On Bahadur Efficiency of Power Divergence Statistics

Peter Harremoës, and Igor Vajda Manuscript submitted February 2010.P. Harremoës is with Copenhagen Business College, Copenhagen, Denmark. I. Vajda is with Institute of Information Theory and Automation, Prague, Czech Republic.

Abstract

It is proved that the information divergence statistic is infinitely more Bahadur efficient than the power divergence statistics of the orders $\alpha>1$ as long as the sequence of alternatives is contiguous with respect to the sequence of null-hypotheses and the the number of observations per bin increases to infinity is not very slow. This improves the former result in Harremoës and Vajda (2008) where the the sequence of null-hypotheses was assumed to be uniform and the restrictions on on the numbers of observations per bin were sharper. Moreover, this paper evaluates also the Bahadur efficiency of the power divergence statistics of the remaining positive orders $0\,<\alpha\leq 1$ . The statistics of these orders are mutually Bahadur-comparable and all of them are more Bahadur efficient than the statistics of the orders $\alpha>1.$ A detailed discussion of the technical definitions and conditions is given, some unclear points are resolved, and the results are illustrated by examples.

Index Terms:

Bahadur efficiency, consistency, power divergence, Rényi divergence.

I Introduction

Problems of detection, classification and identification are often solved by the method of testing statistical hypotheses. Consider signals $Y_{1},Y_{2},...,Y_{n}$ collected from a random source independently at time instants $i=1,2,...,n.$ Signal processing usually requires digitalization based on appropriate quantization. Quantization of the signal space $\mathcal{Y}$ into $k$ disjoint cells (or bins) $\mathcal{Y}_{n1},\mathcal{Y}_{n2},...,\mathcal{Y}_{nk}$ reduces the signals $Y_{1},Y_{2},...,Y_{n}$ into simple $k$ -valued indicators $I_{n}(Y_{1}),I_{n}(Y_{2}),...,I_{n}(Y_{n})$ of their cover cells. Various hypotheses about the data source represented by probability measures $Q_{n}$ on $\mathcal{Y}$ are transformed by the quantization into discrete probability distributions

Q_{n}=\left(q_{n1}=Q(\mathcal{Y}_{n1}),...,q_{nk}=Q(\mathcal{Y}_{nk})\right)

on the quantization cells where for no quantization cell $q_{nj}=0.$ These hypothetical distributions need not be the same as the true distributions $P_{n}=(p_{n1}=P(\mathcal{Y}_{n1}),...,p_{nk}=P(\mathcal{Y}_{nk}))$ . The latter distributions are usually unknown but, by the law of large numbers, they can be approximated by the empirical distributions (vectors of relative cell frequencies)

\hat{P}_{n}=\left(\hat{p}_{n1}=\frac{X_{n1}}{n},...,p_{nk}=\frac{X_{nk}}{n}\right)=\frac{\mathbf{X}_{n}}{n}

(1)

where $X_{nj}$ is the numbers of the signals $Y_{1},Y_{2},...,Y_{n}$ in $\mathcal{Y}_{nj}.$ Formally,

X_{nj}={\textstyle\sum\limits_{i=1}^{n}}1_{\left\{Y_{i}\in\mathcal{Y}_{nj}\right\}}={\textstyle\sum\limits_{i=1}^{n}}1_{\left\{I_{n}(Y_{i})=j\right\}},\text{ \ \ }1\leq j\leq k

(2)

where $1_{A}$ denotes the indicator of the event $A$ . The problem is to decide whether the signals $Y_{1},Y_{2},...,Y_{n}$ are generated by the source $(\mathcal{Y},Q)$ on the basis of the distributions $\hat{P}_{n},Q_{n}$ . A classical method for solving this problem is the method of testing statistical hypotheses in the spirit of Fisher, Neyman and Pearson. In our case the hypothesis is

{\mathcal{H}}:P_{n}=Q_{n}

(3)

and the decision is based either on the likelihood ratio statistic

\hat{T}_{1,n}=2{\textstyle\sum\limits_{j=1}^{k}}X_{nj}\ln\frac{X_{nj}}{nq_{nj}}

(4)

or the Pearson $\chi^{2}$ -statistic

\hat{T}_{2,n}={\textstyle\sum\limits_{j=1}^{k}}\frac{(X_{nj}-nq_{nj})^{2}}{nq_{nj}}

(5)

in the sense that the hypothesis is rejected when the statistic is large, where ”large” depends on the required decision error or risk [1].

It is easy to see (c.f. (13), (14) below) that the classical test statistics (4), (5) are of the form

\hat{T}_{\alpha,n}=2n\hat{D}_{\alpha,n}\overset{\text{def}}{=}2nD_{\alpha}\left(\hat{P}_{n},Q_{n}\right),\quad\alpha\in\{1,2\}

(6)

where $D_{\alpha}\left(P,Q\right)$ for arbitrary $\alpha>0$ and distributions $P=(p_{1},...,p_{k})$ , $Q=(q_{1},...,q_{k})$ denotes the divergence $D_{\phi_{\alpha}}\left(P,Q\right)$ of Csiszár [2] for the power function

\phi_{\alpha}(t)={\frac{t^{\alpha}-\alpha(t-1)-1}{\alpha(\alpha-1)}}\text{ \ when \ \ }\alpha\neq 1

(7)

and

\phi_{1}(t)=\lim_{\alpha\rightarrow 1}\phi_{\alpha}(t)=t\ln t-t+1.

(8)

The power divergences

D_{\alpha}\left(P,Q\right)=\frac{1}{\alpha(\alpha-1)}\left({\textstyle\sum\limits_{j=1}^{k}}p_{j}^{\alpha}q_{j}^{1-\alpha}-1\right)\text{ \ \ }\alpha\neq 1

(9)

or the one-one related Rényi divergences [3]

D_{\alpha}\left(P\|Q\right)=\frac{1}{\alpha-1}\ln{\textstyle\sum\limits_{j=1}^{k}}p_{j}^{\alpha}q_{j}^{1-\alpha}\text{ \ \ }\alpha\neq 1

(10)

with the common information divergence limit

D_{1}\left(P,Q\right)=D_{1}\left(P\|Q\right)={\textstyle\sum\limits_{j=1}^{k}}p_{j}\ln\frac{p_{j}}{q_{j}}

(11)

are often applied in various areas of information theory. In the present context of detection and identification one can mention e.g. the work of Kailath [4] who used the Bhattacharryya distance

B\left(P,Q\right)=-\ln{\textstyle\sum\limits_{j=1}^{k}}\left(p_{j}q_{j}\right)^{1/2}=\frac{1}{2}D_{1/2}\left(P\|Q\right)

which is one-one related to the Hellinger divergence.

In practical applications it is important to use the statistic $\hat{D}_{\alpha_{\text{opt}},n}$ which is optimal in a sufficiently wide class of divergence statistics $\hat{D}_{\alpha,n}$ containing the standard statistical proposals $\hat{D}_{1,n}$ and $\hat{D}_{2,n}$ appearing in (6). We addressed this problem previously [5, 6, 7]. Our solution confirmed the classical statistical result of Quine and Robinson [8] who proved that the likelihood ratio statistic $\hat{D}_{1,n}$ is more efficient in the Bahadur sense than the $\chi^{2}$ -statistic $\hat{D}_{2,n}$ and extended the results of Beirlant et al. [9] and Györfi et al. [10] dealing with Bahadur efficiency of several selected power divergence statistics. Namely, we evaluated the Bahadur efficiencies of the statistics $\hat{D}_{n\alpha}$ in the domain $\alpha\geq 1$ for the numbers $k=k_{n}$ of quantization cells slowly increasing with $n$ when the hypothetical distributions $Q_{n}$ are uniform and the alternative distributions $P_{n}$ are contiguous in the sense that $\lim_{n\rightarrow\infty}D_{\alpha}\left(P_{n},Q_{n}\right)$ exists and identifiable in the sense that this limit is positive. We found that the Bahadur efficiencies decrease with the power parameter in the whole domain $\alpha\geq 1$ . In the present paper we sharpen this result by relaxing conditions on the rate of $k_{n}$ and extend it considerably by admitting non-uniform hypothetical distributions $Q_{n}$ and by evaluating the Bahadur efficiencies also in the domain $0<\alpha\leq 1.$

II Basic model

Let $M(k)$ denote the set of all probability distributions $P=(p_{j}:1\leq j\leq k)$ and

M(k|n)=\left\{P\in M(k):nP\in\{0,1,\ldots\}^{k}\right\}

its subset called the set of types in information theory. We consider hypothetical distributions $Q_{n}=(q_{nj}:1\leq j\leq k)\in M(k)$ restricted by the condition $q_{nj}>0$ and arbitrary alternative distributions $P_{n}=(p_{nj}:1\leq j\leq k)\in M(k).$ The $\{0,1,\ldots\}^{k}$ -valued frequency counts $\mathbf{X}_{n}$ with coordinates introduced in (2) are multinomially distributed in the sense

\mathbf{X}_{n}\sim\mbox{Mult}_{k}(n,P_{n}),n=1,2,\ldots.

(12)

Important components of the model are the empirical distributions $\widehat{P}_{n}\in M(k|n)$ defined by (1). Finally, for arbitrary $P\in M(k)$ and arbitrary $Q\in M(k)$ with positive coordinates we consider the power divergences (9)-(8). For their properties we refer to [11, 12, 13]. In particular, for the empirical and hypothetical distributions $\hat{P}_{n},Q_{n}$ we consider the power divergence statistics $\widehat{D}_{\alpha,n}=D_{\alpha}\left(\hat{P}_{n},Q_{n}\right)$ (c.f. (6))defined by (9), (11) for all $\alpha>0$ .

Example 1

For $\alpha=2,$ $\alpha=1$ and $\alpha=1/2$ we get the special power divergence statistics

$\displaystyle\widehat{D}_{2,n}$	$\displaystyle={\frac{1}{2}}\sum_{j=1}^{n}{\frac{(\widehat{p}_{nj}-q_{nj})^{2}}{q_{nj}}}={\frac{1}{2n}}\hat{T}_{2,n},\vskip 6.0pt plus 2.0pt minus 2.0pt$	(13)
$\displaystyle\widehat{D}_{1,n}$	$\displaystyle=\sum_{j=1}^{n}\widehat{p}_{nj}\ln{\frac{\widehat{p}_{nj}}{q_{nj}}}={\frac{1}{2n}}\hat{T}_{1,n},\vskip 6.0pt plus 2.0pt minus 2.0pt$	(14)
$\displaystyle\widehat{D}_{1/2,n}$	$\displaystyle=2\sum_{j=1}^{n}\left(\widehat{p}_{nj}^{1/2}-q_{nj}^{1/2}\right)^{2}$	(15)

For testing the hypothesis $\mathcal{H}$ of (3) are usually used the re-scaled versions

\widehat{T}_{\alpha,n}=2n\widehat{D}_{\alpha,n}

(16)

distributed under $\mathcal{H}$ asymptotically $\chi^{2}$ with $k-1$ degrees of freedom if $k$ is constant and asymptotically normally if $k=k_{n}$ slowly increases to infinity [14, 15, and references therein] . The statistics (13) and (14) rescaled in this manner were already mentioned in (5) and (4). In (15) is the Hellinger divergence statistics rescaled by $2n$ is known as Freeman–Tukey statistic

\widehat{T}_{1/2,n}=2n\widehat{D}_{1/2,n}=4\sum_{j=1}^{k}(\left(X_{nj}\right)^{1/2}-\left(nq_{nj}\right)^{1/2})^{2}.

(17)

Convention

Unless the hypothesis ${\mathcal{H}}$ is explicitly assumed, the random variables, convergences and asymptotic relations are considered under the alternative ${\mathcal{A}}$ . Further, unless otherwise explicitly stated, the asymptotic relations are considered for $n\longrightarrow\infty$ and the symbols of the type

s_{n}\longrightarrow s\text{ \ \ and \ \ }s_{n}(\mathbf{X}_{n})\overset{p}{\longrightarrow}s

denote the ordinary numerical convergence and the stochastic convergence in probability for $n\longrightarrow\infty$ .

In this paper we consider the following assumptions.

A1:

The number of cells $k=k_{n}\leq n$ of the distributions from $M(k),\,M(k|n)$ depends on the sample size $n$ and increases to infinity. In the rest of the paper the subscript $n$ is suppressed in the symbols containing $k$ .

