数理解析研究所講究録 1308 巻 2003 年 75-83

75
Sequential Point Estimation of the Powers of
an Exponential Scale Parameter

Muktar ALI ( アリ ムクタール)

Graduate School of Science and Technology, Niigata University
Niigata 950-2181, JapanEiichi ISOGAI( 磯貝 英一 )Department of Mathematics, Niigata University

Niigata 950–2181, Japan

1. Introduction

Let $X_{1},X_{2},X_{3}$ , \ldots be independent and identically distributed (i.i.d.) random variables
with the probability density function (p.d.f.)$f_{\mu,\sigma}(x)= \frac{1}{\sigma}\exp(-\frac{x-\mu}{\sigma})I_{(ae\geq\mu)}$,are unknown and $I(A)$ denotes the indicator
where both $\mu\in(-\infty, \infty)$ andfunction of the set . For any given $r\neq 0$ we want to estimate the power of thebased on arandom
be an estimator ofscale parameter . Let, ,of size . Then as aloss function we use the squared error loss
sample. The risk associated with the estimator is given bydefined by$\sigma\in(0, \infty)$$A$$\sigma^{r}$$X_{1}$$\ldots$$X_{n}$$\sigma^{r}$$\delta_{n}=\delta_{n}(X_{1}, \ldots, X_{n})$$n$$\delta_{n}$$L_{n}=(\delta_{||}-\sigma^{r})^{2}$$R_{n}=R_{n}(\delta_{n})=\mathrm{E}(\mathrm{L}\mathrm{n})$. Let $w>0$ be apreassigned error bound for the risk. We want to. In bounded risk pointfind the smallest sample size $n=n_{0}$ which satisfies that
estimation problems the error bound is assumed to be sufficiently small, so that wesuppose that the sample size is sufficiently large. For $n\geq 2$ set$R_{\mathfrak{n}}\leq w$

$w$

$n$

76
$T_{n}= \min\{X_{1}, \ldots,X_{n}\}$We can show that$c_{\mathfrak{n}}\sigma_{n}^{r}$and$\sigma_{n}=(n-1)^{-1}\sum_{\dot{\iota}=1}^{n}(X_{\dot{l}}-T_{n})$.

with $c_{n}=(n-1)^{r}\Gamma(n-1)/\Gamma(r+n-1)$ is the uniformly minimum

variance unbiased estimator of$\sigma^{r}$provided $n> \max\{1,1-r\}$ . We use, however,$\sigma_{n}^{r}$as anestimator of , because is approximately equal tofor large and the calculation
ofis easier than that of. Our goal is to find an asymptotically smallest sample$\sigma^{r}$$\sigma_{n}^{r}$$\sigma_{n}^{r}$$n$$c_{n}\sigma_{n}^{r}$$\mathrm{q}_{\iota}\sigma_{n}^{r}$satisfying that.
Estimation of and are of great importance. For $r=1$ , namely, for the estimationof the standard deviation by , Isogai, Saito and Uno (1999) dealt with this boundedrisk point estimation problem. Minimum risk point estimation problems for $r=1$ wereconsidered by Mukhopadhyay and Ekwo (1987), Ghosh and Mukhopadhyay (1989) andIsogai and Uno (1994). Starr and Woodroofe (1972) treated the same problem for $r=2$ .
Uno and Isogai (2002) proposed afully sequential procedure for the estimation of withnormal scale parameter . For areview one may refer to Mukhopadhyay (1988), Ghoshand Sen (1991) and Ghosh, Mukhopadhyay and Sen (1997). Sometimes, it is of interest
to measure mean
in -units and hence to estimate. For anormal
distribution with mean and varianceboth unknown, Sriram (1990) considered thesequential point estimation problem forby using an estimator of with $=-1$ .
We shall now compute the risk
to find . We can show that forsize$R_{n}=E(\sigma_{n}^{r}-\sigma^{r})^{2}\leq w$$n_{0}$$\sigma^{2}$$\sigma$$\sigma_{n}$$\sigma^{r}$$\sigma$$\mathrm{A}=\mu+\sigma$$\mu\sigma^{-1}$$\sigma$$\sigma^{2}$$\mu$$\mu\sigma^{-1}$$\sigma^{r}$$R_{n}=E(\sigma_{n}^{r}-\sigma^{r})^{2}$$r$$n_{0}$$n> \max\{2,1-2r\}$ .
$R_{n}<\infty$

and$R_{n}=r^{2}\sigma^{2r}n^{-1}+O(n^{-2})$

as$narrow\infty$.

