# An CUSUM Test with Observation-Adjusted Control Limits in Change Detection

Fuquan Tang, Dong Han∗
Department of Statistics, Shanghai Jiao Tong University,
Shanghai 200030, P. R. China.

###### ABSTRACT\\\

In this paper, we not only propose an new optimal sequential test of sum of
logarithmic likelihood ratio (SLR) but also present the CUSUM sequential test
(control chart, stopping time) with the observation-adjusted control limits
(CUSUM-OAL) for monitoring quickly and adaptively the change in distribution
of a sequential observations. Two limiting relationships between the optimal
test and a series of the CUSUM-OAL tests are established. Moreover, we give
the estimation of the in-control and the out-of-control average run lengths
(ARLs) of the CUSUM-OAL test. The theoretical results are illustrated by
numerical simulations in detecting mean shifts of the observations sequence.

††footnotetext: ∗Supported by National Natural Science Foundation of China
(11531001)
∗Corresponding author, E-mail<EMAIL_ADDRESS>
$\,\,MSC\,2010\,subject\,classification$. Primary 62L10; Secondary 62L15

_Keywords_ : Optimal sequential test, CUSUM-OAL test, change detection.

##  1 INTRODUCTION

In order to quickly detect a change in distribution of observations sequence
without exceeding a certain false alarm rate, a great variety of sequential
tests have been proposed, developed and applied to various fields since
Shewhart (1931) proposed a control chart method, see, for example, Siegmund
(1985), Basseville and Nikiforov (1993), Lai (1995, 2001), Stoumbos _et al_.
(2000), Chakraborti _et al_. (2001), Bersimis _et al_. (2007), Montgomery
(2009), Qiu (2014), Tartakovsky _et al_. (2015), Woodall _et al_. (2017),
Bersimis _et al_. (2018) and Chakrabortia and Graham (2019).

One of popular used sequential tests is the following upper-sided CUSUM test
which was proposed by Page (1954).

$T_{C}(c)=\min\\{n\geq 0:\,\,\max_{1\leq k\leq n}\sum_{i=n-k+1}^{n}Z_{i}\geq
c\\},$ (1)

where $c>0$ is a constant control limit,
$Z_{i}=\log[p_{v_{1}}(X_{i})/p_{v_{0}}(X_{i})]$, $p_{v_{0}}(x)$ and
$p_{v_{1}}(x)$ are pre-change and post-change probability density functions
respectively for a sequence of mutually independent observations
$\\{X_{i},\,i\geq 1\\}$, that is, there is a unknown change-point $\tau\geq 1$
such that $X_{1},...,X_{\tau-1}$ have the probability density function
$p_{v_{0}}$, whereas, $X_{\tau},X_{\tau+1},...$ have the probability density
function $p_{v_{1}}$. By the renewal property of the CUSUM test $T_{C}$ we
have $sup_{k\geq 1}\textbf{E}_{k}(T_{C}-k+1|T_{C}\geq
k)=\textbf{E}_{1}(T_{C})$ (see Siegmund 1985, P.25), where
$\textbf{E}_{1}(T_{C})$ is the out-of-control average run length (ARL1),
$\textbf{P}_{k}$ and $\textbf{E}_{k}$ denote the probability and expectation
respectively when the change from $p_{v_{0}}$ to $p_{v_{1}}$ occurs at the
change-point $\tau=k$ for $k\geq 1$.

Though we know that the CUSUM test is optimal under Lorden’s measure (see
Moustakides 1986 and Ritov 1990), the out-of-control ARL1 of the CUSUM test is
not small, especially in detecting small mean shifts ( see Table 1 in Section
4). In other words, the CUSUM test is insensitive in detecting small mean
shifts. Then, how to increase the sensitivity of the CUSUM test ? Note that
the control limit in the CUSUM test is a constant $c$ which does not depend on
the observation samples. Intuitively, if the control limit of the CUSUM test
can become low as the samples mean of the observation sequence increases, then
the alarm time of detecting the increasing mean shifts will be greatly
shortened. Based on this idea, by selecting a decreasing function $g(x)$ we
may define the ( upper-sided ) CUSUM chart $T_{C}(cg)$ with the observation-
adjusted control limits $cg(\hat{Z}_{n})$ ( abbreviated to the CUSUM-OAL chart
) in the following

$T_{C}(cg)=\min\\{n\geq 0:\max_{0\leq k\leq n}\sum_{i=n-k+1}^{n}Z_{i}\geq
cg(\hat{Z}_{n})\\},$ (2)

where $c>0$ is a constant and $\hat{Z}_{n}=\sum_{i=1}^{n}Z_{i}/n$. In other
words, the control limits $cg(\hat{Z}_{n})$ of the CUSUM-OAL test can be
adjusted adaptively according to the observation information
$\\{\hat{Z}_{n}\\}$. Note that the control limits $cg(\hat{Z}_{n})$ may be
negative. In the special case, the CUSUM-OAL chart $T_{C}(cg)$ becomes into
the conventional CUSUM chart $T_{C}(c)$ in (1) when $g\equiv 1$. Similarly, we
can define a down-sided CUSUM-OAL test. In this paper, we consider only the
upper-sided CUSUM-OAL test since the properties of the down-sided CUSUM-OAL
test can be obtained by the similar method.

The main purpose of the present paper is to show the good detection
performance of the CUSUM-OAL test and to give the estimation of its the in-
control and out-of-control ARLs.

The paper is organized as follows. In Section 2, we first present an optimal
SLR sequential test, then define two sequences of the CUSUM-OAL tests and
prove that one of the two sequences of CUSUM-OAL tests converges to the
optimal test, another sequences of CUSUM-OAL tests converges to a combination
of the optimal test and the CUSUM test. The estimation of the in-control and
out-of-control ARLs of the CUSUM-OAL tests and their comparison are given in
Section 3. The detection performances of the three CUSUM-OAL tests and the
conventional CUSUM test are illustrated in Section 4 by comparing their
numerical out-of-control ARLs. Section 5 provides some concluding remarks.
Proofs of the theorems are given in the Appendix.

##  2  AN OPTIMAL SLR TEST, TWO CUSUM-OAL TESTS AND THEIR LIMITING
RELATIONSHIPS

Let $\textbf{P}_{0}$ and $\textbf{E}_{0}$ denote the probability and the
expectation respectively with the probability density $p_{v_{0}}$ when there
is no change for all the time. It is known that
$\textbf{E}_{0}(Z_{1})=\textbf{E}_{v_{0}}(\log[p_{v_{1}}(X_{1})/p_{v_{0}}(X_{1})])<0$
for $p_{v_{1}}(x)\neq p_{v_{0}}(x)$. Without loss of generality, let
$\mu_{0}=\textbf{E}_{0}(Z_{1})<0$.

It follows from Proposition 2.38 in Siegmund (1985) and (5.8)-(5.9) in Chow
_et al_ , P.108) that the following sequence test of sum of logarithmic
likelihood ratio (SLR)

$\displaystyle T_{SLR}$ $\displaystyle=$ $\displaystyle\min\\{n\geq
1:\,\prod_{k=1}^{n}\frac{p_{v_{1}}(X_{k})}{p_{v_{0}}(X_{k})}\,\geq
B\,\\}=\min\\{n\geq 1:\,\sum_{j=1}^{n}Z_{j}\geq c\\}$ (3) $\displaystyle=$
$\displaystyle\min\\{n\geq 1:\,\sum_{j=1}^{n}(Z_{j}-\mu_{0})\geq
c+n|\mu_{0}|\\}$ (4)

for $B>1$, is optimal in the following sense

$\displaystyle\min_{T:\,\,\textbf{P}_{0}(T<\infty)\geq\alpha}\\{\textbf{E}_{1}(T)\\}=\textbf{E}_{1}(T_{SLR})$

for $\textbf{P}_{0}(T_{SLR}<\infty)=\alpha$, where $c=\log B$ and
$0<\alpha<1$.