A2:

The hypothetical distributions $Q_{n}=(q_{nj}>0:1\leq j\leq k)$ are regular in the sense that $\max_{j}q_{nj}\rightarrow 0$ for $n\rightarrow\infty$ and that there exists $\varrho>0$ such that

q_{nj}>\frac{\varrho}{k}\text{ \ for all }1\leq j\leq k\text{\ and }n=1,2,\ldots\text{ .}

(18)

A3 $\alpha$ :

The alternative ${\mathcal{A}}:(P_{n}:n=1,2,\ldots)$ is identifiable in the sense that there exits $0<\Delta_{\alpha}<\infty$ such that

D_{\alpha,n}\overset{def}{=}D_{\alpha}(P_{n},Q_{n})\longrightarrow\Delta_{\alpha}\text{\ \ under }{\mathcal{A}}.

(19)

Under A2

-\ln q_{nj}<\ln\frac{k}{\varrho}\text{ \ \ and \ \ }\ln^{2}q_{nj}<\ln^{2}\frac{k}{\varrho}.

(20)

Further, logical complement to the hypothesis ${\mathcal{H}}$ is the alternative denoted by ${\mathcal{A}}.$ By (3), under ${\mathcal{A}}$ the alternative distributions $P_{n}$ differ from $Q_{n}$ . Assumption A3 $\alpha$ means that the alternative distributions are neither too close to nor too distant from $Q_{n}$ in the sense of $D_{\alpha}$ -divergence for given $\alpha>0$ . Since for all $n=1,2,\ldots$

D_{\alpha,n}=D_{\alpha}(Q_{n},Q_{n})\equiv 0\ \ \text{so\ that }\Delta_{\alpha}=0\text{\ \ under }{\mathcal{H}}

it is clear that the hypothesis ${\mathcal{A}}$ is under A1, A2, A3 $\alpha$ distinguished from the hypothesis ${\mathcal{H}}$ by achieving a positive $D_{\alpha}$ -divergence limit $\Delta_{\alpha}$ . In what follows we use the abbreviated notations

	$\displaystyle\text{{A}}(\alpha)$	$\displaystyle=\left\{\text{{A1},{A2},\thinspace{A3}}\alpha\right\},$		(21)
	$\displaystyle\text{{A}}(\alpha_{1},\alpha_{2})$	$\displaystyle=\left\{\text{{A1},{A2},\thinspace{A3}}\alpha_{1}\text{, {A3}}\alpha_{2}\right\}$		(22)

for the combinations of assumptions.

Definition 2

Under A( $\alpha$ ) we say that the statistic $\widehat{D}_{\alpha,n}$ is consistent with parameter $\Delta_{\alpha}$ appearing in (19) if

\widehat{D}_{\alpha,n}\overset{p}{\longrightarrow}\Delta_{\alpha}\text{ \ \ under }{\mathcal{A}}

(23)

and

\widehat{D}_{\alpha,n}\overset{p}{\longrightarrow}0\text{ \ \ under }{\mathcal{H}}

(24)

i.e. if $\widehat{D}_{\alpha,n}\overset{p}{\longrightarrow}\Delta_{\alpha}$ under both ${\mathcal{A}}$ and ${\mathcal{H}}$ . If (24) is replaced by the stronger condition that the expectation E $\widehat{D}_{\alpha,n}$ tends to zero under ${\mathcal{H}}$ , in symbols

\text{{E}}\left[\left.\widehat{D}_{\alpha,n}\right|{\mathcal{H}}\right]\longrightarrow 0,

(25)

then $\widehat{D}_{\alpha,n}$ is said strongly consistent.

Definition 3

We say that the statistic $\widehat{D}_{\alpha,n}$ is Bahadur stable if there is a continuous function with a Bahadur relative function $\varrho_{\alpha}:$ $]0,\infty[^{2}$ $\rightarrow$ $]0,\infty[$ such that the probability of error function

\mathsf{e}_{\alpha,n}(\Delta)=\mathsf{P}\left(\left.\widehat{D}_{\alpha,n}>\Delta\right|{\mathcal{H}}\right),\text{ \ \ }\Delta>0

(26)

corresponding to the test rejecting ${\mathcal{H}}$ when $\widehat{D}_{\alpha,n}>\Delta$ satisfies for all $\Delta_{1},\Delta_{2}>0$ the relation

\frac{\ln\mathsf{e}_{\alpha,n}(\Delta_{1})}{\ln\mathsf{e}_{\alpha,n}(\Delta_{2})}\longrightarrow\varrho_{\alpha}\left(\Delta_{1},\Delta_{2}\right).

If this condition holds then $\varrho_{\alpha}$ is called the Bahadur relative function.

Obviously, the Bahadur relative functions are multiplicative in the sense

\varrho_{\alpha}\left(\Delta_{1},\Delta_{2}\right)\varrho_{\alpha}\left(\Delta_{2},\Delta_{3}\right)=\varrho_{\alpha}\left(\Delta_{1},\Delta_{3}\right).

Statistics that are Bahadur stable have the nice property that the asymptotic behavior of the error function $\mathsf{e}_{\alpha,n}(\Delta)$ is determined by its behavior for just a single argument $\Delta^{\ast}>0.$ Indeed, if $\widehat{D}_{\alpha,n}$ is Bahadur stable and if we define for a fixed $\Delta^{\ast}>0$ the sequence

c_{\alpha}^{\ast}(n)=-\frac{n}{\ln\mathsf{e}_{\alpha,n}(\Delta^{\ast})}

(27)

then for all $\Delta>0$

-{\frac{c_{\alpha}^{\ast}(n)}{n}}\ln\mathsf{e}_{\alpha,n}(\Delta)\longrightarrow\varrho_{\alpha}\left(\Delta,\Delta^{\ast}\right)\text{ \ \ for all }\Delta>0.

Moreover, if the expressions $\ -{c_{\alpha}(n)/n}\ln\mathsf{e}_{\alpha,n}(\Delta)$ converge for a sequence $c_{\alpha}\left(n\right)$ then the ratio $c_{\alpha}(n)/c_{\alpha}^{\ast}(n)$ tends to a constant.

Motivation of the next definition

Suppose that condition A( $\alpha_{1},\alpha_{2}$ ) holds and denote for each $\alpha\in\{\alpha_{1},\alpha_{2}\}$ and $n=1,2,\ldots$ by $\Delta_{\alpha}+\varepsilon_{\alpha,n}$ the critical value of the statistics $\widehat{D}_{\alpha_{i},n}$ leading to the rejection of $\mathcal{H}$ with a fixed power $0<\mathsf{p}<1$ . In other words, let

\mathsf{p}=\text{{P}}\left(\widehat{D}_{\alpha,n}>\Delta_{\alpha}+\varepsilon_{\alpha,n}\right)\text{ \ \ for all }n=1,2,\ldots\text{ }

where the sequence $\varepsilon_{\alpha,n}=\varepsilon_{\alpha,n}(\mathsf{p})$ depends on the fixed $\mathsf{p}$ . Since the assumed consistency of $\widehat{D}_{\alpha,n}$ implies that $\varepsilon_{\alpha,n}$ tends to zero, the corresponding error probabilities $\mathsf{e}_{\alpha,n}(\Delta_{\alpha}+\varepsilon_{\alpha,n})$ $=\mathsf{P}\left(\left.\widehat{D}_{\alpha,n}>\Delta_{\alpha}+\varepsilon_{\alpha,n}\right|\mathcal{H}\right)$ can be approximated by $\mathsf{e}_{\alpha,n}(\Delta_{\alpha})$ $=\mathsf{P}\left(\left.\widehat{D}_{\alpha,n}>\Delta_{\alpha}\right|\mathcal{H}\right).$ By (33),

-{\frac{c_{\alpha}(n)}{n}}\ln\mathsf{e}_{\alpha,n}(\Delta_{\alpha})\longrightarrow g_{\alpha}(\Delta_{\alpha}).

Hence the error $\mathsf{e}_{\alpha_{1},n}(\Delta_{\alpha_{1}})$ of the statistic $\widehat{D}_{\alpha_{1},n}$ tends to zero with the same exponential rate as $\mathsf{e}_{\alpha_{2},m_{n}}(\Delta_{\alpha_{2}})$ achieved by $\widehat{D}_{\alpha_{2},m_{n}}$ for possibly different sample sizes $m_{n}\neq n$ with the property $m_{n}\longrightarrow\infty$ if the corresponding error exponents

g_{\alpha_{1}}(\Delta_{\alpha_{1}})\frac{n}{c_{\alpha_{1}}(n)}\text{ \ \ and \ \ }g_{\alpha_{2}}(\Delta_{\alpha_{2}})\frac{m_{n}}{c_{\alpha_{2}}(m_{n})}

(28)

tend to infinity with the same rate in the sense

\frac{m_{n}}{c_{\alpha_{2}}(m_{n})}=\frac{g_{\alpha_{1}}(\Delta_{\alpha_{1}})}{g_{\alpha_{2}}(\Delta_{\alpha_{2}})}.\frac{n}{c_{\alpha_{1}}(n)}\left(1+o(1)\right).

(29)

The sample sizes $m_{n}$ and $n$ needed by the statistics $\widehat{D}_{\alpha_{2},n}$ and $\widehat{D}_{\alpha_{1},n}$ to achieve the same rate of convergence of errors are thus mutually related by the formula

{\frac{m_{n}}{n}=}\text{ }\frac{g_{\alpha_{1}}(\Delta_{\alpha_{1}})}{g_{\alpha_{2}}(\Delta_{\alpha_{2}})}.\frac{c_{\alpha_{2}}(m_{n})}{c_{\alpha_{1}}(n)}\left(1+o(1)\right).

(30)

Obviously, the statistic $\widehat{D}_{\alpha_{1},n}$ is asymptotically less or more efficient than $\widehat{D}_{\alpha_{2},n}$ if the ratio $m_{n}/n$ of sample sizes needed to achieve the same rate of convergence of errors to zero tends to a constant larger or smaller than $1$ . This motivates the following definition which refers to the typical convergent situation

{\frac{c_{\alpha_{2}}(m_{n})}{c_{\alpha_{1}}(n)}\longrightarrow}\text{ }c_{\alpha_{2}/\alpha_{1}}\text{ \ \ for some }0\leq c_{\alpha_{2}/\alpha_{1}}\leq\infty.\vskip 6.0pt plus 2.0pt minus 2.0pt

(31)

Definition 4

If there is a continuous function

g_{\alpha}:\left]0,\infty\right[\mathcal{\ }{\mathcal{\rightarrow}}\left]0,\infty\right[

and a sequence $c_{\alpha}(n)$ such that for all $x>0$ the error function

\mathsf{e}_{\alpha,n}(x)=\mathsf{P}\left(\left.D_{\alpha,n}>x\right|\mathcal{H}\right),\text{ \ \ }x>0

(32)

satisfies for all $x>0$ the relation

-{\frac{c_{\alpha}(n)}{n}}\ln\mathsf{e}_{\alpha,n}(x)\longrightarrow g_{\alpha}(x)

(33)

then $g_{\alpha}$ is called Bahadur function of the statistic $D_{\alpha,n}$ generated by $c_{\alpha}(n)$ . If (33) is replaced by the condition

-{\frac{c_{\alpha}(n)}{n}}\ln\mathsf{e}_{\alpha,n}(x+\varepsilon_{n})\longrightarrow g_{\alpha}(x)\text{ \ \ for arbitrary }\varepsilon_{n}\longrightarrow 0

(34)

then the function $g_{\alpha}$ is strongly Bahadur.

Definition 5

Let us assume that A( $\alpha_{1},\alpha_{2}$ ) holds and that for each $\alpha\in\{\alpha_{1},\alpha_{2}\}$ the statistic $\widehat{D}_{n,\alpha}$ is consistent with parameter $\Delta_{\alpha}$ and has a Bahadur function $g_{\alpha}$ generated by a sequence $c_{\alpha}(n)$ such that (31) is satisfied. Then the Bahadur efficiency of $\widehat{D}_{\alpha_{1},n}$ with respect to $\widehat{D}_{\alpha_{2},n}$ is the number from the interval $[0,\infty]$ defined by the formula

\mbox{BE}\left(\widehat{D}_{\alpha_{1},n}\,;\text{ }\widehat{D}_{\alpha_{2},n}\right)={\frac{g_{\alpha_{1}}(\Delta_{\alpha_{1}})}{g_{\alpha_{2}}(\Delta_{\alpha_{2}})}}.c_{\alpha_{2}/\alpha_{1}}.

(35)

Hereafter we shall consider also the slightly modified concept of Bahadur efficiency.

Definition 6

Let in addition to the assumptions of Definition 5, the statistics $\widehat{D}_{\alpha_{1},n},$ $\widehat{D}_{\alpha_{2},n}$ be strongly consistent and the functions $g_{\alpha_{1}},$ $g_{\alpha_{2}}$ strongly Bahadur. Then the Bahadur efficiency (35) is said to be Bahadur efficiency in the strong sense.

