# Bartlett correction of an independence test in a multivariate Poisson model

Rolf Larsson111Department of Mathematics, Uppsala University, P.O.Box 480,
SE-751 06 Uppsala, Sweden. e-mail<EMAIL_ADDRESS>
###### Abstract

We consider a system of dependent Poisson variables, where each variable is
the sum of an independent variate and a common variate. It is the common
variate that creates the dependence.

Within this system, a test of independence may be constructed where the null
hypothesis is that the common variate is identically zero. In the present
paper, we consider the maximum log likelihood ratio test. For this test, it is
well-known that the asymptotic distribution of the test statistic is an equal
mixture of zero and a chi square distribution with one degree of freedom. We
examine a Bartlett correction of this test, in the hope that we will get
better approximation of the nominal size for moderately large sample sizes.

This correction is explicitly derived, and its usefulness is explored in a
simulation study. For practical purposes, the correction is found to be useful
in dimension two, but not in higher dimensions.

##  1 Introduction

Bartlett correction is a simple technique to obtain small sample corrections
of asymptotic distributions, in particular useful for maximum log likelihood
ratio tests. It was introduced in Bartlett (1937). Let the sample size be $n$,
and write $Q_{n}=-2\log\Lambda_{n}$, where $\Lambda_{n}$ is the ratio between
the maximum likelihoods under the null and the alternative hypotheses,
respectively. Denoting the distributional limit of $Q_{n}$ by $Q_{\infty}$,
the simple idea is to write

$Q_{n}\approx\frac{EQ_{n}}{EQ_{\infty}}Q_{\infty},$

where

$EQ_{n}=EQ_{\infty}+n^{-1}R+o(n^{-1}),$

where $E$ is the expectation operator and $o(n^{-1})$ is an error term that
tends to zero faster than $n^{-1}$ as $n\to\infty$. Hence,

$Q_{n}\approx(1+n^{-1}R^{\prime})Q_{\infty},$ (1)

for a constant $R^{\prime}=R/EQ_{\infty}$. Obviously, the approximation in (1)
asymptotically corrects the first moment of the statistic, in the sense that
the error is of smaller order than $n^{-1}$. In fact, for maximum log
likelihood ratio tests, Lawley (1956) shows that under regularity conditions,
all higher moments are corrected in the same way. Hayakawa (1977) went on to
prove that this practice also corrects the fractiles of the distribution.

In standard situations, it is well known that, under the null hypothesis, the
asymptotic distribution of the maximum log likelihood ratio (LR) test
($Q_{\infty}$ above) is $\chi^{2}$ with one degree of freedom, $\chi^{2}(1)$.
However, the present paper deals with a case where the parameter of the null
hypothesis is on the boundary of the parameter space. Here, the standard
result does not apply. Indeed, Self and Liang (1987) have shown that the
corresponding distribution is an equal mixture of a point mass at one and
$\chi^{2}(1)$. If the parameter is $\lambda$ (say), and we test null
hypothesis $H_{0}$: $\lambda=0$ vs $H_{1}$: $\lambda>0$, then the asymptotic
probability that the maximum likelihood estimator $\hat{\lambda}=0$ is $1/2$,
and the asymptotic distribution of $(Q_{n}|\hat{\lambda}>0)$ is $\chi^{2}(1)$.

In the present paper, we will deal with a dependent Poisson model, see e.g.
Karlis (2003). This model is as follows: Assume that $U$,
$X_{1},X_{2},...,X_{m}$ are independent Poisson with parameters $\lambda$,
$\mu_{1},\mu_{2},...,\mu_{m}$, respectively. Then, construct the dependent
Poisson variates $(Y_{1},Y_{2},...,Y_{m})$ through $Y_{j}=U+X_{j}$,
$j=1,2,...,m$. It is clear that the $Y_{j}$ are simultaneously independent if
and only if $\lambda=0$. Hence, it is of great interest to test $H_{0}$:
$\lambda=0$ vs $H_{1}$: $\lambda>0$. Doing this with the LR test is the focus
of the present paper.

One application of such a test can be found in Larsson (2020), where it can be
seen as a building block in the search algorithm for the best discrete factor
analysis model. (There, the best model is defined as the one with the smallest
AIC, but comparing AIC values is equivalent to using the LR test on some
significance level, if the null model is a special case of the alternative
model.)

The rest of the paper unfolds as follows. Section 2 gives the asymptotic
approximation of the expectation of the LR test under the null. In section 3,
Bartlett correction is discussed. Section 4 gives small sample simulation
results, and section 5 concludes.

##  2 Asymptotic approximation of the expectation

Suppose we have a random sample $(y_{i1},y_{i2},...,y_{im})$, $i=1,2,...,n$,
from the distribution of $(Y_{1},Y_{2},...,Y_{m})$ of the dependent Poisson
model by Karlis (2003). We want to test $H_{0}$: $\lambda=0$ (independence) vs
$H_{1}$: $\lambda>0$ (dependence) by using the likelihood ratio test.

The first object is to find the maximum likelihood estimates (MLEs) of the
parameters.

Let $y_{i}=\min(y_{i1},y_{i2},...,y_{im})$ for all $i$. The likelihood is

$\displaystyle L(\lambda,\mu_{1},\mu_{2},...,\mu_{m})$
$\displaystyle=\prod_{i=1}^{n}\sum_{u=0}^{y_{i}}e^{-\lambda}\frac{\lambda^{u}}{u!}\prod_{j=1}^{m}e^{-\mu_{j}}\frac{\mu_{j}^{y_{ij}-u}}{(y_{ij}-u)!}$
$\displaystyle=e^{-n(\lambda+\sum_{j=1}^{m}\mu_{j})}\left(\prod_{j=1}^{m}\mu_{j}^{n\bar{y}_{j}}\right)\prod_{i=1}^{n}\sum_{u=0}^{y_{i}}\frac{1}{u!\prod_{j=1}^{m}(y_{ij}-u)!}\left(\frac{\lambda}{\prod_{j=1}^{m}\mu_{j}}\right)^{u},$
(2)

where $\bar{y}_{j}=n^{-1}\sum_{i=1}^{n}y_{ij}$ for $j=1,2,...,m$.

Let $\hat{\lambda}$ and $\hat{\mu}_{j}$ be the MLEs of $\lambda$ and
$\mu_{j}$, $j=1,2,...,m$. Since it is known that (Karlis, 2003, Larsson, 2020)
$\hat{\lambda}+\hat{\mu}_{j}=\bar{y}_{j}$, the task is to maximize the
likelihood in (2) reformulated as

$L(\lambda)=e^{-n\left\\{\sum_{j=1}^{m}\bar{y}_{j}-(m-1)\lambda\right\\}}\left\\{\prod_{j=1}^{m}(\bar{y}_{j}-\lambda)^{n\bar{y}_{j}}\right\\}\prod_{i=1}^{n}S_{i}(\lambda),$
(3)

where

$S_{i}(\lambda)\stackrel{{\scriptstyle\rm{def}}}{{=}}\sum_{u=0}^{y_{i}}\frac{\xi(\lambda)^{u}}{u!\prod_{j=1}^{m}(y_{ij}-u)!},$
(4)

with

$\xi(\lambda)\stackrel{{\scriptstyle\rm{def}}}{{=}}\frac{\lambda}{\prod_{j=1}^{m}(\bar{y}_{j}-\lambda)}.$
(5)

As usual, it seems to be easier to focus on maximizing the log likelihood,
which via (3) is given as

$\displaystyle l(\lambda)$
$\displaystyle\stackrel{{\scriptstyle\rm{def}}}{{=}}\log\\{L(\lambda)\\}$
$\displaystyle=n\left\\{(m-1)\lambda-\sum_{j=1}^{m}\bar{y}_{j}\right\\}+n\sum_{j=1}^{m}\bar{y}_{j}\log(\bar{y}_{j}-\lambda)+\sum_{i=1}^{n}s_{i}(\lambda),$
(6)

where $s_{i}(\lambda)=\log\\{S_{i}(\lambda)\\}$. The derivative of (6) with
respect to $\lambda$ is

$l^{\prime}(\lambda)=n\left(m-1-\sum_{j=1}^{m}\frac{\bar{y}_{j}}{\bar{y}_{j}-\lambda}\right)+\sum_{i=1}^{n}s_{i}^{\prime}(\lambda).$
(7)

The derivative of $s_{i}(\lambda)$ w.r.t $\lambda$ is, emphasizing the
dependency on the observations
${\boldsymbol{y}_{i}}=(y_{i1},y_{i2},...,y_{im})$,

$s_{i}^{\prime}(\lambda)=\frac{S_{i}^{\prime}(\lambda;{\boldsymbol{y}_{i}})}{S_{i}(\lambda;{\boldsymbol{y}_{i}})},$

where from (4), denoting $I\\{A\\}$ as the indicator of the event $A$,

$\displaystyle S_{i}^{\prime}(\lambda;{\boldsymbol{y}_{i}})$
$\displaystyle\stackrel{{\scriptstyle\rm{def}}}{{=}}\xi^{\prime}(\lambda)\sum_{u=1}^{y_{i}}\frac{\xi(\lambda)^{u-1}}{(u-1)!\prod_{j=1}^{m}(y_{ij}-u)!}I\\{y_{i}>0\\}$
$\displaystyle=\xi^{\prime}(\lambda)\sum_{u=1}^{(y_{i}-1)+1}\frac{\xi(\lambda)^{u-1}}{(u-1)!\prod_{j=1}^{m}\\{(y_{ij}-1-(u-1)\\}!}I\\{y_{i}>0\\}$
$\displaystyle=\xi^{\prime}(\lambda)S_{i}(\lambda;\boldsymbol{y}_{i}-\boldsymbol{1})I\\{y_{i}>0\\},$

where $\boldsymbol{1}=(1,1,...,1)$. Hence, because by (5),

$\xi^{\prime}(\lambda)=\frac{\prod_{j=1}^{m}(\bar{y}_{j}-\lambda)-\lambda\frac{d}{d\lambda}\prod_{j=1}^{m}(\bar{y}_{j}-\lambda)}{\left\\{\prod_{j=1}^{m}(\bar{y}_{j}-\lambda)\right\\}^{2}}=\frac{1+\lambda\sum_{j=1}^{m}(\bar{y}_{j}-\lambda)^{-1}}{\prod_{j=1}^{m}(\bar{y}_{j}-\lambda)},$

it follows that

$s_{i}^{\prime}(\lambda)=\frac{1+\lambda\sum_{j=1}^{m}(\bar{y}_{j}-\lambda)^{-1}}{\prod_{j=1}^{m}(\bar{y}_{j}-\lambda)}\frac{S_{i}(\lambda;\boldsymbol{y}_{i}-\boldsymbol{1})}{S_{i}(\lambda;\boldsymbol{y_{i}})}I\\{y_{i}>0\\}.$

Thus, via (7), and the identity

$\sum_{j=1}^{m}\frac{\bar{y}_{j}}{\bar{y}_{j}-\lambda}=m+\lambda\sum_{j=1}^{m}(\bar{y}_{j}-\lambda)^{-1},$

we have

$\displaystyle l^{\prime}(\lambda)$
$\displaystyle=\left\\{1+\lambda\sum_{j=1}^{m}(\bar{y}_{j}-\lambda)^{-1}\right\\}$
$\displaystyle\cdot\left\\{-n+\frac{1}{\prod_{j=1}^{m}(\bar{y}_{j}-\lambda)}\sum_{i=1}^{n}\frac{S_{i}(\lambda;\boldsymbol{y}_{i}-\boldsymbol{1})}{S_{i}(\lambda;\boldsymbol{y_{i}})}I\\{y_{i}>0\\}\right\\}.$
(8)

To find the MLE $\hat{\lambda}$, we need to solve the equation
$l^{\prime}(\lambda)=0$. In practice, this has to be done by numerical
methods. For our purposes, it is enough to derive an asymptotic approximation.

To this end, we Taylor expand (4) as (here, $y_{ij(2)}=y_{ij}(y_{ij}-1)$
etcetera, and $O$ in fact means order in probability)

$\displaystyle S_{i}(\lambda)=\frac{1}{\prod_{j=1}^{m}y_{ij}!}$
$\displaystyle\left\\{1+\prod_{j=1}^{m}y_{ij}\xi(\lambda)+\frac{\prod_{j=1}^{m}y_{ij(2)}}{2!}\xi(\lambda)^{2}+\frac{\prod_{j=1}^{m}y_{ij(3)}}{3!}\xi(\lambda)^{3}\right.$
$\displaystyle\left.+\frac{\prod_{j=1}^{m}y_{ij(4)}}{4!}\xi(\lambda)^{4}+O(\lambda^{5})\right\\}.$
(9)

Here, Taylor expansion of (5) yields (the sums and products are over $1\leq
j\leq m$, etcetera), introducing $P=\prod_{j}\bar{y}_{j}$,

$\displaystyle\xi(\lambda)$ $\displaystyle=\lambda
P^{-1}\prod_{j=1}^{m}\left(1-\frac{\lambda}{\bar{y}_{j}}\right)^{-1}$
$\displaystyle=\lambda
P^{-1}\prod_{j}\left(1+\frac{\lambda}{\bar{y}_{j}}+\frac{\lambda^{2}}{\bar{y}_{j}^{2}}+...\right)$
$\displaystyle=\lambda
P^{-1}\left\\{1+\sum_{j}\frac{1}{\bar{y}_{j}}\lambda+\left(\sum_{j}\frac{1}{\bar{y}_{j}^{2}}+\sum_{j<k}\frac{1}{\bar{y}_{j}\bar{y}_{k}}\right)\lambda^{2}\right.$
$\displaystyle\left.+\left(\sum_{j}\frac{1}{\bar{y}_{j}^{3}}+\sum_{j<k}\frac{1}{\bar{y}_{j}^{2}\bar{y}_{k}}+\sum_{j<k}\frac{1}{\bar{y}_{j}\bar{y}_{k}^{2}}+\sum_{j<k<l}\frac{1}{\bar{y}_{j}\bar{y}_{k}\bar{y}_{l}}\right)\lambda^{3}+O(\lambda^{4})\right\\},$
(10)

which inserted into (9) implies

$\displaystyle S_{i}(\lambda)$
$\displaystyle=\frac{1}{\prod_{j}y_{ij}!}\left[1+\prod_{j}y_{ij}\left\\{P^{-1}\lambda+\frac{1}{2}a_{2}({\boldsymbol{y}}_{i})P^{-2}\lambda^{2}+\frac{1}{6}a_{3}({\boldsymbol{y}}_{i})P^{-3}\lambda^{3}\right\\}+O(\lambda^{4})\right],$

where

$a_{2}({\boldsymbol{y}}_{i})\stackrel{{\scriptstyle\rm{def}}}{{=}}2P\sum_{j}\bar{y}_{j}^{-1}+\prod_{j}(y_{ij}-1),$
(11)

and

$\displaystyle a_{3}({\boldsymbol{y}}_{i})$
$\displaystyle\stackrel{{\scriptstyle\rm{def}}}{{=}}6P^{2}\left(\sum_{j}\bar{y}_{j}^{-2}+\sum_{j<k}\bar{y}_{j}^{-1}\bar{y}_{k}^{-1}\right)+6P\prod_{j}(y_{ij}-1)\sum_{k}\bar{y}_{k}^{-1}$
$\displaystyle+\prod_{j}(y_{ij}-1)(y_{ij}-2).$ (12)

Now,

$\displaystyle\frac{S_{i}(\lambda;\boldsymbol{y}_{i}-\boldsymbol{1})}{S_{i}(\lambda;\boldsymbol{y}_{i})}=\prod_{j}y_{ij}\cdot$
$\displaystyle\frac{1+\prod_{j}(y_{ij}-1)\left\\{P^{-1}\lambda+\frac{1}{2}a_{2}(\boldsymbol{y}_{i}-\boldsymbol{1})P^{-2}\lambda^{2}++\frac{1}{6}a_{3}(\boldsymbol{y}_{i}-\boldsymbol{1})P^{-3}\lambda^{3}\right\\}+O(\lambda^{4})}{1+\prod_{j}y_{ij}\left\\{P^{-1}\lambda+\frac{1}{2}a_{2}(\boldsymbol{y}_{i})P^{-2}\lambda^{2}+\frac{1}{6}a_{3}(\boldsymbol{y}_{i})P^{-3}\lambda^{3}\right\\}+O(\lambda^{4})},$
(13)

and by Taylor expansion,

$\displaystyle\frac{S_{i}(\lambda;\boldsymbol{y}_{i}-\boldsymbol{1})}{S_{i}(\lambda;\boldsymbol{y}_{i})}$
$\displaystyle=\prod_{j}y_{ij}\left[1+\left\\{\prod_{j}(y_{ij}-1)-\prod_{j}y_{ij}\right\\}P^{-1}\lambda\right.$
$\displaystyle\left.+b_{2}(\boldsymbol{y}_{i})P^{-2}\lambda^{2}+b_{3}(\boldsymbol{y}_{i})P^{-3}\lambda^{3}+O(\lambda^{4})\right],$
(14)

where

$\displaystyle b_{2}(\boldsymbol{y}_{i})$
$\displaystyle\stackrel{{\scriptstyle\rm{def}}}{{=}}\frac{1}{2}\left\\{\prod_{j}(y_{ij}-1)a_{2}(\boldsymbol{y}_{i}-\boldsymbol{1})-\prod_{j}y_{ij}a_{2}(\boldsymbol{y}_{i})\right\\}$
$\displaystyle+\prod_{j}y_{ij}^{2}-\prod_{j}y_{ij}(y_{ij}-1).$ (15)

and

$\displaystyle b_{3}(\boldsymbol{y}_{i})$
$\displaystyle\stackrel{{\scriptstyle\rm{def}}}{{=}}\frac{1}{6}\left\\{\prod_{j}(y_{ij}-1)a_{3}(\boldsymbol{y}_{i}-\boldsymbol{1})-\prod_{j}y_{ij}a_{3}(\boldsymbol{y}_{i})\right\\}$
$\displaystyle-\prod_{j}y_{ij}^{3}+\prod_{j}y_{ij}^{2}a_{2}(\boldsymbol{y}_{i})+\prod_{j}y_{ij}^{2}(y_{ij}-1)$
$\displaystyle-\frac{1}{2}\prod_{j}y_{ij}\left\\{\prod_{j}(y_{ij}-1)a_{2}(\boldsymbol{y}_{i}-\boldsymbol{1})+\prod_{j}y_{ij}a_{2}(\boldsymbol{y}_{i})\right\\}.$
(16)

Moreover, let

$C_{1}\stackrel{{\scriptstyle\rm{def}}}{{=}}n^{-1}\sum_{i=1}^{n}\prod_{j}y_{ij}\left\\{\prod_{j}(y_{ij}-1)-\prod_{j}y_{ij}\right\\},$
(17)
$C_{2}\stackrel{{\scriptstyle\rm{def}}}{{=}}n^{-1}\sum_{i=1}^{n}\prod_{j}y_{ij}b_{2}(\boldsymbol{y}_{i}),$
(18)

and

$C_{3}\stackrel{{\scriptstyle\rm{def}}}{{=}}n^{-1}\sum_{i=1}^{n}\prod_{j}y_{ij}b_{3}(\boldsymbol{y}_{i}),$
(19)

Now, via (14) and (17)-(19),

$\displaystyle S$
$\displaystyle\stackrel{{\scriptstyle\rm{def}}}{{=}}n^{-1}\sum_{i=1}^{n}\frac{S_{i}(\lambda;\boldsymbol{y}_{i}-\boldsymbol{1})}{S_{i}(\lambda;\boldsymbol{y}_{i})}I\\{y_{i}>0\\}$
$\displaystyle=\bar{y}_{11...1}+P^{-1}C_{1}\lambda+P^{-2}C_{2}\lambda^{2}+P^{-3}C_{3}\lambda^{3}+O(\lambda^{4}),$
(20)

where

$\bar{y}_{11...1}\stackrel{{\scriptstyle\rm{def}}}{{=}}n^{-1}\prod_{j}y_{ij}.$
(21)

From (8) and (20), $l^{\prime}(\lambda)=0$ implies
$\prod_{j}(\bar{y}_{j}-\lambda)=S$, where

$\displaystyle\prod_{j}(\bar{y}_{j}-\lambda)=P\prod_{j}\left(1-\frac{\lambda}{\bar{y}_{j}}\right)$
$\displaystyle=P\left(1-\sum_{j}\bar{y}_{j}^{-1}\lambda+\sum_{j<k}\bar{y}_{j}^{-1}\bar{y}_{k}^{-1}\lambda^{2}-\sum_{j<k<l}\bar{y}_{j}^{-1}\bar{y}_{k}^{-1}\bar{y}_{l}^{-1}\lambda^{3}\right)+O(\lambda^{4}),$

and so, via (20),

$0=n^{-1/2}d_{0}+d_{1}\lambda+d_{2}\lambda^{2}+d_{3}\lambda^{3}+O(\lambda^{4}),$
(22)

where

$\displaystyle d_{0}$ $\displaystyle=\sqrt{n}\left(\bar{y}_{11...1}-P\right),$
(23) $\displaystyle d_{1}$
$\displaystyle=P^{-1}C_{1}+P\sum_{j}\bar{y}_{j}^{-1},$ (24) $\displaystyle
d_{2}$
$\displaystyle=P^{-2}C_{2}-P\sum_{j<k}\bar{y}_{j}^{-1}\bar{y}_{k}^{-1},$ (25)
$\displaystyle d_{3}$
$\displaystyle=P^{-3}C_{3}+P\sum_{j<k<l}\bar{y}_{j}^{-1}\bar{y}_{k}^{-1}\bar{y}_{l}^{-1}.$
(26)

It will be shown below that as $n\to\infty$, $d_{0}$, $d_{1}$, $d_{2}$ and
$d_{3}$ are $O_{p}(1)$.

Now, putting $\lambda=n^{-1/2}c_{1}+n^{-1}c_{2}+n^{-3/2}c_{3}+O(n^{-2})$ and
inserting into (22), we get

$\displaystyle 0$
$\displaystyle=n^{-1/2}d_{0}+d_{1}(n^{-1/2}c_{1}+n^{-1}c_{2}+n^{-3/2}c_{3})+d_{2}(n^{-1/2}c_{1}+n^{-1}c_{2})^{2}$
$\displaystyle+d_{3}(n^{-1/2}c_{1})^{3}+O(n^{-2})$
$\displaystyle=n^{-1/2}(d_{0}+d_{1}c_{1})+n^{-1}(d_{1}c_{2}+d_{2}c_{1}^{2})+n^{-3/2}(d_{1}c_{3}+2d_{2}c_{1}c_{2}+d_{3}c_{1}^{3})$
$\displaystyle+O(n^{-2}),$

and by putting all $n$ power coefficients equal to zero, we obtain the root

$\hat{\lambda}=-\frac{n^{-1/2}d_{0}}{d_{1}}\left\\{1+\frac{n^{-1/2}d_{0}d_{2}}{d_{1}^{2}}+\frac{n^{-1}d_{0}^{2}(2d_{2}^{2}-d_{1}d_{3})}{d_{1}^{4}}+O(n^{-3/2})\right\\}.$
(27)

##  3 The log likelihood ratio test

Putting $\lambda=0$ in (2) we get the likelihood under the null of
independence. Then, inserting the MLEs $\hat{\mu}_{j}=\bar{y}_{j}$, we get the
maximized likelihood under $H_{0}$ as

$L_{0}\stackrel{{\scriptstyle\rm{def}}}{{=}}e^{-n\sum_{j=1}^{m}\bar{y}_{j}}\left(\prod_{j=1}^{m}\bar{y}_{j}^{n\bar{y}_{j}}\right)\prod_{i=1}^{n}\prod_{j=1}^{m}(y_{ij}!)^{-1}.$
(28)

Now, define the log likelihood ratio test statistic as

$Q_{n}=-2\log\left(\frac{L_{0}}{L_{1}}\right)=2(\log L_{1}-\log L_{0}),$

where $L_{1}$ is the maximized likelihood under $H_{1}$. In the following, we
will implicitly condition on the event $\hat{\lambda}>0$.

