI,*Universityof NorthernIowa
JACEKWESOLOWSKI,**PolitechnikaWarszawska
Abstract
Themeanandthevarianceof thetimeS(t) spentby a systembelowa randomthreshold
untilt areobtainedwhenthesystemlevelis modelledby thecurrentvalueof a sequence
of independent
andidenticallydistributed
randomvariablesappearing
at theepochsof a
Poissonprocess.Inthecaseof thehomogeneous
Poissonprocess,the
nonhomogeneous
asymptoticdistributionof S(t)/t as t -* oo is derived.
Poissonprocess;
Exceedancestatistics;limit theorems,nonhomogeneous
Keywords:
randomthreshold;
Poissondrivensequenceof observations;
orderstatistics
AMS2000SubjectClassification:
Primary60G55;60G70;60K10
60K37;62E15;62E20
Secondary
1. Introduction
If shocks occurringin time affect the level of an economic, financial,environmental,biological or engineeringsystem, then the proportionof time spent by the system below a threshold
is frequentlyof interest. If Yi denotes the system level duringthe periodbetween the (i - 1)th
and ith shocks and if N(t) counts the numberof shocks during[0, t] for t > 0, thenthe process
Zt = El
Yn1(N(t) = n - 1), t > 0, keeps track of the level of the system. Given any
t > 0, we are interestedin the proportionof time during[0, t] when the process Z = (Zt)t>o is
below the system's threshold.However,thereare situationsin which thereis no practicalway
to identify the system's thresholdwith certainty. Therefore,the thresholdwill be considered
a random variable X and interest will center on S(t)/t, where S(t) denotes the total time
during [0, t] that the process Z falls below X. Assuming that the process N = (N(t))t>o is
a nonhomogeneousPoisson process with N, Y = (Yi)i>l, and X independent,we obtain the
mean and varianceof S(t)/t in Section 2. The choice of the nonhomogeneousPoisson process
to model the time epochs of shocks was first proposed by Esary et al. [4]. We also prove,
in Section 3, that if the shock process N is homogeneous Poisson, then S(t)/t converges in
distributionto G(X), where G denotes the common distributionfunctionof the Yi.
A related but easier scheme of exceedance was proposed by Wesolowski and Ahsanullah
[8]. They investigated the exact and asymptotic distributionsof three statistics connected
with exceeding an independentrandomthresholdin a sequence of independentand identically
distributed(i.i.d.) observations.In particular,they considereda discrete analogueof our S(t).
Some additionaldistributionalpropertiesrelated to the exceedance scheme of [8] have been
recently studiedby Bairamovand Eryilmaz[I], Bairamovand Kotz [2] and Eryilmaz[3].
Received 22 November2000; revision received 4 April 2003.
* Postal address:Departmentof Mathematics,Universityof NorthernIowa, CedarFalls, IA 50614-0506, USA.
** Postal address:WydzialMatematykii Nauk Informacyjnych,PolitechnikaWarszawska,Warszawa00-661, Poland.
Email address:[email protected]
807
808
S. N. U. A. KIRMANI
ANDJ. WESOLOWSKI
2. Mean and variance
AssumingthatY = (Yi)i=l,2,...is a sequenceof i.i.d. observationswith commondistribution
function G, one of the statisticsconsideredin [8] was the numberS*(n) of observationsin a
sampleof size n falling below the randomlevel X, where X is independentof Y. It was proved
therethatthe conditionaldistributionof S*(n) given X is binomialwith parametersn andG(X).
Consequently,
E(S*(n)) = n E(G(X))
and
var(S*(n)) = n E(G(X)G(X)) + n2 var(G(X)),
where G = 1- G.
Here, we study similarcharacteristicsin the case when observationsarriveat the epochs of
a nonhomogeneousPoisson process N = (N(t))t,o which is independentof (Y, X).