Motivation of Definition 6

Let the assumptions of this definition hold then for each $\alpha\in\{\alpha_{1},\alpha_{2}\},and$ $u>0$ the function

L_{\alpha,n}(u)=\text{{P}}\left(\left.{\widehat{T}}_{\alpha,n}-\mathsf{E}\left[\left.{\widehat{T}}_{\alpha,n}\right|{\mathcal{H}}\right]\geq u\right|{\mathcal{H}}\right),\ \text{(cf. \ref{10})}

denotes the level of the error of the statistic

{\widehat{T}}_{\alpha,n}-\mathsf{E}\left[\left.{\widehat{T}}_{\alpha,n}\right|{\mathcal{H}}\right]\equiv 2n\left(\widehat{D}_{\alpha,n}-\mathsf{E}\left[\left.\widehat{D}_{\alpha,n}\right|{\mathcal{H}}\right]\right)\

for critical value $u>0$ . By the assumed strong consistency of $\widehat{D}_{\alpha,n},$

\frac{\text{{E}}\left[\left.\widehat{T}_{\alpha,n}\right|{\mathcal{H}}\right]}{2n}\longrightarrow 0\ \ \ \ \ \mbox{(cf.(\ref{9}))}.

This means that the sequence $c_{\alpha}(n)$ generating the strongly Bahadur $g_{\alpha}$ satisfies for all $t>0$ the relation

-{\frac{c_{\alpha}(n)}{n}}\ln\text{{P}}\left(\left.{\widehat{T}}_{\alpha,n}\geq\text{{E}}\left[\left.{\widehat{T}}_{\alpha,n}\right|{\mathcal{H}}\right]+2nt\right|{\mathcal{H}}\right)\longrightarrow g_{\alpha}(t)\text{ \ \ .}

(cf. (34))

Consequently, by the argument of Quine and Robinson [8, p. 732],

\lim\nolimits_{n}-{\frac{c_{\alpha}(n)}{n}}\ln L_{\alpha,n}({\widehat{T}}_{\alpha,n})\overset{p}{\longrightarrow}g_{\alpha}(\Delta_{\alpha}).

Hence [8], the error level $L_{\alpha_{1},n}({\widehat{T}}_{\alpha_{1},n})$ of the statistic ${\widehat{T}}_{\alpha_{1},n}=2n\widehat{D}_{\alpha_{1},n}$ is asymptotically equivalent to the error level $L_{\alpha_{2},m_{n}}({\widehat{T}}_{\alpha_{2},m_{n}})$ of the statistic ${\widehat{T}}_{\alpha_{2},m_{n}}=2m_{n}\widehat{D}_{\alpha_{2},m_{n}}$ achieved by a sample size $m_{n}$ if the comparability (29) takes place or, in other words, if the sample sizes $n$ and $m_{n}$ are mutually related by (30). In other words, the concept of Bahadur efficiency introduced in this paper coincides under the stronger assumptions of Definition 6 with the Bahadur efficiency of Quine and Robinson [8].

Harremoës and Vajda [5] assumed the same strong consistency as in Definition 6 but introduced the Bahadur efficiency by the slightly different formula

\mbox{BE}\left(\widehat{D}_{\alpha_{1},n}\,;\text{ }\widehat{D}_{\alpha_{2},n}\right)={\frac{g_{\alpha_{1}}(\Delta_{\alpha_{1}})}{g_{\alpha_{2}}(\Delta_{\alpha_{2}})}}.\bar{c}_{\alpha_{2}/\alpha_{1}}

(36)

where¹¹1Due to a missprint, $\alpha_{1}$ and $\alpha_{2}$ were interchanged behind the limit in [5, Eq. 30], but the formula was used in the correct form (36). In the Appendix we prove that the conclusions made on the basis of the original formula (36) hold unchanged under the present precised formula (35).

\bar{c}_{\alpha_{2}/\alpha_{1}}=\lim_{n\longrightarrow\infty}\frac{c_{\alpha_{2}}(n)}{c_{\alpha_{1}}(n)}.

(37)

III Consistency

In this section we study the consistency of the class of power divergence statistics $D_{\alpha}(\widehat{P}_{n},Q_{n}),$ $\alpha>0.$ In the domain $\alpha<0$ this consistency was studied in the particular case of uniform $Q$ by Harremoës and Vajda [6].

Theorem 7

Let distributions $Q_{n}\in M\left(k\right)$ satisfy the assumption A( $\alpha$ ). Assume that $f$ is uniformly continuous. Then the statistic $D_{f}(\widehat{P}_{n},Q_{n})$ is strongly consistent provided

\frac{n}{k}\longrightarrow\infty.

(38)

Proof:

Under $\mathcal{H}$ we have $D_{f}(P_{n},Q_{n})=D_{f}(Q_{n},Q_{n})=0.$ Hence it suffices to prove

\left|\Lambda_{\alpha,n}\right|\overset{p}{\longrightarrow}0\text{ \ \ under both }\mathcal{H}\ \text{and }\mathcal{A}

(39)

for $\Lambda_{\alpha,n}=D_{f}(\widehat{P}_{n},Q_{n})-D_{f}(P_{n},Q_{n})$ . For simplicity we skip the subscript $n$ in the symbols $\widehat{P}_{n},P_{n},$ and $Q_{n}$ , i.e. we substitute

\widehat{P}_{n}=\widehat{P}=(\widehat{p}_{j}:1\leq j\leq k),\text{ \ \ }P_{n}=P=(p_{j}:1\leq j\leq k).

(40)

This leads to the simplified formula $\Lambda_{\alpha,n}=D_{f}(\widehat{P},Q)-D_{f}(P,Q).$ We can without loss of generality assume that $D_{f}(P,Q)$ is constant not only under $\mathcal{H}$ (where the constant is automatically $0$ ) but also under $\mathcal{A}$ (where the assumed detectability implies the convergence $D_{f}(P,Q)\longrightarrow\Delta_{\alpha}$ for $0<\Delta_{\alpha}<\infty$ ). In this asymptotic sense we use the equalities

D_{f}(P,Q)=\frac{\sum q_{j}\left(\frac{p_{j}}{q_{j}}\right)^{\alpha}-1}{\alpha\left(\alpha-1\right)}=\Delta_{\alpha}

(41)

and

\Lambda_{\alpha,n}=D_{f}(\widehat{P},U)-\Delta_{\alpha}\text{.}

(42)

Choose some $0<s<1$ and define

f^{s}\left(t\right)=\left\{\begin{array}[c]{ll}f\left(t\right)&\text{for }t\geq s,\\ f\left(s\right)+f_{+}^{\prime}\left(s\right)\left(t-s\right)&\text{for }0\leq t<s.\end{array}\right.

Then

0\leq f\left(t\right)-f^{s}\left(t\right)\leq f\left(0\right)-f^{s}\left(0\right)

so that (9) implies

0\leq D_{f}\left(P,Q\right)-D_{f^{s}}\left(P,Q\right)\leq f\left(0\right)-f^{s}\left(0\right).

The function $f^{s}$ is Lipschitz with the Lipschitz constant $\lambda=\max\left\{\left|f_{+}^{\prime}\left(s\right)\right|,\left|f^{\prime}\left(\infty\right)\right|\right\}$ i.e. $\left|f\left(t_{1}\right)-f\left(t_{2}\right)\right|\leq\lambda\left|t_{1}-t_{2}\right|$ for all $t_{1},t_{2}\geq 0.$ Then

\left|D_{f^{s}}(\widehat{P}_{n},Q)-D_{f^{s}}(P_{n},Q)\right|\\ =\left|\sum_{j=1}^{k}q_{j}\,f^{s}\left({\frac{\widehat{p}_{j}}{q_{j}}}\right)-\sum_{j=1}^{k}\,q_{j}f^{s}\left({\frac{p_{j}}{q_{j}}}\right)\right|\\ =\sum_{j=1}^{k}q_{j}\,\left|f^{s}\left({\frac{\widehat{p}_{j}}{q_{j}}}\right)-f^{s}\left({\frac{p_{j}}{q_{j}}}\right)\right|\leq\sum_{j=1}^{k}q_{j}\lambda\,\left|{\frac{\widehat{p}_{j}}{q_{j}}}-{\frac{p_{j}}{q_{j}}}\right|\\ =\lambda\sum_{j=1}^{k}\,\left|\widehat{p}_{j}-p_{j}\right|\leq\lambda\left(\sum_{j=1}^{k}\frac{\left(\widehat{p}_{j}-p_{j}\right)^{2}}{p_{j}}\right)^{1/2}

where in the last step we used the Schwarz inequality. Since

\mathsf{E}\left[\left(\widehat{p}_{j}-p_{j}\right)^{2}\right]=p_{j}(1-p_{j})/n\leq\ p_{j}/n

(43)

it holds

\mathsf{E}\left|D_{f^{s}}(\widehat{P}_{n},Q)-D_{f^{s}}(P_{n},Q)\right|\\ \leq\lambda\left(\mathsf{E}\left[\sum_{j=1}^{k}\frac{\left(\widehat{p}_{j}-p_{j}\right)^{2}}{p_{j}}\right]\right)^{1/2}\leq\lambda\left(\frac{k}{n}\right)^{1/2}.

Consequently,

\mathsf{E}\left|D_{f}(\widehat{P}_{n},Q)-D_{f}(P_{n},Q)\right|\\ \leq 2\left(f\left(0\right)-f^{s}\left(0\right)\right)+\lambda\left(k/n\right)^{1/2}

so that under (49)

\underset{n\rightarrow\infty}{\lim\sup\text{ }}\mathsf{E}\left|D_{f}(\widehat{P}_{n},Q_{n})-D_{f}(P_{n},Q_{n})\right|\\ \leq 2\left(f\left(0\right)-f^{s}\left(0\right)\right).

This holds for all $s>0.$ Since $f\left(0\right)-f^{s}\left(0\right)\longrightarrow 0$ for $s\downarrow 0,$ we see that in this case (38) implies (39). ∎

The interpretation of condition 38 is that the mean number of observations per bin should tend to infinity under $\mathcal{H}$ . Note that this condition does not exclude that we will observe empty cells.

Our results are concentrated in Theorem 9 below. Its proof uses the following auxiliary result.

Lemma 8

For $x,y\geq 0$ and $1\leq\alpha\leq 2$ it holds

L_{\alpha}\left(x,y\right)\leq\phi_{\alpha}(y)-\phi_{\alpha}(x)\leq U_{\alpha}\left(x,y\right)

(44)

where

L_{\alpha}\left(x,y\right)=(y-x)\phi_{\alpha}^{\prime}(x)

(45)

and

U_{\alpha}\left(x,y\right)=L_{\alpha}\left(x,y\right)+\frac{1}{\alpha}x^{\alpha-2}\left(y-x\right)^{2}.\vskip 6.0pt plus 2.0pt minus 2.0pt

(46)

Proof:

First assume $1<\alpha<2.$ Since $\frac{1}{\alpha}x^{\alpha-2}\left(y-x\right)^{2}$ is nonnegative, it suffices to prove

\phi_{\alpha}(y)\geq\phi_{\alpha}(x)+\phi_{\alpha}^{\prime}(x)\left(y-x\right)

(47)

and

\phi_{\alpha}(y)\leq\phi_{\alpha}(x)+\phi_{\alpha}^{\prime}(x)\left(y-x\right)+\frac{1}{\alpha}x^{\alpha-2}\left(y-x\right)^{2}.

(48)

But Inequality (47) is evident since the function $y\rightarrow\phi_{\alpha}(y)$ is convex. We shall prove that the function

f\left(y\right)=\\ \phi_{\alpha}\left(y\right)-\left(\phi_{\alpha}\left(x\right)+\phi_{\alpha}^{\prime}(x)\left(y-x\right)+\frac{1}{\alpha}x^{\alpha-2}\left(y-x\right)^{2}\right)

is non-positive. First we observe that $f(0)=f(x)=0$ . By differentiating $f\left(y\right)$ we get

f^{\prime}\left(y\right)={\frac{y^{\alpha-1}-1}{\alpha-1}-}\left(\phi_{\alpha}^{\prime}(x)+\frac{2}{\alpha}x^{\alpha-2}\left(y-x\right)\right)

so that $f^{\prime}\left(x\right)=0.$ Differentiating once more we get

f^{\prime\prime}\left(y\right)=y^{\alpha-2}{-}\frac{2}{\alpha}x^{\alpha-2}.