In particular, if $\textbf{P}_{0}$ is the standard normal distribution with
mean shift $\mu>0$ after change-point, we have $Z_{j}-\mu_{0}=\mu X_{j}$,
where $\mu_{0}=-\mu^{2}/2$. It follows from proposition 4 in Frisén (2003)
that the SLR test $T_{SLR}$ in (4) is also optimal (minimal ARL1) with the
same false alarm probability $\textbf{P}_{0}(T<\tau)$.

It can be seen that the in-control average run length of $T_{SLR}$ is
infinite, that is, ARL${}_{0}=\textbf{E}_{0}(T_{SLR})=\infty$. However, the
minimal ARL1 with finite ARL0 is a widely used optimality criterion in
statistical quality control (see Montgomery, 2009) and detection of abrupt
changes (see Basseville and Nikiforov, 1993). In order to get finite ARL0 for
$T_{SLR}$, we replace the constant control limit $c$ of $T_{SLR}$ in (3) or
(4) with the dynamic control limit $n(\mu_{0}-r)$ and obtain a modified SLR
test $T_{SLR}(r)$ in the following

$T_{SLR}(r)=\min\\{n\geq 1:\,\sum_{j=1}^{n}Z_{j}\geq
n(\mu_{0}-r)\\}=\min\\{n\geq 1:\,\sum_{j=1}^{n}(Z_{j}-\mu_{0})\geq-rn\\}$ (5)

for $r\geq 0$.

For comparison, the in-control $\textbf{ARL}_{0}$ of all candidate sequential
tests are constrained to be equal to the same desired level of type I error,
the test with the lowest out-of-control $\textbf{ARL}_{v}$ has the highest
power or the fastest monitoring (detection) speed.

In the following example 1, the numerical simulations of the out-of-control
ARLs of the CUSUM-OAL tests $T_{C}(cg_{u,0})$ in detecting the mean shifts of
observations with normal distribution will be compared with that of the SLR
tests $T^{*}(r)$ and $T^{*}(0)$, and that of the CUSUM-SLR test
$T_{C}(c)\wedge T^{*}(0):=\min\\{T_{C}(c),\,T^{*}(0)\\}$ in the following
Table 1. These comparisons lead us to guess that there are some limiting
relationships between $T_{C}(cg_{u,r})$ and $T^{*}(r)$, and
$T_{C}(c\widetilde{g_{u}})$ and $T_{C}(c)\wedge T^{*}(0)$, respectively.

Example 1. Let $X_{1},X_{2},....$ be mutually independent following the normal
distribution $N(0,1)$ if there is no change. After the change-point $\tau=1$,
the mean $\textbf{E}_{\mu}(X_{k})$ ( $k\geq 1$ ) will change from $v_{0}=0$ to
$v=0.1,0.25,0.5,0.75,1,1.25,1.5,3$. Here, we let
$p_{v_{0}}(x)=e^{-x^{2}/2}/\sqrt{2\pi}$,
$p_{v_{1}}(x)=e^{-(x-1)^{2}/2}/\sqrt{2\pi}$ and therefore, $Z_{k}=X_{k}-1/2$
for $k\geq 1$, where $v_{1}=1$ is a given reference value which for the CUSUM
test is the magnitude of a shift in the process mean to be detected quickly.
We conducted the numerical simulation based on 1,000,000 repetitions.

The following Table 1 lists the simulation results of the ARLs of the tests
$T_{C}(c)$, $T_{C}(c\widetilde{g_{u}})$ for $u=1,10,10^{2},10^{3},10^{4}$,
$T^{*}(0.0007)$, $T_{C}(c)\wedge T^{*}(0)$ and $T^{*}(0)$ for detecting the
mean shifts, where the mean shift $0.0$ means that there is no change which
corresponds to the in-control ARL0 and all tests have the common
ARL${}_{0}\approx 1000$ except the test $T^{*}(0)$ which has
ARL${}_{0}=\infty$. The values in the parameters are the standard deviations
of the tests.

Table 1. ARLs of $T_{C}(c)$, $T_{C}(c\widetilde{g_{u}})$, $T^{*}(r)$,
$T_{C}(c)\wedge T^{*}(0)$ and $T^{*}(0)$ when $\tau=1$.
---
Tests | u, r |  |  |  | Shifts |  |  |  |
| c | 0.0 | 0.1 | 0.25 | 0.5 | 0.75 | 1.0 | 1.5 | 3.00
$T_{C}(c)$ |  | 1000.59 | 439.00 | 147.72 | 38.91 | 17.32 | 10.50 | 5.82 | 2.61
| c=5.0742 | (993.16) | (431.89) | (140.25) | (31.79) | (11.23) | (5.49) | (2.26) | (0.66)
| $u=1.0$ | 1000.43 | 237.34 | 46.84 | 10.02 | 4.39 | 2.65 | 1.52 | 1.03
| $c=5.6125$ | (1510.55) | (391.64) | (79.68) | (15.37) | (5.72) | (2.90) | (1.10) | (0.14)
| $u=10.0$ | 999.17 | 18.16 | 4.57 | 2.19 | 1.57 | 1.31 | 1.11 | 1.01
| $c=7.7790$ | (4969.50) | (77.77) | (11.06) | (3.01) | (1.45) | (0.86) | (0.38) | (0.04)
| $u=10^{2}$ | 1000.66 | 8.15 | 3.38 | 1.92 | 1.47 | 1.27 | 1.11 | 1.01
| $c=9.97$ | (13951.15) | (29.90) | (6.98) | (2.33) | (1.20) | (0.73) | (0.34) | (0.04)
| $u=10^{3}$ | 1001.49 | 7.56 | 3.28 | 1.89 | 1.46 | 1.26 | 1.11 | 1.01
| $c=11.38$ | (25423.13) | (26.45) | (6.60) | (2.27) | (1.18) | (0.72) | (0.33) | (0.04)
| $u=10^{4}$ | 1001.32 | 7.51 | 3.26 | 1.89 | 1.45 | 1.24 | 1.11 | 1.01
$T_{C}(c\widetilde{g_{u}})$ | $c=11.84$ | (30042.90) | (26.22) | (6.56) | (2.27) | (1.17) | (0.72) | (0.33) | (0.04)
| $r=0.0007$ | 1001 | 7.44 | 3.25 | 1.88 | 1.45 | 1.25 | 1.09 | 1.01
$T^{*}(r)$ |  | (43017) | (26.07) | (6.57) | (2.24) | (1.18) | (0.71) | (0.33) | (0.04)
| $r=0$ | 1001.13 | 7.52 | 3.25 | 1.88 | 1.45 | 1.25 | 1.09 | 1.01
$T_{C}(c)\wedge T^{*}(0)$ | $c=11.9271$ | (31090.71) | (26.88) | ( 6.57) | (2.24) | (1.17) | (0.72) | (0.33) | (0.04)
| $r=0$ | $\infty$ | 7.47 | 3.25 | 1.88 | 1.45 | 1.25 | 1.09 | 1.01
$T^{*}(0)$ |  |  ($\infty$) | (26.24) | (6.54) | (2.27) | (1.18) | (0.72) | (0.33) | (0.04)

From the last row in Table 1, it’s a little surprising that though the ARL0 of
$T^{*}(0)$ is infinite, that is, $\textbf{E}_{0}(T^{*}(0))=\infty$, the
detection speed of $T^{*}(0)$ is faster than that of the CUSUM chart $T_{C}$
for all mean shifts, in particular, for detecting the small mean shift 0.1,
the speed of $T^{*}(0)$ is only 7.47 which is very faster than the speed, 439,
of the CUSUM test. Moreover, both control charts $T^{*}(0.0007)$ and
$T_{C}(11.9271)\wedge T^{*}(0)$ not only have the nearly same detection
performance as $T^{*}(0)$ but also can have the finite in-control ARL0. Note
particularly that when the number $u$ in $\widetilde{g_{u}}$ is taken from $0$
to $1,10,10^{2},10^{3},10^{4}$, the detection speed of
$T_{C}(c\widetilde{g_{u}})$ is getting faster and faster, approaching to that
of $T_{C}(c)\wedge T^{*}(0)$. This inspires us to prove the following
theoretic results.