Via (6) and (28), we have

$\frac{1}{2}n^{-1}Q_{n}=(m-1)\hat{\lambda}+\sum_{j}\bar{y}_{j}\log\left(1-\frac{\hat{\lambda}}{\bar{y}_{j}}\right)+n^{-1}\sum_{i=1}^{n}s_{i}^{*}(\hat{\lambda}),$
(29)

with

$s_{i}^{*}(\hat{\lambda})=\log\left\\{S_{i}(\hat{\lambda})\prod_{j}y_{ij}!\right\\},$
(30)

where, via (9),

$\displaystyle S_{i}(\hat{\lambda})\prod_{j}y_{ij}!$
$\displaystyle=1+\prod_{j}y_{ij}\xi(\hat{\lambda})+\frac{\prod_{j}y_{ij(2)}}{2!}\xi(\hat{\lambda})^{2}+\frac{\prod_{j}y_{ij(3)}}{3!}\xi(\hat{\lambda})^{3}+\frac{\prod_{j}y_{ij(4)}}{4!}\xi(\hat{\lambda})^{4}$
$\displaystyle+O(\hat{\lambda}^{5}).$

Now, (30) and the Taylor expansion
$\log(1+x)=x-x^{2}/2+x^{3}/3-x^{4}/4+O(x^{5})$ yields

$s_{i}^{*}(\hat{\lambda})=\prod_{j}y_{ij}\xi(\hat{\lambda})+\frac{e_{2i}}{2}\xi(\hat{\lambda})^{2}+\frac{e_{3i}}{3}\xi(\hat{\lambda})^{3}+\frac{e_{4i}}{4}\xi(\hat{\lambda})^{4}+O(\hat{\lambda}^{5}),$
(31)

where

$\displaystyle e_{2i}$
$\displaystyle=\prod_{j}y_{ij(2)}-\left(\prod_{j}y_{ij}\right)^{2},$ (32)
$\displaystyle e_{3i}$
$\displaystyle=\frac{1}{2}\prod_{j}y_{ij(3)}-\frac{3}{2}\prod_{j}y_{ij}\prod_{j}y_{ij(2)}+\left(\prod_{j}y_{ij}\right)^{3},$
(33) $\displaystyle e_{4i}$
$\displaystyle=\frac{1}{6}\prod_{j}y_{ij(4)}-\frac{2}{3}\prod_{j}y_{ij}\prod_{j}y_{ij(3)}-\frac{1}{2}\left(\prod_{j}y_{ij(2)}\right)^{2}$
$\displaystyle+2\left(\prod_{j}y_{ij}\right)^{2}\prod_{j}y_{ij(2)}-\left(\prod_{j}y_{ij}\right)^{4}.$
(34)

Moreover, because for $j=1,2,...,m$,

$\log\left(\frac{\bar{y}_{j}-\hat{\lambda}}{\bar{y}_{j}}\right)=\log\left(1-\frac{\hat{\lambda}}{\bar{y}_{j}}\right)=-\frac{\hat{\lambda}}{\bar{y}_{j}}-\frac{\hat{\lambda}^{2}}{2\bar{y}_{j}^{2}}-\frac{\hat{\lambda}^{3}}{3\bar{y}_{j}^{3}}-\frac{\hat{\lambda}^{4}}{4\bar{y}_{j}^{4}}+O(\hat{\lambda}^{5}),$

(29) and (31) yield

$\displaystyle\frac{1}{2}n^{-1}Q_{n}$
$\displaystyle=(m-1)\hat{\lambda}+\sum_{j=1}^{m}\bar{y}_{j}\left(-\frac{\hat{\lambda}}{\bar{y}_{j}}-\frac{\hat{\lambda}^{2}}{2\bar{y}_{j}^{2}}-\frac{\hat{\lambda}^{3}}{3\bar{y}_{j}^{3}}-\frac{\hat{\lambda}^{4}}{4\bar{y}_{j}^{4}}\right)+\bar{y}_{11...1}\xi(\hat{\lambda})$
$\displaystyle+\frac{1}{2}\bar{e}_{2}\xi(\hat{\lambda})^{2}+\frac{1}{3}\bar{e}_{3}\xi(\hat{\lambda})^{3}+\frac{1}{4}\bar{e}_{4}\xi(\hat{\lambda})^{4}+O(\hat{\lambda}^{5}),$
(35)

where $\bar{e}_{k}=n^{-1}\sum_{i}e_{ki}$ for $k=2,3,4$. Here, from (32) and
(17),

$\bar{e}_{2}=C_{1},$ (36)

Let $\overline{\cal Y}(2^{m-1},0^{1})$ be the sum of all products of $m-1$
distinct ${\bar{y}_{j}}^{2}$, etcetera. Then, inserting (10) into (35), we get
in view of (23) that

$\displaystyle\frac{1}{2}n^{-1}Q_{n}$
$\displaystyle=-\hat{\lambda}-\frac{\overline{\cal
Y}(1^{m-1},0^{1})}{2\prod_{j}\bar{y}_{j}}\hat{\lambda}^{2}-\frac{\overline{\cal
Y}(2^{m-1},0^{1})}{3\prod_{j}\bar{y}_{j}^{2}}\hat{\lambda}^{3}-\frac{\overline{\cal
Y}(3^{m-1},0^{1})}{4\prod_{j}\bar{y}_{j}^{3}}\hat{\lambda}^{4}$
$\displaystyle+\bar{y}_{11...1}\left(\frac{1}{\prod_{j}\bar{y}_{j}}\hat{\lambda}+\frac{\overline{\cal
Y}(1^{m-1},0^{1})}{\prod_{j}\bar{y}_{j}^{2}}\hat{\lambda}^{2}+\frac{\overline{\cal
Y}(2^{m-1},0^{1})+\overline{\cal
Y}(2^{m-2},1^{2})}{\prod_{j}\bar{y}_{j}^{3}}\hat{\lambda}^{3}\right.$
$\displaystyle\left.+\frac{\overline{\cal Y}(3^{m-1},0^{1})+\overline{\cal
Y}(3^{m-2},2^{1},1^{1})+\overline{\cal
Y}(3^{m-3},2^{3})}{\prod_{j}\bar{y}_{j}^{4}}\hat{\lambda}^{4}\right)$
$\displaystyle+\frac{1}{2}\bar{e}_{2}\left(\frac{1}{\prod_{j}\bar{y}_{j}}\hat{\lambda}+\frac{\overline{\cal
Y}(1^{m-1},0^{1})}{\prod_{j}\bar{y}_{j}^{2}}\hat{\lambda}^{2}+\frac{\overline{\cal
Y}(2^{m-1},0^{1})+\overline{\cal
Y}(2^{m-2},1^{2})}{\prod_{j}\bar{y}_{j}^{3}}\hat{\lambda}^{3}\right)^{2}$
$\displaystyle+\frac{1}{3}\bar{e}_{3}\left(\frac{1}{\prod_{j}\bar{y}_{j}}\hat{\lambda}+\frac{\overline{\cal
Y}(1^{m-1},0^{1})}{\prod_{j}\bar{y}_{j}^{2}}\hat{\lambda}^{2}\right)^{3}+\frac{1}{4}\bar{e}_{4}\left(\frac{1}{\prod_{j}\bar{y}_{j}}\hat{\lambda}\right)^{4}+O(\hat{\lambda}^{5})$
$\displaystyle=\frac{n^{-1/2}d_{0}}{\prod_{j}\bar{y}_{j}}\hat{\lambda}+\frac{d_{32}}{2\prod_{j}\bar{y}_{j}^{2}}\hat{\lambda}^{2}+\frac{d_{33}}{3\prod_{j}\bar{y}_{j}^{3}}\hat{\lambda}^{3}+\frac{d_{34}}{4\prod_{j}\bar{y}_{j}^{4}}\hat{\lambda}^{4}+O(\hat{\lambda}^{5}),$

where by some simplifications,

$\displaystyle d_{32}$ $\displaystyle=(2\bar{y}_{11...1}-P)\overline{\cal
Y}(1^{m-1},0^{1})+\bar{e}_{2},$ (37) $\displaystyle d_{33}$
$\displaystyle=(3\bar{y}_{11...1}-P)\overline{\cal
Y}(2^{m-1},0^{1})+3\bar{y}_{11...1}\overline{\cal Y}(2^{m-2},1^{2})$
$\displaystyle+3\overline{\cal Y}(1^{m-1},0^{1})\bar{e}_{2}+\bar{e}_{3},$ (38)
$\displaystyle d_{34}$ $\displaystyle=(4\bar{y}_{11...1}-P)\overline{\cal
Y}(3^{m-1},0^{1})$ $\displaystyle+4\bar{y}_{11...1}\left\\{\overline{\cal
Y}(3^{m-2},2^{1},1^{1})+\overline{\cal Y}(3^{m-3},2^{3})\right\\}$
$\displaystyle+2\left\\{\overline{\cal Y}(1^{m-1},0^{1})^{2}+2\overline{\cal
Y}(2^{m-1},0^{1})+2\overline{\cal Y}(2^{m-2},1^{2})\right\\}\bar{e}_{2}$
$\displaystyle+4\overline{\cal Y}(1^{m-1},0^{1})\bar{e}_{3}+\bar{e}_{4}.$ (39)

Hence, by (27),

$\displaystyle\frac{1}{2}Q$
$\displaystyle=\frac{d_{0}^{2}}{P(-d_{1})}\left\\{1+\frac{n^{-1/2}d_{0}d_{2}}{d_{1}^{2}}+\frac{n^{-1}d_{0}^{2}(2d_{2}^{2}-d_{1}d_{3})}{d_{1}^{4}}\right\\}$
$\displaystyle+\frac{d_{32}d_{0}^{2}}{2P^{2}d_{1}^{2}}\left\\{1+\frac{n^{-1/2}d_{0}d_{2}}{d_{1}^{2}}+\frac{n^{-1}d_{0}^{2}(2d_{2}^{2}-d_{1}d_{3})}{d_{1}^{4}}\right\\}^{2}$
$\displaystyle+\frac{n^{-1/2}d_{33}d_{0}^{3}}{3P^{3}(-d_{1})^{3}}\left(1+\frac{n^{-1/2}d_{0}d_{2}}{d_{1}^{2}}\right)^{3}+\frac{n^{-1}d_{34}d_{0}^{4}}{4P^{4}d_{1}^{4}}+O(n^{-3/2})$
$\displaystyle=f_{0}+n^{-1/2}f_{1}+n^{-1}f_{2}+O(n^{-3/2}),$ (40)

where

$\displaystyle f_{0}$
$\displaystyle=\frac{d_{0}^{2}}{2P^{2}d_{1}^{2}}(-2Pd_{1}+d_{32}),$ (41)
$\displaystyle f_{1}$
$\displaystyle=\frac{d_{0}^{3}}{3P^{3}d_{1}^{4}}(-3P^{2}d_{1}d_{2}+3Pd_{2}d_{32}-d_{33}d_{1}),$
(42) $\displaystyle f_{2}$
$\displaystyle=\frac{d_{0}^{4}}{4P^{4}d_{1}^{6}}(-8P^{3}d_{1}d_{2}^{2}+10P^{2}d_{2}^{2}d_{32}-4Pd_{33}d_{1}d_{2}+d_{34}d_{1}^{2}$
$\displaystyle\qquad\qquad+4P^{3}d_{1}^{2}d_{3}-4P^{2}d_{1}d_{3}d_{32}).$ (43)

In order to proceed from here, for $j=1,2,...,m$ write

$\bar{y}_{j}=\mu_{j}+n^{-1/2}r_{j},$ (44)

where

$r_{j}=\sqrt{n}(\bar{y}_{j}-\mu_{j}).$ (45)

At first, we use (44) and the expansion $(1+x)^{-1}=1-x+x^{2}+O(x^{3})$ to get

$\displaystyle\bar{y}_{j}^{-1}$
$\displaystyle=(\mu_{j}+n^{-1/2}r_{j})^{-1}=\mu_{j}^{-1}\left(1+n^{-1/2}\frac{r_{j}}{\mu_{j}}\right)^{-1}$
$\displaystyle=\mu_{j}^{-1}\left(1-n^{-1/2}\frac{r_{j}}{\mu_{j}}+n^{-1}\frac{r_{j}^{2}}{\mu_{j}^{2}}\right)+O(n^{-3/2}).$
(46)

Now, defining $p_{1}=\prod_{j}\mu_{j}$, we have because $E(P)=p_{1}$ that, via
(44),

$\displaystyle P$
$\displaystyle=p_{1}\prod_{j}\left(1+n^{-1/2}\frac{r_{j}}{\mu_{j}}\right)$
$\displaystyle=p_{1}\left\\{1+n^{-1/2}\sum_{j}\frac{r_{j}}{\mu_{j}}+n^{-1}\sum_{j<k}\frac{r_{j}r_{k}}{\mu_{j}\mu_{k}}+O(n^{-3/2})\right\\},$
(47)

implying

$\displaystyle P^{-1}$
$\displaystyle=p_{1}^{-1}\left[1-n^{-1/2}\sum_{j}\frac{r_{j}}{\mu_{j}}+n^{-1}\left\\{\left(\sum_{j}\frac{r_{j}}{\mu_{j}}\right)^{2}-\sum_{j<k}\frac{r_{j}r_{k}}{\mu_{j}\mu_{k}}\right\\}+O(n^{-3/2})\right]$
$\displaystyle=p_{1}^{-1}\left\\{1-n^{-1/2}\sum_{j}\frac{r_{j}}{\mu_{j}}+n^{-1}\sum_{j\geq
k}\frac{r_{j}r_{k}}{\mu_{j}\mu_{k}}+O(n^{-3/2})\right\\}.$ (48)

To tackle $d_{1}=P^{-1}C_{1}+P\sum_{j}\bar{y}_{j}^{-1}$ (cf (24)), we write

$C_{1}=C_{10}+n^{-1/2}C_{11},\quad C_{11}=\sqrt{n}(C_{1}-C_{10}),$ (49)

where $C_{10}=E(C_{1})$. Now, write

$-d_{1}=d_{10}+n^{-1/2}d_{11}+n^{-1}d_{12}+O(n^{-3/2}),$ (50)

where from (24), for $m>2$,

$-d_{10}=p_{1}^{-1}C_{10}+p_{1}s_{1},$ (51)

letting $s_{1}\stackrel{{\scriptstyle\rm{def}}}{{=}}\sum_{j}\mu_{j}^{-1}$.

Now, from (24) and (46)-(49),

$\displaystyle d_{1}$
$\displaystyle=p_{1}^{-1}\left(1-n^{-1/2}\sum_{j}\frac{r_{j}}{\mu_{j}}+n^{-1}\sum_{j\geq
k}\frac{r_{j}r_{k}}{\mu_{j}\mu_{k}}\right)(C_{10}+n^{-1/2}C_{11})$
$\displaystyle+p_{1}\left(1+n^{-1/2}\sum_{j}\frac{r_{j}}{\mu_{j}}+n^{-1}\sum_{j<k}\frac{r_{j}r_{k}}{\mu_{j}\mu_{k}}\right)\sum_{j}\mu_{j}^{-1}\left(1-n^{-1/2}\frac{r_{j}}{\mu_{j}}+n^{-1}\frac{r_{j}^{2}}{\mu_{j}^{2}}\right)$
$\displaystyle+O(n^{-3/2}),$

from which it follows that

$\displaystyle d_{11}$
$\displaystyle=p_{1}^{-1}\left(C_{10}\sum_{j}\frac{r_{j}}{\mu_{j}}-C_{11}\right)-p_{1}\left(\sum_{j}\mu_{j}^{-1}\sum_{j}\frac{r_{j}}{\mu_{j}}-\sum_{j}\frac{r_{j}}{\mu_{j}^{2}}\right),$
(52) $\displaystyle d_{12}$
$\displaystyle=p_{1}^{-1}\left(C_{11}\sum_{j}\frac{r_{j}}{\mu_{j}}-C_{10}\sum_{j\geq
k}\frac{r_{j}r_{k}}{\mu_{j}\mu_{k}}\right)$ $\displaystyle-
p_{1}\left(\sum_{j}\mu_{j}^{-1}\sum_{j<k}\frac{r_{j}r_{k}}{\mu_{j}\mu_{k}}-\sum_{j}\frac{r_{j}}{\mu_{j}}\sum_{j}\frac{r_{j}}{\mu_{j}^{2}}+\sum_{j}\frac{r_{j}^{2}}{\mu_{j}^{3}}\right).$
(53)

Moreover, from (50),

$(-d_{1})^{-1}=d_{10}^{-1}\left(1-n^{-1/2}\frac{d_{11}}{d_{10}}+n^{-1}\frac{d_{11}^{2}-d_{10}d_{12}}{d_{10}^{2}}\right)+O(n^{-3/2}).$
(54)

Now, write (41) as

$f_{0}=f_{01}+f_{02},$ (55)

where

$\displaystyle f_{01}$ $\displaystyle=\frac{d_{0}^{2}}{P(-d_{1})},$ (56)
$\displaystyle f_{02}$
$\displaystyle=\frac{d_{0}^{2}d_{32}}{2P^{2}d_{1}^{2}}.$ (57)

Next, introduce the notation $\dot{y}_{ij}=y_{ij}-\mu_{j}$ for $j=1,2,...m$.
Moreover, let $\bar{\dot{y}}_{j}=n^{-1}\sum_{i}\dot{y}_{ij}$,
$\bar{\dot{y}}_{110...0}=n^{-1}\sum_{i}\dot{y}_{1j}y_{2j}$ and so on. It
follows from (23) that, for $m>2$,

$\displaystyle d_{0}$
$\displaystyle=\sqrt{n}(\bar{y}_{11...1}-\bar{y}_{1}\bar{y}_{2}\cdots\bar{y}_{m})$
$\displaystyle=n^{-1/2}\sum_{i}(\mu_{1}+\dot{y}_{i1})\cdots(\mu_{m}+\dot{y}_{im})-n^{1/2}(\mu_{1}+\bar{\dot{y}}_{1})\cdots(\mu_{m}+\bar{\dot{y}}_{m})$
$\displaystyle=\sqrt{n}(\bar{\dot{y}}_{11...1}+\mu_{1}\bar{\dot{y}}_{01...1}+...+\mu_{m}\bar{\dot{y}}_{1...10}+...$
$\displaystyle+\mu_{1}\mu_{2}\cdots\mu_{m-2}\bar{\dot{y}}_{0...011}+...+\mu_{3}\cdots\mu_{m}\bar{\dot{y}}_{110...0})$
$\displaystyle-\sqrt{n}(\mu_{1}\mu_{2}\cdots\mu_{m-2}\bar{\dot{y}}_{m-1}\bar{\dot{y}}_{m}+...+\mu_{3}\cdots\mu_{m}\bar{\dot{y}}_{1}\bar{\dot{y}}_{2})$
$\displaystyle-\sqrt{n}(\mu_{1}\mu_{2}\cdots\mu_{m-3}\bar{\dot{y}}_{m-2}\bar{\dot{y}}_{m-1}\bar{\dot{y}}_{m}+...+\mu_{4}\cdots\mu_{m}\bar{\dot{y}}_{1}\bar{\dot{y}}_{2}\bar{\dot{y}}_{3})+...$
$\displaystyle=d_{00}-n^{-1/2}(r^{2},\mu^{m-2})-n^{-1}(r^{3},\mu^{m-3})+O(n^{-3/2}),$
(58)

where for $k=2,3$ (note that from (45), $r_{j}=\sqrt{n}\bar{\dot{y}}_{j}$)

$(r^{k},\mu^{m-k})=\mu_{1}\mu_{2}\cdots\mu_{m-k}r_{m-k+1}\cdots
r_{m}+...+\mu_{k+1}\cdots\mu_{m}r_{1}\cdots r_{k},$ (59)

and

$d_{00}=\sqrt{n}\bar{\dot{y}}_{11...1}+(r^{\otimes(m-1)},\mu^{1})+...+(r^{\otimes
2},\mu^{m-2}),$ (60)

with, for $k=1,2,...,m-2$,

$(r^{\otimes(m-k)},\mu^{k})=\sqrt{n}(\mu_{1}\mu_{2}\cdots\mu_{k}\bar{\dot{y}}_{0...01...1}+...+\mu_{m-k+1}\cdots\mu_{m}\bar{\dot{y}}_{1...10...0}),$
(61)

where the $\bar{\dot{y}}$ terms have indices with $k$ zeros. It is seen that
$d_{00}$ and the $(r^{k},\mu^{m-k})$ are $O(1)$.

If $m=2$, it follows that $d_{0}=d_{00}-n^{-1/2}r_{1}r_{2}$, where
$d_{00}=\sqrt{n}\bar{\dot{y}}_{11}$.