Thus $f^{\prime\prime}(y)>0$ for $y<x_{\alpha}\overset{def}{=}\left(\alpha/2\right)^{\frac{1}{2-\alpha}}x$ and $f^{\prime\prime}(y)<0$ for $y>x_{\alpha}.$ Since $x_{\alpha}<x$ and $f(y)$ is concave on $[x_{\alpha},1]$ , it is maximized on this interval at $y=x$ where $f(x)=0$ . Thus $f\left(y\right)\leq 0$ on this interval and in particular $f(x_{\alpha})\leq 0$ . This together with $f(0)=0$ and the convexity of $f\left(y\right)$ on the interval $[0,$ $x_{\alpha}]$ implies $f\left(y\right)\leq 0$ for $y\in\left[0,x\right]$ . This completes the proof of the non-positivity of $f\left(y\right)$ , i.e. the proof of (48). The cases $\alpha=2$ and $\alpha=1$ follow by continuity. ∎

The main result of this section is the following theorem.

Theorem 9

Let distributions $Q_{n}\in M\left(k\right)$ satisfy the assumption A( $\alpha$ ). Then the statistic $D_{\alpha}(\widehat{P}_{n},Q_{n})$ is strongly consistent provided

\rule{-19.91692pt}{0.0pt}0<\alpha\leq 2\text{ \ \ \ \ and \ \ \ \ }{\frac{n}{k}}\longrightarrow\infty\vskip-8.53581pt

(49)

and consistent provided

\rule{5.69054pt}{0.0pt}\alpha>2\rule{28.45274pt}{0.0pt}\text{\ \ \ and\ \ \ \ \ }{\frac{n}{k\log k}}\longrightarrow\infty.\vskip 6.0pt plus 2.0pt minus 2.0pt

(50)

Proof:

We shall use the same notation as in the proof of Theorem 7. In the proof we treat separately the cases

	i	$\displaystyle:0<\alpha<1,\quad\text{ }$
	ii	$\displaystyle:1<\alpha\leq 2,\quad\text{ }$
	iii	$\displaystyle:\alpha=1,$
	iv	$\displaystyle:\alpha>2.$

Case i ( $0<\alpha<1$ )

This follows from Theorem 7 because $x\rightarrow\phi_{\alpha}\left(x\right)$ is uniformly continuous.

Case ii ( $1<\alpha\leq 2$ )

Here we get from (42)

\Lambda_{\alpha,n}=\sum_{j=1}^{k}q_{j}\left(\phi_{\alpha}\left(\frac{\widehat{p}_{j}}{q_{j}}\right)-\phi_{\alpha}\left(\frac{p_{j}}{q_{j}}\right)\right)

(51)

so that Lemma 8 implies

\sum_{j=1}^{k}q_{j}L_{\alpha}\left(\frac{p_{j}}{q_{j}},\frac{\widehat{p}_{j}}{q_{j}}\right)\leq\Lambda_{\alpha,n}\leq\\ \\ \sum_{j=1}^{k}q_{j}L_{\alpha}\left(\frac{p_{j}}{q_{j}},\frac{\widehat{p}_{j}}{q_{j}}\right)+\sum_{j=1}^{k}q_{j}\frac{1}{\alpha}\left(\frac{p_{j}}{q_{j}}\right)^{\alpha-2}\left(\frac{\widehat{p}_{j}}{q_{j}}-\frac{p_{j}}{q_{j}}\right)^{2}

and

\left|\Lambda_{\alpha,n}\right|\leq\left|\sum_{j=1}^{k}(\widehat{p}_{j}-p_{j})\phi_{\alpha}^{\prime}\left(\frac{p_{j}}{q_{j}}\right)\right|+\sum_{j=1}^{k}\frac{p_{j}^{\alpha-2}}{q_{j}^{\alpha-1}}\frac{\left(\widehat{p}_{j}-p_{j}\right)^{2}}{\alpha}.

We take the mean and get

E\left|\Lambda_{\alpha,n}\right|\leq\\ \mathsf{E}\left|\sum_{j=1}^{k}(\widehat{p}_{j}-p_{j})\phi_{\alpha}^{\prime}\left(\frac{p_{j}}{q_{j}}\right)\right|+\sum_{j=1}^{k}\frac{p_{j}^{\alpha-2}}{\alpha q_{j}^{\alpha-1}}\mathsf{E}\left[\left(\widehat{p}_{j}-p_{j}\right)^{2}\right]

The terms on the right hand side are treated separately.

	$\displaystyle\sum_{j=1}^{k}\frac{p_{j}^{\alpha-2}}{q_{j}^{\alpha-1}}\frac{\mathsf{E}\left[\left(\widehat{p}_{j}-p_{j}\right)^{2}\right]}{\alpha}$	$\displaystyle=\sum_{j=1}^{k}\frac{p_{j}^{\alpha-2}}{q_{j}^{\alpha-1}}\frac{\mathsf{E}\left[\left(n\widehat{p}_{j}-np_{j}\right)^{2}\right]}{\alpha n^{2}}$
		$\displaystyle=\sum_{j=1}^{k}\frac{p_{j}^{\alpha-2}}{q_{j}^{\alpha-1}}\frac{np_{j}\left(1-p_{j}\right)}{\alpha n^{2}}$
		$\displaystyle\leq\frac{1}{\alpha n}\sum_{j=1}^{k}\frac{p_{j}^{\alpha-1}}{\left(\frac{\rho}{k}\right)^{\alpha-1}}$
		$\displaystyle\leq\frac{k^{\alpha-1}}{\alpha n\rho^{\alpha-1}}\sum_{j=1}^{k}p_{j}^{\alpha-1}.$

The function $P\rightarrow\sum_{j=1}^{k}p_{j}^{\alpha-1}$ is concave so it attains its maximum for $P=\left(1/k,1/k,\cdots,1/k\right).$ Therefore

	$\displaystyle\sum_{j=1}^{k}\frac{p_{j}^{\alpha-2}}{q_{j}^{\alpha-1}}\frac{\mathsf{E}\left[\left(\widehat{p}_{j}-p_{j}\right)^{2}\right]}{\alpha}$	$\displaystyle\leq\frac{k^{\alpha-1}}{\alpha n\rho^{\alpha-1}}k\left(\frac{1}{k}\right)^{\alpha-1}$
		$\displaystyle=\frac{1}{\alpha\rho^{\alpha-1}}\frac{k}{n}.$

Next we bound the first term.

\mathsf{E}\left|\sum_{j=1}^{k}(\widehat{p}_{j}-p_{j})\phi_{\alpha}^{\prime}\left(\frac{p_{j}}{q_{j}}\right)\right|\\ \leq\mathsf{E}\left[\left(\sum_{j=1}^{k}(\widehat{p}_{j}-p_{j})\phi_{\alpha}^{\prime}\left(\frac{p_{j}}{q_{j}}\right)\right)^{2}\right]^{1/2}\\ =\left(\sum_{i,j=1}^{k}Cov\left(\widehat{p}_{i},\widehat{p}_{j}\right)\phi_{\alpha}^{\prime}\left(\frac{p_{i}}{q_{i}}\right)\phi_{\alpha}^{\prime}\left(\frac{p_{j}}{q_{j}}\right)\right)^{1/2}\\ =\frac{1}{n}\left(\sum_{i,j=1}^{k}Cov\left(n\widehat{p}_{i},n\widehat{p}_{j}\right)\phi_{\alpha}^{\prime}\left(\frac{p_{i}}{q_{i}}\right)\phi_{\alpha}^{\prime}\left(\frac{p_{j}}{q_{j}}\right)\right)^{1/2}\\ =\frac{1}{n}\left(\begin{array}[c]{c}\sum_{i=1}^{k}Var\left(n\widehat{p}_{i}\right)\left(\phi_{\alpha}^{\prime}\left(\frac{p_{i}}{q_{i}}\right)\right)^{2}\\ +\sum_{i\neq j}Cov\left(n\widehat{p}_{i},n\widehat{p}_{j}\right)\phi_{\alpha}^{\prime}\left(\frac{p_{i}}{q_{i}}\right)\phi_{\alpha}^{\prime}\left(\frac{p_{j}}{q_{j}}\right)\end{array}\right)^{1/2}.

This equals

\frac{1}{n}\left(\begin{array}[c]{c}\sum_{i=1}^{k}np_{i}\left(1-p_{i}\right)\left(\phi_{\alpha}^{\prime}\left(\frac{p_{i}}{q_{i}}\right)\right)^{2}\\ +\sum_{i\neq j}np_{i}p_{j}\phi_{\alpha}^{\prime}\left(\frac{p_{i}}{q_{i}}\right)\phi_{\alpha}^{\prime}\left(\frac{p_{j}}{q_{j}}\right)\end{array}\right)^{1/2}\\ \leq\frac{1}{n^{1/2}}\left(\begin{array}[c]{c}\sum_{i=1}^{k}p_{i}\left(\phi_{\alpha}^{\prime}\left(\frac{p_{i}}{q_{i}}\right)\right)^{2}\\ +\sum_{i,j}p_{i}p_{j}\phi_{\alpha}^{\prime}\left(\frac{p_{i}}{q_{i}}\right)\phi_{\alpha}^{\prime}\left(\frac{p_{j}}{q_{j}}\right)\end{array}\right)^{1/2}.

This can be bounded as

\frac{1}{n^{1/2}}\left(\begin{array}[c]{c}\sum_{i=1}^{k}p_{i}\left(\phi_{\alpha}^{\prime}\left(\frac{p_{i}}{q_{i}}\right)\right)^{2}\\ +\left(\sum_{i}p_{i}\phi_{\alpha}^{\prime}\left(\frac{p_{i}}{q_{i}}\right)\right)^{2}\end{array}\right)^{1/2}\\ \leq\left(\frac{2}{n}\sum_{i=1}^{k}p_{i}\left(\phi_{\alpha}^{\prime}\left(\frac{p_{i}}{q_{i}}\right)\right)^{2}\right)^{1/2}.

These bounds can be combined into

\mathsf{E}\left|\Lambda_{\alpha,n}\right|\leq\frac{1}{\alpha\rho^{\alpha-1}}\frac{k}{n}+\left(\frac{2}{n}\sum_{i=1}^{k}p_{i}\left(\phi_{\alpha}^{\prime}\left(\frac{p_{i}}{q_{i}}\right)\right)^{2}\right)^{1/2}.

(52)

Under (49) the first term tends to zero as $n\rightarrow\infty.$ The last term does the same, which is seen from the inequalities

\frac{2}{n}\sum_{i=1}^{k}p_{i}\left(\frac{\left(\frac{p_{i}}{q_{i}}\right)^{\alpha-1}-1}{\alpha-1}\right)^{2}\\ \leq\frac{2}{n}\sum_{i=1}^{k}p_{i}\frac{\left(\frac{p_{i}}{q_{i}}\right)^{2\alpha-2}+1}{\left(\alpha-1\right)^{2}}\\ =\frac{2}{n\left(\alpha-1\right)^{2}}\left(\sum_{i=1}^{k}q_{i}\left(\frac{p_{i}}{q_{i}}\right)^{\alpha}\left(\frac{p_{i}}{q_{i}}\right)^{\alpha-1}+1\right)\\ \leq\frac{2}{n\left(\alpha-1\right)^{2}}\sum_{i=1}^{k}q_{i}\left(\frac{p_{i}}{q_{i}}\right)^{\alpha}\left(\frac{1}{\rho/k}\right)^{\alpha-1}+\frac{2}{n\left(\alpha-1\right)^{2}}\\ =\frac{k^{\alpha-1}}{n}\frac{2}{\left(\alpha-1\right)^{2}\rho^{\alpha-1}}\sum_{i=1}^{k}q_{i}\left(\frac{p_{i}}{q_{i}}\right)^{\alpha}+\frac{2}{n\left(\alpha-1\right)^{2}}\\ =\frac{k^{\alpha-1}}{n}\frac{2\left(\alpha\left(\alpha-1\right)\Delta+1\right)}{\left(\alpha-1\right)^{2}\rho^{\alpha-1}}+\frac{2}{n\left(\alpha-1\right)^{2}}.

Case iii ( $\alpha=1$ )

For $\alpha=1$ in Inequality 52 we get

\mathsf{E}\left|\Lambda_{1,n}\right|\leq\frac{k}{n}+\left(\frac{2}{n}\sum_{i=1}^{k}p_{i}\left(\ln\frac{p_{i}}{q_{i}}\right)^{2}\right)^{1/2}.