Theorem 2. Let $\tau=1$ and $\\{X_{k},k\geq 1\\}$ be an i.i.d. observations
sequence with $\mu_{0}=\textbf{E}_{0}(Z_{1})<0$. Then

$\displaystyle\lim_{u\to\infty}T_{C}(cg_{u,r})=T^{*}(r),\,\,\,\,\,\,\,\,\lim_{u\to\infty}T_{C}(c\widetilde{g_{u}})=T_{C}(c)\wedge
T^{*}(0).$ (6)

Theorem 2 shows that when the constant control limit $c$ of the CUSUM test
$T_{C}(c)$ is replaced with the observation-adjusted control limits
$\\{cg_{u,r}(\hat{Z}_{n})\\}$ and $\\{c\widetilde{g_{u}}(\hat{Z}_{n})\\}$
respectively, the corresponding two CUSUM-OAL tests $\\{T_{C}(cg_{u,r})\\}$
and $\\{T_{C}(c\widetilde{g_{u}})\\}$ will converge to the optimal SLR test
$T^{*}(r)$ and the CUSUM-SLR test $T_{C}(c)\wedge T^{*}(0)$ as $u\to\infty$,
respectively. In other words, the fastest alarm times that
$\\{T_{C}(cg_{u,r})\\}$ and $\\{T_{C}(c\widetilde{g_{u}})\\}$ can be reached
are $T^{*}(r)$ and $T_{C}(c)\wedge T^{*}(0)$, respectively.

Remark 2. Since $T_{C}(cg_{0,r})=T_{C}(c\widetilde{g_{0}})=T_{C}(c)$ when
$u=0$, it follows that both $\\{T_{C}(cg_{u,r}):\,u\geq 0\\}$ and
$\\{T_{C}(c\widetilde{g_{u}}):\,u\geq 0\\}$ can be seen as two ”long bridges”
connecting $T_{C}(c)$ and $T^{*}(r)$, and $T_{C}(c)$ and $T_{C}(c)\wedge
T^{*}(0)$, respectively.

##  3 ESTIMATION AND COMPARISON OF ARL OF THE CUSUM-OAL TEST

In this section we will give an estimation of the ARLs of the following CUSUM-
OAL test that can be written as

$T_{C}(cg)=\min\\{n\geq 1:\max_{1\leq k\leq n}\sum_{i=n-k+1}^{n}Z_{i}\geq
cg(\hat{Z}_{n}(ac))\\},$ (7)

where $g(.)$ is a decreasing function,
$\hat{Z}_{n}(ac)=\frac{1}{j}\sum_{i=n-j+1}^{n}Z_{i},j=\min\\{n,[ac]\\}$ for
$ac\geq 1,a>0,c>0$, and $[x]$ denotes the smallest integer greater than or
equal to $x$. Here $\hat{Z}_{n}(ac)$ is a sliding average of the statistics,
$Z_{i},n-j+1\leq i\leq n$, which will become
$\hat{Z}_{n}=\frac{1}{n}\sum_{i=1}^{n}Z_{i}$ when $a=\infty$.

Next we discuss on the the post-change probability distribution in order to
estimate the ARLs of $T_{C}(cg)$.

Usually we rarely know the post-change probability distribution $P_{v}$ of the
observation process before it is detected. But the possible change domain and
its boundary (including the size and form of the boundary) about $v$ may be
determined by engineering knowledge, practical experience or statistical data.
So we may assume that the region of parameter space $V$ and a probability
distribution $Q$ on $V$ are known. If we have no prior knowledge of the
possible value of $v$ after the change time $\tau$, we may assume that $v$
occurs equally on $V$, that is, the probability distribution $Q$ is an equal
probability distribution (or uniform distribution ) on $V$. For example, let
$P_{v}$ be the normal distribution and $v=(\mu,\sigma)$, where $\mu$ and
$\sigma$ denote the mean and standard deviation respectively, we can take the
set
$V=\\{(\mu,\sigma):\mu_{1}\leq\mu\leq\mu_{2},0<\sigma_{1}\leq\sigma\leq\sigma_{2}\\}$
and $Q$ is subject to the uniform distribution $U(V)$ on $V$ if $v$ occurs
equally on $V$, where the numbers $\mu_{1},\mu_{2},\sigma_{1}$ and
$\sigma_{2}$ are known. It means that we know the domain of the possible post-
change distributions, $P_{v},v\in V$, i.e., the boundary $\partial{V}$ of the
parameter space $V$ is known.

Next we shall divide the parameter space $V$ into three subsets $V^{+}$,
$V^{0}$ and $V^{-}$ by the Kullback-Leibler information distance. Let

$\displaystyle
V^{-}=\\{v:E_{v}(Z_{1})<0\\},\,\,\,\,\,\,V^{0}=\\{v:E_{v}(Z_{1})=0\\},\,\,\,\,\,\,V^{+}=\\{v:E_{v}(Z_{1})>0\\}$

where $E_{v}(Z_{1})=I(P_{v}|P_{v_{0}})-I(P_{v}|P_{v_{1}})$ and

$\displaystyle
I(P_{v}|P_{v_{0}})=E_{v}(\log[\frac{P_{v}(X_{1})}{P_{v_{0}}(X_{1})}]),\,\,\,\,\,\,\,\,I(P_{v}|P_{v_{1}})=E_{v}(\log[\frac{P_{v}(X_{1})}{P_{v_{1}}(X_{1})}])$

are two Kullblak-Leibler information distances between $P_{v}$, $P_{v_{0}}$
and $P_{v}$, $P_{v_{1}}$. Since $I(p|q)=0$ if and only if $p=q$, where $p$ and
$q$ are two probability measures, it follows that
$E_{v_{0}}(Z_{1})=-I(P_{v_{0}}|P_{v_{1}})$, and therefore, $v_{0}\in V^{-}$
when $P_{v_{1}}\neq P_{v_{0}}$. When $v\in V^{-}$, i.e.,
$I(P_{v}|P_{v_{0}})<I(P_{v}|P_{v_{1}})$, it means that $P_{v}$ is closer to
$P_{v_{0}}$ than to $P_{v_{1}}$ according to the Kullblak-Leibler information
distance. There is a similar explanation for $v\in V^{+}$ or $\in V^{0}$.

Suppose the post-change distribution $P_{v}$ and the function $g(x)$ satisfy
the following conditions:
(I) The probability $P_{v}$ is not a point mass at $E_{v}(Z_{1})$ and
$P_{v}(Z_{1}>0)>0$.
(II) The moment-generating function $h_{v}(\theta)=E_{v}(e^{\theta Z_{1}})$
satisfies $h_{v}(\theta)<\infty$ for some $\theta>0$.
(III) The function $g(x)$ is decreasing, its second order derivative function
$g^{\prime\prime}(x)$ is continuous and bounded, and there is a positive
number $x^{*}$ such that $g(x^{*})=0$.

Let $\tilde{Z}_{1}=Z_{1}+|g^{\prime}(\mu)|(Z_{1}-\mu)/a$,
$\tilde{h}(\theta)=\textbf{E}_{v}(e^{\theta\tilde{Z}_{1}})$ and

$\displaystyle
H_{v}(\theta)=\frac{au}{g(\mu)}\ln\tilde{h}_{v}(\theta)+(1-\frac{au}{g(\mu)})\ln
h_{v}(\theta)$

for $\theta\geq 0$, where $E_{v}(Z_{1})=\mu<0$, $a\leq g(\mu)/u$ and
$u=H^{\prime}_{v}(\theta^{*}_{v})$, where $\theta^{*}_{v}>0$ satisfies
$H_{v}(\theta^{*}_{v})=0$. Note that $H_{v}(\theta)$ is a convex function and
$H^{\prime}_{v}(0)=\mu<0$. It follows that there is a unique positive number
$\theta^{*}_{v}>0$ such that $H_{v}(\theta^{*}_{v})=0$.