Furthermore, (56), (48) and (54) yield

$\displaystyle f_{01}$
$\displaystyle=\frac{d_{0}^{2}}{p_{1}d_{10}}\left(1-n^{-1/2}\sum_{j}\frac{r_{j}}{\mu_{j}}+n^{-1}\sum_{j\geq
k}\frac{r_{j}r_{k}}{\mu_{j}\mu_{k}}\right)$
$\displaystyle\left(1-n^{-1/2}\frac{d_{11}}{d_{10}}+n^{-1}\frac{d_{11}^{2}-d_{10}d_{12}}{d_{10}^{2}}\right)+O(n^{-3/2})$
$\displaystyle=\frac{d_{0}^{2}}{p_{1}d_{10}}(1+n^{-1/2}s_{011}+n^{-1}s_{012}),$
(62)

where

$\displaystyle s_{011}$
$\displaystyle=-\left(\sum_{j}\frac{r_{j}}{\mu_{j}}+\frac{d_{11}}{d_{10}}\right),$
(63) $\displaystyle s_{012}$ $\displaystyle=\sum_{j\geq
k}\frac{r_{j}r_{k}}{\mu_{j}\mu_{k}}+\frac{d_{11}}{d_{10}}\sum_{j}\frac{r_{j}}{\mu_{j}}+\frac{d_{11}^{2}-d_{10}d_{12}}{d_{10}^{2}}.$
(64)

To calculate the expectation of (62), we at first have, via (58),

$\displaystyle E(d_{0}^{2})$
$\displaystyle=E(d_{00}^{2})-2n^{-1/2}E\\{d_{00}(r^{2},\mu^{m-2})\\}$
$\displaystyle+n^{-1}\left[E\\{(r^{2},\mu^{m-2})^{2}\\}-2E\\{d_{00}(r^{3},\mu^{m-3})\\}\right]+O(n^{-3/2}),$
(65)

where, from (60),

$\displaystyle E(d_{00}^{2})$
$\displaystyle=nE(\bar{\dot{y}}_{11...1}^{2})+E\left\\{(r^{\otimes(m-1)},\mu^{1})^{2}\right\\}+...+E\left\\{(r^{\otimes
2},\mu^{m-2})^{2}\right\\}$
$\displaystyle+2\sqrt{n}E\left\\{\bar{\dot{y}}_{11...1}(r^{\otimes(m-1)},\mu^{1})\right\\}+...+E\left\\{(r^{\otimes
3},\mu^{m-3})(r^{\otimes 2},\mu^{m-2})\right\\}.$ (66)

Here, because all $E(\dot{y}_{ij})=0$,

$\displaystyle nE(\bar{\dot{y}}_{11...1}^{2})$
$\displaystyle=n^{-1}E\left\\{\left(\sum_{i}\dot{y}_{i1}\cdots\dot{y}_{im}\right)^{2}\right\\}$
$\displaystyle=n^{-1}\sum_{i}\sum_{j}E(\dot{y}_{i1}\cdots\dot{y}_{im}\dot{y}_{j1}\cdots\dot{y}_{jm})$
$\displaystyle=n^{-1}\sum_{i}E(\dot{y}_{i1}^{2})\cdots
E(\dot{y}_{im}^{2})=\mu_{1}\cdots\mu_{m}=p_{1}.$ (67)

A similar calculation shows that $E(\bar{\dot{y}}_{I}\bar{\dot{y}}_{J})=0$ for
index sets $I\neq J$. As a consequence, we have for $k=1,...,m-2$ that, from
(61),

$\displaystyle E\left\\{(r^{\otimes(m-k)},\mu^{k})^{2}\right\\}$
$\displaystyle=nE\left\\{(\mu_{1}\mu_{2}\cdots\mu_{k}\bar{\dot{y}}_{0...01...1}+...+\mu_{m-k+1}\cdots\mu_{m}\bar{\dot{y}}_{1...10...0})^{2}\right\\}$
$\displaystyle=n\left\\{\mu_{1}^{2}\cdots\mu_{k}^{2}E(\bar{\dot{y}}_{0...01...1}^{2})+...+\mu_{m-k+1}^{2}\cdots\mu_{m}^{2}E(\bar{\dot{y}}_{1...10...0}^{2})\right\\},$

where as in (67),

$nE(\bar{\dot{y}}_{1...10...0}^{2})=\mu_{1}\cdots\mu_{m-k}$

etcetera, and it follows that

$E\left\\{(r^{\otimes(m-k)},\mu^{k})^{2}\right\\}=p_{1}\sum_{j_{1}<...<j_{k}}\mu_{j_{1}}\cdots\mu_{j_{k}}.$
(68)

Moreover, as above,

$\sqrt{n}E\left\\{\bar{\dot{y}}_{11...1}(r^{\otimes(m-1)},\mu^{1})\right\\}=nE\left\\{\bar{\dot{y}}_{11...1}(\mu_{1}\bar{\dot{y}}_{01...1}+...+\mu_{m}\bar{\dot{y}}_{1...10})\right\\}=0,$

etcetera, and so, via (66)-(68) and (51), we find that

$E(d_{00}^{2})=p_{1}d_{10}.$ (69)

As for the next term in (65), we have

$\displaystyle n^{-1/2}E\\{d_{00}(r^{2},\mu^{m-2})\\}$
$\displaystyle=n^{-1/2}E\left[\left\\{\sqrt{n}\bar{\dot{y}}_{11...1}+(r^{\otimes
m-1},\mu^{1})+...+(r^{\otimes 2},\mu^{m-2})\right\\}(r^{2},\mu^{m-2})\right],$
(70)

where

$\displaystyle E\left\\{\bar{\dot{y}}_{11...1}(r^{2},\mu^{m-2})\right\\}$
$\displaystyle=E\left\\{\bar{\dot{y}}_{11...1}(\mu_{1}\cdots\mu_{m-2}r_{m-1}r_{m}+...+\mu_{3}\cdots\mu_{m}r_{1}r_{2})\right\\}$
$\displaystyle=\mu_{1}\cdots\mu_{m-2}E(\bar{\dot{y}}_{11...1}r_{m-1}r_{m})+...+\mu_{3}\cdots\mu_{m}E(\bar{\dot{y}}_{11...1}r_{1}r_{2}).$

Here,

$E(\bar{\dot{y}}_{11...1}r_{1}r_{2})=n^{-2}\sum_{i}\sum_{j}\sum_{k}E(\dot{y}_{i1}\cdots\dot{y}_{im}\dot{y}_{j1}\dot{y}_{k2}),$

If $m>2$ this is zero. For $m=2$, it is

$n^{-2}\sum_{i}E(\dot{y}_{i1}^{2}\dot{y}_{i2}^{2})=n^{-1}\mu_{1}\mu_{2}.$

In this case, this is the only contribution to (70). For $m>2$, we also have
terms like

$\displaystyle E\left\\{(r^{\otimes
2},\mu^{m-2})(r^{2},\mu^{m-2})\right\\}=n^{3/2}E$
$\displaystyle\left\\{(\mu_{1}\cdots\mu_{m-2}\bar{\dot{y}}_{0...011}+...+\mu_{3}\cdots\mu_{m}\bar{\dot{y}}_{110...0})\right.$
$\displaystyle\left.(\mu_{1}\cdots\mu_{m-2}\bar{\dot{y}}_{m-1}\bar{\dot{y}}_{m}+...+\mu_{3}\cdots\mu_{m}\bar{\dot{y}}_{1}\bar{\dot{y}}_{2})\right\\},$

where to evaluate the right-hand side, we need terms like

$\displaystyle E(\bar{\dot{y}}_{110...0}\bar{\dot{y}}_{1}\bar{\dot{y}}_{2})$
$\displaystyle=n^{-3}\sum_{i}\sum_{j}\sum_{k}E(\dot{y}_{i1}\dot{y}_{i2}\dot{y}_{j1}\dot{y}_{k2})=n^{-2}\mu_{1}\mu_{2},$
$\displaystyle E(\bar{\dot{y}}_{110...0}\bar{\dot{y}}_{1}\bar{\dot{y}}_{3})$
$\displaystyle=n^{-3}\sum_{i}\sum_{j}\sum_{k}E(\dot{y}_{i1}\dot{y}_{i2}\dot{y}_{j1}\dot{y}_{k3})=0,$

etcetera. It follows that

$\displaystyle E\left\\{(r^{\otimes 2},\mu^{m-2})(r^{2},\mu^{m-2})\right\\}$
$\displaystyle=n^{-1/2}(\mu_{1}^{2}\cdots\mu_{m-2}^{2}\mu_{m-1}\mu_{m}+...+\mu_{3}^{2}\cdots\mu_{m}^{2}\mu_{1}\mu_{2})$
$\displaystyle=n^{-1/2}p_{1}\sum_{j_{1}<...<j_{m-2}}\mu_{j_{1}}\cdots\mu_{j_{m-2}}.$
(71)

Similarly,

$\displaystyle E\left\\{(r^{\otimes
3},\mu^{m-3})(r^{2},\mu^{m-2})\right\\}=n^{3/2}E$
$\displaystyle\left\\{(\mu_{1}\cdots\mu_{m-3}\bar{\dot{y}}_{0...0111}+...+\mu_{4}\cdots\mu_{m}\bar{\dot{y}}_{1110...0})\right.$
$\displaystyle\left.(\mu_{1}\cdots\mu_{m-2}\bar{\dot{y}}_{m-1}\bar{\dot{y}}_{m}+...+\mu_{3}\cdots\mu_{m}\bar{\dot{y}}_{1}\bar{\dot{y}}_{2})\right\\},$

where

$E(\bar{\dot{y}}_{1110...0}\bar{\dot{y}}_{1}\bar{\dot{y}}_{2})=n^{-3}\sum_{i}\sum_{j}\sum_{k}E(\dot{y}_{i1}\dot{y}_{i2}\dot{y}_{i3}\dot{y}_{j1}\dot{y}_{k2})=0,$

etcetera, implying

$E\left\\{(r^{\otimes 3},\mu^{m-3})(r^{2},\mu^{m-2})\right\\}=0.$

In this manner, (70) and (71) imply

$\displaystyle n^{-1/2}E\left\\{d_{00}(r^{2},\mu^{m-2})\right\\}$
$\displaystyle=n^{-1/2}E\left\\{(r^{\otimes
2},\mu^{m-2})(r^{2},\mu^{m-2})\right\\}$
$\displaystyle=n^{-1}p_{1}\sum_{j_{1}<...<j_{m-2}}\mu_{j_{1}}\cdots\mu_{j_{m-2}}.$
(72)

To tackle the $n^{-1}$ terms of (65), we have

$E\left\\{(r^{2},\mu^{m-2})^{2}\right\\}=n^{2}E\left\\{(\mu_{1}\cdots\mu_{m-2}\bar{\dot{y}}_{m-1}\bar{\dot{y}}_{m}+...+\mu_{3}\cdots\mu_{m}\bar{\dot{y}}_{1}\bar{\dot{y}}_{2})^{2}\right\\},$

where

$\displaystyle E(\bar{\dot{y}}_{1}^{2}\bar{\dot{y}}_{2}^{2})$
$\displaystyle=n^{-4}\sum_{i}\sum_{j}\sum_{k}\sum_{l}E(\dot{y}_{i1}\dot{y}_{j1}\dot{y}_{k2}\dot{y}_{l2})=n^{-2}\mu_{1}\mu_{2}+O(n^{-3}),$
$\displaystyle E(\bar{\dot{y}}_{1}^{2}\bar{\dot{y}}_{2}\bar{\dot{y}}_{3})$
$\displaystyle=n^{-4}\sum_{i}\sum_{j}\sum_{k}\sum_{l}E(\dot{y}_{i1}\dot{y}_{j1}\dot{y}_{k2}\dot{y}_{l3})=0,$

and so on, implying

$\displaystyle E\left\\{(r^{2},\mu^{m-2})^{2}\right\\}$
$\displaystyle=\mu_{1}^{2}\cdots\mu_{m-2}^{2}\mu_{m-1}\mu_{m}+...+\mu_{3}^{2}\cdots\mu_{m}^{2}\mu_{1}\mu_{2}$
$\displaystyle=p_{1}\sum_{j_{1}<...<j_{m-2}}\mu_{j_{1}}\cdots\mu_{j_{m-2}}.$
(73)

Moreover, from (60),

$\displaystyle E\\{d_{00}(r^{3},\mu^{m-3})\\}$
$\displaystyle=E\left[\left\\{\sqrt{n}\bar{\dot{y}}_{11...1}+(r^{\otimes
m-1},\mu^{1})+...+(r^{\otimes 2},\mu^{m-2})\right\\}(r^{3},\mu^{m-3})\right],$

where as above,

$\displaystyle E\left\\{(r^{\otimes 2},\mu^{m-2})(r^{3},\mu^{m-3})\right\\}$
$\displaystyle=n^{2}E\left\\{(\mu_{1}\cdots\mu_{m-2}\bar{\dot{y}}_{0...011}+...+\mu_{3}\cdots\mu_{m}\bar{\dot{y}}_{110...0})\right.$
$\displaystyle\left.(\mu_{1}\cdots\mu_{m-3}\bar{\dot{y}}_{m-2}\bar{\dot{y}}_{m-1}\bar{\dot{y}}_{m}+...+\mu_{4}\cdots\mu_{m}\bar{\dot{y}}_{1}\bar{\dot{y}}_{2}\bar{\dot{y}}_{3})\right\\}$
$\displaystyle=0,$

and so on. The only term that breaks this pattern is

$\displaystyle E\left\\{(r^{\otimes 3},\mu^{m-3})(r^{3},\mu^{m-3})\right\\}$
$\displaystyle=n^{2}E\left\\{(\mu_{1}\cdots\mu_{m-3}\bar{\dot{y}}_{0...0111}+...+\mu_{4}\cdots\mu_{m}\bar{\dot{y}}_{1110...0})\right.$
$\displaystyle\left.(\mu_{1}\cdots\mu_{m-3}\bar{\dot{y}}_{m-2}\bar{\dot{y}}_{m-1}\bar{\dot{y}}_{m}+...+\mu_{4}\cdots\mu_{m}\bar{\dot{y}}_{1}\bar{\dot{y}}_{2}\bar{\dot{y}}_{3})\right\\},$

where nonzero contributions arise from terms like

$E(\bar{\dot{y}}_{1110...0}\bar{\dot{y}}_{1}\bar{\dot{y}}_{2}\bar{\dot{y}}_{3})=n^{-4}\sum_{i}\sum_{j}\sum_{k}\sum_{l}E({\dot{y}}_{i1}{\dot{y}}_{i2}{\dot{y}}_{i3}{\dot{y}}_{j1}{\dot{y}}_{k2}{\dot{y}}_{l3})=n^{-3}\mu_{1}\mu_{2}\mu_{3}.$

But this means that

$E\\{d_{00}(r^{3},\mu^{m-3})\\}=O(n^{-1}),$

which may be neglected. Thus, (65), (69), (72) and (73) imply

$E(d_{0}^{2})=p_{1}\left(d_{10}-n^{-1}\sum_{j_{1}<...<j_{m-2}}\mu_{j_{1}}\cdots\mu_{j_{m-2}}\right)+O(n^{-3/2}),$
(74)

for $m>2$, and the equation is true also for $m=2$ if the sum is interpreted
as 1.

Next, we go on with $E(d_{0}^{2}s_{011})$. To this end, using (58) (cf (65)),

$\displaystyle E(d_{0}^{2}r_{1})$
$\displaystyle=E(d_{00}^{2}r_{1})-2n^{-1/2}E\\{d_{00}(r^{2},\mu^{m-2})r_{1}\\}$
$\displaystyle+n^{-1}\left[E\\{(r^{2},\mu^{m-2})^{2}r_{1}\\}-2E\\{d_{00}(r^{3},\mu^{m-3})r_{1}\\}\right]+O(n^{-3/2}),$
(75)

where, from (60) (cf (66)),

$\displaystyle E(d_{00}^{2}r_{1})$
$\displaystyle=nE(\bar{\dot{y}}_{11...1}^{2}r_{1})+E\left\\{(r^{\otimes(m-1)},\mu^{1})^{2}r_{1}\right\\}+...+E\left\\{(r^{\otimes
2},\mu^{m-2})^{2}r_{1}\right\\}$
$\displaystyle+2\sqrt{n}E\left\\{\bar{\dot{y}}_{11...1}(r^{\otimes(m-1)},\mu^{1})r_{1}\right\\}+...+E\left\\{(r^{\otimes
3},\mu^{m-3})(r^{\otimes 2},\mu^{m-2})r_{1}\right\\}.$ (76)

Here, in the usual manner,

$\displaystyle nE(\bar{\dot{y}}_{11...1}^{2}r_{1})$
$\displaystyle=n^{3/2}E(\bar{\dot{y}}_{11...1}^{2}\bar{\dot{y}}_{1})$
$\displaystyle=n^{-3/2}\sum_{i}\sum_{j}\sum_{k}E(\dot{y}_{i1}\cdots\dot{y}_{im}\dot{y}_{j1}\cdots\dot{y}_{jm}\dot{y}_{k1})$
$\displaystyle=n^{-3/2}\sum_{i}E(\dot{y}_{i1}^{3})E(\dot{y}_{i2}^{2})\cdots
E(\dot{y}_{im}^{2})$ $\displaystyle=n^{-1/2}p_{1},$ (77)

since the third central moment of a ${\rm Po}(\mu)$ variate is $\mu$.
Moreover, we have for $k=1,...,m-2$ that

$\displaystyle E\left\\{(r^{\otimes(m-k)},\mu^{k})^{2}r_{1}\right\\}$
$\displaystyle=n^{3/2}E\left\\{(\mu_{1}\mu_{2}\cdots\mu_{k}\bar{\dot{y}}_{0...01...1}+...+\mu_{m-k+1}\cdots\mu_{m}\bar{\dot{y}}_{1...10...0})^{2}\bar{\dot{y}}_{1}\right\\}$
$\displaystyle=n^{3/2}\left\\{\mu_{1}^{2}\cdots\mu_{k}^{2}E(\bar{\dot{y}}_{0...01...1}^{2}\bar{\dot{y}}_{1})+...+\mu_{m-k+1}^{2}\cdots\mu_{m}^{2}E(\bar{\dot{y}}_{1...10...0}^{2}\bar{\dot{y}}_{1})\right\\},$

because as above, cross terms disappear in the last step. Here, the
expectations are nonzero only if the first index in $\bar{\dot{y}}_{I}$ is
one, and in such cases we e.g. have

$\displaystyle E(\bar{\dot{y}}_{1...10...0}^{2}\bar{\dot{y}}_{1})$
$\displaystyle=n^{-3}\sum_{i}\sum_{j}\sum_{k}E(\dot{y}_{i1}\cdots\dot{y}_{i,m-k}\dot{y}_{j1}\cdots\dot{y}_{j,m-k}\dot{y}_{k1})$
$\displaystyle=n^{-2}\mu_{1}\cdots\mu_{m-k}.$ (78)

Hence,

$\displaystyle E\left\\{(r^{\otimes(m-k)},\mu^{k})^{2}r_{1}\right\\}$
$\displaystyle=n^{3/2}\left\\{\mu_{2}^{2}\cdots\mu_{k+1}^{2}E(\bar{\dot{y}}_{10...01...1}^{2}\bar{\dot{y}}_{1})+...+\mu_{m-k+1}^{2}\cdots\mu_{m}^{2}E(\bar{\dot{y}}_{1...10...0}^{2}\bar{\dot{y}}_{1})\right\\}$
$\displaystyle=n^{-1/2}\left(\mu_{2}^{2}\cdots\mu_{k+1}^{2}\mu_{1}\mu_{k+1}\cdots\mu_{m}+...+\mu_{m-k+1}^{2}\cdots\mu_{m}^{2}\mu_{1}\cdots\mu_{m-k}\right)$
$\displaystyle=n^{-1/2}p_{1}\sum_{1<j_{1}<...<j_{k}}\mu_{j_{1}}\cdots\mu_{j_{k}}.$
(79)

For the usual reasons, there is only one more term that contributes to (76),
and this term is

$\displaystyle E\left\\{(r^{\otimes 3},\mu^{m-3})(r^{\otimes
2},\mu^{m-2})r_{1}\right\\}$
$\displaystyle=n^{3/2}E\left\\{(\mu_{1}\cdots\mu_{m-3}\bar{\dot{y}}_{0...0111}+...+\mu_{4}\cdots\mu_{m}\bar{\dot{y}}_{1110...0})\right.$
$\displaystyle\left.(\mu_{1}\cdots\mu_{m-2}\bar{\dot{y}}_{0...011}+...+\mu_{3}\cdots\mu_{m}\bar{\dot{y}}_{110...0})\bar{\dot{y}}_{1}\right\\}.$

The nonzero expectations that arise from here are of the type

$\displaystyle
E(\bar{\dot{y}}_{1110...0}\bar{\dot{y}}_{0110...0}\bar{\dot{y}}_{1})$
$\displaystyle=n^{-3}\sum_{i}\sum_{j}\sum_{k}E(\dot{y}_{i1}\dot{y}_{i2}\dot{y}_{i3}\dot{y}_{j2}\dot{y}_{j3}\dot{y}_{k1})$
$\displaystyle=n^{-3}\sum_{i}E(\dot{y}_{i1}^{2})E(\dot{y}_{i2}^{2})E(\dot{y}_{i3}^{2})=n^{-2}\mu_{1}\mu_{2}\mu_{3},$

implying

$\displaystyle E\left\\{(r^{\otimes 3},\mu^{m-3})(r^{\otimes
2},\mu^{m-2})r_{1}\right\\}$
$\displaystyle=n^{3/2}\left\\{\mu_{1}\mu_{4}^{2}\cdots\mu_{m}^{2}E(\bar{\dot{y}}_{1110...0}\bar{\dot{y}}_{0110...0}\bar{\dot{y}}_{1})+...+\mu_{1}\mu_{2}^{2}\cdots\mu_{m-2}^{2}E(\bar{\dot{y}}_{10...011}\bar{\dot{y}}_{0...011}\bar{\dot{y}}_{1})\right\\}$
$\displaystyle=n^{-1/2}(\mu_{1}\mu_{4}^{2}\cdots\mu_{m}^{2}\mu_{1}\mu_{2}\mu_{3}+...+\mu_{1}\mu_{2}^{2}\cdots\mu_{m-2}^{2}\mu_{1}\mu_{m-1}\mu_{m})$
$\displaystyle=n^{-1/2}p_{1}\mu_{1}\sum_{1<j_{1}<...<j_{m-3}}\mu_{j_{1}}\cdots\mu_{j_{m-3}}.$

Hence, via (76) and (79),

$E(d_{00}^{2}r_{1})=n^{-1/2}p_{1}\xi_{1},$ (80)

where

$\displaystyle\xi_{1}$
$\displaystyle=1+\sum_{1<j_{1}}\mu_{j_{1}}+...+\sum_{1<j_{1}<...<j_{m-2}}\mu_{j_{1}}\cdots\mu_{j_{m-2}}$
$\displaystyle+\mu_{1}\sum_{1<j_{1}<...<j_{m-3}}\mu_{j_{1}}\cdots\mu_{j_{m-3}}.$
(81)

The other terms in (75) may be neglected. For example, we have from (60) that

$\displaystyle E\left\\{d_{00}(r^{2},\mu^{m-2})r_{1}\right\\}$
$\displaystyle=E\left[\left\\{\sqrt{n}\bar{\dot{y}}_{11...1}+(r^{\otimes(m-1)},\mu^{1})+...+(r^{\otimes
2},\mu^{m-2})\right\\}(r^{2},\mu^{m-2})r_{1}\right],$

where

$\displaystyle
E\left\\{(r^{\otimes(m-k)},\mu^{k})(r^{2},\mu^{m-2})r_{1}\right\\}$
$\displaystyle=n^{2}E\left\\{(\mu_{1}\cdots\mu_{k}\bar{\dot{y}}_{0...01...1}+...+\mu_{m-k+1}\cdots\mu_{m}\bar{\dot{y}}_{1...10...0})\right.$
$\displaystyle\left.(\mu_{1}\cdots\mu_{m-2}\bar{\dot{y}}_{m-1}\bar{\dot{y}}_{m}+...+\mu_{3}\cdots\mu_{m}\bar{\dot{y}}_{1}\bar{\dot{y}}_{2})\bar{\dot{y}}_{1}\right\\},$

with e.g.