Using $\ln p_{i}\leq 0$ we find that last term on the right satisfies the relations

	$\displaystyle\frac{1}{n}\sum_{j=1}^{k}p_{i}\left(\ln^{2}p_{i}-2\ln p_{i}\ln q_{i}+\ln^{2}q_{i}\right)$		(53)
	$\displaystyle=\frac{1}{n}\sum_{j=1}^{k}p_{i}\ln^{2}p_{i}-\frac{2}{n}\sum_{j=1}^{k}p_{i}\ln p_{i}\ln q_{i}+\frac{1}{n}\sum_{j=1}^{k}p_{i}\ln^{2}q_{i}$
	$\displaystyle\leq\frac{1}{n}\sum_{j=1}^{k}p_{i}\ln^{2}p_{i}+\frac{\ln^{2}\frac{k}{\varrho}}{n}\sum_{j=1}^{k}p_{i}\leq\frac{1}{n}\sum_{j=1}^{k}p_{i}\ln^{2}p_{i}+\frac{\ln^{2}\frac{k}{\varrho}}{n}.$

The function $x\rightarrow x\ln^{2}x$ is concave in the interval $\left[0;e^{-1}\right]$ and convex in the interval $\left[e^{-1};1\right].$ Therefore we we can apply the method of [16] to verify that $\sum_{i=1}^{k}p_{i}\ln^{2}p_{i}$ attains its maximum for a mixture of uniform distributions on $k$ points and on subset of $k-1$ of these points. Thus

\frac{1}{n}\sum_{i=1}^{k}p_{i}\ln^{2}p_{j}\leq\frac{1}{n}\sum_{i=1}^{k}\frac{1}{k-1}\ln^{2}\left(\frac{1}{k}\right)\\ =\frac{k\ln^{2}k}{n\left(k-1\right)}\leq\frac{2\ln^{2}k}{n}

(54)

and we can conclude that under (49) the first term in (53) tends to zero as $n$ tends to infinity. Obviously, under (49) also the second term in (51) tends to zero so that the desired relation (39) holds.

Case iv ( $\alpha>2$ )

By A2,

	$\displaystyle D_{\alpha}(P,Q)$	$\displaystyle=\frac{1}{\alpha(\alpha-1)}\left(\sum_{j=1}^{k}p_{j}^{\alpha}q_{j}^{1-\alpha}-1\right)$
		$\displaystyle\geq\frac{1}{\alpha(\alpha-1)}\left(\left(\frac{k}{\varrho}\right)^{\alpha-1}\sum_{j=1}^{k}p_{j}^{\alpha}-1\right)$

so that

\sum_{j=1}^{k}p_{j}^{\alpha}\leq\left(\alpha(\alpha-1)\Delta+1\right)\left(\frac{\varrho}{k}\right)^{\alpha-1}

(55)

where we replaced $D_{\alpha}(P,Q)$ by $\Delta=\Delta_{\alpha}$ in the sense of (41). Further, by the Taylor formula

\widehat{p}_{j}^{\alpha}=p_{j}^{\alpha}+\alpha\,p_{j}^{\alpha-1}(\widehat{p}_{j}-p_{j})+{\frac{\alpha(\alpha-1)}{2}}\,\xi_{j}^{\alpha-2}(\widehat{p}_{j}-p_{j})^{2}

(56)

where $\xi_{j}$ is between $p_{j}$ and $\widehat{p}_{j}.$ We shall look for a highly probable upper bound on $\widehat{p}_{j}$ . Choose any $b>1$ and consider the random event

E_{nj}(b)=\{\widehat{p}_{j}\geq b\max\left\{p_{j},q_{j}\right\}\}.

We shall prove that under (50) it holds

\boldsymbol{\pi}_{n}(b)\overset{def}{=}\mathsf{P}\left(\cup_{j}E_{nj}(b)\right)\longrightarrow 0.

(57)

The components $X_{j}=X_{nj}$ of the observation vector $\boldsymbol{X}_{n}$ defined in Section 1 are approximately Poisson distributed, Po $\left(np_{j}\right),$ so that

\mathsf{P}\left(\widehat{p}_{j}\geq b\max\left\{p_{j},q_{j}\right\}\right)=\mathsf{P}\left(X_{j}\geq nb\max\left\{p_{j},q_{j}\right\}\right)\\ \leq\exp\{-D_{1}\left(\text{Po}\left(b\max\left\{np_{j},nq_{j}\right\}\right),\text{ Po}\left(np_{j}\right)\right)\}

for the divergence $D_{1}(P,Q)$ defined by (9)-(8) with $P,Q$ replaced by the corresponding Poisson distributions. But

D_{1}\left(\text{Po}\left(bnq_{j}\right),\text{ Po}\left(np_{j}\right)\right)=np_{j}\phi_{1}\left(b\right)

(58)

for the logarithmic function $\phi_{1}\geq 0$ introduced in (7). Since for all $0\leq p_{j},q_{j}\leq 1$

\phi_{1}\left(\frac{b\max\left\{p_{j},q_{j}\right\}_{j}}{p_{j}}\right)\geq\phi_{1}\left(b\right)>1\text{ \ for }b>1,\text{\ }

it holds

D_{1}\left(\text{Po}\left(b\max\left\{np_{j},nq_{j}\right\}\right),\text{ Po}\left(np_{j}\right)\right)\\ \geq D_{1}\left(\text{Po}\left(bnq_{j}\right),\text{ Po}\left(nq_{j}\right)\right).

Consequently,

$\displaystyle\boldsymbol{\pi}_{n}(b)$	$\displaystyle\leq\sum_{j}\mathsf{P}\left(\widehat{p}_{j}\geq b\max\left\{p_{j},q_{j}\right\}\right)$
	$\displaystyle\leq\sum_{j}\exp\{-D_{1}\left(\text{Po}\left(b\max\left\{np_{j},nq_{j}\right\}\right),\text{ Po}\left(np_{j}\right)\right)\}$
	$\displaystyle\leq\sum_{j}\exp\left\{-D_{1}\left(\text{Po}\left(bnq_{j}\right),\text{ Po}\left(nq_{j}\right)\right)\right\}$
	$\displaystyle=\sum_{j}\exp\left\{-nq_{j}\phi_{1}\left(b\right)\right\}\text{ \ \ \ \ (cf. (\ref{Poisson}))}$
	$\displaystyle\leq k\exp\left\{-n\frac{\varrho}{k}\phi_{1}\left(b\right)\right\}=k^{1-\frac{n}{k\log k}\varrho\phi_{1}\left(b\right)}.$	(59)

Assumption (50) implies that the exponent in (59) tends to $-\infty$ so that (57) holds. Therefore it suffices to prove (39) under the condition that for all sufficiently large $n$ the random events $\cup_{j}E_{nj}(b)$ fail to take place, i.e. that

\widehat{p}_{j}<b\max\left\{p_{j},q_{j}\right\}\text{ \ \ for all \ }1\leq j\leq k.

(60)

Let us start with the fact that under (60) it holds $\xi_{j}\leq\left\{bp_{j},bq_{j}\right\}$ and then

\xi_{j}^{\alpha-2}\leq\left(\max\left\{bp_{j},bq_{j}\right\}\right)^{\alpha-2}\leq b^{\alpha-2}p_{j}^{\alpha-2}+b^{\alpha-2}\frac{\varrho^{\alpha-2}}{k^{\alpha-2}}.

(61)

Applying this in the Taylor formula (56) we obtain

	$\displaystyle\left\|\widehat{p}_{j}^{\alpha}-p_{j}^{\alpha}\right\|$	$\displaystyle\leq\alpha\,p_{j}^{\alpha-1}\left\|\widehat{p}_{j}-p_{j}\right\|$
		$\displaystyle+{\frac{\alpha(\alpha-1)b^{\alpha-2}}{2}}\,\left(p_{j}^{\alpha-2}+\frac{\varrho^{\alpha-2}}{k^{\alpha-2}}\right)(\widehat{p}_{j}-p_{j})^{2}.$

Hence under (60) we get from (51) and Lemma 8

\left|\Lambda_{\alpha,n}\right|\leq\frac{k^{\alpha-1}}{\alpha\left(\alpha-1\right)}\sum_{j=1}^{n}\alpha\,p_{j}^{\alpha-1}\left|\widehat{p}_{j}-p_{j}\right|\\ +\frac{k^{\alpha-1}}{\alpha\left(\alpha-1\right)}\sum_{j=1}^{n}{\frac{\alpha(\alpha-1)b^{\alpha-2}}{2}}\,\left(p_{j}^{\alpha-2}+\frac{\varrho^{\alpha-2}}{k^{\alpha-2}}\right)(\widehat{p}_{j}-p_{j})^{2}.

(62)

Applying (55) and using Jensen’s inequality and the expectation bound (43), we upper bound $\mathbf{E}\left|\Lambda_{\alpha,n}\right|$ by

	$\displaystyle\frac{\left(\alpha\left(\alpha-1\right)\Delta+1\right)^{1/2}}{\alpha\left(\alpha-1\right)}\left(\frac{\sum p_{j}^{\alpha-1}}{k^{1-\alpha}n}\right)^{1/2}$
	$\displaystyle+{\frac{b^{\alpha-2}k^{\alpha-1}}{2}}\sum_{j=1}^{k}\left(p_{j}^{\alpha-2}+\frac{\varrho^{\alpha-2}}{k^{\alpha-2}}\right)\mathsf{E}\left[(\widehat{p}_{j}-p_{j})^{2}\right]\vskip 6.0pt plus 2.0pt minus 2.0pt$
	$\displaystyle\leq\frac{\left(\alpha\left(\alpha-1\right)\Delta+1\right)^{1/2}}{\alpha\left(\alpha-1\right)}\left(\frac{\sum p_{j}^{\alpha-1}}{k^{1-\alpha}n}\right)^{1/2}$
	$\displaystyle+{\frac{b^{\alpha-2}k^{\alpha-1}}{2}}\sum_{j=1}^{k}\left(p_{j}^{\alpha-2}+\frac{\varrho^{\alpha-2}}{k^{\alpha-2}}\right)\frac{p_{j}}{n}\vskip 6.0pt plus 2.0pt minus 2.0pt$
	$\displaystyle=\frac{\left(\alpha\left(\alpha-1\right)\Delta+1\right)^{1/2}}{\alpha\left(\alpha-1\right)}\left(\frac{\sum p_{j}^{\alpha-1}}{k^{1-\alpha}n}\right)^{1/2}$		(63)
	$\displaystyle+{\frac{b^{\alpha-2}}{2}}\frac{k^{\alpha-1}\sum_{j=1}^{k}p_{j}^{\alpha-1}}{n}+\frac{b^{\alpha-2}\varrho^{\alpha-2}}{2}\frac{k}{n}.$

Obviously, under (60) the desired relation (39) holds if the assumption (50) implies the convergence

\frac{\sum p_{j}^{\alpha-1}}{k^{1-\alpha}n}\rightarrow 0.

(64)

However, by Schwarz inequality and (55),

\sum_{j=1}^{k}\,p_{j}^{\alpha-1}=\sum_{j=1}^{k}p_{j}\,\left(p_{j}^{\alpha-1}\right)^{(\alpha-2)/(\alpha-1)}\vskip 6.0pt plus 2.0pt minus 2.0pt\\ \leq\,\left(\sum_{j=1}^{k}p_{j}p_{j}^{\alpha-1}\right)^{(\alpha-2)/(\alpha-1)}=\left(\sum_{j=1}^{k}p_{j}^{\alpha}\right)^{(\alpha-2)/(\alpha-1)}\vskip 6.0pt plus 2.0pt minus 2.0pt\\ \leq\left(\left(\alpha\left(\alpha-1\right)\Delta+1\right)\frac{\varrho^{\alpha-1}}{k^{\alpha-1}}\right)^{(\alpha-2)/(\alpha-1)}\\ =\frac{\varrho^{\alpha-2}\left(\alpha\left(\alpha-1\right)\Delta+1\right)^{(\alpha-2)/(\alpha-1)}}{k^{\alpha-2}}

so that the validity of (39) under (50) is obvious and the proof is complete.

∎

Condition 50 is stronger than Condition 38 and implies that for any fixed number $a>0$ eventually any bins will contain more than $a$ observations.

IV Bahadur efficiency

In this section we study the Bahadur efficiency in the class of power divergence statistics $\hat{D}_{\alpha,n}=D_{\alpha}(\hat{P}_{n},Q_{n})$ , $\alpha>0$ . As before, we use the simplified notations

P_{n}=P,\text{ }Q_{n}=Q\text{ \ \ and \ }k_{n}=k.