Note that the following function

$\displaystyle\Theta(x)=\theta(\frac{1}{x})-xH_{v}(\theta(\frac{1}{x}))-2\theta^{*}_{v}$

satisfies that $\Theta(1/u)=\theta(u)-2\theta^{*}_{v}=-\theta^{*}_{v}$, where
$\theta(u)=\theta^{*}_{v}$,

$\displaystyle\Theta^{\prime}(x)=-\frac{1}{x^{2}}\theta^{\prime}(\frac{1}{x})-H_{v}(\theta(\frac{1}{x}))+\frac{1}{x}\theta^{\prime}(\frac{1}{x})H_{v}^{\prime}(\theta(\frac{1}{x}))=-H_{v}(\theta(\frac{1}{x}))$

and therefore,
$\Theta^{\prime}(\theta(u))=-H(\theta(u))=-H(\theta^{*}_{v})=0$,
$\Theta^{\prime}(\theta(1/x))>0$ for $x>1/u$ and
$\Theta^{\prime}(\theta(1/x))<0$ for $x>1/u$. Hence, there exists a positive
number $b$ defined in (LABEL:f08).

It can be seen, the main part of $\textbf{ARL}_{v}(T_{c}(g))$ will be an
exponential function, square function, and linear function of $c$ when the
process $\\{Z_{k}:k\geq 0\\}$ has no change or a ”small change”, a ”medium
change” and a ”large change” from $P_{v_{0}}$ to $P_{v}$, respectively. Here,
the ”small change” ($v\in V^{-}$) means that $P_{v}$ is closer to $P_{v_{0}}$
than to $P_{v_{1}}$, i.e., $I(P_{v}|P_{v_{0}})<I(P_{v}|P_{v_{1}})$, and the
”large change” is just the opposite. The ”medium change” ($v\in V^{0}$)
corresponds to $I(P_{v}|P_{v_{0}})=I(P_{v}|P_{v_{1}})$.

In this paper, we will use another method to prove Theorem 3 since Wald’s
identity and the martingale method do not hold or can not work for showing the
ARLs estimation of the test $T_{c}(g)$ when $g$ is not constant.

Next we compare the detection performance of the CUSUM-OAL test
($\textbf{ARL}_{v}(T_{c^{\prime}}(g))$) with that of the CUSUM test
($\textbf{ARL}_{v}(T_{C}(c))$) by using (LABEL:f11) in Theorem 4.1.

Let $\textbf{ARL}_{v_{0}}(T_{c^{\prime}}(g))=\textbf{ARL}_{v_{0}}(T_{C}(c))$
for large $c^{\prime}$ and $c$. We have
$c=c^{\prime}\theta_{v_{0}}^{*}g(\mu_{0})+o(1)$. Hence

$\displaystyle\textbf{ARL}_{v}(T_{C}(c))>\textbf{ARL}_{v}(T_{c}(g))$

for $s^{*}_{v}\theta^{*}_{v_{0}}>g(\mu)\theta^{*}_{v}/g(\mu_{0})$ when
$\mu_{0}<\mu<0$ and for $\theta^{*}_{v_{0}}>g(\mu)/g(\mu_{0})$ when $\mu\geq
0$. This means that $\textbf{ARL}_{v}(T_{c}(g))$ can be smaller than
$\textbf{ARL}_{v}(T_{C}(c))$ as long as $g(\mu)/g(\mu_{0})$ is small for all
$\mu>\mu_{0}$.

##  4  NUMERICAL SIMULATION AND A REAL EXAMPLE ILLUSTRATION

###  4.1 Numerical Simulation of ARLs for $\tau$$\geq 1$

By the simulation results of ARLs in Table 1, we see that the detection
performance of $T^{*}(r)$, $T_{C}(c)\wedge T^{*}(0)$, $T^{*}(0)$ and
$T_{C}(c\widetilde{g_{u}})$ for large $u$ is much better than that of the
conventional CUSUM test $T_{C}$ for $\tau=1$.

The following Table 2 illustrates the simulation values of
$\textbf{E}_{\tau_{i},v}$ and $\textbf{J}_{ACE}$ of nine tests in detecting
two mean shifts $v=0.1$ and $v=1$ after six change-points, $\tau_{i},\,1\leq
i\leq 6$ with ARL${}_{0}(T)=\textbf{E}_{0}(T)\approx 500$.

Table 2. Simulation of $\textbf{E}_{\tau_{i},v}$ and $\textbf{J}_{ACE}$ for
detecting two mean shifts $v=0.1,\,v=1$.
---
Tests | v | $\underset{\tau_{0}=0}{\textbf{E}_{0,v}}$ | $\underset{\tau_{1}=1}{\textbf{E}_{1,v}}$ | $\underset{\tau_{2}=10}{\textbf{E}_{10,v}}$ | $\underset{\tau_{3}=50}{\textbf{E}_{50,v}}$ | $\underset{\tau_{4}=100}{\textbf{E}_{100,v}}$ | $\underset{\tau_{5}=150}{\textbf{E}_{150,v}}$ | $\underset{\tau_{6}=200}{\textbf{E}_{200,v}}$ | $\textbf{J}_{ACE}$ |
|  | _parameter_ | _c=4.3867_ |  |  |  |  |  |  |
| v=0.1 | 498.55 | 247.25 | 241.73 | 243.52 | 243.87 | 244.23 | 243.28 | 243.98 |
$T_{C}(c)$ |  | (493.25) | (241.76) | (243.46) | (263.86) | (288.53) | (314.22) | (339.73) |  |
| v=1 | 498.55 | 9.15 | 8.48 | 8.46 | 8.46 | 8.46 | 8.46 | 8.58 |
|  | (493.25) | (5.00) | (5.13) | (5.87) | (6.77) | (7.63) | (8.49) |  |
|  | _parameter_ | _c=6.5839_ |  |  |  |  |  |  |
| v=0.1 | 498.11 | 8.06 | 41.52 | 95.15 | 136.72 | 167.20 | 193.34 | 107.00 |
$T_{C}(c\widetilde{g}_{100})$ |  | (6540.43) | (29.29) | (184.62) | (501.30) | (781.77) | (1013.44) | (1225.66) |  |
| v=1 | 498.11 | 1.25 | 4.15 | 9.03 | 12.38 | 14.69 | 16.60 | 9.68 |
|  | (6540.43) | (0.73) | (10.64) | (34.72) | (56.19) | (72.86) | (88.02) |  |
|  | _parameter_ | _r=0.00137_ |  |  |  |  |  |  |
| v=0.1 | 499.43 | 7.44 | 38.03 | 87.34 | 123.01 | 150.69 | 172.79 | 96.55 |
$T^{*}(r)$ |  | (16588.89) | (26.07) | (168.31) | (473.17) | (734.40) | (966.69) | (1164.92) |  |
| v=1 | 499.43 | 1.25 | 4.11 | 9.02 | 12.73 | 15.51 | 17.86 | 10.08 |
|  | (16588.89) | (0.71) | (10.72) | (36.12) | (61.00) | (82.62) | (102.32) |  |
|  | _parameter_ | _c=10.4889_ |  |  |  |  |  |  |
$T_{C}(c)$ | v=0.1 | 499.58 | 7.50 | 38.30 | 88.61 | 125.16 | 152.58 | 179.39 | 98.59 |
$\wedge$ |  | (10932.09) | (26.43) | (170.16) | (476.22) | (746.80) | (966.95) | (1202.64) |  |
$T^{*}(0)$ | v=1 | 499.58 | 1.25 | 4.11 | 9.04 | 12.10 | 13.88 | 15.10 | 9.25 |
|  | (10932.09) | (0.72) | (10.70) | (35.90) | (56.80) | (71.61) | (83.38) |  |
| v=0.1 | $\infty$ | 7.50 | 38.58 | 88.64 | 125.00 | 152.92 | 177.42 | 98.34 |
|  |  ($\infty$) | (26.54) | (170.58) | (475.69) | (743.32) | (972.55) | (1191.50) |  |
$T^{*}(0)$ | v=1 | $\infty$ | 1.25 | 4.12 | 9.09 | 12.77 | 15.60 | 18.02 | 10.14 |
|  |  ($\infty$) | (0.72) | (10.75) | (36.28) | (60.67) | (82.23) | (101.86) |  |
|  | _parameter_ | _m=50_ | _c=5.8093_ | _p=1_ |  |  |  |  |
| v=0.1 | 500.87 | 164.18 | 251.68 | 266.00 | 266.51 | 266.07 | 266.44 | 246.81 |
|  | (686.03) | (239.82) | (348.33) | (377.68) | (399.06) | (418.45) | (438.60) |  |
| v=1 | 500.87 | 2.68 | 7.34 | 9.41 | 9.34 | 9.35 | 9.36 | 7.91 |
|  | (686.03) | (2.94) | (7.25) | (9.45) | (10.04) | (10.70) | (11.36) |  |
|  | _parameter_ | _m=30_ | _c=6.7701_ | _p=1_ |  |  |  |  |
| v=0.1 | 500.64 | 174.61 | 259.52 | 264.36 | 264.87 | 264.67 | 264.55 | 248.76 |
|  | (666.64) | (244.09) | (345.62) | (366.62) | (388.06) | (407.65) | (428.29) |  |
| v=1 | 500.64 | 2.80 | 8.07 | 9.36 | 9.37 | 9.36 | 9.36 | 8.05 |
|  | (666.64) | (3.18) | (7.68) | (8.82) | (9.52) | (10.18) | (10.84) |  |
|  | _parameter_ | _m=10_ | _c=26.3031_ | _p=1_ |  |  |  |  |
$T_{C}(ch_{m})$ | v=0.1 | 500.12 | 218.56 | 288.10 | 287.93 | 287.44 | 287.59 | 287.68 | 276.22 |
|  | (613.17) | (276.99) | (356.17) | (375.19) | (398.04) | (422.13) | (447.11) |  |
| v=1 | 500.12 | 3.77 | 9.35 | 9.31 | 9.30 | 9.30 | 9.30 | 8.39 |
|  | (613.17) | (4.83) | (7.44) | (8.02) | (8.73) | (9.47) | (10.23) |  |
| v=0.1 | 500.38 | 6.89 | 32.94 | 58.76 | 46.86 | 7.57 | 88.88 | 40.32 |
|  | (15373.79) | (19.02) | (116.99) | (248.01) | (223.84) | (130.62) | (2208.53) |  |
$T^{*}_{M}$ | v=1 | 500.38 | 1.25 | 4.11 | 9.07 | 12.79 | 1.25 | 9.12 | 6.27 |
|  | (15373.79) | (0.72) | (10.72) | (36.27) | (60.88) | (6.60) | (175.62) |  |