The mainobjectof ourinterestis the process Z = (Z(t))t>o which keeps trackof the current
Yj as follows:
00
Ynl(N(t) = n - 1),
Z(t) =
t > 0.
n=1
Denote by S(t) the time spent by the process Z below the level X up to the time t, and by
Wi, i = 0, 1,..., the interarrivaltimes of the process N, that is, Wi = Vi+l - Vi, where
Vi = inf{t > 0 : N(t) = i} is the ith epoch of N, i = 0, 1, 2.... Then S(t) has the form
N(t)-I
S(t) =
- i=
N(t)-I
Wi l(Yi+ <X) + (t
-
E
Wi) l(YN(t)+, <X)
l(N(t) > 1)
i=
+ t 1(YN(t)+l < X) 1(N(t) = 0)
N(t)-I
= ( E Wi[1(Yi+ < X) - I(YN(t)+I <
X)])
i=O
+ t l(YN(t)+l < X).
(N(t) > 1)
(1)
Though we are unable to derive the exact distributionof S(t), the first two moments are
computableand similarin form to the correspondingresultsin [8].
Proposition 1. In the model definedabove,for any t > 0,
E(S(t)) = tE(G(X)),
var(S(t)) =
(2)
(t) E(G(X)G(X)) + t2 var(G(X)),
(3)
where
X(t) = 2
P((y
Ny)
N(x))dxdy < t2.
<x<y<t
Before provingthe above proposition,we presenta result on orderstatistics which will be
used lateron in the proof.
809
Timespentbelowa randomthreshold
Lemma 1. Let X:n, ..., Xn:nbe orderstatisticsfroman i.i.d. samplewithdistributionfunction
F having support[0, a]. Then
(Xi:n - Xi-i:n)2 + (a - Xn:n)2 = 2
E X1:n +
-
-
i=2
(F(x) + F(y))n dx dy. (4)
/
<x<y<a
Proof Observe first that, for any square integrablerandom variable X with distribution
function F having support[0, a],
E(X2) = 2
yF(y) dy = 2 f
J~O
F(y) dx dy.
JJO~<x<y<a
Consequently,
E((a - X)2) = 2<x<ya
F(a - y) dx dy
=2 O<x<y<a
=2r
F(x)dxdy.
JJo<x<y<a
We proceedby inductionwith respectto n. For n = 1, the resulthas just been proved. Denote
the left-handside of (4) by Ln. Then
n
- (
Ln = E E( X2
+ >(Xi:n
i=2
-
Xil:n)2
- Xn:n)2
Xn:n) + (a
It is known (see, for instance, [6, Chapter4]) that the conditionaldistributionof (Xl:n,....
Xn-l:n) given Xn:n = x is the same as the joint distribution of order statistics from an i.i.d.
sample of size n - 1 based on the distributionfunction
GX(u)=
GF(u)=
u) foru <x,
F(x)
foru > x.
1
Then, by the inductionassumption,it follows that
Ln = E(2
= 2
[GXn:n
+ Gxn,:(y)]-1 dxdy + (a - Xn:n)2)
(x) +
<x<y<Xn:n
y
[Gu(x) + G(y)]n-l
(
If1
= 21
(n
<x<y<a
dFn:n(u) + Fn:n(x)) dx dy
y
O<x<y<a
[t + F(x) - F(y)]"-1 dt + Fn(x)) dx dy,
F(y)
which immediatelyimplies the result.
ANDJ. WESOLOWSKI
N. U. A. KIRMANI
810
~~~~~~~~~S.
810
Proof of Proposition1. By the independenceproperties,we have
oc n
E(S(t))
1
< X)
E(Wi [1(Y,I?
=L
1(Y,,+ < X)] IN(t) = n) P(N(t) =n)
-
n=I i=O
00
+ tLEE(1(Yn+i < X) I N(t) = n) P(N(t) =n)
n=O
I
oo n
=L
E(Wj I N(t)
n)[E(1(Y1?l < X))
=
-
E(1(Yn+l < X))] P(N(t)
=n)
n=1 i=0
+ tE (G (X)).