The results are concentrated in Theorem 13 below. Its proof is based on the following lemmas. The first two of them make use of the Rényi divergences of orders $\alpha>0$

	$\displaystyle D_{\alpha}\left(P\\|Q\right)$	$\displaystyle={\frac{1}{\alpha-1}}\ln\sum_{j=1}^{k}p_{j}^{\alpha}q_{j}^{1-\alpha},\ \$
	$\displaystyle D_{1}\left(P\\|Q\right)$	$\displaystyle=\lim_{\alpha\rightarrow 1}D_{\alpha}\left(P\\|Q\right)=D\left(P\\|Q\right)$

where $D\left(P\|Q\right)$ is the classical information divergence denoted above by $D_{1}\left(P,Q\right)$ . There is a monotone relationship between the Rényi and power divergences given by the formula

	$\displaystyle D_{\alpha}\left(P\\|Q\right)$	$\displaystyle={\frac{1}{\alpha-1}}\ln\left(1+\alpha\left(\alpha-1\right)D_{\alpha}\left(P,Q\right)\right),\text{ \ \ }$		(65)
	$\displaystyle D_{1}\left(P\\|Q\right)$	$\displaystyle=D_{1}\left(P,Q\right).$		(66)

Lemma 10

Let $P$ and $Q$ be probability vectors on the set $\mathcal{X}$ . If $\alpha<\beta$ then

D_{\alpha}\left(P\|Q\right)\leq D_{\beta}\left(P\|Q\right).

with equality if and only there exists a subset $A\subseteq\mathcal{X}$ such that $P=Q\left(\cdot\mid A\right).\vskip 6.0pt plus 2.0pt minus 2.0pt$

Proof:

By Jensen’s inequality

	$\displaystyle D_{\alpha}\left(P\\|Q\right)$	$\displaystyle={\frac{1}{\alpha-1}}\ln\sum_{j=1}^{k}p_{j}^{\alpha}q_{j}^{1-\alpha}$
		$\displaystyle={\frac{1}{\alpha-1}}\ln\sum_{j=1}^{k}p_{j}\left(\left(\frac{p_{j}}{q_{j}}\right)^{\beta-1}\right)^{\frac{\alpha-1}{\beta-1}}$
		$\displaystyle\leq{\frac{1}{\alpha-1}}\ln\left(\sum_{j=1}^{k}p_{j}\left(\frac{p_{j}}{q_{j}}\right)^{\beta-1}\right)^{\frac{\alpha-1}{\beta-1}}$
		$\displaystyle\leq{\frac{1}{\beta-1}}\ln\sum_{j=1}^{k}p_{j}\left(\frac{p_{j}}{q_{j}}\right)^{\beta-1}$
		$\displaystyle=D_{\beta}\left(P\\|Q\right).$

The equality takes place if and only if $\left(\frac{p_{j}}{q_{j}}\right)^{\beta-1}$ is constant $P$ -almost surely. Therefore $\frac{p_{j}}{q_{j}}$ is constant on the support of $P$ that we shall denote $A.$ Now $P$ equals $Q$ conditioned on $A.$ ∎

Lemma 11

Let $0<\alpha\leq 1.$ If

{\frac{\,n}{k\ln n}}\longrightarrow\infty.

(67)

and $q_{\max}\rightarrow 0$ as $n\rightarrow\infty$ then the statistic $\widehat{D}_{\alpha,n}$ is Bahadur stable and consistent and the constant sequence generates the Bahadur function

g_{\alpha}(\Delta)=\\ \left\{\begin{array}[c]{ll}\displaystyle{\frac{\ln\left(1+\alpha(\alpha-1)\,\Delta\right)}{\alpha-1}},\quad\Delta>0&\mbox{when}\ 0<\alpha<1\vskip 6.0pt plus 2.0pt minus 2.0pt\\ \displaystyle\lim_{\alpha\rightarrow 1}g_{\alpha}(\Delta)=\Delta,\quad\text{\ \ \ \ \ }\Delta>0&\mbox{when}\ \alpha=1.\vskip 6.0pt plus 2.0pt minus 2.0pt\end{array}\right.\vskip 6.0pt plus 2.0pt minus 2.0pt

(68)

Proof:

Let us first consider $0<\alpha<1$ . The minimum of $D_{1}(P,Q)$ given $D_{\alpha}(P\|Q)\geq\Delta$ is lower bounded by $\Delta.$ Let $\varepsilon>0$ be given. If $q_{\max}$ is sufficiently small there exist sets $A_{-}\subseteq A_{+}$ such that

-\ln Q\left(A_{+}\right)\leq\Delta\leq-\ln Q\left(A_{-}\right)\leq\Delta+\varepsilon.

Let $P_{s}$ denote the mixture $\left(1-s\right)Q\left(\cdot\mid A_{+}\right)+sQ\left(\cdot\mid A_{-}\right).$ Then $s\rightarrow D_{\alpha}\left(P_{s}\|Q\right)$ is a continuous function satisfying

	$\displaystyle D_{\alpha}\left(P_{0}\\|Q\right)$	$\displaystyle\leq\Delta,$
	$\displaystyle D_{\alpha}\left(P_{1}\\|Q\right)$	$\displaystyle\geq\Delta.$

In particular there exist $s\in\left[0,1\right]$ such that $D_{\alpha}\left(P_{s}\|Q\right)=\Delta.$ For this $s$ we have

D_{1}\left(P_{s},Q\right)\\ \leq\left(1-s\right)D_{1}\left(Q\left(\cdot\mid A_{+}\right),Q\right)+sD_{1}\left(Q\left(\cdot\mid A_{-}\right),Q\right)\\ =\left(1-s\right)\left(-\ln Q\left(A_{+}\right)\right)+s\left(-\ln Q\left(A_{-}\right)\right)\\ \leq\left(1-s\right)\Delta+s\left(\Delta+\varepsilon\right)=\Delta+\varepsilon.

Hence

\Delta\leq\inf D_{1}\left(P,Q\right)\leq\Delta+\varepsilon

where the infimum is taken over all $P$ satisfying $D_{\alpha}\left(P\|Q\right)=\Delta$ and where $n$ is sufficiently large. This holds for all $\varepsilon>0$ so the Bahadur function of the statistic $D_{\alpha}\left(\hat{P}\|Q\right)$ is $g\left(\Delta\right)=\Delta.$ The Bahadur function of the power divergence statistics $D_{\alpha}\left(\hat{P},Q\right)$ can be calculated using Equality 65. ∎

Lemma 12

Let $\alpha>1.$ If assumptions A( $\alpha$ ) holds for for the uniform distributions $Q_{n}=U$ and the sequence

c_{\alpha}(n)={\frac{\,k^{(\alpha-1)/\alpha}}{\ln k}}

(69)

satisfies the condition

{\frac{n}{c_{\alpha}(n)\,k\ln n}}\longrightarrow\infty

(70)

then the statistic $\widehat{D}_{\alpha,n}=D_{\alpha}(\hat{P}_{n},Q_{n})$ is consistent and the sequence (69) generates the Bahadur function

g_{\alpha}(\Delta)=\left(\alpha(\alpha-1)\,\Delta\right)^{1/\alpha},\quad\Delta>0.\vskip 6.0pt plus 2.0pt minus 2.0pt

(71)

Proof:

If the sequence (69) satisfies (70) then Theorem 1 implies the consistency of $\widehat{D}_{\alpha,n}$ . Formula (71) was already mentioned in Example 2 above with a reference to Harremoës and Vajda [5]). ∎

Theorem 13

Let the assumption A( $\alpha_{1},\alpha_{2}$ ) hold where $0<\alpha_{1}<\alpha_{2}.$ If

{\frac{k\ln n}{n}}\longrightarrow 0

(72)

then the statistics

	$\displaystyle\widehat{D}_{\alpha_{1},n}$	$\displaystyle=D_{\alpha_{1}}(\hat{P}_{n},Q_{n}),$
	$\displaystyle\widehat{D}_{\alpha_{2},n}$	$\displaystyle=D_{\alpha_{2}}(\hat{P}_{n},Q_{n})$

satisfy the relation

\mbox{BE}\left(\widehat{D}_{\alpha_{1},n};\widehat{D}_{\alpha_{2},n}\right)\\ =\left\{\begin{array}[c]{ll}\displaystyle{\frac{\alpha_{2}-1}{\alpha_{1}-1}}\cdot{\frac{\ln\left(1+\alpha_{1}(\alpha_{1}-1)\Delta_{\alpha_{1}}\right)\,}{\ln\left(1+\alpha_{2}(\alpha_{2}-1)\,\Delta_{\alpha_{2}}\right)}}&\mbox{for}\ \alpha_{2}<1\vskip 6.0pt plus 2.0pt minus 2.0pt\vskip 6.0pt plus 2.0pt minus 2.0pt\vskip 6.0pt plus 2.0pt minus 2.0pt\\ \displaystyle{\frac{1}{\alpha_{1}-1}}\cdot{\frac{\ln\left(1+\alpha_{1}(\alpha_{1}-1)\Delta_{\alpha_{1}}\right)\,}{\Delta_{\alpha_{2}}}}&\mbox{for}\ \alpha_{2}=1.\vskip 6.0pt plus 2.0pt minus 2.0pt\end{array}\right.

(73)

{\frac{k^{2-1/\alpha_{2}}\ln n}{n}}\longrightarrow 0

(74)

then the statistics $\widehat{D}_{\alpha_{1},n}=D_{\alpha_{1}}(\hat{P}_{n},U)$ and $\widehat{D}_{\alpha_{2},n}=D_{\alpha_{2}}(\hat{P}_{n},U)$ satisfy the relation

\mbox{BE}\left(\widehat{D}_{\alpha_{1},n};\widehat{D}_{\alpha_{2},n}\right)=\infty\text{ \ \ \ }\mbox{for}\text{ }\alpha_{2}>1.\vskip 6.0pt plus 2.0pt minus 2.0pt

(75)

Proof:

By Lemma 11, the assumptions of Definition 5 hold. The first assertion follows directly from Definition 3 since, by Lemma 11,

\frac{g_{\alpha_{1}}(\Delta_{\alpha_{1}})}{g_{\alpha_{2}}(\Delta_{\alpha_{2}})}=\\ \left\{\begin{array}[c]{ll}\displaystyle{\frac{\alpha_{2}-1}{\alpha_{1}-1}}\cdot{\frac{\ln\left(1+\alpha_{1}(\alpha_{1}-1)\Delta_{\alpha_{1}}\right)}{\ln\left(1+\alpha_{2}(\alpha_{2}-1)\,\Delta_{\alpha_{2}}\right)}}&\mbox{when}\ \alpha_{2}<1\vskip 6.0pt plus 2.0pt minus 2.0pt\vskip 6.0pt plus 2.0pt minus 2.0pt\vskip 6.0pt plus 2.0pt minus 2.0pt\\ \displaystyle{\frac{1}{\alpha_{1}-1}}\cdot{\frac{\ln\left(1+\alpha_{1}(\alpha_{1}-1)\Delta_{\alpha_{1}}\right)}{\Delta_{\alpha_{2}}}}&\mbox{when}\ \alpha_{2}=1.\vskip 6.0pt plus 2.0pt minus 2.0pt\end{array}\right.

(76)

The second assertion was for $\alpha_{1}=1$ deduced in Section 2 from the lemmas presented there. The argument was based on the fact that $c_{\alpha_{1}}(n)=1$ for $\alpha_{1}=1$ . But $c_{\alpha}(n)=1$ for all $0<\alpha\leq 1$ so that extension from $\alpha_{1}=1$ to $0<\alpha_{1}<1$ is straightforward. ∎

Example 14

Let

P_{n}=\left(p_{nj}\overset{def}{=}\frac{1_{\left\{1\leq j\leq k/2\right\}}}{\left\lfloor k/2\right\rfloor}\right),\text{ \ \ }n=1,2,\ldots

(77)

where $1_{A}$ is the indicator function, $\left\lfloor\cdot\right\rfloor$ stands for the integer part (floor function) and, as before,

U=\left(u_{j}\overset{def}{=}1/k:1\leq j\leq k\right).