Ignoring the order term above, we can find the asymptotically smallest sample size
satisfying that $R_{n}\leq w$ . Suppose$r^{2}\sigma^{2r}n^{-1}\leq w$

For simplicity,

or equivalently,$n \geq\frac{r^{2}\sigma^{2r}}{w}=n^{*}$(say).

$n_{0}$(1.1)

is assumed to be an integer. Then $n_{0}=n^{*}$ is the asymptotically best
fixed sample size if is known. Unfortunately, the asymptotically best fixed sample size
procedure
cannot be used since is unknown. Further there is no fixed sample size$n^{*}$$\sigma$$n_{0}$$\sigma$procedure satisfying our condition. Thus we need to find asequential sampling rule.

In Section 2we shall propose afully sequential procedure for this estimation problem
and give two theorems concerning the second order approximation to its average sample

77 77
size and risk associated with our procedure. We shall also consider aclass of sequential
estimators derived on the basis of the idea of bias-correction and compare them from thepoint of view of risk. In Section 3we shall provide brief simulation results.

2. Results
In this section we shall propose afully sequential procedure $N$ motivated by the formin (1.1) and give two theorems concerning the second order approximation to itsof$n^{*}$. We shall also consider aas
average sample size $E(N)$ and risk
, based on the idea ofclass of sequential estimators, including the ordinary estimatorbias-correction. The comparison will be made from the point of view of risk. It will turnout that we can find an appropriate sequential estimator to reduce the risk associatedwith the ordinary one.
In this paper we propose the stopping rule defined by$warrow \mathrm{O}$$R_{N}=E(\sigma_{N}^{r}-\sigma^{r})^{2}$$\sigma_{N}^{r}$$N=N_{w}(r)= \inf\{n\geq m:n\geq\frac{r^{2}\sigma_{n}^{2r}}{w}l_{n}\}$

(2.1)

,

where $m$ is astarting sample size satisfying that $m> \max\{2,1-2r\}$ andpositive function of onsuch that$x$$l_{l}$

is agiven$(0, \infty)$$l_{x}=1+ \frac{l_{0}}{x}+o(\frac{1}{x})$as$xarrow \mathrm{o}\mathrm{o}$with aconstant$l_{0}$.

. Once the sampling stops
We can show that $P(Nw(r) <\infty)=1$ for all $w>0$ and. Then the risk associated with is given by
bystage, we estimate
at the. The following two theorems are concerned with the second orderapproximation to the average sample size and risk.
$r$$N\mathrm{t}\mathrm{h}$$\sigma^{r}$$\neq 0$$\sigma_{N}^{r}$$\sigma_{N}^{r}$$R_{N}=E(\sigma_{N}^{r}-\sigma^{r})^{2}$

Theorem 2.1. If $m>m_{1}(r)$ , then as$warrow \mathrm{O}$$E(N)=n^{*}+\rho+l_{0}-r(2r+1)+o(1)$ ,where

if $r>0$$<0$$1-2r$ if$1+6r$$m_{1}(r)=\{$$f$and$\rho$

is acertain constant with

$0\leq\rho\leq\pm 2$ $+2r^{2}$.

78
If $m>m_{2}(r)$ , then as

Theorem 2.2.

$warrow \mathrm{O}$$n^{*}( \frac{R_{N}}{w}-1)=11r^{2}+8r+1+\frac{3}{4}(r-1)^{2}-\rho-l_{0}+o(1)$

,

where
$\max\{1+1\mathrm{O}r, 7+8r\}$

if

$r>0$

$7-14r$if$r<0$ .

$m_{2}(r)=\{$(i) If we take an arbitrary constantRemark 2.1.

$l_{0}$

such that$l_{0}>11r^{2}+8r+1+ \frac{3}{4}(r-1)^{2}-\rho$

,

(2.2)

then from Theorem 2.2 we have that $R_{N}<w$ for sufficiently small $w>0$ . Thus ourcondition on the risk is satisfied for sufficiently small .
$w$(ii) Theorem 2.1 of Isogai, Saito and Uno (1999) with $a=0$ and $b=1$ is the same as

Theorems 2.1 and 2.2 with $r=1$ except for the condition on the starting sample size.
The difference of this condition is caused by the fact that this paper deals with all powers.