$E(\bar{\dot{y}}_{1...10...0}\bar{\dot{y}}_{1}^{2}\bar{\dot{y}}_{2})=n^{-4}\sum_{i}\sum_{j}\sum_{k}\sum_{l}E(\dot{y}_{i1}\cdots\dot{y}_{i,m-k}\dot{y}_{j1}\dot{y}_{k1}\dot{y}_{l2})=O(n^{-3}),$

implying that

$E\left\\{(r^{\otimes m-k},\mu^{k})(r^{\otimes
2},\mu^{m-2})r_{1}\right\\}=O(n^{-1}).$

Hence, from (80), we have

$E(d_{0}^{2}r_{1})=n^{-1/2}p_{1}\xi_{1}+O(n^{-1}).$ (82)

In the following, for all $j=2,...,m$ we will define $\xi_{j}$ analogously as
$\xi_{1}$.

To tackle $E(d_{0}^{2}C_{11})$, it follows as above that it is of the same
order as $E(d_{00}^{2}C_{11})$. Here, from (60),

$\displaystyle d_{00}$
$\displaystyle=\sqrt{n}(\bar{\dot{y}}_{11...1}+\mu_{1}\bar{\dot{y}}_{01...1}+...+\mu_{m}\bar{\dot{y}}_{1...10}+...$
$\displaystyle+\mu_{1}\cdots\mu_{m-2}\bar{\dot{y}}_{0...011}+...+\mu_{3}\cdots\mu_{m}\bar{\dot{y}}_{110...0}),$
(83)

and by (17) and (49), we may write $C_{11}=\sqrt{n}(C_{1}-C_{10})$ where

$C_{1}=n^{-1}\sum_{i}C_{i1},$ (84)

with

$C_{i1}=\prod_{j}y_{ij}\left\\{\prod_{j}(y_{ij}-1)-\prod_{j}y_{ij}\right\\}.$

Next, analogously define

$X=\prod_{j}X_{j}\left\\{\prod_{j}(X_{j}-1)-\prod_{j}X_{j}\right\\},$ (85)

where $X_{j}$ is a Poisson variate with parameter $\mu_{j}$ and all $X_{j}$
are simultaneously independent.

Let

$\displaystyle U_{i}$
$\displaystyle={\dot{y}}_{i1}\cdots{\dot{y}}_{im}+\mu_{1}{\dot{y}}_{i2}\cdots{\dot{y}}_{im}+...+\mu_{m}{\dot{y}}_{i1}\cdots{\dot{y}}_{i,m-1}+...$
$\displaystyle+\mu_{1}\cdots\mu_{m-2}{\dot{y}}_{i,m-1}{\dot{y}}_{im}+...+\mu_{3}\cdots\mu_{m}{\dot{y}}_{i1}{\dot{y}}_{i2}.$
(86)

Now, it follows from (83) and (84) that

$E(d_{00}^{2}C_{11})=n^{-3/2}\sum_{i}\sum_{j}\sum_{k}E\\{U_{i}U_{j}(C_{k1}-C_{10})\\}.$

Here, the expectation terms are nonzero only if $i=j=k$, and so, with
$\dot{X}=X-C_{10}$,

$E(d_{00}^{2}C_{11})=n^{-1/2}E\left({\dot{S}}^{2}\dot{X}\right),$ (87)

where with $\dot{X}_{j}=X_{j}-\mu_{j}$ for all $j$,

$\displaystyle\dot{S}$
$\displaystyle={\dot{X}}_{1}\cdots{\dot{X}}_{m}+\mu_{1}{\dot{X}}_{2}\cdots{\dot{X}}_{m}+...+\mu_{m}{\dot{X}}_{1}\cdots{\dot{X}}_{m-1}+...$
$\displaystyle+\mu_{1}\cdots\mu_{m-2}{\dot{X}}_{m-1}{\dot{X}}_{m}+...+\mu_{3}\cdots\mu_{m}{\dot{X}}_{1}{\dot{X}}_{2}.$
(88)

The expectation in (87) may be worked out as a function of
$\mu_{1},...,\mu_{m}$, but we refrain to do this until later. Now, from (52)
we get

$\displaystyle E(d_{0}^{2}d_{11})$
$\displaystyle=p_{1}^{-1}\left\\{C_{10}\sum_{j}\frac{E(d_{0}^{2}r_{j})}{\mu_{j}}-E(d_{0}^{2}C_{11})\right\\}$
$\displaystyle-
p_{1}\left\\{\sum_{j}\mu_{j}^{-1}\sum_{j}\frac{E(d_{0}^{2}r_{j})}{\mu_{j}}-\sum_{j}\frac{E(d_{0}^{2}r_{j})}{\mu_{j}^{2}}\right\\},$

where we get $E(d_{0}^{2}r_{j})$ from (82) and $E(d_{0}^{2}C_{11})$ from (87),
so that

$\displaystyle E(d_{0}^{2}d_{11})$
$\displaystyle=n^{-1/2}\left\\{C_{10}\sum_{j}\mu_{j}^{-1}\xi_{j}-p_{1}^{-1}E\left({\dot{S}}^{2}\dot{X}\right)\right.$
$\displaystyle\left.-p_{1}^{2}\left(s_{1}\sum_{j}\mu_{j}^{-1}\xi_{j}-\sum_{j}\mu_{j}^{-2}\xi_{j}\right)\right\\}+O(n^{-1}),$
(89)

where we have introduced

$s_{k}=\sum_{j}\mu_{j}^{-k},$

for $k=1,2,...$.

Then in turn, via (63) and (82), we get

$\displaystyle E(d_{0}^{2}s_{011})$
$\displaystyle=-\sum_{j}\frac{E(d_{0}^{2}r_{j})}{\mu_{j}}-\frac{E(d_{0}^{2}d_{11})}{d_{10}}$
$\displaystyle=-n^{-1/2}\left[p_{1}\sum_{j}\mu_{j}^{-1}\xi_{j}+d_{10}^{-1}\left\\{C_{10}\sum_{j}\mu_{j}^{-1}\xi_{j}-p_{1}^{-1}E\left({\dot{S}}^{2}\dot{X}\right)\right.\right.$
$\displaystyle\left.\left.-p_{1}^{2}\left(s_{1}\sum_{j}\mu_{j}^{-1}\xi_{j}-\sum_{j}\mu_{j}^{-2}\xi_{j}\right)\right\\}\right]+O(n^{-1}).$
(90)

To go on with $E(d_{0}^{2}s_{012})$, we at first note that, from (83),

$E(d_{00}^{2}r_{1}^{2})=n^{-2}\sum_{i}\sum_{j}\sum_{k}\sum_{l}E\\{U_{i}U_{j}{\dot{y}}_{k1}{\dot{y}}_{l1}\\}.$

The expectation inside of the sum is nonzero not only for $i=j=k=l$ (which
creates a $O(n^{-1})$ term) but also for $i=j$ and $k=l$, where $i$ and $k$
are not necessarily equal. It follows as above that

$E(d_{00}^{2}r_{1}^{2})=E(\dot{S}^{2})E(\dot{X}_{1}^{2})+O(n^{-1})=\mu_{1}E(\dot{S}^{2})+O(n^{-1}).$
(91)

Similarly, it is seen that $E(d_{00}^{2}r_{1}r_{2})=O(n^{-1})$, since in this
case, such a ’cross product term’ cannot arise.

Next, to calculate $E(d_{00}^{2}d_{11}r_{1})$, we get as above that, with
$\dot{X}=X-C_{10}$ (observe that $E(\dot{S}X_{j})=0$ for all $j$),

$\displaystyle E(d_{00}^{2}C_{11}r_{1})$
$\displaystyle=n^{-2}\sum_{i}\sum_{j}\sum_{k}\sum_{l}E\\{U_{i}U_{j}(C_{k1}-C_{10}){\dot{y}}_{l1}\\}$
$\displaystyle=E(\dot{S}^{2})E(\dot{X}\dot{X}_{1})+O(n^{-1}).$ (92)

Thus, via (52) and (91),

$\displaystyle E(d_{00}^{2}d_{11}r_{1})$
$\displaystyle=p_{1}^{-1}\left\\{C_{10}\frac{E(d_{00}^{2}r_{1}^{2})}{\mu_{1}}-E(d_{00}^{2}C_{11}r_{1})\right\\}$
$\displaystyle-
p_{1}\left\\{\sum_{j}\mu_{j}^{-1}\frac{E(d_{00}^{2}r_{1}^{2})}{\mu_{1}}-\frac{E(d_{00}^{2}r_{1}^{2})}{\mu_{1}^{2}}\right\\}$
$\displaystyle=p_{1}^{-1}E({\dot{S}}^{2})\left\\{C_{10}-E(\dot{X}\dot{X}_{1})\right\\}-p_{1}E({\dot{S}}^{2})(s_{1}-\mu_{1}^{-1})$
$\displaystyle+O(n^{-1}).$ (93)

Next, (52) and simplifications yield

$\displaystyle d_{11}^{2}$
$\displaystyle=p_{1}^{-2}C_{11}^{2}+2DC_{11}\sum_{j}\frac{r_{j}}{\mu_{j}}+p_{1}^{2}D^{2}\left(\sum_{j}\frac{r_{j}}{\mu_{j}}\right)^{2}$
$\displaystyle-2C_{11}\sum_{j}\frac{r_{j}}{\mu_{j}^{2}}+p_{1}^{2}\left(\sum_{j}\frac{r_{j}}{\mu_{j}^{2}}\right)^{2}-2p_{1}^{2}D\sum_{j}\frac{r_{j}}{\mu_{j}}\sum_{j}\frac{r_{j}}{\mu_{j}^{2}},$

where

$D=s_{1}-p_{1}^{-2}C_{10}=p_{1}^{-1}d_{10}+2s_{1},$ (94)

following from the fact that, via (50) and (51),

$-d_{10}=p_{1}^{-1}C_{10}+p_{1}s_{1}.$ (95)

To find $E(d_{00}^{2}d_{11}^{2})$, we need

$\displaystyle E(d_{00}^{2}C_{11}^{2})$
$\displaystyle=n^{-2}\sum_{i}\sum_{j}\sum_{k}\sum_{l}E\\{U_{i}U_{j}(C_{k1}-C_{10})(C_{l1}-C_{10})\\}$
$\displaystyle=E(\dot{S}^{2})E({\dot{X}}^{2})+2\left\\{E(\dot{S}\dot{X})\right\\}^{2}+O(n^{-1}).$
(96)

Now, together with (92) and (91), we get

$\displaystyle E(d_{00}^{2}d_{11}^{2})$
$\displaystyle=p_{1}^{-2}E(d_{00}^{2}C_{11}^{2})+2D\sum_{j}\mu_{j}^{-1}E(d_{00}^{2}C_{11}r_{j})+p_{1}^{2}D^{2}\sum_{j}\mu_{j}^{-2}E(d_{00}^{2}r_{j}^{2})$
$\displaystyle-2\sum_{j}\mu_{j}^{-2}E(d_{00}^{2}C_{11}r_{j})+p_{1}^{2}\sum_{j}\mu_{j}^{-4}E(d_{00}^{2}r_{j}^{2})-2p_{1}^{2}D\sum_{j}\mu_{j}^{-3}E(d_{00}^{2}r_{j}^{2})$
$\displaystyle=2p_{1}^{-2}\left\\{E(\dot{S}\dot{X})\right\\}^{2}+E(\dot{S}^{2})\left\\{p_{1}^{-2}E({\dot{X}}^{2})+2DA_{1}+p_{1}^{2}D^{2}s_{1}\right.$
$\displaystyle\left.-2A_{2}+p_{1}^{2}s_{3}-2p_{1}^{2}Ds_{2}\right\\}+O(n^{-1/2}),$
(97)

where

$A_{1}=\sum_{j}\mu_{j}^{-1}E(\dot{X}\dot{X}_{j}),\quad
A_{2}=\sum_{j}\mu_{j}^{-2}E(\dot{X}\dot{X}_{j}).$

Similarly, from (53), (91) and (92),

$\displaystyle E(d_{00}^{2}d_{12})$
$\displaystyle=p_{1}^{-1}\left\\{\sum_{j}\mu_{j}^{-1}E(d_{00}^{2}C_{11}r_{j})-C_{10}\sum_{j}\mu_{j}^{-2}E(d_{00}^{2}r_{j}^{2})\right\\}+O(n^{-1})$
$\displaystyle=p_{1}^{-1}E(\dot{S}^{2})\left(A_{1}-C_{10}s_{1}\right)+O(n^{-1}).$
(98)

Next, (64) implies

$\displaystyle E(d_{00}^{2}s_{012})$
$\displaystyle=\sum_{j}\mu_{j}^{-2}E(d_{00}^{2}r_{j}^{2})+d_{10}^{-1}\sum_{j}\mu_{j}^{-1}E(d_{00}^{2}d_{11}r_{j})$
$\displaystyle+d_{10}^{-2}E(d_{00}^{2}d_{11}^{2})-d_{10}^{-1}E(d_{00}^{2}d_{12})+O(n^{-1}),$

and via (91), (93), (94), (97), (98) and simplifications,

$E(d_{00}^{2}s_{012})=2d_{10}^{-2}p_{1}^{-2}\left\\{E(\dot{S}\dot{X})\right\\}^{2}+E({\dot{S}}^{2})F+O(n^{-1}),$
(99)

with

$\displaystyle F$
$\displaystyle=p_{1}^{-2}d_{10}^{-2}E({\dot{X}}^{2})+4d_{10}^{-2}s_{1}A_{1}-2d_{10}^{-2}A_{2}+p_{1}d_{10}^{-1}(s_{1}^{2}-s_{2})$
$\displaystyle+p_{1}^{2}d_{10}^{-2}(4s_{1}^{3}-4s_{1}s_{2}+s_{3}).$ (100)

Now, from (62), we get

$E(f_{01})=p_{1}^{-1}d_{10}^{-1}\left\\{E(d_{00}^{2})+n^{-1/2}E(d_{00}^{2}s_{011})+n^{-1}E(d_{00}^{2}s_{012})\right\\},$
(101)

where we get $E(d_{00}^{2})$ from (74), $E(d_{00}^{2}s_{011})$ from (90) and
$E(d_{00}^{2}s_{012})$ from (99).

Next, (57), (48) and (54) imply

$f_{02}=\frac{d_{0}^{2}d_{32}}{2p_{1}^{2}d_{10}^{2}}G,$ (102)

where

$\displaystyle G=$
$\displaystyle\left(1-n^{-1/2}\sum_{j}\frac{r_{j}}{\mu_{j}}+n^{-1}\sum_{j\geq
k}\frac{r_{j}r_{k}}{\mu_{j}\mu_{k}}\right)^{2}$
$\displaystyle\cdot\left(1-n^{-1/2}\frac{d_{11}}{d_{10}}+n^{-1}\frac{d_{11}^{2}-d_{10}d_{12}}{d_{10}^{2}}\right)^{2}+O(n^{-3/2})$
$\displaystyle=$ $\displaystyle 1-2n^{-1/2}G_{1}+n^{-1}G_{2}+O(n^{-3/2}),$
(103)

with

$\displaystyle G_{1}$
$\displaystyle=\sum_{j}\frac{r_{j}}{\mu_{j}}+\frac{d_{11}}{d_{10}},$ (104)
$\displaystyle G_{2}$ $\displaystyle=2\sum_{j\geq
k}\frac{r_{j}r_{k}}{\mu_{j}\mu_{k}}+\left(\sum_{j}\frac{r_{j}}{\mu_{j}}\right)^{2}+4\frac{d_{11}}{d_{10}}\sum_{j}\frac{r_{j}}{\mu_{j}}+2\frac{d_{11}^{2}-d_{10}d_{12}}{d_{10}^{2}}+\frac{d_{11}^{2}}{d_{10}^{2}}.$
(105)

We get $d_{32}$ from (37), where (cf (58))

$\displaystyle 2\bar{y}_{11...1}-P=2\bar{y}_{11...1}-\prod_{j}\bar{y}_{j}$
$\displaystyle=2n^{-1}\sum_{i}(\mu_{1}+\dot{y}_{i1})\cdots(\mu_{m}+\dot{y}_{im})-(\mu_{1}+\bar{\dot{y}}_{1})\cdots(\mu_{m}+\bar{\dot{y}}_{m})$
$\displaystyle=2p_{1}+2(\bar{\dot{y}}_{11...1}+\mu_{1}\bar{\dot{y}}_{01...1}+...+\mu_{m}\bar{\dot{y}}_{1...10}+...$
$\displaystyle+\mu_{1}\mu_{2}\cdots\mu_{m-2}\bar{\dot{y}}_{0...011}+...+\mu_{3}\cdots\mu_{m}\bar{\dot{y}}_{110...0})$
$\displaystyle+2(\mu_{1}\cdots\mu_{m-1}\bar{\dot{y}}_{m}+...+\mu_{2}\cdots\mu_{m}\bar{\dot{y}}_{1})$
$\displaystyle-
p_{1}-(\mu_{1}\cdots\mu_{m-1}\bar{\dot{y}}_{m}+...+\mu_{2}\cdots\mu_{m}\bar{\dot{y}}_{1})$
$\displaystyle-(\mu_{1}\cdots\mu_{m-2}\bar{\dot{y}}_{m-1}\bar{\dot{y}}_{m}+...+\mu_{3}\cdots\mu_{m}\bar{\dot{y}}_{1}\bar{\dot{y}}_{2})-...$
$\displaystyle=p_{1}+n^{-1/2}\left\\{2d_{00}+(r,\mu^{m-1})\right\\}-n^{-1}(r^{2},\mu^{m-2})+O(n^{-3/2}),$
(106)

and

$\displaystyle\overline{\cal Y}(1^{m-1},0^{1})$
$\displaystyle=\bar{y}_{01...1}+...+\bar{y}_{1...10}$
$\displaystyle=n^{-1}\sum_{i}(\mu_{2}+\dot{y}_{i2})\cdots(\mu_{m}+\dot{y}_{im})+...$
$\displaystyle+n^{-1}\sum_{i}(\mu_{1}+\dot{y}_{i1})\cdots(\mu_{m-1}+\dot{y}_{i,m-1})$
$\displaystyle=\mu_{2}\cdots\mu_{m}+\mu_{2}\cdots\mu_{m-1}\bar{\dot{y}}_{m}+...+\mu_{3}\cdots\mu_{m}\bar{\dot{y}}_{2}+...$
$\displaystyle+\mu_{2}\bar{\dot{y}}_{00...11}+...+\mu_{m}\bar{\dot{y}}_{01...10}+\bar{\dot{y}}_{01...1}+...$
$\displaystyle+\mu_{1}\cdots\mu_{m-1}+\mu_{1}\cdots\mu_{m-2}\bar{\dot{y}}_{m-1}+...+\mu_{2}\cdots\mu_{m-1}\bar{\dot{y}}_{1}+...$
$\displaystyle+\mu_{1}\bar{\dot{y}}_{01...10}+...+\mu_{m-1}\bar{\dot{y}}_{1...100}+\bar{\dot{y}}_{1...10}$
$\displaystyle=p_{1}\sum_{j}\mu_{j}^{-1}+n^{-1/2}H,$ (107)

where, with $\dot{r}_{12}=\sqrt{n}\bar{\dot{y}}_{110...0}$ ectetera,

$\displaystyle H$
$\displaystyle=\sum_{j_{1}<...<j_{m-2}}\mu_{j_{1}}\cdots\mu_{j_{m-2}}\sum_{k\notin(j_{1},...,j_{m-2})}r_{k}+...$
$\displaystyle+\sum_{j}\mu_{j}\sum_{k_{1}<...<k_{m-2},\forall k_{i}\neq
j}\dot{r}_{k_{1},...,k_{m-2}}+\sum_{j}\sum_{k_{1}<...<k_{m-1},\forall
k_{i}\neq j}\dot{r}_{k_{1},...,k_{m-1}}.$ (108)

Moreover, since by (36) and (49),

$\bar{e}_{2}=C_{1}=C_{10}+n^{-1/2}C_{11},$

we get from (37), (106) and (107) that

$\displaystyle d_{32}$
$\displaystyle=\left[p_{1}+n^{-1/2}\left\\{2d_{00}+(r,\mu^{m-1})\right\\}-n^{-1}(r^{2},\mu^{m-2})\right]$
$\displaystyle\cdot\left(p_{1}s_{1}+n^{-1/2}H\right)+C_{10}+n^{-1/2}C_{11}+O(n^{-3/2})$
$\displaystyle=d_{320}+n^{-1/2}d_{321}+n^{-1}d_{322}+O(n^{-3/2}),$ (109)

where

$\displaystyle d_{320}$ $\displaystyle=p_{1}^{2}s_{1}+C_{10},$ (110)
$\displaystyle d_{321}$
$\displaystyle=p_{1}H+p_{1}s_{1}\left\\{2d_{00}+(r,\mu^{m-1})\right\\}+C_{11},$
(111) $\displaystyle d_{322}$
$\displaystyle=H\left\\{2d_{00}+(r,\mu^{m-1})\right\\}-p_{1}s_{1}(r^{2},\mu^{m-2}).$
(112)

Hence, inserting (103) and (109) into (102), we get

$f_{02}=\frac{d_{0}^{2}}{2p_{1}^{2}d_{10}^{2}}(d_{320}+n^{-1/2}s_{021}+n^{-1}s_{022})+O(n^{-3/2}),$
(113)

where

$\displaystyle s_{021}$ $\displaystyle=d_{321}-2d_{320}G_{1},$ (114)
$\displaystyle s_{022}$ $\displaystyle=d_{322}-2d_{321}G_{1}+d_{320}G_{2}.$
(115)

Now, observing that from (110) and (95),

$d_{320}=-p_{1}d_{10},$ (116)

we get via (74) that

$\displaystyle E(d_{0}^{2}d_{320})$
$\displaystyle=-p_{1}^{2}d_{10}^{2}\left(1-n^{-1}d_{10}^{-1}\sum_{j_{1}<...<j_{m-2}}\mu_{j_{1}}\cdots\mu_{j_{m-2}}\right)$
$\displaystyle+O(n^{-3/2}).$ (117)

To tackle $E(d_{0}^{2}d_{321})$, we at first need $E(d_{0}^{2}H)$. To this
end, we have from (83), (86) and (88) that

$E(d_{00}^{2}H)=n^{-3/2}\sum_{i}\sum_{j}\sum_{l}E(U_{i}U_{j}H_{l})=n^{-1/2}E(\dot{S}^{2}Z),$

where, from (108),

$\displaystyle H_{l}$
$\displaystyle=\sum_{j_{1}<...<j_{m-2}}\mu_{j_{1}}\cdots\mu_{j_{m-2}}\sum_{k\notin(j_{1},...,j_{m-2})}\dot{y}_{lk}+...$
$\displaystyle+\sum_{j}\mu_{j}\sum_{k_{1}<...<k_{m-2},\forall k_{i}\neq
j}\dot{y}_{l,k_{1}}\cdots\dot{y}_{l,k_{m-2}}$
$\displaystyle+\sum_{j}\sum_{k_{1}<...<k_{m-1},\forall k_{i}\neq
j}\dot{y}_{l,k_{1}}\cdots\dot{y}_{l,k_{m-1}},$ (118)

and

$\displaystyle Z$
$\displaystyle=\sum_{j_{1}<...<j_{m-2}}\mu_{j_{1}}\cdots\mu_{j_{m-2}}\sum_{k\notin(j_{1},...,j_{m-2})}\dot{X}_{k}+...$
$\displaystyle+\sum_{j}\mu_{j}\sum_{k_{1}<...<k_{m-2},\forall k_{i}\neq
j}\dot{X}_{k_{1}}\cdots\dot{X}_{k_{m-2}}$
$\displaystyle+\sum_{j}\sum_{k_{1}<...<k_{m-1},\forall k_{i}\neq
j}\dot{X}_{k_{1}}\cdots\dot{X}_{k_{m-1}}.$ (119)

Via (58), it follows that $E(d_{0}^{2}H)$ asymptotically behaves like
$E(d_{00}^{2}H)$, i.e.