Then for $\ \alpha\neq 0\vskip 6.0pt plus 2.0pt minus 2.0pt,1$

	$\displaystyle D_{\alpha}(P_{n},U)$	$\displaystyle={\frac{\sum_{1}^{k}u_{j}\left(\left(p_{nj}/u_{j}\right)^{\alpha}-\alpha\left(p_{nj}/u_{j}-1\right)-1\right)}{\alpha(\alpha-1)}}$
		$\displaystyle={\frac{\sum_{1}^{k}p_{nj}^{\alpha}u_{j}^{1-\alpha}+\sum_{1}^{k}(p_{nj}-u_{j})-\sum_{1}^{k}u_{j}}{\alpha(\alpha-1)}}\vskip 6.0pt plus 2.0pt minus 2.0pt$
		$\displaystyle={\frac{k^{\alpha-1}\sum_{1}^{\left\lfloor k/2\right\rfloor}\left\lfloor k/2\right\rfloor^{-\alpha}-1}{\alpha(\alpha-1)}}\hskip 79.6678pt\vskip 6.0pt plus 2.0pt minus 2.0pt$
		$\displaystyle={\frac{k^{\alpha-1}\left\lfloor k/2\right\rfloor/\left\lfloor k/2\right\rfloor^{\alpha}-1}{\alpha(\alpha-1)}}\vskip 6.0pt plus 2.0pt minus 2.0pt$
		$\displaystyle={\frac{\left(k/\left\lfloor k/2\right\rfloor\right)^{\alpha-1}-1}{\alpha(\alpha-1)}}.$

Therefore the identifiably condition (19) takes on the form

D_{\alpha}(P_{n},U)\longrightarrow\\ \left\{\begin{array}[c]{ll}\displaystyle{\frac{2^{\alpha-1}-1}{\alpha(\alpha-1)}}\overset{def}{=}\Delta_{\alpha}&,\mbox{if}\ \alpha>0,\ \alpha\neq 1\vskip 6.0pt plus 2.0pt minus 2.0pt\vskip 6.0pt plus 2.0pt minus 2.0pt\\ \displaystyle\ln 2\ \overset{def}{=}\Delta_{1}&\mbox{if}\ \alpha=1.\end{array}\right.

If $0<\alpha\leq 1$ then Lemma 12 implies

g_{\alpha}(\Delta)=\ln\left(1+\alpha(\alpha-1)\,\Delta\right)/(\alpha-1)

when $0<\alpha<1$ and $g_{1}(\Delta)=\Delta$ when $\alpha=1$ . If moreover (73) then under the alternative (77)

	$\displaystyle\frac{g_{\alpha}(\Delta_{\alpha})}{g_{1}(\Delta_{1})}$	$\displaystyle={\frac{\ln\left(1+\alpha(\alpha-1)\,\frac{2^{\alpha-1}-1}{\alpha(\alpha-1)}\right)}{(\alpha-1)\ln 2}}\vskip 6.0pt plus 2.0pt minus 2.0pt$
		$\displaystyle={\frac{\ln\left(1+2^{\alpha-1}-1\right)}{(\alpha-1)\ln 2}}=1.$

Hence, by Definition 4, the likelihood ratio statistic $\widehat{D}_{1,n}$ is as Bahadur efficient as any $\widehat{D}_{\alpha,n}$ with $0<\alpha<1$ . If $\alpha>1$ then Lemma 12 implies

{\frac{g_{\alpha}(\Delta_{\alpha})}{g_{1}(\Delta_{1})}}={\frac{(2^{\alpha-1}-1)^{1/\alpha}}{\ln 2}}>1.

However, contrary to this prevalence of $g_{\alpha}(\Delta_{\alpha})\$ over $g_{1}(\Delta_{1})$ , Theorem 13 implies that $\widehat{D}_{1,n}$ is infinitely more Bahadur efficient than $\widehat{D}_{\alpha,n}$ .

Example 15

Let us now consider the truncated geometric distribution

P_{n}=(p_{n1},\ldots,p_{nk})=c_{k}(p)(1,p,\ldots,p^{k})

with parameter $p=p_{n}\in]0,1[$ . Since

1+p+p^{2}+\ldots={\frac{1}{1-p}}\text{ \ \ and \ \ }p^{k+1}+p^{k+2}+\ldots={\frac{p^{k+1}}{1-p},}

it holds

1+p+\cdots+p^{k}={\frac{1}{1-p}}-{\frac{p^{k+1}}{1-p}}={\frac{1-p^{k+1}}{1-p}=}\frac{1}{c_{k}(p)}.

Hence for all $\alpha\neq 0,1$

	$\displaystyle\alpha(\alpha-1)\,D_{\alpha,n}+1$	$\displaystyle={\frac{1}{k}}\sum_{j=0}^{k}\left({\frac{p_{nj}}{1/k}}\right)^{\alpha}$
		$\displaystyle={\frac{1}{k}}\sum_{j=1}^{k}{\frac{k^{\alpha}(1-p)^{\alpha}p^{\alpha j}}{(1-p^{k+1})^{\alpha}}}\vskip 6.0pt plus 2.0pt minus 2.0pt$
		$\displaystyle={\frac{\left(k(1-p)\right)^{\alpha}}{k(1-p^{k+1})^{\alpha}}}\sum_{j=0}^{k}(p^{\alpha})^{j}\vskip 6.0pt plus 2.0pt minus 2.0pt$
		$\displaystyle={\frac{\left(k(1-p)\right)^{\alpha}}{k(1-p^{k+1})^{\alpha}}}\cdot{\frac{1-p^{\alpha(k+1)}}{1-p^{\alpha}}}\vskip 6.0pt plus 2.0pt minus 2.0pt$
		$\displaystyle={\frac{\left(k(1-p)\right)^{\alpha}}{k(1-p^{\alpha})}}\cdot{\frac{1-p^{\alpha(k+1)}}{(1-p^{k+1})^{\alpha}}}.$

In the particular case $p=1-x/k$ for $x\neq 0$ fixed we get $k(1-p)=x$ and

	$\displaystyle k(1-p^{\alpha})\$	$\displaystyle=\ k\left(1-\left(1-{\frac{\alpha x}{k}}+o\left({\frac{x}{k}}\right)\right)\right)\longrightarrow\alpha x,\vskip 6.0pt plus 2.0pt minus 2.0pt$
	$\displaystyle p^{\alpha(k+1)}$	$\displaystyle=\left(1-{\frac{x}{k}}\right)^{\alpha(k+1)}\longrightarrow e^{-x\alpha},\vskip 6.0pt plus 2.0pt minus 2.0pt$
	$\displaystyle p^{k+1}$	$\displaystyle=\left(1-{\frac{x}{k}}\right)^{k+1}\longrightarrow e^{-x}.$

Therefore

	$\displaystyle\alpha(\alpha-1)\,D_{\alpha,n}+1$	$\displaystyle={\frac{x^{\alpha}}{k\left({\frac{\alpha x}{k}}+o\left({\frac{x}{k}}\right)\right)}}\cdot{\frac{1-e^{-x\alpha}}{(1-e^{-x})^{\alpha}}}\vskip 6.0pt plus 2.0pt minus 2.0pt$
		$\displaystyle={\frac{x^{\alpha}}{\alpha x+o(x)}}\cdot{\frac{e^{x\alpha}-1}{(e^{x}-1)^{\alpha}}}.$

Consequently,

\alpha(\alpha-1)\,\Delta_{\alpha}+1={\frac{x^{\alpha-1}}{\alpha}}\cdot{\frac{e^{x\alpha}-1}{(e^{x}-1)^{\alpha}}}

i.e.,

\Delta_{\alpha}=\frac{x^{\alpha-1}(e^{x\alpha}-1)-\alpha(e^{x}-1)^{\alpha}}{\alpha^{2}(\alpha-1)(e^{x}-1)^{\alpha}}\text{ \ \ for }\alpha\neq 0,1.

By the L’Hospital rule,

	$\displaystyle\Delta_{1}$	$\displaystyle=\ln\frac{x}{e(e^{x}-1)}+\frac{xe^{x}}{e^{x}-1}~\text{,}$
	$\displaystyle\Delta_{0}$	$\displaystyle=\frac{\ln(e^{x}-1)-\ln x}{2}~.$

From here one can deduce that if $x\rightarrow 0$ then

\Delta_{\alpha}\longrightarrow 0\text{ \ \ for all \ \ }\alpha\in\mathbb{R}.

If $x=1$ then

\Delta_{\alpha}={\frac{e^{\alpha}-1-(e-1)^{\alpha}}{\alpha^{2}(\alpha-1).(e-1)^{\alpha}}}\text{ \ \ for }\alpha\neq 0,1

and

	$\displaystyle\Delta_{1}$	$\displaystyle=\frac{1-(e-1)\ln(e-1)}{e-1}=0.035,\ \ \ \$
	$\displaystyle\Delta_{0}$	$\displaystyle=\frac{\ln(e-1)}{2}=0.271.$

Using Lemma 2 and Theorem 2 in a similar manner as in the previous example, we find that here $\widehat{D}_{1,n}$ is more Bahadur efficient as any $\widehat{D}_{\alpha,n}$ with $0<\alpha<\infty,$ $\alpha\neq 1.$

V Contiguity

In this paper we proved that the statistics $\hat{D}_{\alpha,n}$ of orders $\alpha>1$ are less Bahadur efficient than those of the orders $0<\alpha\leq 1$ and that the latter are mutually comparable in the Bahadur sense. One may have expected $\hat{D}_{1,n}$ to be much more Bahadur efficient than $\hat{D}_{\alpha,n}$ for $\ 0<\alpha<1.$ In order to understand why this is not the case we have to examine somewhat closer the assumptions of our theory.

Recall that given a sequence of pairs of probability measures $\left(P_{n},Q_{n}\right)_{n\in\mathbb{N}},$ $\left(P_{n}\right)_{n\in\mathbb{N}}$ is said to be contiguous with respect to $\left(Q_{n}\right)_{n\in\mathbb{N}}$ if $Q_{n}\left(A_{n}\right)\rightarrow 0$ for $n\rightarrow\infty$ implies $P_{n}\left(A_{n}\right)\rightarrow 0$ for $n\rightarrow\infty$ and any sequence of sets $\left(A_{n}\right)_{n\in\mathbb{N}}.$ When $\left(P_{n}\right)_{n\in\mathbb{N}}$ is contiguous with respect to $\left(Q_{n}\right)_{n\in\mathbb{N}}$ we write $P_{n}\vartriangleleft Q_{n}.$ Let $P$ and $Q$ be probability measures on the same set $\mathcal{X}$ and let $\left(\mathcal{F}_{n}\right)_{n\in\mathbb{N}}$ be an increasing sequence of finite sub- $\sigma$ -algebras on $\mathcal{X}$ that generates the full $\sigma$ -algebra on $\mathcal{X}.$ If $P_{n}=P_{\mid\mathcal{F}_{n}}$ and $Q_{n}=Q_{\mid\mathcal{F}_{n}}$ then $P_{n}\vartriangleleft Q_{n}$ if and only if $P\ll Q$ where $\ll$ denotes absolute continuity. For completeness we give the proof of the following simple proposition.

Proposition 16

Let $\left(P_{n},Q_{n}\right)_{n\in\mathbb{N}}$ denote a sequence of pairs of probability measures and assume that the sequence $D_{1}\left(P_{n},Q_{n}\right)$ is bounded. Then $P_{n}\vartriangleleft Q_{n}.\vskip 6.0pt plus 2.0pt minus 2.0pt$

Proof:

Assume that the proposition is false. Then there exist $\varepsilon>0$ and a subsequence of sets $\left(A_{n_{k}}\right)_{k\in\mathbb{N}}$ such that $Q_{n_{k}}\left(A_{n_{k}}\right)\rightarrow 0$ for $k\rightarrow\infty$ and $P_{n_{k}}\left(A_{n_{k}}\right)\geq\varepsilon$ for all $k\in\mathbb{N}.$ ∎

In general, a large power $\alpha$ makes the power divergence $D_{a}\left(P,Q\right)$ sensitive to large values of $dP/dQ.$ Therefore the statistics $\widehat{D}_{\alpha,n}$ with large $\alpha$ should be used when the sequence of alternatives $P_{n}$ may not be contiguous with respect to the sequence of hypotheses $Q_{n}.$ Conversely, a small power $\alpha$ makes $D_{a}\left(P,Q\right)$ sensitive to small values of $dP/dQ.$ Therefore $\widehat{D}_{\alpha,n}$ with small $\alpha$ should be used when the sequence of hypotheses $Q_{n}$ is not contiguous with respect to the sequence alternatives $P_{n}.$ Our conditions guarantee $P_{n}\vartriangleleft Q_{n}$ but not the reversed contiguity $Q_{n}\vartriangleleft P_{n}.$ We see that a substantial modification of the conditions is needed in order to guarantee that $\hat{D}_{1,n}$ dominates the divergence statistcs $\hat{D}_{\alpha,n}$ of the orders $\ 0<\alpha<1$ in the Bahadur sense.

VI Appendix: Relations to previous results

As mentioned at the end of Section II, Harremoës and Vajda [5] assumed the same strong consistency as in Definition 4+ but introduced the Bahadur efficiency by the formula (36). The next four lemmas help to clarify the relation between this and the present precised concept of Bahadur efficiency (35).

Under the assumptions of Definition 4, [5] considered the following conditions.

C1:: The limit $\bar{c}_{\alpha_{2}/\alpha_{1}}$ considered in (37) exists.
C2:: Both statistics $\widehat{D}_{\alpha_{i},n}$ are strongly consistent and both functions $g_{\alpha_{i}}$ are strongly Bahadur.