Farther the methods of the proofs are different.

We shall here evaluate the bias of$\sigma_{N}^{r}$.

Proposition 2.1. If $m>m_{3}(r)$ , then as$warrow \mathrm{O}$$E( \sigma_{N}^{r})-\sigma^{r}=-\frac{1}{2}\mathrm{s}.\mathrm{g}\mathrm{n}(r)(3r+1)(n^{*})^{-1/2}w^{1/2}+o(w)$,wheresign(r) $=\{$

1if $r>0$
$<0$
-1 if
$f$

and

$m_{3}(r)=\{$$\max\{1+6r,3+3r\}$

if $r>0$$3-5r$

if$r<0$ .

Taking Proposition 2.1 into account, we consider aclass of sequential estimators$(-\infty, \infty)\}$for$\sigma^{r}$defined by$\sigma_{N}^{r}(k)=\sigma_{N}^{r}+k$ $N^{-1/2}w^{1/2}$.

Then we get the following proposition concerning the bias of$\sigma_{N}^{r}(k)$.

$\{\sigma_{N}^{r}(k),k$$\in$

79 79Proposition 2.2.

If $m>m_{3}(r)$ , then as$warrow \mathrm{O}$.

$E( \sigma_{N}^{r}(k))=\sigma^{r}+\{k-\frac{1}{2}\mathrm{s}\mathrm{i}\mathrm{g}\mathrm{n}(r)(3r+1)\}(n^{*})^{-1/2}w^{1/2}+o(w)$For k$= \frac{1}{2}\mathrm{s}\mathrm{i}\mathrm{g}\mathrm{n}(r)(3r+1)$,$\mathrm{a}\mathrm{r}\mathrm{N}(\mathrm{k})$is asecond-0rder asymptotically unbiased estimator.

We shall now compare the risk ofwith that of$\sigma_{N}^{r}(k)$Theorem 2.3. If $m>m_{2}(r)$ , then asLet k$= \frac{1}{2}\mathrm{s}\mathrm{i}\mathrm{g}\mathrm{n}(r)(5r+1)$$R_{N}(k)<R_{N}$. Let$R_{N}(k)=E(\sigma_{N}^{r}(k)-\sigma^{r})^{2}$.

$warrow \mathrm{O}$$\frac{n^{*}}{w}(R_{N}(k)-R_{N})=k^{2}$

Remark 2.2$\sigma_{N}^{r}$-sign(r)(5r+1)k+0(1).
for r$\neq 0$. Then we havefor sufficiently small w $>0$ if r$\neq-\frac{1}{5}$.

Thus bias-correction is asymptotically effective to reduce the risk for all $r\neq 0$ with using
which is not asecond-0rder asymptotically unbiased$\sigma_{N}^{r}+\frac{1}{2}\mathrm{s}\mathrm{i}\mathrm{g}\mathrm{n}(r)(5r+1)N^{-1/2}w^{1/2}$estimator.

3. Simulation Results
We shall give brief simulation results which are based on 100,000 repetitions. We
choose the constant satisfying the inequality in (2.2) in Tables 1-3. In Tables 4
we choose such that the average sample size $E(N)$ approximately equals the optimal, we use
one . Since we do not know any approximate value of between 0and$\mathrm{m}\mathrm{d}$$5$$l_{0}$$l_{0}$$n^{*}$$\rho$

ashere $\rho=0$ or
the stopping rule $N$ in (2.1)$\rho=\frac{1}{2}+2r^{2}$$\rho$$\frac{1}{2}+2r^{2}$. Prom these simulation results we might need to improve

80
Table 1.

$\rho=0$,$l_{0}>11r^{2}+8r+1+ \frac{3}{4}(r-1)^{2}-\rho$$r=-1$

$r=1$

$r=2$

$w=0.01$$w=0.01$$w=0.04$$m=22$$m=16$$m=24$$l_{0}=8$$l_{0}=21$$l_{0}=62$

$k=2$

$k=3$

108.630260
0.991965
1.011400
0.963153
0.934459

116.346200
0.983436
1.011567
0.984520
0.921439-6.308015$n^{*}=100$$\mu=0$, $\sigma=1$$l_{n}=1+l_{0}/n$$E(N)$$E(\sigma_{N}^{r})$$E(\sigma_{N}^{r}(k))$$R_{N}/w$$R_{N}(k)/w$$\underline{n^{*}(R_{N}(k)-R_{N})/w- 2.869400}$