  * •

Tablenotes:_The parameters for $T^{*}_{M}$ are k1=1, k2=150,
$r_{1}=5.2*10^{-5}$, $r_{2}=1.1*10^{-5}$, and the expectation and standard
deviation in both cases are 1717.06 with 13459.80 and 3918.33 with 16893.25,
respectively._

###  4.2 A Real Example

##  5 CONCLUSION

The contributions of this paper can be summarized to the following three
aspects.

(1) We present the optimal test $T^{*}(r)$ under generalized ARL0 with finite
$\textbf{E}_{0}(T^{*}(r))$.

(2) To enhance the sensitivity of the CUSUM test for detecting the
distribution change we propose a CUSUM-type test with a real-time observation
control limit (CUSUM-OAL). Numerical simulations show that the out-control
ARLs of the CUSUM-OAL tests are significantly smaller than the out-control
ARLs of the CUSUM test. But the CUSUM-OAL tests have bigger standard
deviations than the CUSUM test in the in-control state. Moreover, we obtain
the estimations of the ARLs of the CUSUM-OAL tests. Both theoretical
estimations and numerical simulations show that the CUSUM-OAL tests perform
much better than the CUSUM test when $\tau=1$.

Appendix : Proofs of Theorems

Proof of Theorem 1. Let $v\in V^{-}$. We first prove that

$e^{cg(\mu)\theta^{*}_{v}(1+o(1))}/bc\leq E_{v}(T_{c}(g))\leq
cu^{-1}g(\mu)e^{cg(\mu)\theta^{*}_{v}(1+o(1))}$ (A. 1)

for a large $c$.

Next we first prove the upward inequality of (36). Let $m_{1}=\left\lceil
cu^{-1}g(\mu)\right\rceil$, $m_{k}=\left\lceil km_{1}\right\rceil$ for $k\geq
0$ and $m=\left\lceil
tm_{1}\exp\\{cg(\mu)\theta^{*}_{v}(1+o(1))\\}\right\rceil$ for $t>0$, where
$\left\lceil x\right\rceil$ denotes the smallest integer greater than or equal
to $x$. Without loss of generality, the number $\left\lceil x\right\rceil$
will be replaced by $x$ in the following when $x$ is large. It follows that

$\displaystyle P_{v}(T_{c}(g)>m)$ $\displaystyle=$ $\displaystyle
P_{v}(\sum_{i=n-k+1}^{n}Z_{i}<cg(\hat{Z}_{n}(ac)),\ \ \ 1\leq k\leq n,1\leq
n\leq m)$ (A. 2) $\displaystyle\leq$ $\displaystyle
P_{v}(\sum_{i=m_{j}-m_{1}+1}^{m_{j}}Z_{i}<cg(\hat{Z}_{m_{j}}(ac)),\,\,1\leq
j\leq m/m_{1})$ $\displaystyle=$
$\displaystyle[P_{v}(\sum_{i=1}^{m_{1}}Z_{i}<cg(\hat{Z}_{m_{1}}(ac))]^{m/m_{1}}$

for a large $c$, where
$\hat{Z}_{m_{j}}(ac))=(ac)^{-1}\sum_{i=m_{j}-ac+1}^{m_{j}}Z_{i}$ and the last
quality holds since the events

$\displaystyle\\{\sum_{i=m_{j}-m_{1}+1}^{m_{j}}Z_{i}<cg(\hat{Z}_{m_{j}}(ac))\\},$

$1\leq j\leq m/m_{1}$, are mutually independent and have an identity
distribution. Since $\hat{Z}_{m_{1}}(ac)-\mu\to 0(a.s.)$ and
$ac(\hat{Z}_{m_{1}}(ac)-\mu)^{2}\Rightarrow\chi^{2}$ ($\chi^{2}$
-distribution) as $c\to\infty$, it follows that

$\displaystyle
g(\hat{Z}_{m_{1}}(ac))=g(\mu)+g^{\prime}(\mu)(\hat{Z}_{m_{1}}(ac)-\mu)+O(1/c)$

and

$\displaystyle
P_{v}(\sum_{i=1}^{m_{1}}Z_{i}<cg(\hat{Z}_{m_{1}}(ac))=P_{v}(\sum_{i=m_{1}-ac+1}^{m_{1}}\tilde{Z}_{i}+\sum_{i=1}^{m_{1}-ac}Z_{i}<c(g(\mu)+O(1/c)))$

for a large $c$, where
$\tilde{Z}_{i}=Z_{i}+a^{-1}|g^{\prime}(\mu)|(Z_{i}-\mu)$. Let

$\displaystyle\tilde{h}_{v}(\theta)=E_{v}(e^{\theta\tilde{Z}_{1}}),\,\,\,h_{v}(\theta)=E_{v}(e^{\theta
Z_{1}}).$

and

$\displaystyle
H_{v}(\theta)=\frac{au}{g(\mu)}\ln\tilde{h}_{v}(\theta)+(1-\frac{au}{g(\mu)})\ln
h_{v}(\theta).$

Note that $H_{v}(\theta)$ is a convex function and $H^{\prime}_{v}(0)=\mu<0$.
This means that there is a unique positive number $\theta^{*}_{v}>0$ such that
$H_{v}(\theta^{*}_{v})=0$. Let $u=H^{\prime}_{v}(\theta^{*}_{v}).$ It follows
from (A.9) that