$E(d_{0}^{2}H)=n^{-1/2}E(\dot{S}^{2}Z)+O(n^{-1}).$ (120)

Moreover, from (58),

$E(d_{0}^{2}d_{00})=E(d_{00}^{3})-2n^{-1/2}E\left\\{d_{00}^{2}(r^{2},\mu^{m-2})\right\\}+O(n^{-1}),$
(121)

where via (60),

$\displaystyle E(d_{00}^{3})$
$\displaystyle=n^{3/2}E(\bar{\dot{y}}_{11...1}^{3})+E\left\\{(r^{\otimes(m-1)},\mu^{1})^{3}\right\\}+...+E\left\\{(r^{\otimes
2},\mu^{m-2})^{3}\right\\}$
$\displaystyle+3nE\left\\{\bar{\dot{y}}_{11...1}^{2}(r^{\otimes(m-1)},\mu^{1})\right\\}+...+3E\left\\{(r^{\otimes
3},\mu^{m-3})(r^{\otimes 2},\mu^{m-2})^{2}\right\\}.$ (122)

Here, as before, the ’cross product terms’ are zero, while

$\displaystyle n^{3/2}E(\bar{\dot{y}}_{11...1}^{3})$
$\displaystyle=n^{-3/2}E\left\\{\left(\sum_{i}\dot{y}_{i1}\cdots\dot{y}_{im}\right)^{3}\right\\}$
$\displaystyle=n^{-3/2}\sum_{i}\sum_{j}\sum_{k}E(\dot{y}_{i1}\cdots\dot{y}_{im}\dot{y}_{j1}\cdots\dot{y}_{jm}\dot{y}_{k1}\cdots\dot{y}_{km})$
$\displaystyle=n^{-3/2}\sum_{i}E(\dot{y}_{i1}^{3})\cdots
E(\dot{y}_{im}^{3})=n^{-1/2}\mu_{1}\cdots\mu_{m}=n^{-1/2}p_{1}$ (123)

and

$\displaystyle E\left\\{(r^{\otimes(m-k)},\mu^{k})^{3}\right\\}$
$\displaystyle=n^{3/2}E\left\\{(\mu_{1}\mu_{2}\cdots\mu_{k}\bar{\dot{y}}_{0...01...1}+...+\mu_{m-k+1}\cdots\mu_{m}\bar{\dot{y}}_{1...10...0})^{3}\right\\}$
$\displaystyle=n^{3/2}\left\\{\mu_{1}^{3}\cdots\mu_{k}^{3}E(\bar{\dot{y}}_{0...01...1}^{3})+...+\mu_{m-k+1}^{3}\cdots\mu_{m}^{3}E(\bar{\dot{y}}_{1...10...0}^{3})\right\\},$

where as in (123),

$n^{3/2}E(\bar{\dot{y}}_{1...10...0}^{3})=n^{-1/2}\mu_{1}\cdots\mu_{m-k}$

etcetera, and it follows that

$E\left\\{(r^{\otimes(m-k)},\mu^{k})^{3}\right\\}=n^{-1/2}p_{1}\sum_{j_{1}<...<j_{k}}\mu_{j_{1}}\cdots\mu_{j_{k}}.$
(124)

Conclusively, inserting (123) and (124) into (122), we find from (51) that

$E(d_{00}^{3})=n^{-1/2}p_{1}d_{10}.$ (125)

Next, (60) implies

$\displaystyle E\left\\{d_{00}^{2}(r^{2},\mu^{m-2})\right\\}$
$\displaystyle=E\left[\left\\{\sqrt{n}\bar{\dot{y}}_{11...1}+(r^{\otimes
m-1},\mu^{1})+...+(r^{\otimes
2},\mu^{m-2})\right\\}^{2}(r^{2},\mu^{m-2})\right]$
$\displaystyle=nE\left\\{\bar{\dot{y}}_{11...1}^{2}(r^{2},\mu^{m-2})\right\\}+E\left\\{(r^{\otimes
m-1},\mu^{1})^{2}(r^{2},\mu^{m-2})\right\\}+...$
$\displaystyle+E\left\\{(r^{\otimes
2},\mu^{m-2})^{2}(r^{2},\mu^{m-2})\right\\},$ (126)

since ’cross terms’ have expectation zero. Here,

$\displaystyle E\left\\{\bar{\dot{y}}_{11...1}^{2}(r^{2},\mu^{m-2})\right\\}$
$\displaystyle=E\left\\{\bar{\dot{y}}_{11...1}^{2}(\mu_{1}\cdots\mu_{m-2}r_{m-1}r_{m}+...+\mu_{3}\cdots\mu_{m}r_{1}r_{2})\right\\}$
$\displaystyle=\mu_{1}\cdots\mu_{m-2}E(\bar{\dot{y}}_{11...1}^{2}r_{m-1}r_{m})+...+\mu_{3}\cdots\mu_{m}E(\bar{\dot{y}}_{11...1}^{2}r_{1}r_{2}).$

Here,

$nE(\bar{\dot{y}}_{11...1}^{2}r_{1}r_{2})=n^{-2}\sum_{i}\sum_{j}\sum_{k}\sum_{l}E(\dot{y}_{i1}\cdots\dot{y}_{im}\dot{y}_{j1}\cdots\dot{y}_{jm}\dot{y}_{k1}\dot{y}_{l2})=O(n^{-1})$

(in fact, this is zero for $m>2$), and it is similarly seen that all other
terms of (126) are also $O(n^{-1})$. Hence,
$E\left\\{d_{00}^{2}(r^{2},\mu^{m-2})\right\\}=O(n^{-1})$, and so, via (121)
and (125), we have

$E(d_{0}^{2}d_{00})=n^{-1/2}p_{1}d_{10}+O(n^{-3/2}).$ (127)

Furthermore, we similarly have

$\displaystyle E\left\\{d_{00}^{2}(r,\mu^{m-1})\right\\}$
$\displaystyle=E\left[\left\\{\sqrt{n}\bar{\dot{y}}_{11...1}+(r^{\otimes
m-1},\mu^{1})+...+(r^{\otimes 2},\mu^{m-2})\right\\}^{2}(r,\mu^{m-1})\right]$
$\displaystyle=nE\left\\{\bar{\dot{y}}_{11...1}^{2}(r,\mu^{m-1})\right\\}+E\left\\{(r^{\otimes
m-1},\mu^{1})^{2}(r,\mu^{m-1})\right\\}+...$
$\displaystyle+E\left\\{(r^{\otimes 2},\mu^{m-2})^{2}(r,\mu^{m-1})\right\\},$
(128)

where

$\displaystyle E\left\\{\bar{\dot{y}}_{11...1}^{2}(r,\mu^{m-1})\right\\}$
$\displaystyle=E\left\\{\bar{\dot{y}}_{11...1}^{2}(\mu_{1}\cdots\mu_{m-1}r_{m}+...+\mu_{2}\cdots\mu_{m}r_{1})\right\\}$
$\displaystyle=\mu_{1}\cdots\mu_{m-1}E(\bar{\dot{y}}_{11...1}^{2}r_{m})+...+\mu_{2}\cdots\mu_{m}E(\bar{\dot{y}}_{11...1}^{2}r_{1}),$

where (77) yields

$nE(\bar{\dot{y}}_{11...1}^{2}r_{1})=n^{-1/2}p_{1}.$

It follows that

$nE\left\\{\bar{\dot{y}}_{11...1}^{2}(r,\mu^{m-1})\right\\}=n^{-1/2}p_{1}\sum_{j_{1}<...<j_{m-1}}\mu_{j_{1}}\cdots\mu_{j_{m-1}}.$
(129)

Moreover, for $k=1,...,m-2$, we have

$\displaystyle E\left\\{(r^{\otimes m-k},\mu^{k})^{2}(r,\mu^{m-1})\right\\}$
$\displaystyle=n^{3/2}E\left\\{(\mu_{1}\cdots\mu_{k}\bar{\dot{y}}_{0...01...1}+...+\mu_{m-k+1}\cdots\mu_{m}\bar{\dot{y}}_{1...10...0})^{2}\right.$
$\displaystyle\left.(\mu_{1}\cdots\mu_{m-1}\bar{\dot{y}}_{m}+...+\mu_{2}\cdots\mu_{m}\bar{\dot{y}}_{1})\right\\}$

where from (78),

$n^{3/2}E(\bar{\dot{y}}_{1...10...0}^{2}\bar{\dot{y}}_{1})=n^{-1/2}\mu_{1}\cdots\mu_{m-k}.$

This type of results imply

$\displaystyle E\left\\{(r^{\otimes m-k},\mu^{k})^{2}(r,\mu^{m-1})\right\\}$
$\displaystyle=n^{-1/2}\left(\mu_{2}\cdots\mu_{m}\sum_{1<l_{1}<...<l_{k}}\mu_{l_{1}}^{2}\cdots\mu_{l_{k}}^{2}\prod_{k\notin(l_{1},...,l_{k})}\mu_{k}\right.$
$\displaystyle\left.+...+\mu_{1}\cdots\mu_{m-1}\sum_{l_{1}<...<l_{k}<m}\mu_{l_{1}}^{2}\cdots\mu_{l_{k}}^{2}\prod_{k\notin(l_{1},...,l_{k})}\mu_{k}\right)$
$\displaystyle=n^{-1/2}p_{1}\sum_{j_{1}<...<j_{m-1}}\mu_{j_{1}}\cdots\mu_{j_{m-1}}\sum_{l_{1}<...<l_{k},\forall
l_{j}\in(j_{1},...,j_{m-1})}\mu_{l_{1}}\cdots\mu_{l_{k}}.$

Hence, via (128) and (129),

$\displaystyle E\left\\{d_{00}^{2}(r,\mu^{m-1})\right\\}$
$\displaystyle=n^{-1/2}p_{1}\sum_{j_{1}<...<j_{m-1}}\mu_{j_{1}}\cdots\mu_{j_{m-1}}\left(1+\sum_{k=1}^{m-2}\sum_{l_{1}<...<l_{k},\forall
l_{j}\in(j_{1},...,j_{m-1})}\mu_{l_{1}}\cdots\mu_{l_{k}}\right)$
$\displaystyle=n^{-1/2}\eta.$

Here, an alternative expression more suitable for future calculations is

$\eta=p_{1}^{2}\sum_{j}\mu_{j}^{-1}\left\\{\prod_{k\neq
j}(1+\mu_{k})-\prod_{k\neq j}\mu_{k}\right\\}.$ (130)

In view of (58), this also means that

$E\left\\{d_{0}^{2}(r,\mu^{m-1})\right\\}=n^{-1/2}\eta+O(n^{-1}),$ (131)

because in the usual manner, it can be seen that

$E\\{d_{00}(r^{2},\mu^{m-2})(r,\mu^{m-1})\\}=O(n^{-1}).$

Now, from (111),

$\displaystyle E(d_{0}^{2}d_{321})$
$\displaystyle=p_{1}E(d_{0}^{2}H)+p_{1}s_{1}\left[2E(d_{0}^{2}d_{00})+E\\{d_{0}^{2}(r,\mu^{m-1})\\}\right]+E(d_{0}^{2}C_{11}),$

and via (120), (127), (131), (87) and (58), we get

$\displaystyle E(d_{0}^{2}d_{321})$
$\displaystyle=n^{-1/2}\left\\{p_{1}E(\dot{S}^{2}Z)\right.$
$\displaystyle\left.+p_{1}s_{1}(2p_{1}d_{10}+\eta)+E(\dot{S}^{2}\dot{X})\right\\}+O(n^{-1}).$
(132)

Next, via (104), (82) and (89), we have

$\displaystyle E(d_{0}^{2}G_{1})$
$\displaystyle=\sum_{j}\mu_{j}^{-1}E(d_{0}^{2}r_{j})+d_{10}^{-1}E(d_{0}^{2}d_{11})$
$\displaystyle=n^{-1/2}\left[p_{1}\sum_{j}\mu_{j}^{-1}\xi_{j}+d_{10}^{-1}\left\\{C_{10}\sum_{j}\mu_{j}^{-1}\xi_{j}-p_{1}^{-1}E\left({\dot{S}}^{2}\dot{X}\right)\right.\right.$
$\displaystyle\left.\left.-p_{1}^{2}\left(s_{1}\sum_{j}\mu_{j}^{-1}\xi_{j}-\sum_{j}\mu_{j}^{-2}\xi_{j}\right)\right\\}\right]+O(n^{-1}).$

Here, from (95), we have

$C_{10}=-p_{1}d_{10}-p_{1}^{2}s_{1},$

which yields

$\displaystyle E(d_{0}^{2}G_{1})$
$\displaystyle=-n^{-1/2}d_{10}^{-1}\left\\{p_{1}^{2}\left(2s_{1}\sum_{j}\mu_{j}^{-1}\xi_{j}-\sum_{j}\mu_{j}^{-2}\xi_{j}\right)+p_{1}^{-1}E\left(\dot{S}^{2}\dot{X}\right)\right\\}$
$\displaystyle+O(n^{-1}).$

Hence, together with (132), (111) and (116), we have

$\displaystyle E(d_{0}^{2}s_{021})$
$\displaystyle=E(d_{0}^{2}d_{321})+2p_{1}d_{10}E(d_{0}^{2}G_{1})$
$\displaystyle=n^{-1/2}\left\\{p_{1}E(\dot{S}^{2}Z)+p_{1}s_{1}(2p_{1}d_{10}+\eta)+E(\dot{S}^{2}\dot{X})\right.$
$\displaystyle\left.-2p_{1}^{3}\left(2s_{1}\sum_{j}\mu_{j}^{-1}\xi_{j}-\sum_{j}\mu_{j}^{-2}\xi_{j}\right)-2E(\dot{S}^{2}\dot{X})\right\\}+O(n^{-1}).$
(133)

To find $E(d_{0}^{2}s_{022})$, then in view of (115), we at first need
$E(d_{0}^{2}d_{322})$. In view of (112), we start with $E(d_{0}^{3}H)$. From
(83), (86), (88), (118) and (119),

$E(d_{00}^{3}H)=n^{-2}\sum_{i}\sum_{j}\sum_{k}\sum_{l}E(U_{i}U_{j}U_{k}H_{l})=3E(\dot{S}^{2})E(\dot{S}Z)+O(n^{-1}),$

and because by (58), we have

$E(d_{0}^{2}d_{00}H)=3E(\dot{S}^{2})E(\dot{S}Z)+O(n^{-1}).$ (134)

Similarly, we get

$\displaystyle E\\{d_{00}^{2}H(r,\mu^{m-1})\\}$
$\displaystyle=n^{-2}\sum_{i}\sum_{j}\sum_{k}\sum_{l}E\\{U_{i}U_{j}H_{k}(\mu_{1}\cdots\mu_{m-1}\dot{y}_{lm}+...+\mu_{2}\cdots\mu_{m}\dot{y}_{l1})\\}$
$\displaystyle=E(\dot{S}^{2})E(ZV),$

where

$V=\mu_{1}\cdots\mu_{m-1}\dot{X}_{m}+...+\mu_{2}\cdots\mu_{m}\dot{X}_{1},$
(135)

implying

$E\\{d_{0}^{2}H(r,\mu^{m-1})\\}=E(\dot{S}^{2})E(ZV)+O(n^{-1}).$ (136)

Moreover,

$\displaystyle E\\{d_{00}^{2}(r^{2},\mu^{m-2})\\}$
$\displaystyle=n^{-2}\sum_{i}\sum_{j}\sum_{k}\sum_{l}E\\{U_{i}U_{j}(\mu_{1}\cdots\mu_{m-2}\dot{y}_{k,m-1}\dot{y}_{l,m}+...+\mu_{2}\cdots\mu_{m}\dot{y}_{k1}\dot{y}_{l2})\\}$
$\displaystyle=O(n^{-1}),$

implying $E\\{d_{0}^{2}(r^{2},\mu^{m-2})\\}=O(n^{-1})$. Hence, via (112),
(134) and (136), we get

$E(d_{0}^{2}d_{322})=E(\dot{S}^{2})\\{6E(\dot{S}Z)+E(ZV)\\}+O(n^{-1}).$ (137)

As for $E(d_{0}^{2}d_{321}G_{1})$, we start with $E(d_{0}^{2}d_{321}r_{1})$,
and to this end, we have in the usual manner

$\displaystyle E(d_{00}^{2}Hr_{1})$
$\displaystyle=n^{-2}\sum_{i}\sum_{j}\sum_{k}\sum_{l}E(U_{i}U_{j}H_{k}\dot{y}_{l1})$
$\displaystyle=E(\dot{S}^{2})E(Z\dot{X}_{1})+O(n^{-1}),$ (138)
$E(d_{00}^{3}r_{1})=n^{-2}\sum_{i}\sum_{j}\sum_{k}\sum_{l}E(U_{i}U_{j}U_{k}\dot{y}_{l1})=0,$
(139)

and

$\displaystyle E\\{d_{00}^{2}(r,\mu^{m-1})r_{1}\\}$
$\displaystyle=n^{-2}\sum_{i}\sum_{j}\sum_{k}\sum_{l}E\\{U_{i}U_{j}(\mu_{1}\cdots\mu_{m-1}\dot{y}_{km}+...+\mu_{2}\cdots\mu_{m}\dot{y}_{m1})\dot{y}_{l1}\\}$
$\displaystyle=p_{1}E(\dot{S}^{2})+O(n^{-1}),$ (140)

which together with (92) and (111) yield

$\displaystyle E(d_{0}^{2}d_{321}r_{1})$
$\displaystyle=E(\dot{S}^{2})\\{p_{1}E(Z\dot{X}_{1})+p_{1}^{2}s_{1}+E(\dot{X}\dot{X}_{1})\\}+O(n^{-1})$
$\displaystyle=E(\dot{S}^{2})\\{p_{1}E(\tilde{Z}\dot{X}_{1})+p_{1}^{2}s_{1}\\},$
(141)

where $\tilde{Z}=Z+p_{1}^{-1}\dot{X}$. Similarly, to find
$E(d_{0}^{2}d_{321}d_{11})$, we need, by (52),

$\displaystyle E(d_{00}^{2}Hd_{11})$
$\displaystyle=p_{1}^{-1}\left\\{C_{10}\sum_{j}\mu_{j}^{-1}E(d_{00}^{2}Hr_{j})-E(d_{00}^{2}HC_{11})\right\\}$
$\displaystyle-
p_{1}\left\\{s_{1}\sum_{j}\mu_{j}^{-1}E(d_{00}^{2}Hr_{j})-\sum_{j}\mu_{j}^{-2}E(d_{00}^{2}Hr_{j})\right\\}.$
(142)

Here, we have in the usual manner that

$\displaystyle E(d_{00}^{2}HC_{11})$
$\displaystyle=n^{-2}\sum_{i}\sum_{j}\sum_{k}\sum_{l}E\\{U_{i}U_{j}H_{k}\dot{(}C_{l1}-C_{10})\\}$
$\displaystyle=E(\dot{S}^{2})E(Z\dot{X})+2E(\dot{S}Z)E(\dot{S}\dot{X})+O(n^{-1}),$
(143)

implying via (138) and (142) that

$\displaystyle E(d_{0}^{2}Hd_{11})$
$\displaystyle=p_{1}^{-1}E(\dot{S}^{2})\left\\{C_{10}\sum_{j}\mu_{j}^{-1}E(Z\dot{X}_{j})-E(Z\dot{X})\right\\}$
$\displaystyle-2p_{1}^{-1}E(\dot{S}Z)E(\dot{S}\dot{X})$ $\displaystyle-
p_{1}E(\dot{S}^{2})\left\\{s_{1}\sum_{j}\mu_{j}^{-1}E(Z\dot{X}_{j})-\sum_{j}\mu_{j}^{-2}E(Z\dot{X}_{j})\right\\}$
$\displaystyle+O(n^{-1}).$ (144)

Similarly, via (52) and (139), we have

$E(d_{00}^{3}d_{11})=-p_{1}^{-1}E(d_{00}^{3}C_{11}),$

where

$\displaystyle E(d_{00}^{3}C_{11})$
$\displaystyle=n^{-2}\sum_{i}\sum_{j}\sum_{k}\sum_{l}E\\{U_{i}U_{j}U_{k}(C_{l1}-C_{10})\\}$
$\displaystyle=3E(\dot{S}^{2})E(\dot{S}\dot{X})+O(n^{-1}),$ (145)

and so,

$E(d_{0}^{2}d_{00}d_{11})=-3p_{1}^{-1}E(\dot{S}^{2})E(\dot{S}\dot{X})+O(n^{-1}).$
(146)

To find $E\\{d_{00}^{2}(r,\mu^{m-1})d_{11}\\}$, we need (cf (135) and (136))

$\displaystyle E\\{d_{00}^{2}(r,\mu^{m-1})C_{11}\\}$
$\displaystyle=n^{-2}\sum_{i}\sum_{j}\sum_{k}\sum_{l}E\\{U_{i}U_{j}(\mu_{1}\cdots\mu_{m-1}\dot{y}_{km}+...+\mu_{2}\cdots\mu_{m}\dot{y}_{k1})(C_{l1}-C_{10})\\}$
$\displaystyle=E(\dot{S}^{2})E(V\dot{X})+O(n^{-1}),$

Together with (52) and (140), this gives

$\displaystyle E\\{d_{00}^{2}(r,\mu^{m-1})d_{11}\\}$
$\displaystyle=p_{1}^{-1}E(\dot{S}^{2})\left\\{C_{10}p_{1}s_{1}-E(V\dot{X})\right\\}$
$\displaystyle-p_{1}^{2}E(\dot{S}^{2})(s_{1}^{2}-s_{2})+O(n^{-1}).$

and because from (95), $C_{10}=-p_{1}d_{10}-p_{1}^{2}s_{1}$ which implies

$\displaystyle E\\{d_{00}^{2}(r,\mu^{m-1})d_{11}\\}$
$\displaystyle=E(\dot{S}^{2})\left\\{-p_{1}d_{10}s_{1}-p_{1}^{-1}E(V\dot{X})-2p_{1}^{2}s_{1}^{2}+p_{1}^{2}s_{2}\right\\}$
$\displaystyle+O(n^{-1}).$ (147)