Theformula(2) follows sinceE(1(Y?i+I< X)) -E (1(Yn + < X)) = E(G (X)) -E (G (X)) = 0.
In orderto find the varianceof S(t), we firstcomputeits second conditionalmomentgiven
N(t)
=
n
0:
>
nlI
E(S2 (t)
I N (t)
n)
=
N(t) =n) E((Ii+ I
: EW2I
=
In+)1)2
i-O
n-I
+ >jE(Wi Wj
N(t)
=n)
-
E((Ii+1
-
In?1)(Ij-+l
I+)
i#i
n-I
+ 2t
2...E(W1
N(t)
E(In?1 (Ij?I _ In?+,)) ? t2 E(In+1),
=n)
i-O
where Ij
=
1(Yj
<
X) for j
E
EIil-
-(I
Observethat
2.
=1,
I1
1)2)
-
0
2 E(G (X)Gij(X)),
0 <i, j
In+j)) = E(G(X)G(X)),
In,(jlE(I+I(I+I-
=
In+,))
<i
0
- E(G(X)G(X)),
<i
n,
<
<
n,i :Aj
<
n.
Consequently,
E(S2(t) I N(t)
=n)
E
E(G(X) G(X)) E((t
?
t2
-
V
I N(t)
=n)
+E
:w7
N (t)
=n
E(G2(X))
for n > 0. Setting Eformulaalso holdsforn
Hence, for any t > 0,
=
=O0and recallingthat Vo =0 almost surely,we find that the above
) tE(G
E (X)).
N(t)
Osince,by directcomputation,E(S2(t) N
N(t)-1
var(S (t))
=
E(G (X) G(X)) E((t
-
-
? E(
VN(t))
~
,
var(G(X))
N (t)
=E (G(X) G(X)) E VI?LV
i -2
_1)2+(t-_VN(t))2
+t2var(G(X)).
Timespentbelowa randomthreshold
811
Now to get (3) it suffices to compute X(t). In orderto do that, we recall (see, for instance,
[7, Theorem 12.2.1]) that the conditional distributionof the random vector (V1,..., VN(t))
given N(t) = n is equal to the joint distributionof the orderstatistics of a randomsample of
size n from the distribution
0,
x<0,
^(x)
OA
H(x)=
-<x <t,
A(t)'
1,
t <x,
where A is the mean value function of the Poisson process. Now applying Lemma 1 and
changingthe orderof integration,we obtainthat
N(t)
(-E
E(Ev2+
V2+E(Vi-Vi_
i=2
k
= 2
y|
V,_1)2 +(t-V
+ (t)2
n)2 N(t) )
dx dy
E([H(x) + H(y)]N())
<x<y<t
= 2|
+ H(y) - 1]) dx dy
exp(A(t)[H(x)
yt
<x<y<t
= 2
exp(-[A(y)
- A(x)]) dx dy.
<x<y<t
Remark 1. ObservethatE(S(t)) does not dependon the parametersof the drivingprocess N.
On the other hand, var(S(t)) dependson the intensity A of the Poisson process and increases
in t at most as a quadraticfunction.
Remark 2. If N is a homogeneous Poisson process with mean value function A(t) = Xt,
where X is a positive constant,then
t-Y
(t
e-xw dw dy =
X(t) = 2f J
2
(e-t
- 1 + Xt).
In case N is a nonhomogeneousPoisson process with A(t) = log(l + t),
rt rt-Y
(t)=
1
14 y
+y
+y+w
2
dwdy = t + -
2
log(l
+ t).
Note that X(t)/t2 -> 0 as t -- oo when N is homogeneous,but not when A(t) = log(l + t).