Table 2. r $=-1$ ,$n^{*}=40$$\mu=0$,$l_{0}=5$$m=22$ ,$E(N)$$E(\sigma_{N}^{r})$$E(\sigma_{N}^{r}(k))$$R_{N}/w$$R_{N}(k)/w$

$k$

$=2$

$\rho=\frac{1}{2}+2r^{2}=2.5$,$k$$=5.5$

137.330710
0.951511
1.048849

0.9491650.800041-14.912360$l_{0}>11r^{2}+8r+1+ \frac{3}{4}(r-1)^{2}-\rho=4.5$$\sigma=0.5$$\sigma=1$$\sigma=2$$=0.1$$w=0.025$$w=0.00625$$\sigma^{r}=2$$\sigma^{r}=1$$\sigma^{r}=0.5$

45.752540
1.965109
2.061243
0.918271

45.700810
0.982012
1.030103
0.914489

0.865381

0.859792

45.738710
0.491141
0.515186
0.925572
0.870948$eo$$\underline{n^{*}(R_{N}(k)-R_{N})/w-2.115595-2.187878-2.184982}$

81
Table 3. r $=1$ ,$\rho=\frac{1}{2}+2r^{2}=2.5$,$l_{0}>11r^{2}+8r+1+ \frac{3}{4}(r-1)^{2}-\rho=17.5$$\sigma=0.5$$\sigma=1$$\sigma=2$$l_{0}=18$$w=0.00625$$w=0.025$$w=0.1$$k=3$$\sigma^{r}=0.5$$\sigma^{r}=1$$\sigma^{r}=2$52.049260
0.481075
0.514786
1.0048480.870552

52.088870
0.962627
1.030025
1.002715
0.871007-5.268332

52.200480
1.928253
2.062843
0.984277
0.862504-4.870901$n^{*}=40$$\mu=0$,$m=16$ ,$E(N)$$E(\sigma_{N}^{r})$$E(\sigma_{N}^{r}(k))$$R_{N}/w$$R_{N}(k)/w$$\underline{n^{*}(R_{N}(k)-R_{N})/w-5.371849}$

Table 4. $r=-1$ ,$\rho=\frac{1}{2}+2r^{2}=2.5$,$l_{0}\leq r(2r+1)-\rho=-1.5$$\sigma=0.5$$\sigma=1$$\sigma=2$$l_{0}=-2$$w=0.1$$w=0.025$$w=0.00625$$=2$$\sigma^{r}=2$$\sigma^{r}=1$$\sigma^{r}=0.5$

39.236610
1.951794
2.056594
1.034014
0.942045-3.678757

39.303210
0.976822
1.029155
1.018080
0.931146-3.477360

39.232770
0.487953
0.514149
1.026389
0.935053-3.653437$n^{*}=40$$\mu=0$,$m=22$ ,$k$$E(N)$$E(\sigma_{N}^{r})$$E(\sigma_{N}^{r}(k))$$R_{N}/w$$R_{N}(k)/w$$n^{*}(R_{N}(k)-R_{N})/w$

81

82
Table 5. r $=1$ ,$n^{*}=40$$\mu=0$ ,$l_{0}=0$$m=16$ ,$k=3$$E(N)$$E(\sigma_{N}^{r})$$E(\sigma_{N}^{r}(k))$$R_{N}/w$$R_{N}(k)/w$$\rho=\frac{1}{2}+2r^{2}=2.5$,$l_{0}\leq \mathrm{r}(2\mathrm{r}+1)-\rho=0.5$$\sigma=0.5$

a $=1$$\sigma=2$$w=0.00625$$w=0.025$$w=0.1$$\sigma^{r}=0.5$$\sigma^{r}=1$$\sigma^{r}=2$37.647620
0.468557
0.509244
1.470935
1.107491

37.596930
0.936478
1.017909
1.470930
1.103551

37.616820
1.873344
2.036156
1.473031
1.106660-14.695145-14.654831$\underline{n^{*}(R_{N}(k)-R_{N})/w14.537765}$

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