As for $E(d_{00}^{2}C_{11}d_{11})$, we have

$\displaystyle E(d_{00}^{2}C_{11}^{2})$
$\displaystyle=n^{-2}\sum_{i}\sum_{j}\sum_{k}\sum_{l}E\\{U_{i}U_{j}(C_{k1}-C_{10})(C_{l1}-C_{10})\\}$
$\displaystyle=E(\dot{S}^{2})E(\dot{X}^{2})+2\\{E(\dot{S}\dot{X})\\}^{2}+O(n^{-1}),$
(148)

implying via (92) and (52) that

$\displaystyle E(d_{00}^{2}C_{11}d_{11})$
$\displaystyle=p_{1}^{-1}E(\dot{S}^{2})\left\\{C_{10}\sum_{j}\mu_{j}^{-1}E(\dot{X}\dot{X}_{j})-E(\dot{X}^{2})\right\\}-2p_{1}^{-1}\left\\{E(\dot{S}\dot{X})\right\\}^{2}$
$\displaystyle-
p_{1}E(\dot{S}^{2})\left\\{s_{1}\sum_{j}\mu_{j}^{-1}E(\dot{X}\dot{X}_{j})-\sum_{j}\mu_{j}^{-2}E(\dot{X}\dot{X}_{j})\right\\}+O(n^{-1}).$

This, together with (111), (144), (146), (147), (95) and simplifications
yields

$\displaystyle E(d_{0}^{2}d_{321}d_{11})$
$\displaystyle=p_{1}E(d_{0}^{2}Hd_{11})+p_{1}s_{1}\left[2E(d_{0}^{2}d_{00}d_{11})+E\\{d_{0}^{2}(r,\mu^{m-1})d_{11}\\}\right]$
$\displaystyle+E(d_{0}^{2}C_{11}d_{11})$
$\displaystyle=E(\dot{S}^{2})\left\\{C_{10}\sum_{j}\mu_{j}^{-1}E(\tilde{Z}\dot{X}_{j})-E(\tilde{Z}\dot{X})\right\\}-2E(\dot{S}\tilde{Z})E(\dot{S}\dot{X})$
$\displaystyle-
p_{1}^{2}E(\dot{S}^{2})\left\\{s_{1}\sum_{j}\mu_{j}^{-1}E(\tilde{Z}\dot{X}_{j})-\sum_{j}\mu_{j}^{-2}E(\tilde{Z}\dot{X}_{j})\right\\}$
$\displaystyle+s_{1}E(\dot{S}^{2})\left\\{-6E(\dot{S}\dot{X})-d_{10}p_{1}^{2}s_{1}-E(V\dot{X})-2p_{1}^{3}s_{1}^{2}+p_{1}^{3}s_{2}\right\\}$
$\displaystyle+O(n^{-1})$
$\displaystyle=-2E(\dot{S}\tilde{Z})E(\dot{S}\dot{X})+E(\dot{S}^{2})\left\\{(-p_{1}d_{10}-2p_{1}^{2}s_{1})\sum_{j}\mu_{j}^{-1}E(\tilde{Z}\dot{X}_{j})\right.$
$\displaystyle-E(\tilde{Z}\dot{X})+p_{1}^{2}\sum_{j}\mu_{j}^{-2}E(\tilde{Z}\dot{X}_{j})-6s_{1}E(\dot{S}\dot{X})$
$\displaystyle\left.-d_{10}p_{1}^{2}s_{1}^{2}-s_{1}E(V\dot{X})-2p_{1}^{3}s_{1}^{3}+p_{1}^{3}s_{1}s_{2}\right\\}+O(n^{-1}).$
(149)

Hence, using (104) and (141),

$\displaystyle E(d_{0}^{2}d_{321}G_{1})$
$\displaystyle=\sum_{j}\mu_{j}^{-1}E(d_{0}^{2}d_{321}r_{j})+d_{10}^{-1}E(d_{0}^{2}d_{321}d_{11})$
$\displaystyle=E(\dot{S}^{2})\left\\{p_{1}\sum_{j}\mu_{j}^{-1}E(\tilde{Z}\dot{X}_{j})+p_{1}^{2}s_{1}^{2}\right\\}+d_{10}^{-1}E(d_{0}^{2}d_{321}d_{11})$
$\displaystyle+O(n^{-1})$
$\displaystyle=-2d_{10}^{-1}E(\dot{S}\tilde{Z})E(\dot{S}\dot{X})+d_{10}^{-1}E(\dot{S}^{2})\left\\{-2p_{1}^{2}s_{1}\sum_{j}\mu_{j}^{-1}E(\tilde{Z}\dot{X}_{j})-E(\tilde{Z}\dot{X})\right.$
$\displaystyle\left.+p_{1}^{2}\sum_{j}\mu_{j}^{-2}E(\tilde{Z}\dot{X}_{j})-6s_{1}E(\dot{S}\dot{X})-s_{1}E(V\dot{X})-2p_{1}^{3}s_{1}^{3}+p_{1}^{3}s_{1}s_{2}\right\\}$
$\displaystyle+O(n^{-1}).$ (150)

The next term needed is $E(d_{0}^{2}G_{2})$. To this end, we have in the usual
manner that

$E(d_{00}^{2}r_{1}r_{2})=n^{-2}\sum_{i}\sum_{j}\sum_{k}\sum_{l}E(U_{i}U_{j}\dot{y}_{k1}\dot{y}_{l2})=O(n^{-1}).$
(151)

Moreover, from (52), (91), (92) and (151),

$\displaystyle E(d_{0}^{2}d_{11}r_{1})$
$\displaystyle=p_{1}^{-1}\left\\{C_{10}\mu_{1}^{-1}E(d_{0}^{2}r_{1}^{2})-E(d_{0}^{2}C_{11}r_{1})\right\\}$
$\displaystyle-
p_{1}\left\\{s_{1}\mu_{1}^{-1}E(d_{0}^{2}r_{1}^{2})-\mu_{1}^{-2}E(d_{0}^{2}r_{1}^{2})\right\\}+O(n^{-1})$
$\displaystyle=E(\dot{S}^{2})\left[p_{1}^{-1}\left\\{C_{10}-E(\dot{X}\dot{X}_{1})\right\\}-p_{1}\left(s_{1}-\mu_{1}^{-1}\right)\right]+O(n^{-1}).$
(152)

Now, via (105) and (151),

$\displaystyle E(d_{0}^{2}G_{2})$
$\displaystyle=3\sum_{j}\mu_{j}^{-2}E(d_{00}^{2}r_{j}^{2})+4d_{10}^{-1}\sum_{j}\mu_{j}^{-1}E(d_{00}^{2}d_{11}r_{j})$
$\displaystyle+3d_{10}^{-2}E(d_{00}^{2}d_{11}^{2})-2d_{10}^{-1}E(d_{00}^{2}d_{12})+O(n^{-1}),$

and from (91), (152), (97), (98) and simplifications, we get

$\displaystyle E(d_{0}^{2}G_{2})$
$\displaystyle=3E(\dot{S}^{2})s_{1}+d_{10}^{-1}E(\dot{S}^{2})\left\\{6p_{1}^{-1}s_{1}C_{10}-6p_{1}^{-1}A_{1}\right.$
$\displaystyle\left.-4p_{1}(s_{1}^{2}-s_{2})\right\\}+3d_{10}^{-2}E(d_{00}^{2}d_{11}^{2})+O(n^{-1}).$
(153)

Moreover, (94) and (97) yield

$\displaystyle d_{10}^{-2}E(d_{00}^{2}d_{11}^{2})$
$\displaystyle=2d_{10}^{-2}p_{1}^{-2}\left\\{E(\dot{S}\dot{X})\right\\}^{2}+E(\dot{S}^{2})\left\\{d_{10}^{-2}p_{1}^{-2}E(\dot{X}^{2})+2d_{10}^{-1}p_{1}^{-1}A_{1}\right.$
$\displaystyle+4d_{10}^{-2}s_{1}A_{1}+s_{1}+4d_{10}^{-1}p_{1}s_{1}^{2}+4d_{10}^{-2}p_{1}^{2}s_{1}^{3}-2d_{10}^{-2}A_{2}+d_{10}^{-2}p_{1}^{2}s_{3}$
$\displaystyle\left.-2d_{10}^{-1}p_{1}s_{2}-4d_{10}^{-2}p_{1}^{2}s_{1}s_{2}\right\\}+O(n^{-1/2}),$

and together with (153), (95) and simplifications, we find

$\displaystyle E(d_{0}^{2}G_{2})$
$\displaystyle=6d_{10}^{-2}p_{1}^{-2}\left\\{E(\dot{S}\dot{X})\right\\}^{2}+E(\dot{S}^{2})\left\\{6s_{1}+6d_{10}^{-1}p_{1}^{-1}s_{1}C_{10}+8d_{10}^{-1}p_{1}s_{1}^{2}\right.$
$\displaystyle-2d_{10}^{-1}p_{1}s_{2}+3d_{10}^{-2}p_{1}^{-2}E(\dot{X}^{2})+12d_{10}^{-2}s_{1}A_{1}+12d_{10}^{-2}p_{1}^{2}s_{1}^{3}$
$\displaystyle\left.-6d_{10}^{-2}A_{2}+3d_{10}^{-2}p_{1}^{2}s_{3}-12d_{10}^{-2}p_{1}^{2}s_{1}s_{2}\right\\}+O(n^{-1/2})$
$\displaystyle=6d_{10}^{-2}p_{1}^{-2}\left\\{E(\dot{S}\dot{X})\right\\}^{2}+E(\dot{S}^{2})\left\\{2d_{10}^{-1}p_{1}(s_{1}^{2}-s_{2})\right.$
$\displaystyle+3d_{10}^{-2}p_{1}^{-2}E(\dot{X}^{2})+12d_{10}^{-2}s_{1}A_{1}+12d_{10}^{-2}p_{1}^{2}s_{1}^{3}$
$\displaystyle\left.-6d_{10}^{-2}A_{2}+3d_{10}^{-2}p_{1}^{2}s_{3}-12d_{10}^{-2}p_{1}^{2}s_{1}s_{2}\right\\}+O(n^{-1/2}).$
(154)

Now, by (115), (137), (150), (116) and (154),

$\displaystyle E(d_{0}^{2}s_{022})$
$\displaystyle=E(d_{0}^{2}d_{322})-2E(d_{0}^{2}d_{321}G_{1})+d_{320}E(d_{0}^{2}G_{2})$
$\displaystyle=E(\dot{S}^{2})\\{6E(\dot{S}Z)+E(ZV)\\}+4d_{10}^{-1}E(\dot{S}\tilde{Z})E(\dot{S}\dot{X})$
$\displaystyle-2d_{10}^{-1}E(\dot{S}^{2})\left\\{-2p_{1}^{2}s_{1}\sum_{j}\mu_{j}^{-1}E(\tilde{Z}\dot{X}_{j})-E(\tilde{Z}\dot{X})\right.$
$\displaystyle\left.+p_{1}^{2}\sum_{j}\mu_{j}^{-2}E(\tilde{Z}\dot{X}_{j})-6s_{1}E(\dot{S}\dot{X})-s_{1}E(V\dot{X})-2p_{1}^{3}s_{1}^{3}+p_{1}^{3}s_{1}s_{2}\right\\}$
$\displaystyle-6d_{10}^{-1}p_{1}^{-1}\left\\{E(\dot{S}\dot{X})\right\\}^{2}$
$\displaystyle-
d_{10}^{-1}E(\dot{S}^{2})\left\\{2d_{10}p_{1}^{2}(s_{1}^{2}-s_{2})+3p_{1}^{-1}E(\dot{X}^{2})+12p_{1}s_{1}A_{1}+12p_{1}^{3}s_{1}^{3}\right.$
$\displaystyle\left.-6p_{1}A_{2}+3p_{1}^{3}s_{3}-12p_{1}^{3}s_{1}s_{2}\right\\}+O(n^{-1/2}),$

and using $\tilde{Z}=Z+p_{1}^{-1}\dot{X}$, defining for $k=1,2$,

$\tilde{A}_{k}=\sum_{j}\mu_{j}^{-k}E(\tilde{Z}\dot{X}_{j})-3p_{1}^{-1}A_{k}=\sum_{j}\mu_{j}^{-k}E\\{(Z-2p_{1}^{-1}\dot{X})\dot{X}_{j}\\},$
(155)

we get after some simplification

$\displaystyle E(d_{0}^{2}s_{022})$
$\displaystyle=E(\dot{S}^{2})\\{6E(\dot{S}Z)+E(ZV)\\}+4d_{10}^{-1}E(\dot{S}Z)E(\dot{S}\dot{X})$
$\displaystyle-2d_{10}^{-1}p_{1}^{-1}\left\\{E(\dot{S}\dot{X})\right\\}^{2}-d_{10}^{-1}E(\dot{S}^{2})\tilde{F}+O(n^{-1/2}),$
(156)

where

$\displaystyle\tilde{F}$
$\displaystyle=-4p_{1}^{2}s_{1}\tilde{A}_{1}-2E(Z\dot{X})+p_{1}^{-1}E(\dot{X}^{2})+2p_{1}^{2}\tilde{A}_{2}$
$\displaystyle-12s_{1}E(\dot{S}\dot{X})-2s_{1}E(V\dot{X})+2d_{10}p_{1}^{2}(s_{1}^{2}-s_{2})$
$\displaystyle+p_{1}^{3}(8s_{1}^{3}-10s_{1}s_{2}+3s_{3}).$ (157)

To get a final expression for $E(f_{0})=E(f_{01})+E(f_{02})$ (see (55)), then
in view of (101), (113) and (116), we need

$\left(p_{1}^{-1}d_{10}^{-1}+\frac{1}{2}p_{1}^{-2}d_{10}^{-2}d_{320}\right)E(d_{0}^{2})=\frac{1}{2}p_{1}^{-1}d_{10}^{-1}E(d_{0}^{2}),$

and via (74), this is

$\displaystyle\left(p_{1}^{-1}d_{10}^{-1}+\frac{1}{2}p_{1}^{-2}d_{10}^{-2}d_{320}\right)E(d_{0}^{2})$
$\displaystyle=\frac{1}{2}-\frac{1}{2}n^{-1}d_{10}^{-1}\sum_{j_{1}<...<j_{m-2}}\mu_{j_{1}}\cdots\mu_{j_{m-2}}+O(n^{-2}).$
(158)

Moreover, via (90), (133) and simplifications,

$\displaystyle
n^{1/2}p_{1}^{-2}d_{10}^{-2}\left\\{p_{1}d_{10}E(d_{0}^{2}s_{011})+\frac{1}{2}E(d_{0}^{2}s_{021})\right\\}$
$\displaystyle=p_{1}^{-2}d_{10}^{-2}\left\\{\frac{1}{2}E(\dot{S}^{2}\dot{X})+\frac{1}{2}p_{1}E(\dot{S}^{2}Z)+p_{1}^{2}s_{1}d_{10}+\frac{1}{2}p_{1}s_{1}\eta\right\\}$
$\displaystyle+O(n^{-1/2}).$ (159)

Furthermore, via (99), (156) and a little simplification, we get

$\displaystyle
p_{1}^{-2}d_{10}^{-2}\left\\{p_{1}d_{10}E(d_{0}^{2}s_{012})+\frac{1}{2}E(d_{0}^{2}s_{022})\right\\}$
$\displaystyle=p_{1}^{-2}d_{10}^{-2}\left[p_{1}^{-1}d_{10}^{-1}\left\\{E(\dot{S}\dot{X})\right\\}^{2}+3E(\dot{S}^{2})E(\dot{S}Z)+\frac{1}{2}E(\dot{S}^{2})E(ZV)\right.$
$\displaystyle\left.+2d_{10}^{-1}E(\dot{S}Z)E(\dot{S}\dot{X})+d_{10}^{-1}E(\dot{S}^{2})\left\\{p_{1}d_{10}^{2}F-\frac{1}{2}\tilde{F}\right\\}\right]+O(n^{-1/2}),$
(160)

where via (100), (157) and simplification,

$\displaystyle p_{1}d_{10}^{2}F-\frac{1}{2}\tilde{F}$
$\displaystyle=\frac{1}{2}p_{1}^{-1}E(\dot{X}^{2})+E(Z\dot{X})+6s_{1}E(\dot{S}\dot{X})+s_{1}E(V\dot{X})$
$\displaystyle+2p_{1}^{2}s_{1}\sum_{j}\mu_{j}^{-1}E(Z\dot{X}_{j})-p_{1}^{2}\sum_{j}\mu_{j}^{-2}E(Z\dot{X}_{j})$
$\displaystyle+p_{1}^{3}s_{1}s_{2}-\frac{1}{2}p_{1}^{3}s_{3}.$ (161)

Hence, we may conclude from (55), (101), (113) and (158)-(161) that

$E(f_{0})=\frac{1}{2}+n^{-1}\rho_{0}+O(n^{-3/2}),$ (162)

where

$\displaystyle\rho_{0}$
$\displaystyle=-\frac{1}{2}d_{10}^{-1}\sum_{j_{1}<...<j_{m-2}}\mu_{j_{1}}\cdots\mu_{j_{m-2}}+p_{1}^{-2}d_{10}^{-2}\cdot$
$\displaystyle\left[\frac{1}{2}E(\dot{S}^{2}\dot{X})+\frac{1}{2}p_{1}E(\dot{S}^{2}Z)+p_{1}^{2}s_{1}d_{10}+\frac{1}{2}p_{1}s_{1}\eta\right.$
$\displaystyle+p_{1}^{-1}d_{10}^{-1}\left\\{E(\dot{S}\dot{X})\right\\}^{2}+3E(\dot{S}^{2})E(\dot{S}Z)+\frac{1}{2}E(\dot{S}^{2})E(ZV)$
$\displaystyle+2d_{10}^{-1}E(\dot{S}Z)E(\dot{S}\dot{X})+d_{10}^{-1}E(\dot{S}^{2})\cdot$
$\displaystyle\left\\{\frac{1}{2}p_{1}^{-1}E(\dot{X}^{2})+E(Z\dot{X})+6s_{1}E(\dot{S}\dot{X})+s_{1}E(V\dot{X})\right.$
$\displaystyle+2p_{1}^{2}s_{1}\sum_{j}\mu_{j}^{-1}E(Z\dot{X}_{j})-p_{1}^{2}\sum_{j}\mu_{j}^{-2}E(Z\dot{X}_{j})$
$\displaystyle\left.\left.+p_{1}^{3}s_{1}s_{2}-\frac{1}{2}p_{1}^{3}s_{3}\right\\}\right].$
(163)

To get further, it is time to evaluate the expectation terms in (163). As for
$\dot{S}$, it turns out easiest to replace the formula in (88) by the
equivalent

$\displaystyle\dot{S}$ $\displaystyle=\prod_{j}X_{j}-\dot{X}_{1}\prod_{j\neq
1}\mu_{j}-...-\dot{X}_{m}\prod_{j\neq m}\mu_{j}-\prod_{j}\mu_{j}$
$\displaystyle=\prod_{j}X_{j}-p_{1}\left(1+\sum_{j}\mu_{j}^{-1}\dot{X}_{j}\right).$
(164)

This gives

$\displaystyle E(\dot{S}^{2})$
$\displaystyle=\prod_{j}E(X_{j}^{2})+p_{1}^{2}\left\\{1+\sum_{j}\mu_{j}^{-2}E(\dot{X}_{j}^{2})\right\\}$
$\displaystyle-2p_{1}\left\\{\prod_{j}E(X_{j})+\sum_{j}\mu_{j}^{-1}E\left(\dot{X}_{j}\prod_{k}X_{k}\right)\right\\}$
$\displaystyle=p_{1}\prod_{j}(1+\mu_{j})-p_{1}^{2}(1+s_{1})$
$\displaystyle=p_{1}d_{10},$ (165)

comparing to (51) and

$C_{10}=E(X)=p_{1}\left\\{p_{1}-\prod_{j}(1+\mu_{j})\right\\},$ (166)

in the last step.

Moreover, because $X$ in (85) is derived from $C_{1}$ in (17), we may write

$X=\prod_{j}X_{j}\left\\{\prod_{j}(X_{j}-1)-\prod_{j}X_{j}\right\\}=\prod_{j}X_{j}(X_{j}-1)-\prod_{j}X_{j}^{2},$
(167)

To calculate $E(\dot{X}^{2})=E(X^{2})-C_{10}^{2}$ (recall that $C_{10}=E(X)$),
we need

$E(X^{2})=\prod_{j}E\left\\{X_{j}^{2}(X_{j}-1)^{2}\right\\}+\prod_{j}E(X_{j}^{4})-2\prod_{j}E\left\\{X_{j}^{3}(X_{j}-1)\right\\},$

where knowledge of noncentral Poisson moments and simplifications leads to

$E(X^{2})=p_{1}^{2}p_{241}+p_{1}p_{1761}-2p_{1}^{2}p_{452},$

letting

$p_{abc}\stackrel{{\scriptstyle\rm{def}}}{{=}}\prod_{j}(a+b\mu_{j}+c\mu_{j}^{2}+d\mu_{j}^{3}),$
(168)

for integers $a,b,c,d$, where $p_{a,b,c,0}=p_{abc}$, etcetera.. Via (166), it
follows that

$\displaystyle E(\dot{X}^{2})$
$\displaystyle=p_{1}\left\\{p_{1}p_{241}+p_{1761}-2p_{1}p_{452}-p_{1}(p_{1}-p_{11})^{2}\right\\}.$
(169)

Next, we work out $E(\dot{S}\dot{X})=E(\dot{S}X)$. To this end, (164) and
(167) yield