3. Asymptotic distribution
It was shown in [8] that S*(n)/n converges in distributionto G(X), S*(n)/n -n G(X) as
n -> oo. Here the formulaefor E(S(t)) and var(S(t)) suggest that a similarresult cannot be
true unless X(t)/t2 convergesto 0 as t -> oo. At the same time it is reasonableto conjecture
that S(t)/t
-
G(X) if X (t)/t2 -
0. We are not able to answer this question in full generality;
however,in the case of the homogeneousPoisson process, the answeris affirmative.
Proposition 2. If N is a homogeneousPoissonprocess, then
S(t)
-S
t
G(X) o as t -- o<.
S. N. U. A. KIRMANIAND J. WESOLOWSKI
812
Proof Let Ij = 1(Yj < X) for j = 1, 2,.. .and rewrite (1) as S(t) = S1 (t+
where
S2(t)+ S3(t,
N(t)-
Sl(t) =
1) E
Wjlj+l,
j=o
1(N(t)
S2(t) = (t - VN(t))IN(t)+I
(N(t) > 1)
and
S3(t) = tlN(t)+l 1(N(t) = 0).
Clearly, S2(t) < t - VN(t). But the distribution of t - VN(t) is known; see, for instance, [5].
Thus, for any e > 0 and t sufficiently large,
-
Pt
- 0 as t Consequently, S2(t)/t
Further, for any E > 0,
VN(t) >)
=e-xt
oo.
S3 (t)
= P(IN(t)+l 1(N(t) = 0) >
)
< P(N(t) = 0)
=
e -Xt
as t -- oo;
-0
0.
hence S3(t)/t
Observe now that l(N(t) > 1) -> 1 as t -> oo. Consequently, to prove that S (t)/t G(X), it suffices to prove that
N(t)-I
-
E
j=0
Wjj+il
G(X).
This will be done in two steps.
Denoting by X the intensity of the Poisson process, we will first prove that
LtJ-I
j=o
W
w 1j+l
G(X)
where L-Jis the integer-part function. Note that the sequence (Wj Ij+l)j=o, ... is conditionally
i.i.d. given X and E(WjIj+l I X) = G(X)/X. Then, by the law of large numbers, for any
real z,
LJL-1
E exp(
E
WjIj+
j=o
(L
-EE(exP
tJ
l
iz
Lw
Lxt j-
/ +,) E
)
) -ast
-* oc.
813
Timespent below a randomthreshold
Secondly, we will show that, as t -+ oo,
[AXt
N(t)-l
(
- E
W,I
Wjij+i)p o0.
j=0
j=0
To this end, observe that
N(t)
___=
L[tJ
N(t) At p
-- 1
At [LtJ
as t --oo.
Let
< N(t) < (l +
AE = {(1 -e)LtJ
)LXAtJ for any E >
and
= E(G(X))2-E(G(X))
X2
Then, for sufficiently large t, P(A)) > 1 e. Thus,
a2 = var(Woi)
lAtJ-1
N(t)-l
p(_
WjIj+l-
E
E
j=0
LtJ-1
N(t)-l
<P(
> E1
WjIj+l
j=0
WjIj+l
E
-
E
WjIj+I
j =0
j =0
N(t)-1
LAtJ-1
> tel
n A)
t
n A,
+E.
But
E
P(
j =0
WjjI+1- E
>L wj+l
j =0
max{N(t), LtJ)-1
=P(
WIj+l > t?l n A.)
E
j=min(N(t), LXtJ)
KL(1+)
L-AJJ - 1
<P(
E
wjIj+i>
j=L(1-e)L),tJJ
(5[ E P 7J j>
t? }n Ae)
)
j=O
<
2eXta2
-2 E t2
0
as t -+oo.
This completes the proof of the theorem.
Remark 3. Observe that, in the case of nonrandom threshold, the convergence in Proposition 2
holds in probability.
Acknowledgement
The authors are indebted to the referee for helpful remarks, particularly one which led to
Lemma 1.
814
S. N. U. A. KIRMANI AND J. WESOLOWSKI
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