$\displaystyle
P_{v}(\sum_{i=m_{1}-ac+1}^{m_{1}}\tilde{Z}_{i}+\sum_{i=1}^{m_{1}-ac}Z_{i}\geq
c(g(\mu)+O(1/c)))$
$\displaystyle=P_{v}(\sum_{i=m_{1}-ac+1}^{m_{1}}\tilde{Z}_{i}+\sum_{i=1}^{m_{1}-ac}Z_{i}\geq
m_{1}u(1+O(1/c)))$ $\displaystyle\geq\exp\\{-m_{1}(\theta
u^{\prime}-H_{v}(\theta)+\frac{1}{m_{1}}\log(F^{m_{1}}_{\theta}(m_{1}u^{\prime})-F^{m_{1}}_{\theta}(m_{1}u))+o(1))\\}$
$\displaystyle=\exp\\{-cg(\mu)(\theta\frac{u^{\prime}}{u}-\frac{1}{u}H_{v}(\theta)+o(1))\\}$

for a large $c$. Taking $\theta\searrow\theta^{*}_{v}$ and $u^{\prime}\searrow
u$, we have

$\displaystyle
P_{v}(\sum_{i=m_{1}-ac+1}^{m_{1}}\tilde{Z}_{i}+\sum_{i=1}^{m_{1}-ac}Z_{i}<c(g(\mu)+O(1/c)))\leq
1-\exp\\{-cg(\mu)\theta^{*}_{v}(1+o(1))\\}$

for a large $c$. Thus, by (A.11) we have

$\displaystyle
P_{v}(T_{c}(g)>m)\leq[1-\exp\\{-cg(\mu)\theta^{*}_{v}(1+o(1))\\}]^{m/m_{1}}\to
e^{-t}.$ (A. 3)

as $c\to\infty$. By the properties of exponential distribution, we have

$\displaystyle E_{v}(T_{c}(g)))\leq
cu^{-1}g(\mu)e^{cg(\mu)\theta^{*}_{v}(1+o(1))}$

for a large $c$.

To prove the downward inequality of (A.10), let

$\displaystyle U_{m}$ $\displaystyle=$
$\displaystyle\\{\sum_{i=n-k+1}^{n}Z_{i}<cg(\hat{Z}_{n}(ac)),\;\;1\leq k\leq
ac-1,\,bc\leq n\leq m\\}$ $\displaystyle V_{m}$ $\displaystyle=$
$\displaystyle\\{\sum_{i=n-k+1}^{n}Z_{i}<cg(\hat{Z}_{n}(ac)),\;\;ac\leq k\leq
bc-1,\,bc\leq n\leq m\\}$ $\displaystyle W_{m}$ $\displaystyle=$
$\displaystyle\\{\sum_{i=n-k+1}^{n}Z_{i}<cg(\hat{Z}_{n}(ac)),\;\;bc\leq k\leq
n,\;bc\leq n\leq m\\}$ $\displaystyle S_{bc}$ $\displaystyle=$
$\displaystyle\\{\sum_{i=n-k+1}^{n}Z_{i}<cg(\hat{Z}_{n}(ac)),\;\;1\leq k\leq
n,\;1\leq n\leq bc-1\\},$

where $b$ is defined in (LABEL:f08) and without loss of generality, we assume
that $b>a$. Obviously, $\\{T_{c}(g)>m\\}=U_{m}V_{m}W_{m}S_{bc}$.

Let $k=xcg(\mu)$. By Chebyshev’s inequality, we have

$\displaystyle P_{v}(\sum_{i=n-k+1}^{n}Z_{i}<cg(\hat{Z}_{n}(ac)))$
$\displaystyle=$ $\displaystyle
P_{v}\Big{(}\sum_{i=n-k+1}^{n}\tilde{Z}_{i}+\sum_{i=n-ac+1}^{n-k}\tilde{\tilde{Z}}_{i}<cg(\mu)(1+o(1))\Big{)}$
$\displaystyle\geq$ $\displaystyle
1-\exp\\{-cg(\mu)(\theta-x\tilde{H}_{v}(\theta)+o(1))\\}$

for $1\leq k\leq ac-1,\,bc\leq n\leq m$, where
$\tilde{\tilde{Z}}_{i}=-g^{\prime}(\mu)(Z_{i}-\mu)/a$ and

$\displaystyle\tilde{H}_{v}(\theta)=\ln\tilde{h}_{v}(\theta)+(\frac{ac}{k}-1)\ln\hat{h}_{v}(\theta),\,\,\,\,\,\hat{h}_{v}(\theta)=E_{v}(e^{\theta\tilde{\tilde{Z}}_{i}}).$

Since $\tilde{H}_{v}(\theta)$ and $H_{v}(\theta)$ are two convex functions and

$\displaystyle\tilde{H}^{\prime}_{v}(0)-H^{\prime}_{v}(0)=0,$
$\displaystyle\tilde{H}^{\prime\prime}_{v}(0)-H^{\prime\prime}_{v}(0)=\sigma^{2}[(1+\frac{g^{\prime}(\mu)}{a})^{2}+(\frac{ac}{k}-1)-\frac{au}{g(\mu)}(1-\frac{g^{\prime}(\mu)}{a})^{2}+\frac{au}{g(\mu)}-1]>0,$

it follows that $\tilde{\theta}_{v}^{*}\geq\theta_{v}^{*}$, where
$\tilde{\theta}_{v}^{*}$ and $\theta_{v}^{*}$ satisfy
$\tilde{H}_{v}(\tilde{\theta}_{v}^{*})=H_{v}(\theta_{v}^{*})=0$. Hence

$\displaystyle
P_{v}(\sum_{i=n-k+1}^{n}\tilde{Z}_{i}+\sum_{i=n-ac+1}^{n-k}\tilde{\tilde{Z}}_{i}$
$\displaystyle<cg(\mu)(1+o(1)))$ $\displaystyle\geq
1-\exp\\{-cg(\mu)\theta_{v}^{*}(1+o(1))\\}$ (A. 4)

for $1\leq k\leq ac-1,\,bc\leq n\leq m$. Similarly, we can get

$\displaystyle
P_{v}(\sum_{i=n-ac+1}^{n}\tilde{Z}_{i}+\sum_{i=n-k+1}^{n-ac}Z_{i}$
$\displaystyle<cg(\mu)(1+o(1)))$ $\displaystyle\geq
1-\exp\\{-cg(\mu)\theta_{v}^{*}(1+o(1))\\}$ (A. 5)

for $ac\leq k\leq bc-1,\,bc\leq n\leq m$, and

$\displaystyle
P_{v}(\sum_{i=n-ac+1}^{n}\tilde{Z}_{i}+\sum_{i=n-k+1}^{n-ac}Z_{i}$
$\displaystyle<$ $\displaystyle cg(\mu)(1+o(1)))$ (A. 6) $\displaystyle\geq$
$\displaystyle 1-\exp\\{-2cg(\mu)\theta_{v}^{*}(1+o(1))\\}$

for $bc\leq k\leq n,\,bc\leq n\leq m$.

Let $m=tcg(\mu)\theta_{v}^{*}/bc$ for $t>0$. By (A.13), (A.14), (A.15) and
Theorem 5.1 in Esary, Proschan and Walkup (1967) we have

$\displaystyle P_{v}(U_{m}V_{m})$ $\displaystyle\geq$
$\displaystyle\prod_{n=bc}^{m}\prod_{k=1}^{bc-1}P_{v}(\sum_{i=n-k+1}^{n}\tilde{Z}_{i}-\frac{g^{\prime}(\mu)}{a}\sum_{i=n-ac+1}^{n-k}(Z_{i}-\mu)<cg(\mu)(1+o(1)))$
$\displaystyle\geq$
$\displaystyle[1-\exp\\{-cg(\mu)\theta_{v}^{*}(1+o(1))\\}]^{bcm}\to e^{-t}$

and

$\displaystyle P_{v}(W_{m})$ $\displaystyle\geq$
$\displaystyle\prod_{n=bc}^{m}\prod_{k=bc}^{n}P_{v}(\sum_{i=n-ac+1}^{n}\tilde{Z}_{i}+\sum_{i=n-k+1}^{n-ac}Z_{i}<cg(\mu)(1+o(1)))$
$\displaystyle\geq$
$\displaystyle[1-\exp\\{-2cg(\mu)\theta_{v}^{*}(1+o(1))\\}]^{(m-bc)^{2}}\to 1$

as $c\to+\infty$.