$\displaystyle E(\dot{S}\dot{X})$
$\displaystyle=\prod_{j}E\left\\{X_{j}^{2}(X_{j}-1)\right\\}-\prod_{j}E(X_{j}^{3})$
$\displaystyle-
p_{1}\left[\prod_{j}E\left\\{X_{j}(X_{j}-1)\right\\}+\sum_{j}\mu_{j}^{-1}E\left\\{\dot{X}_{j}\prod_{k}X_{k}(X_{k}-1)\right\\}\right]$
$\displaystyle+p_{1}\left\\{\prod_{j}E(X_{j}^{2})+\sum_{j}\mu_{j}^{-1}E\left(\dot{X}_{j}\prod_{k}X_{k}^{2}\right)\right\\},$

and it follows that

$\displaystyle E(\dot{S}\dot{X})$
$\displaystyle=p_{1}\\{p_{1}p_{21}-p_{131}-p_{1}^{2}-2p_{1}^{2}s_{1}$
$\displaystyle+p_{1}p_{11}+p_{1}\sum_{j}\mu_{j}^{-1}(1+2\mu_{j})\prod_{k\neq
j}(1+\mu_{k})\\}$
$\displaystyle=p_{1}(p_{1}p_{21}-p_{131}-p_{1}^{2}-2p_{1}^{2}s_{1}+p_{1}p_{11}+p_{1}p_{11}s_{1;11}^{(12)}),$
(170)

where

$s_{a;bc}^{(de)}\stackrel{{\scriptstyle\rm{def}}}{{=}}\sum_{j}\mu_{j}^{-a}(b+c\mu_{j})^{-1}(d+e\mu_{j}),$
(171)

for integers $a,b,c,d,e$, etcetera. Next, we use (164) to write

$\displaystyle\dot{S}^{2}$
$\displaystyle=\prod_{j}X_{j}^{2}-2p_{1}\left(\prod_{j}X_{j}+\sum_{j}\mu_{j}^{-1}\dot{X}_{j}\prod_{k}X_{k}\right)$
$\displaystyle+p_{1}^{2}\left(1+2\sum_{j}\mu_{j}^{-1}\dot{X}_{j}+\sum_{j}\sum_{k}\mu_{j}^{-1}\mu_{k}^{-1}\dot{X}_{j}\dot{X}_{k}\right),$
(172)

which via (167) yields

$\displaystyle E(\dot{S}^{2}X)$
$\displaystyle=\prod_{j}E\\{X_{j}^{3}(X_{j}-1)\\}-\prod_{j}E(X_{j}^{4})$
$\displaystyle-2p_{1}\left[\prod_{j}E\\{X_{j}^{2}(X_{j}-1)\\}-\prod_{j}E(X_{j}^{3})\right]$
$\displaystyle-2p_{1}\left[\sum_{j}\mu_{j}^{-1}E\left\\{\dot{X}_{j}\prod_{k}X_{k}^{2}(X_{k}-1)\right\\}-\sum_{j}\mu_{j}^{-1}E\left(\dot{X}_{j}\prod_{k}X_{k}^{3}\right)\right]$
$\displaystyle+p_{1}^{2}\left[\prod_{j}E\\{X_{j}(X_{j}-1)\\}-\prod_{j}E(X_{j}^{2})\right]$
$\displaystyle+2p_{1}^{2}\left[\sum_{j}\mu_{j}^{-1}E\left\\{\dot{X}_{j}\prod_{k}X_{k}(X_{k}-1)\right\\}-\sum_{j}\mu_{j}^{-1}E\left(\dot{X}_{j}\prod
X_{k}^{2}\right)\right]$
$\displaystyle+p_{1}^{2}\left[\sum_{j}\sum_{k}\mu_{j}^{-1}\mu_{k}^{-1}E\left\\{\dot{X}_{j}\dot{X}_{k}\prod_{l}X_{l}(X_{l}-1)\right\\}\right.$
$\displaystyle\left.-\sum_{j}\sum_{k}\mu_{j}^{-1}\mu_{k}^{-1}E\left(\dot{X}_{j}\dot{X}_{k}\prod_{l}X_{l}^{2}\right)\right],$

and inserting the required moment formulae, we obtain

$\displaystyle E(\dot{S}^{2}X)$
$\displaystyle=p_{1}(p_{1}p_{451}-p_{1761})-2p_{1}^{2}(p_{1}p_{21}-p_{131})$
$\displaystyle-2p_{1}^{2}\left\\{p_{1}\sum_{j}\mu_{j}^{-1}(4+3\mu_{j})\prod_{k\neq
j}(2+\mu_{k})\right.$
$\displaystyle\left.-\sum_{j}\mu_{j}^{-1}(1+6\mu_{j}+3\mu_{j}^{2})\prod_{k\neq
j}(1+3\mu_{k}+\mu_{k}^{2})\right\\}+p_{1}^{3}(p_{1}-p_{11})$
$\displaystyle+2p_{1}^{3}\left\\{2p_{1}s_{1}-\sum_{j}\mu_{j}^{-1}(1+2\mu_{j})\prod_{k\neq
j}(1+\mu_{k})\right\\}$
$\displaystyle+p_{1}^{3}\left\\{p_{1}s_{1}(4s_{1}+1)-\sum_{j}\mu_{j}^{-2}(1+5\mu_{j}+\mu_{j}^{2})\prod_{k\neq
j}(1+\mu_{k})\right.$ $\displaystyle\left.-\sum_{j\neq
k}\mu_{j}^{-1}\mu_{k}^{-1}(1+2\mu_{j})(1+2\mu_{k})\prod_{l\notin(j,k)}(1+\mu_{l})\right\\}$
$\displaystyle=p_{1}(p_{1}p_{451}-p_{1761})-2p_{1}^{2}(p_{1}p_{21}-p_{131})-2p_{1}^{2}(p_{1}p_{21}s_{1;21}^{(43)}-p_{131}s_{1;131}^{(163)})$
$\displaystyle+p_{1}^{3}(p_{1}-p_{11})+2p_{1}^{3}(2p_{1}s_{1}-p_{11}s_{1;11}^{(12)})$
$\displaystyle+p_{1}^{3}\\{p_{1}s_{1}(4s_{1}+1)-p_{11}s_{2;11}^{(151)}-p_{11}(s_{1;11}^{(12)})^{2}+p_{11}s_{2;121}^{(144)}\\}.$
(173)

Moreover, by (165) and (166),

$\displaystyle E(\dot{S}^{2}\dot{X})$
$\displaystyle=E(\dot{S}^{2}X)-C_{10}E(\dot{S}^{2})$
$\displaystyle=E(\dot{S}^{2}X)-p_{1}^{2}d_{10}\left\\{p_{1}-\prod_{j}(1+\mu_{j})\right\\}$
$\displaystyle=E(\dot{S}^{2}X)-p_{1}^{2}d_{10}(p_{1}-p_{11}).$ (174)

Next, by (135) we may write

$V=p_{1}\sum_{j}\mu_{j}^{-1}\dot{X}_{j},$ (175)

and via (167), we find

$\displaystyle E(V\dot{X})=E(VX)$
$\displaystyle=p_{1}\left[\sum_{j}\mu_{j}^{-1}E\left\\{\dot{X}_{j}\prod_{k}X_{k}(X_{k}-1)\right\\}-\sum_{j}\mu_{j}^{-1}E\left(\dot{X}_{j}\prod_{k}X_{k}^{2}\right)\right]$
$\displaystyle=p_{1}^{2}\left\\{2p_{1}s_{1}-\sum_{j}\mu_{j}^{-1}(1+2\mu_{j})\prod_{k\neq
j}(1+\mu_{k})\right\\}$
$\displaystyle=p_{1}^{2}(2p_{1}s_{1}-p_{11}s_{1;11}^{(12)}).$ (176)

We are left with the terms involving $Z$. To attack these terms, observe that
(119) may be reformulated as

$Z=\sum_{j}\prod_{k\neq j}X_{k}-p_{1}s_{1}.$ (177)

From (177), we find

$E(Z\dot{X}_{1})=\sum_{j}\prod_{k\neq j}E(\dot{X}_{1}X_{k})=\sum_{j\neq
1}\prod_{k\neq j}\mu_{k}=\sum_{j\neq
1}\mu_{j}^{-1}p_{1}=p_{1}(s_{1}-\mu_{1}^{-1}),$

which implies

$\sum_{j}\mu_{j}^{-1}E(Z\dot{X}_{j})=p_{1}(s_{1}^{2}-s_{2}),$ (178)

and

$\sum_{j}\mu_{j}^{-2}E(Z\dot{X}_{j})=p_{1}(s_{1}s_{2}-s_{3}),$ (179)

Moreover, (175) and (177) yield

$\displaystyle E(ZV)$
$\displaystyle=p_{1}\sum_{j}\mu_{j}^{-1}\sum_{k}E\left(\dot{X}_{j}\prod_{l\neq
k}X_{l}\right)=p_{1}\sum_{j}\mu_{j}^{-1}\sum_{k\neq
j}E\left(\dot{X}_{j}\prod_{l\neq k}X_{l}\right)$
$\displaystyle=p_{1}\sum_{j}\mu_{j}^{-1}\sum_{k\neq j}\prod_{l\neq
k}\mu_{l}=p_{1}^{2}\sum_{j}\mu_{j}^{-1}\sum_{k\neq
j}\mu_{k}^{-1}=p_{1}^{2}(s_{1}^{2}-s_{2}).$ (180)

Furthermore, we get via (167), (166) and (177) that

$\displaystyle E(Z\dot{X})=E(ZX)$ $\displaystyle=\sum_{j}E\left\\{\prod_{k\neq
j}X_{k}\prod_{l}X_{l}(X_{l}-1)\right\\}-\sum_{j}E\left(\prod_{k\neq
j}X_{k}\prod_{l}X_{l}^{2}\right)-p_{1}s_{1}E(X)$
$\displaystyle=\sum_{j}E\left\\{X_{j}(X_{j}-1)\right\\}\prod_{k\neq
j}E\left\\{X_{k}^{2}(X_{k}-1)\right\\}$
$\displaystyle-\sum_{j}E(X_{j}^{2})\prod_{k\neq j}E(X_{k}^{3})-p_{1}s_{1}E(X)$
$\displaystyle=p_{1}^{2}\sum_{j}\prod_{k\neq
j}(2+\mu_{k})-p_{1}\sum_{j}(1+\mu_{j})\prod_{k\neq j}(1+3\mu_{k}+\mu_{k}^{2})$
$\displaystyle-p_{1}^{2}s_{1}(p_{1}-p_{11})$
$\displaystyle=p_{1}^{2}p_{21}s_{21}-p_{1}p_{131}s_{0;131}^{(11)}-p_{1}^{2}s_{1}(p_{1}-p_{11}).$
(181)

Next, we may observe that from (164) and (175), we have

$\dot{S}=\prod_{j}X_{j}-p_{1}-V,$ (182)

and so, because $E(Z)=0$,

$E(\dot{S}Z)=E\left(Z\prod_{j}X_{j}\right)-E(ZV),$ (183)

where via (177),

$\displaystyle E\left(Z\prod_{j}X_{j}\right)$
$\displaystyle=\sum_{j}E\left(\prod_{l}X_{l}\prod_{k\neq
j}X_{k}\right)-p_{1}s_{1}E\left(\prod_{j}X_{j}\right)$
$\displaystyle=\sum_{j}E(X_{j})\prod_{k\neq j}E(X_{k}^{2})-p_{1}^{2}s_{1}$
$\displaystyle=p_{1}\sum_{j}\prod_{k\neq
j}(1+\mu_{k})-p_{1}^{2}s_{1}=p_{1}p_{11}s_{0;11}-p_{1}^{2}s_{1}.$ (184)

Thus, (180), (183) and (184) imply

$E(\dot{S}Z)=p_{1}p_{11}s_{0;11}-p_{1}^{2}(s_{1}+s_{1}^{2}-s_{2}).$ (185)

Next, (182) yields, because $E(Z)=0$,

$\displaystyle E(\dot{S}^{2}Z)$ $\displaystyle=E\left(Z\prod
X_{j}^{2}\right)-2p_{1}E\left(Z\prod X_{j}\right)-2E\left(ZV\prod
X_{j}\right)$ $\displaystyle+2p_{1}E(ZV)+E(ZV^{2}).$ (186)

Here, by (177),

$\displaystyle E\left(Z\prod_{j}X_{j}^{2}\right)$
$\displaystyle=\sum_{j}E\left(\prod_{l}X_{l}^{2}\prod_{k\neq
j}X_{k}\right)-p_{1}s_{1}E\left(\prod_{j}X_{j}^{2}\right)$
$\displaystyle=\sum_{j}E(X_{j}^{2})\prod_{k\neq
j}E(X_{k}^{3})-p_{1}^{2}s_{1}\prod_{j}(1+\mu_{j})$
$\displaystyle=p_{1}\sum_{j}(1+\mu_{j})\prod_{k\neq
j}(1+3\mu_{k}+\mu_{k}^{2})-p_{1}^{2}s_{1}\prod_{j}(1+\mu_{j})$
$\displaystyle=p_{1}p_{131}s_{0;131}^{(11)}-p_{1}^{2}p_{11}s_{1}.$ (187)

Moreover, via (175) and (177),

$\displaystyle E\left(ZV\prod_{j}X_{j}\right)$
$\displaystyle=\sum_{j}E\left(V\prod_{q}X_{q}\prod_{k\neq
j}X_{k}\right)-p_{1}s_{1}E\left(V\prod_{j}X_{j}\right)$
$\displaystyle=p_{1}\left\\{\sum_{j}\sum_{l}\mu_{l}^{-1}E\left(\dot{X}_{l}\prod_{q}X_{q}\prod_{k\neq
j}X_{k}\right)-p_{1}s_{1}\sum_{k}\mu_{k}^{-1}E\left(\dot{X}_{k}\prod_{j}X_{j}\right)\right\\}$
$\displaystyle=p_{1}\left\\{\sum_{j}\mu_{j}^{-1}E(\dot{X}_{j}X_{j})\prod_{k\neq
j}E(X_{k}^{2})+\sum_{j\neq l}\mu_{l}^{-1}E(X_{j})\prod_{k\neq
j}E(\dot{X}_{k}X_{k}^{2})\right.$
$\displaystyle\left.-p_{1}s_{1}\sum_{k}\mu_{k}^{-1}E(\dot{X}_{k}X_{k})\prod_{j\neq
k}E(X_{j})\right\\}$
$\displaystyle=p_{1}^{2}\left\\{\sum_{j}\mu_{j}^{-1}\prod_{k\neq
j}(1+\mu_{k})+\sum_{j\neq l}\mu_{l}^{-1}\prod_{k\neq
j}(1+2\mu_{k})-p_{1}s_{1}^{2}\right\\}$
$\displaystyle=p_{1}^{2}\\{p_{11}s_{1;11}+p_{12}(s_{1}s_{12}-s_{1;12})-p_{1}s_{1}^{2}\\}.$
(188)

Furthermore, (177) gives

$E(ZV^{2})=\sum_{j}E\left(V^{2}\prod_{k\neq
j}X_{k}\right)-p_{1}s_{1}E(V^{2}),$ (189)

where from (175),

$\displaystyle E(V^{2})$
$\displaystyle=p_{1}^{2}\sum_{j}\sum_{k}\mu_{j}^{-1}\mu_{k}^{-1}E(\dot{X}_{j}\dot{X}_{k})=p_{1}^{2}\sum_{j}\mu_{j}^{-2}E(\dot{X}_{j}^{2})=p_{1}^{2}s_{1},$
(190)

and

$\displaystyle\sum_{j}E\left(V^{2}\prod_{k\neq j}X_{k}\right)$
$\displaystyle=p_{1}^{2}\sum_{j}\sum_{q}\sum_{l}\mu_{q}^{-1}\mu_{l}^{-1}E\left(\dot{X}_{q}\dot{X}_{l}\prod_{k\neq
j}X_{k}\right)$ $\displaystyle=p_{1}^{2}(v_{1}+v_{2}),$ (191)

where we get the term with $q=l$ as

$\displaystyle v_{1}$
$\displaystyle=\sum_{j}\sum_{l}\mu_{l}^{-2}E\left(\dot{X}_{l}^{2}\prod_{k\neq
j}X_{k}\right)$
$\displaystyle=\sum_{j}\mu_{j}^{-2}E(\dot{X}_{j}^{2})\prod_{k\neq
j}E(X_{k})+\sum_{j\neq
l}\mu_{l}^{-2}E(\dot{X}_{l}^{2}X_{l})\prod_{k\notin(j,l)}E(X_{k})$
$\displaystyle=p_{1}s_{2}+p_{1}\sum_{j\neq
l}\mu_{l}^{-2}\mu_{j}^{-1}(1+\mu_{l})=p_{1}s_{2}+p_{1}(s_{1}s_{2}^{(11)}-s_{3}^{(11)}),$
(192)

and for $q\neq l$ in (191) we have

$\displaystyle v_{2}$ $\displaystyle=\sum_{j}\sum_{q\neq
l}\mu_{q}^{-1}\mu_{l}^{-1}E\left(\dot{X}_{q}\dot{X}_{l}\prod_{k\neq
j}X_{k}\right)$ $\displaystyle=\sum_{j}\sum_{q\notin(j,l),j\neq
l}\mu_{q}^{-1}\mu_{l}^{-1}E(\dot{X}_{q}X_{q})E(\dot{X}_{l}X_{l})\prod_{k\notin(q,l)}E(X_{k})$
$\displaystyle=p_{1}\sum_{j}\sum_{q\notin(j,l),j\neq
l}\mu_{q}^{-1}\mu_{l}^{-1}=p_{1}(m-2)(s_{1}^{2}-s_{2}).$ (193)

Hence, (189)-(193) together yield

$E(ZV^{2})=p_{1}^{3}\left\\{(m-3)(s_{1}^{2}-s_{2})+s_{1}s_{2}^{(11)}-s_{3}^{(11)}\right\\}.$
(194)

Thus, from (186), (187), (184), (188), (180), (194) and some simplification,

$\displaystyle E(\dot{S}^{2}Z)$
$\displaystyle=p_{1}p_{131}s_{0;131}^{(11)}-p_{1}^{2}\\{p_{11}(s_{1}+2s_{0;11}+2s_{1;11})+2p_{12}(s_{1}s_{12}-s_{1;12})\\}$
$\displaystyle+p_{1}^{3}\\{2s_{1}+2s_{1}^{2}+(m-1)(s_{1}^{2}-s_{2})+s_{1}s_{2}^{(11)}-s_{3}^{(11)}\\}.$
(195)

We may now insert (165), (178), (179) and (180) into (163) to obtain, after
small simplifications,

$\displaystyle\rho_{0}$
$\displaystyle=-\frac{1}{2}d_{10}^{-1}\sum_{j_{1}<...<j_{m-2}}\mu_{j_{1}}\cdots\mu_{j_{m-2}}+p_{1}^{-2}d_{10}^{-2}\cdot$
$\displaystyle\left[\frac{1}{2}E(\dot{S}^{2}\dot{X})+\frac{1}{2}p_{1}E(\dot{S}^{2}Z)+p_{1}^{2}s_{1}d_{10}+\frac{1}{2}p_{1}s_{1}\eta\right.$
$\displaystyle+p_{1}^{-1}d_{10}^{-1}\left\\{E(\dot{S}\dot{X})\right\\}^{2}+3p_{1}d_{10}E(\dot{S}Z)+\frac{1}{2}p_{1}^{3}d_{10}(s_{1}^{2}-s_{2})$
$\displaystyle+2d_{10}^{-1}E(\dot{S}Z)E(\dot{S}\dot{X})+p_{1}\cdot$
$\displaystyle\left\\{\frac{1}{2}p_{1}^{-1}E(\dot{X}^{2})+E(Z\dot{X})+6s_{1}E(\dot{S}\dot{X})+s_{1}E(V\dot{X})\right.$
$\displaystyle\left.\left.+2p_{1}^{3}s_{1}(s_{1}^{2}-s_{2})+\frac{1}{2}p_{1}^{3}s_{3}\right\\}\right],$
(196)

where

$\sum_{j_{1}<...<j_{m-2}}\mu_{j_{1}}\cdots\mu_{j_{m-2}}=p_{1}\sum_{j<k}\mu_{j}^{-1}\mu_{k}^{-1}=\frac{1}{2}p_{1}(s_{1}^{2}-s_{2})$
(197)

and from (130),

$\eta=p_{1}^{2}(p_{11}s_{1;11}-p_{1}s_{2}).$ (198)

Next, we focus on $f_{1}$. We have from (228) that

$f_{1}d_{0}^{-3}=\frac{d_{2}}{P^{2}d_{1}^{4}}D_{1}+D_{2},$ (199)

where

$D_{1}=P(-d_{1})+d_{32},$ (200)

and

$D_{2}=\frac{d_{33}}{3P^{3}(-d_{1})^{3}}.$ (201)

Now, from (47), (50), (109), (116) and (200), we have

$\displaystyle D_{1}$
$\displaystyle=p_{1}d_{10}\left(1+n^{-1/2}\sum_{j}\frac{r_{j}}{\mu_{j}}\right)\left(1+n^{-1/2}\frac{d_{11}}{d_{10}}\right)+d_{320}+n^{-1/2}d_{321}+O(n^{-1})$
$\displaystyle=n^{-1/2}\left(p_{1}d_{10}\sum_{j}\mu_{j}^{-1}r_{j}+p_{1}d_{11}+d_{321}\right)+O(n^{-1}),$
(202)

and with

$d_{33}=d_{330}+n^{-1/2}d_{331}+O(n^{-1}),$ (203)

we have by (201) that

$\displaystyle D_{2}$
$\displaystyle=\frac{1}{3}p_{1}^{-3}d_{10}^{-3}d_{330}\left(1+n^{-1/2}\sum_{j}\frac{r_{j}}{\mu_{j}}\right)^{-3}\left(1+n^{-1/2}\frac{d_{11}}{d_{10}}\right)^{-3}\left(1+n^{-1/2}\frac{d_{331}}{d_{330}}\right)$
$\displaystyle+O(n^{-1})$
$\displaystyle=\frac{1}{3}p_{1}^{-3}d_{10}^{-3}d_{330}\left\\{1+n^{-1/2}\left(-3\sum_{j}\mu_{j}^{-1}r_{j}-3d_{10}^{-1}d_{11}+d_{330}^{-1}d_{331}\right)\right\\}$
$\displaystyle+O(n^{-1}).$ (204)

Since $D_{1}$ is of order $n^{-1/2}$, we only need to keep track of the main
term (limit) of $d_{2}$. From (25), this limit is

$d_{20}=p_{1}^{-2}C_{20}-p_{1}\sum_{j<k}\mu_{j}^{-1}\mu_{k}^{-1}=p_{1}^{-2}C_{20}-\frac{1}{2}p_{1}(s_{1}^{2}-s_{2}),$
(205)

where $C_{20}$ is the limit of (18). In fact, (15) and (18) yield

$\displaystyle C_{20}$
$\displaystyle=E\left(\prod_{j}X_{j}\left[p_{1}s_{1}\left\\{\prod_{j}(X_{j}-1)-\prod_{j}X_{j}\right\\}\right.\right.$
$\displaystyle+\frac{1}{2}\prod_{j}(X_{j}-1)(X_{j}-2)-\frac{1}{2}\prod_{j}X_{j}(X_{j}-1)$
$\displaystyle\left.\left.+\prod_{j}X_{j}^{2}-\prod_{j}X_{j}(X_{j}-1)\right]\right)$
$\displaystyle=p_{1}s_{1}\prod_{j}E\\{X_{j}(X_{j}-1)\\}-p_{1}s_{1}\prod_{j}E(X_{j}^{2})$
$\displaystyle+\frac{1}{2}\prod_{j}E\\{X_{j}(X_{j}-1)(X_{j}-2)\\}-\frac{3}{2}\prod_{j}E\\{X_{j}^{2}(X_{j}-1)\\}$
$\displaystyle+\prod_{j}E(X_{j}^{3}),$

It follows that

$\displaystyle C_{20}$
$\displaystyle=p_{1}^{3}s_{1}-p_{1}^{2}s_{1}\prod_{j}(1+\mu_{j})+\frac{1}{2}p_{1}^{3}$
$\displaystyle-\frac{3}{2}p_{1}^{2}\prod_{j}(2+\mu_{j})+p_{1}\prod_{j}(1+3\mu_{j}+\mu_{j}^{2})$
$\displaystyle=p_{1}^{3}s_{1}-p_{1}^{2}p_{11}s_{1}+\frac{1}{2}p_{1}^{3}-\frac{3}{2}p_{1}^{2}p_{21}+p_{1}p_{131},$

which via (205) yields

$d_{20}=p_{1}s_{1}-p_{11}s_{1}+\frac{1}{2}p_{1}-\frac{3}{2}p_{21}+p_{1}^{-1}p_{131}-\frac{1}{2}p_{1}(s_{1}^{2}-s_{2}).$
(206)

Now, by (199) and (202),

$\displaystyle f_{1}d_{0}^{-3}$
$\displaystyle=n^{-1/2}\frac{d_{20}}{p_{1}^{2}d_{10}^{4}}\left(p_{1}d_{10}\sum_{j}\mu_{j}^{-1}r_{j}+p_{1}d_{11}+d_{321}\right)$
$\displaystyle+D_{2}+O(n^{-1}).$ (207)

To tackle $D_{2}$, we at first need $d_{330}$, the limit of $d_{33}$. Here, we
find that from (17) and (32), the limit of $\bar{e}_{2}$ is $C_{10}$, given in
(166). Analogously, define $e_{30}$ as the limit of $\bar{e}_{3}$, cf (33).
This yields

$\displaystyle e_{30}$
$\displaystyle=\frac{1}{2}\prod_{j}E\\{X_{j}(X_{j}-1)(X_{j}-2)-\frac{3}{2}\prod_{j}E\\{X_{j}^{2}(X_{j}-1)\\}+\prod_{j}E(X_{j}^{3})$
$\displaystyle=\frac{1}{2}p_{1}^{3}-\frac{3}{2}p_{1}^{2}\prod_{j}(2+\mu_{j})+p_{1}\prod_{j}(1+3\mu_{j}+\mu_{j}^{2})$
$\displaystyle=\frac{1}{2}p_{1}^{3}-\frac{3}{2}p_{1}^{2}p_{21}+p_{1}p_{131}.$
(208)

Then, letting $M(2^{m-1},0^{1})$ be the limit of $\overline{\cal
Y}(2^{m-1},0^{1})$ etcetera, (38) yields

$\displaystyle d_{330}$
$\displaystyle=2p_{1}M(2^{m-1},0^{1})+3p_{1}M(2^{m-2},1^{2})$
$\displaystyle+3M(1^{m-1},0^{1})C_{10}+e_{30}.$ (209)

Here,

$\displaystyle M(2^{m-1},0^{1})$
$\displaystyle=\sum_{j_{1}<...<j_{m-1}}\prod_{k}\mu_{j_{k}}^{2}=p_{1}^{2}s_{2},$
(210) $\displaystyle M(2^{m-2},1^{2})$
$\displaystyle=p_{1}\sum_{j_{1}<...<j_{m-2}}\prod_{k}\mu_{j_{k}}=p_{1}^{2}\sum_{j<k}\mu_{j}^{-1}\mu_{k}^{-1}$
$\displaystyle=\frac{1}{2}p_{1}^{2}(s_{1}^{2}-s_{2}),$ (211) $\displaystyle
M(1^{m-1},0^{1})$ $\displaystyle=p_{1}s_{1}.$ (212)

Inserting this together with (166) and (208) into (209), we get

$d_{330}=p_{1}^{3}\left(\frac{3}{2}s_{1}^{2}+\frac{1}{2}s_{2}\right)+3p_{1}^{2}s_{1}(p_{1}-p_{11})+\frac{1}{2}p_{1}^{3}-\frac{3}{2}p_{1}^{2}p_{21}+p_{1}p_{131}.$
(213)

To find $d_{331}$, we need the expansions

$\bar{e}_{2}=C_{10}+n^{-1/2}C_{11}+O(n^{-1})$ (214)

and

$\bar{e}_{3}=e_{30}+n^{-1/2}e_{31}+O(n^{-1}),$ (215)

where $e_{30}$ is in (208). Moreover, we write e.g.