Lemma 17

Let the assumptions of Definition 5 hold. Under C1 the Bahadur efficiency (36) coincides with the present Bahadur efficiency (35). If moreover C2 holds then (36) is the Bahadur efficiency in the strong sense.

Proof:

The first assertion is clear from (36) and (35). Under C2 the assumptions of Definition 3+ hold. Hence the second assertion follows from Definition 6. ∎

Lemma 18

Let the assumptions of Definition 3 hold and let $b(\alpha):\mathcal{I}\longrightarrow]0,1[$ be increasing and $d_{\alpha}:\mathcal{I}\longrightarrow]0,\infty[$ arbitrary function on an interval $\mathcal{I}$ covering $\{\alpha_{1},\alpha_{2}\}$ . If for every $\alpha\in\{\alpha_{1},\alpha_{2}\}$ the sequence $c_{\alpha}(n)$ generating the Bahadur function $g_{\alpha}$ satisfies the asymptotic condition

c_{\alpha}(n)=n^{b(\alpha)}(d_{\alpha}+o(1))

(78)

then (31) holds for $c_{\alpha_{2}/\alpha_{1}}=\infty$ and condition C1 is satisfied.

Proof:

Under (78) it suffices to prove that (31) holds for $c_{\alpha_{2}/\alpha_{1}}=\infty,$ i.e.

\lim_{n\longrightarrow\infty}\frac{c_{\alpha_{2}}(m_{n})}{c_{\alpha_{1}}(n)}=\infty

(79)

for $m_{n}$ defined by (29). By (78),

c_{\alpha_{2}}(m_{n})=m_{n}^{b(\alpha_{2})}(d_{\alpha_{2}}+o(1))

and

c_{\alpha_{1}}(n)=n^{b(\alpha_{1})}(d_{\alpha_{1}}+o(1))

so that (29) implies

m_{n}^{1-b(\alpha_{2})}=n^{1-b(\alpha_{1})}(\gamma\delta+o(1))

for the finite positive constants

\delta=\frac{d_{\alpha_{1}}}{d_{\alpha_{2}}}\text{ \ \ and \ \ }\gamma={\frac{g_{\alpha_{1}}(\Delta_{\alpha_{1}})}{g_{\alpha_{2}}(\Delta_{\alpha_{2}})}.}

Hence (30) implies

	$\displaystyle\frac{c_{\alpha_{2}}(m_{n})}{c_{\alpha_{1}}(n)}$	$\displaystyle=\frac{m_{n}}{n}(\gamma^{-1}+o(1))\vskip 6.0pt plus 2.0pt minus 2.0pt$
		$\displaystyle=\frac{n^{\frac{1-b(\alpha_{1})}{1-b(\alpha_{2})}}}{n}((\gamma\delta)^{\frac{1}{1-b(\alpha_{2})}}\gamma^{-1}+o(1))\vskip 6.0pt plus 2.0pt minus 2.0pt$
		$\displaystyle=n^{\frac{b(\alpha_{2})-b(\alpha_{1})}{1-b(\alpha_{2})}}(\gamma^{\frac{b(\alpha_{2})}{1-b(\alpha_{2})}}\delta^{\frac{1}{1-b(\alpha_{2})}}+o(1))$

so that (79) holds. ∎

Lemma 19

Let the assumptions of Definition 5 hold and let for every $\alpha\in\{\alpha_{1},\alpha_{2}\}$ the sequence $c_{\alpha}(n)$ generating the Bahadur function $g_{\alpha}$ satisfy the asymptotic condition

c_{\alpha}(n)=\frac{\alpha n^{b(\alpha)}}{\ln n}

(80)

for some increasing function $b(\alpha):\mathcal{I}\longrightarrow]0,1[$ on an interval $\mathcal{I}$ covering $\{\alpha_{1},\alpha_{2}\}$ . Then (31) holds for $c_{\alpha_{2}/\alpha_{1}}=\infty$ and condition C1 is satisfied.

Proof:

Similarly as before, it suffices to prove the relation (79) for $m_{n}$ defined by (29). By (80),

c_{\alpha_{2}}(m_{n})=\frac{\alpha_{2}m_{n}^{b(\alpha_{2})}}{\ln m_{n}}\text{ \ \ and \ \ }c_{\alpha_{1}}(n)=\frac{\alpha_{1}n^{b(\alpha_{1})}}{\ln n}

so that (29) implies

\frac{\alpha_{2}m_{n}^{1-b(\alpha_{2})}}{\ln m_{n}}=\frac{\alpha_{1}n^{1-b(\alpha_{1})}}{\ln n}(\gamma+o(1))

for the same $\gamma$ as in the previous proof. Since $1-b(\alpha_{2})<1-b(\alpha_{1})$ , this implies the asymptotic relation

\frac{m_{n}}{n}\longrightarrow\infty{.}

(81)

Similarly as in the previous proof, we get from (30)

\frac{c_{\alpha_{2}}(m_{n})}{c_{\alpha_{1}}(n)}=\frac{m_{n}}{n}(\gamma^{-1}+o(1))\vskip 6.0pt plus 2.0pt minus 2.0pt=\\ \left(\frac{\alpha_{1}\ln m_{n}}{\alpha_{2}\ln n}\right)^{\frac{1}{1-b(\alpha_{2})}}n^{\frac{b(\alpha_{2})-b(\alpha_{1})}{1-b(\alpha_{2})}}\left(\begin{array}[c]{c}\gamma^{\frac{b(\alpha_{2})}{1-b(\alpha_{2})}}\\ +o(1)\end{array}\right)\vskip 6.0pt plus 2.0pt minus 2.0pt\vskip 6.0pt plus 2.0pt minus 2.0pt\\ >n^{\frac{b(\alpha_{2})-b(\alpha_{1})}{1-b(\alpha_{2})}}(\gamma^{\frac{b(\alpha_{2})}{1-b(\alpha_{2})}}+o(1)).

Therefore the desired relation (79) holds. ∎

Lemma 20

Let the assumptions of Definition 3 hold and let for every $\alpha\in\{\alpha_{1},\alpha_{2}\}$ the sequence $c_{\alpha}(n)$ generating the Bahadur function $g_{\alpha}$ satisfy the asymptotic condition

c_{\alpha}(n)=\frac{\alpha k^{b(\alpha)}}{\ln k}

(82)

where $k=k_{n}\longrightarrow\infty$ is the sequence considered above and $b(\alpha):\mathcal{I}\longrightarrow]0,\infty[$ is increasing on an interval $\mathcal{I}$ covering $\{\alpha_{1},\alpha_{2}\}$ . Then (31) holds for $c_{\alpha_{2}/\alpha_{1}}=\infty$ and condition C1 is satisfied.

Proof:

It suffices to apply Lemma 17 to the sequences

c_{\alpha_{1}}(k)=\frac{\alpha_{1}k^{b(\alpha_{1})}}{\ln k}\text{ \ \ and \ \ }c_{\alpha_{2}}(m_{k})=\frac{\alpha_{2}m_{k}^{b(\alpha_{2})}}{\ln m_{k}}

for $m_{k}$ defined by the condition

\frac{m_{k}}{c_{\alpha_{2}}(m_{k})}=\frac{g_{\alpha_{1}}(\Delta_{\alpha_{1}})}{g_{\alpha_{2}}(\Delta_{\alpha_{2}})}.\frac{k}{c_{\alpha_{1}}(k)}\left(1+o(1)\right)\text{ \ \ (cf. (\ref{15})).}

(83)

∎

Example 21

Let assumptions of Definition 5 hold for $\alpha_{1}=1$ and $\alpha_{2}=\alpha>1,$ and let

{\frac{k^{b(\alpha)+1}\ln n}{n}}\longrightarrow 0\quad\mbox{for}\ \ \ b(\alpha)=(\alpha-1)/\alpha.

(84)

By [5, Eq. 51, 76 and 79] and (84) the sequences

c_{1}(n)=1\quad\mbox{and}\quad c_{\alpha}(n)={\frac{\alpha k^{b(\alpha)}}{\ln k}}

(85)

generate the Bahadur functions

g_{1}(\Delta)=\Delta\quad\mbox{and}\quad g_{\alpha}(\Delta)=\left(\alpha(\alpha-1)\,\Delta\right)^{1/\alpha},\text{ \ \ }\Delta>0.

(86)

Here we cannot apply Lemma 18 since $c_{1}(n)$ is not special case of $c_{\alpha}(n)$ for $\alpha=1.$ An alternative direct approach can be based on the observation that (29) cannot hold if $\lim\inf_{n}m_{n}<\infty.$ In the opposite case $m_{n}\rightarrow\infty$ obviously implies

c_{\alpha/1}\overset{def}{=}\lim\nolimits_{n}{\frac{c_{\alpha}(m_{n})}{c_{1}(n)}}=\infty

so that C1 holds with $\bar{c}_{\alpha_{2}/\alpha_{1}}\equiv c_{\alpha/1}=\infty$ . Hence Lemma 1 implies that the Bahadur efficiency $\mbox{BE}\left(\widehat{D}_{1,n}\,;\,\widehat{D}_{\alpha,n}\right)=\infty$ obtained previously by Harremoës and Vajda [5, Eq. 81] coincides with the Bahadur efficiency of $\widehat{D}_{1,n}$ with respect to $\widehat{D}_{\alpha,n}$ in the present precised sense of (35). Under stronger condition on $k$ than (84), Harremoës and Vajda established also the strong consistency of the statistics $\widehat{D}_{1,n}$ and $\widehat{D}_{\alpha,n}$ . One can verify that (86) are strongly Bahadur functions so that C2 holds as well. Hence, as argued by Lemma 3, we deal here with the Bahadur efficiency in the strong sense.

Example 22

Let assumptions of Definition 5 hold for $\alpha_{1}>1$ and let the function $b(\alpha)$ be defined by (84) for all $\alpha\geq 1$ . Harremoës and Vajda (2008) proved that if the sequence $k$ satisfies the condition (84) with $\alpha=\alpha_{2}$ then for all $\alpha\in\{\alpha_{1},\alpha_{2}\}$ the function $g_{\alpha}(\Delta)$ given by the second formula in (86) is Bahadur function of the statistics $\widehat{D}_{\alpha,n}$ generated by the sequences $c_{\alpha}(n)$ from the second formula in (85). Thus in this case the assumptions of Lemma 18 hold. From Lemmas 20 and 17 we conclude that the Bahadur efficiency

\mbox{BE}\left(\widehat{D}_{\alpha_{1},n}\,;\,\widehat{D}_{\alpha_{2},n}\right)=\infty\quad\mbox{for all}\ \text{\ \ }0<\alpha_{1}<\alpha_{2}<\infty

obtained in [5, Eq. 81] coincides with the Bahadur efficiency in the present precise sense. Similarly as in the previous example, we can arrive to the conclusion that this is the Bahadur efficiency in the strong sense.

Acknowledgement. This research was supported by the European Network of Excellence and by the GAČR grant 202/10/0618.

References

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	$\displaystyle D_{\alpha}\left(P\\|Q\right)$	$\displaystyle={\frac{1}{\alpha-1}}\ln\sum_{j=1}^{k}p_{j}^{\alpha}q_{j}^{1-\alpha},\ \$
	$\displaystyle D_{1}\left(P\\|Q\right)$	$\displaystyle=\lim_{\alpha\rightarrow 1}D_{\alpha}\left(P\\|Q\right)=D\left(P\\|Q\right)$

On Bahadur Efficiency of Power Divergence Statistics

Abstract

Index Terms:

I Introduction

II Basic model

Example 1

Convention

Definition 2

Definition 3

Motivation of the next definition

Definition 4

Definition 5

Definition 6

Motivation of Definition 6

III Consistency

Theorem 7

Proof:

Lemma 8

Proof:

Theorem 9

Proof:

Case i (0<α<10<\alpha<1)

Case ii (1<α≤21<\alpha\leq 2)

Case iii (α=1\alpha=1)

Case iv (α>2\alpha>2)

IV Bahadur efficiency

Lemma 10

Proof:

Lemma 11

Proof:

Lemma 12

Proof:

Theorem 13

Proof:

Example 14

Example 15

V Contiguity

Proposition 16

Proof:

VI Appendix: Relations to previous results

Lemma 17

Proof:

Lemma 18

Proof:

Lemma 19

Proof:

Lemma 20

Proof:

Example 21

Example 22

References

Case i ( $0<\alpha<1$ )

Case ii ( $1<\alpha\leq 2$ )

Case iii ( $\alpha=1$ )

Case iv ( $\alpha>2$ )