Finally,

$\displaystyle P_{v}(S_{bc})$ $\displaystyle\geq$ $\displaystyle
P_{v}(\sum_{i=n-k+1}^{n}Z_{i}<cg_{0},\;\;1\leq k\leq n,\;1\leq n\leq bc-1)$
$\displaystyle\geq$
$\displaystyle\prod_{n=1}^{bc-1}\prod_{k=1}^{n}(1-\exp\\{-cg_{0}\theta+k\ln
h_{v}(\theta)\\})$ $\displaystyle\geq$
$\displaystyle[1-\exp\\{-cg_{0}\theta_{0}\\}]^{(bc)^{2}}\to 1$

as $c\to+\infty$, where $\theta_{0}>0$ satisfies $h_{v}(\theta_{0})=1$.

Thus

$\displaystyle P_{v}(T_{c}(g)>m)=P_{v}(U_{m}V_{m}W_{m}S_{bc})\searrow e^{-t}.$

as $c\to\infty$. This implies that

$\displaystyle E_{v}(T_{c}(g))\geq e^{cg(\mu)\theta^{*}_{v}(1+o(1))}/bc$

for a large $c$. This completes the proof of (A.10).

Let $v\in V^{0}$.

Let $m_{1}=(cg(0))^{2}/\sigma^{2}$. It follows that

$\displaystyle E_{v}(T_{c}(g))$ $\displaystyle=$
$\displaystyle\sum_{n=0}^{\infty}P_{v}(T_{c}(g)>n)$ $\displaystyle\leq$
$\displaystyle
m_{1}+\sum_{n=m_{1}}^{2m_{1}}P_{v}(T_{c}(g)>n)+...+\sum_{n=km_{1}}^{(k+1)m_{1}}P_{v}(T_{c}(g)>n)+....$
$\displaystyle\leq$ $\displaystyle
m_{1}[1+\sum_{k=1}^{\infty}P_{v}(T_{c}(g)>km_{1})].$

Note that

$\displaystyle P_{v}(T_{c}(g)>km_{1})\leq
P_{v}\Big{(}\sum_{i=(j-1)m_{1}+1}^{jm_{1}}Z_{i}+\sum_{i=jm_{1}-ac+1}^{jm_{1}}Z^{\prime}_{i}<cg(0)(1+o(1)),\,\,1\leq
j\leq k\Big{)}$ $\displaystyle=$
$\displaystyle[P_{v}\Big{(}\sum_{i=1}^{m_{1}}Z_{i}+\sum_{i=m_{1}-ac+1}^{m_{1}}Z^{\prime}_{i}<cg(0)(1+o(1))\Big{)}]^{k}$
$\displaystyle=$
$\displaystyle[P_{v}\Big{(}\sum_{i=1}^{m_{1}-ac}Z_{i}+\sum_{i=m_{1}-ac+1}^{m_{1}}(1+A)Z_{i}\,<cg(0)(1+o(1))\Big{)}]^{k}$
$\displaystyle=$
$\displaystyle[P_{v}\Big{(}\frac{\sum_{i=1}^{m_{1}-ac}Z_{i}}{cg(0)}+\frac{\sum_{i=m_{1}-ac+1}^{m_{1}}(1+A)Z_{i}}{cg(0)}\,<(1+o(1))\Big{)}]^{k}$

for a large $c$, where $A=|g^{\prime}(0)|/a$, and

$\displaystyle\frac{\sum_{i=1}^{m_{1}-ac}Z_{i}}{cg(0)}\Rightarrow X\sim
N(0,1),\,\,\,\,\,\,\,\,\frac{\sum_{i=m_{1}-ac+1}^{m_{1}}(1+A)Z_{i}}{cg(0)}\to
0$

as $c\to\infty$. Thus

$\displaystyle E_{v}(T_{c}(g))\leq
m_{1}[1+\sum_{k=1}^{\infty}[\Phi(1+o(1))]^{k}]=\frac{(cg(0))^{2}}{\sigma^{2}}\frac{(1+o(1))}{1-\Phi(1)},$

where $\Phi(.)$ is the standard normal distribution.

Let $m_{2}=(cg(0))^{2}/(8\sigma^{2}\ln c)$. Note that

$\displaystyle\prod_{n=1}^{ac}\prod_{k=1}^{n}P_{v}\Big{(}\sum_{i=n-k+1}^{n}Z_{i}+c|g^{\prime}(0)|n^{-1}\sum_{i=1}^{n}Z_{i}<cg(0)(1+o(1))\Big{)}$
$\displaystyle\geq[P_{v}\Big{(}\sum_{i=1}^{ac}Z_{i}+\sum_{i=1}^{ac}Z^{\prime}_{i}<cg(0)(1+o(1))\Big{)}]^{(ac)^{2}}=(1+o(1))(1-\Phi(\frac{\sqrt{c}g(0)}{\sqrt{a}(1+A)}))^{(ac)^{2}}\to
1$

as $c\to\infty$, since

$\displaystyle
P_{v}(\frac{\sum_{i=1}^{ac}Z_{i}+\sum_{i=1}^{ac}Z^{\prime}_{i}}{\sigma(1+A)\sqrt{ac}})\Rightarrow
X\sim N(0,1)$

as $c\to\infty$. It follows that

$\displaystyle E_{v}(T_{c}(g))\geq\sum_{n=0}^{m_{2}}P_{v}(T_{c}(g)>n)\geq
m_{2}P_{v}(T_{c}(g)>m_{2})$ $\displaystyle\geq$ $\displaystyle
m_{2}(1+o(1))\prod_{n=ac+1}^{m_{1}}\prod_{k=1}^{n}P_{v}\Big{(}\sum_{i=n-k+1}^{n}Z_{i}+\sum_{i=n-ac+1}^{n}Z^{\prime}_{i}<cg(0)(1+o(1))\Big{)}$
$\displaystyle\geq$ $\displaystyle
m_{2}(1+o(1))[P_{v}\Big{(}\sum_{i=1}^{m_{2}}Z_{i}+\sum_{i=m_{2}-ac+1}^{m_{2}}Z^{\prime}_{i}<cg(0)(1+o(1))\Big{)}]^{m^{2}_{2}}$
$\displaystyle=$ $\displaystyle
m_{2}(1+o(1))[P_{v}\Big{(}\frac{\sum_{i=1}^{m_{2}}Z_{i}+\sum_{i=m_{2}-ac+1}^{m_{2}}Z^{\prime}_{i}}{\sqrt{m_{2}}\sigma}<\sqrt{8\ln
c}(1+o(1))\Big{)}]^{m^{2}_{2}}$ $\displaystyle=$ $\displaystyle
m_{2}(1+o(1))[\Phi(\sqrt{8\ln
c})]^{m^{2}_{2}}=m_{2}(1+o(1))[1-\frac{1}{c^{4}\sqrt{8\ln c}}]^{m^{2}_{2}}\to
m_{2}(1+o(1))$

as $c\to\infty$, where the third inequality comes from Theorem 5.1 in Esary,
Proschan and Walkup (1967). Thus, we have

$\displaystyle\frac{(cg(0))^{2}}{8\sigma^{2}\ln c}(1+o(1))\leq
E_{v}(T_{c}(g))\leq\frac{(cg(0))^{2}}{\sigma^{2}(1-\Phi(1))}(1+o(1)).$

Let $v\in V^{+}$ and let

$\displaystyle T_{0}=\min\\{n:\sum_{i=1}^{n}Z_{i}+\frac{ac}{(ac)\wedge
n}\sum_{i=n-(ac)\wedge n+1}^{n}Z^{\prime}_{i}\geq c\\}.$

The uniform integrability of $\\{T_{c}(g)/c\\}$ for $c\geq 1$, follows from
the well-known uniform integrability of $\\{T_{0}/c\\}$ (see Gut (1988)).

By the Strong Large Number Theorem we have

$\displaystyle\mu$ $\displaystyle=$
$\displaystyle\lim_{n\to\infty}\frac{\sum_{i=1}^{n}Z_{i}}{n}$ $\displaystyle=$
$\displaystyle\lim_{n\to\infty}\max_{1\leq j\leq
n}\frac{1}{n}[\sum_{i=j}^{n}Z_{i}+g^{\prime}(\mu)a^{-1}\sum_{i=n-ac+1}^{n}(Z_{i}-\mu)]$

Note that $T_{c}(g)\to\infty$ as $c\to\infty$,

$\displaystyle\max_{1\leq j\leq
T_{c}(g)}[\sum_{i=j}^{T_{c}(g)}Z_{i}+g^{\prime}(\mu)a^{-1}\sum_{i=T_{c}(g)-ac+1}^{T_{c}(g)}(Z_{i}-\mu)]\geq
cg(\mu)(1+o(1)),$

and

$\displaystyle\max_{1\leq j\leq
T_{c}(g)-1}[\sum_{i=j}^{T_{c}(g)-1}Z_{i}+g^{\prime}(\mu)a^{-1}\sum_{i=T_{c}(g)-ac}^{T_{c}(g)-1}(Z_{i}-\mu)]\leq
cg(\mu)(1+o(1)).$

It follows that

$\displaystyle\mu\longleftarrow\max_{1\leq j\leq
T_{c}(g)}\frac{1}{T_{c}(g)}[\sum_{i=j}^{T_{c}(g)}Z_{i}+g^{\prime}(\mu)a^{-1}\sum_{i=T_{c}(g)-ac+1}^{T_{c}(g)}(Z_{i}-\mu)]$
$\displaystyle\geq$ $\displaystyle\frac{cg(\mu)(1+o(1))}{T_{c}(g)}$
$\displaystyle\geq$ $\displaystyle\max_{1\leq j\leq
T_{k}-1}\frac{1}{T_{c}(g)-1}[\sum_{i=j}^{T_{c}(g)-1}Z_{i}+g^{\prime}(\mu)a^{-1}\sum_{i=T_{c}(g)-ac}^{T_{c}(g)-1}(Z_{i}-\mu)]\longrightarrow\mu$

as $c\to\infty$. By the uniform integrability of $\\{T_{c}(g)/c\\}$ and using
Theorem A.1.1 in Gut’s book (1988), we have

$\displaystyle E_{v}(T_{c}(g))=(1+o(1))\frac{cg(\mu)}{\mu}$

for a large $c$. This completes the proof of Theorem 2.

Proof of Theorem 4. Since $g(x)<0$ for $x>a^{*}$, $a^{*}\leq\mu^{*}$ and
$\mu^{*}\geq 0$, it follows that

$\displaystyle
P_{v}\Big{(}m\hat{Z}_{m}<cg(\hat{Z}_{m}),\,\,\,\hat{Z}_{m}>a^{*}\Big{)}\leq
P_{v}(\hat{Z}_{m}<\mu^{*})$

and

$\displaystyle P_{v}(T_{c}(g)>m)$ $\displaystyle=$ $\displaystyle
P_{v}\Big{(}\sum_{i=n-k+1}^{n}Z_{i}<cg(\hat{Z}_{n}),\,\,\,1\leq k\leq
n,\,\,1\leq n\leq m\Big{)}\leq
P_{v}\Big{(}m\hat{Z}_{m}<cg(\hat{Z}_{m})\Big{)}$ $\displaystyle=$
$\displaystyle P_{v}\Big{(}m\hat{Z}_{m}<cg(\hat{Z}_{m}),\,\,\hat{Z}_{m}\leq
a^{*}\Big{)}+P_{v}\Big{(}m\hat{Z}_{m}<cg(\hat{Z}_{m}),\,\,\,\hat{Z}_{m}>a^{*}\Big{)}$
$\displaystyle\leq$ $\displaystyle 2P_{v}(\hat{Z}_{m}<\mu^{*}).$

Furthermore,

$\displaystyle P_{v}(\hat{Z}_{m}<\mu^{*})$ $\displaystyle=$ $\displaystyle
P_{v}(\sum_{i}^{m}-Z_{i}>-m\mu^{*})=P_{v}(\sum_{i}^{m}(\mu-
Z_{i})>m(\mu-\mu^{*}))$ $\displaystyle=$ $\displaystyle
P_{v}(e^{\theta\sum_{i}^{m}(\mu-Z_{i})}>e^{\theta m(\mu-\mu^{*})})\leq
e^{-m[\theta(\mu-\mu^{*})-\ln M(\theta)]},$

where $M(\theta)=E_{v}(e^{\theta(\mu-Z_{1})})$ and the last inequality follows
from Chebychev’s inequality. Note that $h(\theta)=\theta(\mu-\mu^{*})-\ln
M(\theta)$ attains its maximum value
$h(\theta^{*})=\theta^{*}(\mu-\mu^{*})-\ln M(\theta^{*})>0$ at
$\theta=\theta^{*}>0$, where $h^{\prime}(\theta^{*})=0$. So,

$\displaystyle E_{v}(T_{c}(g))=1+\sum_{m=1}^{\infty}P_{v}(T_{c}(g)>m)\leq
1+2\sum_{m=1}^{\infty}e^{-m[\theta^{*}(\mu-\mu^{*})-\ln
M(\theta^{*})]}=\frac{e^{\theta^{*}(\mu-\mu^{*})-\ln
M(\theta^{*})}+1}{e^{\theta^{*}(\mu-\mu^{*})-\ln M(\theta^{*})}-1}.$

Let $k>1$. It follows that

$\displaystyle E_{vk}(T_{c}(g)-k+1)^{+}$ $\displaystyle=$
$\displaystyle\sum_{m=1}^{\infty}P_{vk}(T_{c}(g)>m+k-1,T_{c}(g)>k-1)$
$\displaystyle\leq$
$\displaystyle(a_{0}+1)(k-1)P_{0}(T_{c}(g)>k-1)+\sum_{m\geq(a_{0}+1)(k-1)}^{\infty}P_{vk}(T_{c}(g)>m+k-1).$

Similarly, we have

$\displaystyle P_{vk}(T_{c}(g)>m+k-1)$ $\displaystyle=$ $\displaystyle
P_{vk}\Big{(}\sum_{i=n-k+1}^{n}Z_{i}<cg(\hat{Z}_{n}),\,\,\,1\leq k\leq
n,\,\,1\leq n\leq m+k-1\Big{)}\leq 2P_{vk}(\hat{Z}_{m+k-1}<\mu^{*})$
$\displaystyle=$ $\displaystyle 2P_{vk}\Big{(}\sum_{i=k-1}^{m+k-1}(\mu-
Z_{i})+\sum_{i=1}^{k-1}(\mu_{0}-Z_{i})>m(\mu-\mu^{*})+(k-1)(\mu_{0}-\mu^{*})\Big{)}$
$\displaystyle\leq$ $\displaystyle
2\exp\\{-m\Big{(}\theta^{*}(\mu-\mu^{*})-\ln
M(\theta^{*})+\frac{k-1}{m}[\mu_{0}-\mu^{*}-\ln
M_{0}(\theta^{*})]\Big{)}\\}\leq e^{-mb}$

for $m\geq(a_{0}+1)(k-1)$, since

$\displaystyle\theta^{*}(\mu-\mu^{*})-\ln
M(\theta^{*})+\frac{k-1}{m}[\mu_{0}-\mu^{*}-\ln M_{0}(\theta^{*})]\geq b$

for $m\geq(a_{0}+1)(k-1)$. Thus,

$\displaystyle E_{vk}(T_{c}(g)-k+1)^{+}$ $\displaystyle\leq$
$\displaystyle(a_{0}+1)(k-1)P_{0}(T_{c}(g)\geq
k)+2\sum_{m\geq(a_{0}+1)(k-1)}^{\infty}e^{-mb}$ $\displaystyle\leq$
$\displaystyle(a_{0}+1)(k-1)P_{0}(T_{c}(g)>\geq
k)+\frac{2e^{-(a_{0}+1)(k-1)b}}{1-e^{-b}}.$

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