$\overline{\cal
Y}(2^{m-1},0^{1})=M(2^{m-1},0^{1})+n^{-1/2}\overline{R}(2^{m-1},0^{1}).$

Next, analogous to (106), we have

$3\bar{y}_{11...1}-P=2p_{1}+n^{-1/2}\\{3d_{00}+2(r,\mu^{m-1})\\}+O(n^{-1}),$

and similarly,

$\bar{y}_{11...1}=p_{1}+n^{-1/2}\\{d_{00}+(r,\mu^{m-1})\\}+O(n^{-1}),$

implying via (38) and (210)-(212) that

$\displaystyle d_{331}$
$\displaystyle=M(2^{m-1},0^{1})\\{3d_{00}+2(r,\mu^{m-1})\\}+2p_{1}\overline{R}(2^{m-1},0^{1})$
$\displaystyle+3M(2^{m-2},1^{2})\\{d_{00}+(r,\mu^{m-1})\\}+3p_{1}\overline{R}(2^{m-2},1^{2})$
$\displaystyle+3M(1^{m-1},0^{1})C_{11}+3C_{10}R(1^{m-1},0^{1})+e_{31}$
$\displaystyle=p_{1}^{2}s_{2}\\{3d_{00}+2(r,\mu^{m-1})\\}+2p_{1}\overline{R}(2^{m-1},0^{1})$
$\displaystyle+\frac{3}{2}p_{1}^{2}(s_{1}^{2}-s_{2})\\{d_{00}+(r,\mu^{m-1})\\}+3p_{1}\overline{R}(2^{m-2},1^{2})$
$\displaystyle+3p_{1}s_{1}C_{11}+3C_{10}R(1^{m-1},0^{1})+e_{31}.$ (216)

To find $E(d_{0}^{3}D_{2})$, we need to calculate

$\displaystyle E(d_{0}^{3}r_{1})$
$\displaystyle=E(d_{00}^{3}r_{1})+O(n^{-1/2})$
$\displaystyle=n^{-2}\sum_{i}\sum_{j}\sum_{k}\sum_{l}E(U_{i}U_{j}U_{k}\dot{y}_{l1})+O(n^{-1/2})$
$\displaystyle=O(n^{-1/2}).$ (217)

Moreover, by (146) and (58),

$E(d_{0}^{3}d_{11})=-3p_{1}^{-1}E(\dot{S}^{2})E(\dot{S}\dot{X})+O(n^{-1/2}).$
(218)

To get $E(d_{0}^{3}d_{321})$, we find in the usual manner

$\displaystyle E(d_{0}^{3}H)$
$\displaystyle=n^{-2}\sum_{i}\sum_{j}\sum_{k}\sum_{l}E(U_{i}U_{j}U_{k}H_{l})+O(n^{-1/2})$
$\displaystyle=3E(\dot{S}^{2})E(\dot{S}Z)+O(n^{-1/2}),$ (219) $\displaystyle
E(d_{0}^{3}d_{00})$
$\displaystyle=n^{-2}\sum_{i}\sum_{j}\sum_{k}\sum_{l}E(U_{i}U_{j}U_{k}U_{l})+O(n^{-1/2})$
$\displaystyle=3\\{E(\dot{S}^{2})\\}^{2}+O(n^{-1/2}),$ (220)

and

$\displaystyle E\\{d_{0}^{3}(r,\mu^{m-1}\\}$
$\displaystyle=n^{-2}\sum_{i}\sum_{j}\sum_{k}\sum_{l}E(U_{i}U_{j}U_{k}V_{l})+O(n^{-1/2})$
$\displaystyle=O(n^{-1/2}),$ (221)

which together with (145) inserted into (111) yields

$\displaystyle E(d_{0}^{3}d_{321})$
$\displaystyle=p_{1}E(d_{0}^{3}H)+p_{1}s_{1}\left[2E(d_{0}^{3}d_{00})+E\\{d_{0}^{3}(r,\mu^{m-1})\\}\right]+E(d_{0}^{3}C_{11})$
$\displaystyle=3E(\dot{S}^{2})\left\\{p_{1}E(\dot{S}Z)+2p_{1}s_{1}E(\dot{S}^{2})+E(\dot{S}\dot{X})\right\\}+O(n^{-1/2}).$
(222)

To get $E(d_{0}^{3}D_{2})$, we in addition need $E(d_{0}^{3}d_{331})$, and to
that end, we must calculate terms like $E\\{d_{0}^{3}R(1^{m-1},0^{1})\\}$.
However, because

$\overline{\cal Y}(1^{m-1},0^{1})=\bar{y}_{1...10}+...+\bar{y}_{01...1},$

where e.g.

$\bar{y}_{1...10}=\mu_{1}\cdots\mu_{m-1}+n^{-1/2}(\mu_{1}\cdots\mu_{m-2}r_{m-1}+...+\mu_{2}\cdots\mu_{m-1}r_{1})+O(n^{-1}),$

it follows that

$\displaystyle R(1^{m-1},0^{1})$
$\displaystyle=\mu_{1}\cdots\mu_{m-2}r_{m-1}+...+\mu_{2}\cdots\mu_{m-1}r_{1}+...$
$\displaystyle+\mu_{2}\cdots\mu_{m-1}r_{m}+...+\mu_{3}\cdots\mu_{m}r_{2}+O(n^{-1}),$

and so, it follows from (217) that

$E\\{d_{0}^{3}R(1^{m-1},0^{1})\\}=O(n^{-1/2}).$

It is similarly seen that $E\\{d_{0}^{3}\overline{R}(2^{m-1},0^{1})\\}$ and
$E\\{d_{0}^{3}\overline{R}(2^{m-2},1^{2})\\}$ are also $O(n^{-1/2})$.

However, the term $E(d_{0}^{3}e_{31})$ is not that simple, but we can perform
the same ’trick’ as with the corresponding term for $e_{2}$,
$E(d_{0}^{3}C_{11})$. Indeed, as in (145), we have

$\displaystyle E(d_{0}^{3}e_{31})$
$\displaystyle=n^{-2}\sum_{i}\sum_{j}\sum_{k}\sum_{l}E(U_{i}U_{j}U_{k}e_{3l})+O(n^{-1/2})$
$\displaystyle=3E(\dot{S}^{2})E(\dot{S}\epsilon_{3})+O(n^{-1/2}),$ (223)

where via (33),

$\epsilon_{3}=\frac{1}{2}\prod_{j}X_{j(3)}-\frac{3}{2}\prod_{j}X_{j}\prod_{j}X_{j(2)}+\left(\prod_{j}X_{j}\right)^{3},$
(224)

where as before, $X_{j}$ is a Poisson variate with parameter $\mu_{j}$ and all
$X_{j}$ are simultaneously independent. Moreover, $X_{j(2)}=X_{j}(X_{j}-1)$
and
$X_{j(3)}=X_{j}(X_{j}-1)(X_{j}-2)$. Moreover, we find in the usual manner that

$\displaystyle E(\dot{S}\epsilon_{3})$
$\displaystyle=\frac{1}{2}\prod_{j}E\\{X_{j}^{2}(X_{j}-1)(X_{j}-2)\\}-\frac{3}{2}\prod_{j}E\\{X_{j}^{3}(X_{j}-1)\\}+\prod_{j}E(X_{j}^{4})$
$\displaystyle-
p_{1}\left[\frac{1}{2}\prod_{j}E\\{X_{j}(X_{j}-1)(X_{j}-2)\\}-\frac{3}{2}\prod_{j}E\\{X_{j}^{2}(X_{j}-1)\\}+\prod_{j}E(X_{j}^{3})\right.$
$\displaystyle+\frac{1}{2}\sum_{j}\mu_{j}^{-1}E\left\\{\dot{X}_{j}\prod_{k}X_{k}(X_{k}-1)(X_{k}-2)\right\\}$
$\displaystyle\left.-\frac{3}{2}\sum_{j}\mu_{j}^{-1}E\left\\{\dot{X}_{j}\prod_{k}X_{k}^{2}(X_{k}-1)\right\\}+\sum_{j}\mu_{j}^{-1}E\left(\dot{X}_{j}\prod_{k}X_{k}^{3}\right)\right]$
$\displaystyle=\frac{1}{2}p_{1}^{3}\prod_{j}(3+\mu_{j})-\frac{3}{2}p_{1}^{2}\prod_{j}(4+5\mu_{j}+\mu_{j}^{2})$
$\displaystyle+p_{1}\prod_{j}(1+7\mu_{j}+6\mu_{j}^{2}+\mu_{j}^{3})-\frac{1}{2}p_{1}^{4}+\frac{3}{2}p_{1}^{3}\prod_{j}(2+\mu_{j})$
$\displaystyle-
p_{1}^{2}\prod_{j}(1+3\mu_{j}+\mu_{j}^{2})-\frac{3}{2}p_{1}^{4}s_{1}+\frac{3}{2}p_{1}^{3}\sum_{j}\mu_{j}^{-1}(4+3\mu_{j})\prod_{k\neq
j}(2+\mu_{k})$ $\displaystyle-
p_{1}^{2}\sum_{j}\mu_{j}^{-1}(1+6\mu_{j}+3\mu_{j}^{2})\prod_{k\neq
j}(1+3\mu_{k}+\mu_{k}^{2})$
$\displaystyle=\frac{1}{2}p_{1}^{3}p_{31}-\frac{3}{2}p_{1}^{2}p_{451}+p_{1}p_{1761}-\frac{1}{2}p_{1}^{4}+\frac{3}{2}p_{1}^{3}p_{21}-p_{1}^{2}p_{131}-\frac{3}{2}p_{1}^{4}s_{1}$
$\displaystyle+\frac{3}{2}p_{1}^{3}p_{21}s_{1;21}^{(43)}-p_{1}^{2}p_{131}s_{1;131}^{(163)}.$
(225)

Now, from (216),

$\displaystyle E(d_{0}^{3}d_{331})$
$\displaystyle=p_{1}^{2}s_{2}\left[3E(d_{0}^{3}d_{00})+2E\\{d_{0}^{3}(r,\mu^{m-1}\\}\right]$
$\displaystyle+\frac{3}{2}p_{1}^{2}(s_{1}^{2}-s_{2})\left[E(d_{0}^{3}d_{00})+E\\{d_{0}^{3}(r,\mu^{m-1}\\}\right]$
$\displaystyle+3p_{1}s_{1}E(d_{0}^{3}C_{11})+E(d_{0}^{3}e_{31})+O(n^{-1/2}),$

and (220), (221), (145) and (224) yield

$\displaystyle E(d_{0}^{3}d_{331})$
$\displaystyle=3E(\dot{S}^{2})\left\\{3p_{1}^{2}s_{2}E(\dot{S}^{2})\right.+\frac{3}{2}p_{1}^{2}(s_{1}^{2}-s_{2})E(\dot{S}^{2})$
$\displaystyle\left.+3p_{1}s_{1}E(\dot{S}\dot{X})+E(\dot{S}\epsilon_{3})\right\\}+O(n^{-1/2}).$
(226)

Now, via (204) and (217),

$\displaystyle E(d_{0}^{3}D_{2})$
$\displaystyle=\frac{1}{3}p_{1}^{-3}d_{10}^{-3}d_{330}\left[E(d_{0}^{3})\right.$
$\displaystyle\left.+n^{-1/2}\left\\{-3d_{10}^{-1}E(d_{0}^{3}d_{11})+d_{330}^{-1}E(d_{0}^{3}d_{331})\right\\}\right]+O(n^{-1}),$

and inserting (127), (218), (226) and (165), we find

$\displaystyle E(d_{0}^{3}D_{2})$
$\displaystyle=n^{-1/2}\frac{1}{3}p_{1}^{-3}d_{10}^{-3}\left[\left\\{p_{1}d_{10}+9E(\dot{S}\dot{X})\right\\}d_{330}\right.$
$\displaystyle+3p_{1}d_{10}\left\\{\frac{3}{2}p_{1}^{3}d_{10}(s_{1}^{2}+s_{2})\right.$
$\displaystyle\left.\left.+3p_{1}s_{1}E(\dot{S}\dot{X})+E(\dot{S}\epsilon_{3})\right\\}\right]+O(n^{-1}),$
(227)

where $d_{330}$ is given in (213), $E(\dot{S}X)$ is given in (170) and
$E(\dot{S}\epsilon_{3})$ is given in (225).

Now, we have from (207) and (217) that

$E(f_{1})=n^{-1/2}\frac{d_{20}}{p_{1}^{2}d_{10}^{4}}\left\\{p_{1}E(d_{0}^{3}d_{11})+E(d_{0}^{3}d_{321})\right\\}+E(d_{0}^{3}D_{2})+O(n^{-1}),$

and (218), (222) and a little simplification yields

$\displaystyle E(f_{1})$
$\displaystyle=n^{-1/2}\frac{d_{20}}{p_{1}d_{10}^{4}}3E(\dot{S}^{2})\left\\{E(\dot{S}Z)+2s_{1}E(\dot{S}^{2})\right\\}$
$\displaystyle+E(d_{0}^{3}D_{2})+O(n^{-1}),$

and by (165) and (227), we obtain

$\displaystyle n^{1/2}E(f_{1})$
$\displaystyle=3d_{20}d_{10}^{-3}\left\\{E(\dot{S}Z)+2p_{1}d_{10}s_{1}\right\\}$
$\displaystyle+\frac{1}{3}p_{1}^{-2}d_{10}^{-2}d_{330}+3p_{1}^{-3}d_{10}^{-3}E(\dot{S}\dot{X})d_{330}$
$\displaystyle+\frac{3}{2}p_{1}d_{10}^{-1}(s_{1}^{2}+s_{2})+3p_{1}^{-1}d_{10}^{-2}s_{1}E(\dot{S}\dot{X})+p_{1}^{-2}d_{10}^{-2}E(\dot{S}\epsilon_{3})$
$\displaystyle+O(n^{-1/2}).$ (228)

Finally, we evaluate $f_{2}$. To this end, only $E(d_{0}^{4})$ and limits of
the terms in (43) are needed. At first, in the usual manner we have

$\displaystyle E(d_{0}^{4})$
$\displaystyle=n^{-2}\sum_{i}\sum_{j}\sum_{k}\sum_{l}E(U_{i}U_{j}U_{k}U_{l})+O(n^{-1/2})$
$\displaystyle=3\\{E(\dot{S}^{2})\\}^{2}+O(n^{-1/2}).$ (229)

Hence, from (43), (116) and (165), letting $d_{30}$ be the limit of $d_{3}$,

$\displaystyle E(f_{2})$
$\displaystyle=\frac{3\\{E(\dot{S}^{2})\\}^{2}}{4p_{1}^{4}d_{10}^{6}}$
$\displaystyle(8p_{1}^{3}d_{10}d_{20}^{2}+10p_{1}^{2}d_{20}^{2}d_{320}+4p_{1}d_{330}d_{10}d_{20}+d_{340}d_{10}^{2}$
$\displaystyle+4p_{1}^{3}d_{10}^{2}d_{30}+4p_{1}^{2}d_{10}d_{30}d_{320})+O(n^{-1/2})$
$\displaystyle=\frac{3}{4p_{1}^{2}d_{10}^{3}}(-2p_{1}^{3}d_{20}^{2}+4p_{1}d_{330}d_{20}+d_{340}d_{10})$
$\displaystyle+O(n^{-1/2}),$ (230)

Here, $d_{20}$ is given by (206), $d_{330}$ is in (213), and similarly,
$d_{340}$ is obtained as the limit of $d_{34}$ in (39) as

$\displaystyle d_{340}$ $\displaystyle=3p_{1}M(3^{m-1},0^{1})$
$\displaystyle+4p_{1}\left\\{M(3^{m-2},2^{1},1^{1})+M(3^{m-3},2^{3})\right\\}$
$\displaystyle+2\left\\{M(1^{m-1},0^{1})^{2}+2M(2^{m-1},0^{1})+2M(2^{m-2},1^{2})\right\\}C_{10}$
$\displaystyle+4M(1^{m-1},0^{1})e_{30}+e_{40},$ (231)

where $e_{40}$ is the limit of (34). Here, we get in the usual manner

$\displaystyle M(3^{m-1},0^{1})$ $\displaystyle=p_{1}^{3}s_{3},$ (232)
$\displaystyle M(3^{m-2},2^{1},1^{1})$
$\displaystyle=p_{1}^{3}\sum_{j}\mu_{j}^{-1}\sum_{k\neq
j}\mu_{k}^{-2}=p_{1}^{3}(s_{1}s_{2}-s_{3}),$ (233) $\displaystyle
M(3^{m-3},2^{3})$
$\displaystyle=p_{1}^{3}\sum_{j_{1}<j_{2}<j_{3}}\prod_{k}\mu_{j_{k}}^{-1}=\frac{1}{6}p_{1}^{3}(s_{1}^{3}-3s_{1}s_{2}+2s_{3}),$
(234)

and inserting (232)-(234) together with (210)-(212) into (231) and
simplifying, we find

$\displaystyle d_{340}$
$\displaystyle=p_{1}^{4}\left(\frac{2}{3}s_{1}^{3}+2s_{1}s_{2}+\frac{1}{3}s_{3}\right)+2p_{1}^{2}(2s_{1}^{2}+s_{2})C_{10}$
$\displaystyle+4p_{1}s_{1}e_{30}+e_{40}.$ (235)

Here, (34) implies

$\displaystyle e_{40}$
$\displaystyle=\frac{1}{6}\prod_{j}E\\{X_{j}(X_{j}-1)(X_{j}-2)(X_{j}-3)\\}-\frac{2}{3}\prod_{j}E\\{X_{j}^{2}(X_{j}-1)(X_{j}-2)\\}$
$\displaystyle-\frac{1}{2}\prod_{j}E\\{X_{j}^{2}(X_{j}-1)^{2}\\}+2\prod_{j}E\\{X_{j}^{3}(X_{j}-1)\\}-\prod_{j}E(X_{j}^{4})$
$\displaystyle=\frac{1}{6}p_{1}^{4}-\frac{2}{3}p_{1}^{3}\prod_{j}(3+\mu_{j})-\frac{1}{2}p_{1}^{2}\prod_{j}(2+4\mu_{j}+\mu_{j}^{2})$
$\displaystyle+2p_{1}^{2}\prod_{j}(4+5\mu_{j}+\mu_{j}^{2})-p_{1}\prod_{j}(1+7\mu_{j}+6\mu_{j}^{2}+\mu_{j}^{3})$
$\displaystyle=\frac{1}{6}p_{1}^{4}-\frac{2}{3}p_{1}^{3}p_{31}-\frac{1}{2}p_{1}^{2}p_{241}+2p_{1}^{2}p_{451}-p_{1}p_{1761}.$
(236)

Hence, inserting (166), (208) and (236) into (235), we get

$\displaystyle d_{340}$
$\displaystyle=p_{1}^{4}\left(\frac{2}{3}s_{1}^{3}+2s_{1}s_{2}+\frac{1}{3}s_{3}\right)+2p_{1}^{3}(2s_{1}^{2}+s_{2})(p_{1}-p_{11})$
$\displaystyle+2p_{1}^{4}s_{1}-6p_{1}^{3}p_{21}s_{1}+4p_{1}^{2}p_{131}s_{1}+\frac{1}{6}p_{1}^{4}-\frac{2}{3}p_{1}^{3}p_{31}-\frac{1}{2}p_{1}^{2}p_{241}$
$\displaystyle+2p_{1}^{2}p_{451}-p_{1}p_{1761}.$ (237)

Hence, by inserting (206), (213) and (237) into (230), we get $E(f_{2})$. One
final simplification is to write

$\displaystyle q_{1}$
$\displaystyle=p_{1}s_{1}-p_{11}s_{1}-\frac{1}{2}p_{1}(s_{1}^{2}-s_{2}),$
(238) $\displaystyle q_{2}$
$\displaystyle=\frac{1}{2}p_{1}-\frac{3}{2}p_{21}+p_{1}^{-1}p_{131},$ (239)
$\displaystyle q_{3}$
$\displaystyle=p_{1}^{3}\left(\frac{3}{2}s_{1}^{2}+\frac{1}{2}s_{2}\right)+3p_{1}^{2}s_{1}(p_{1}-p_{11}),$
(240)

so that by (206) and (213),

$\displaystyle d_{20}$ $\displaystyle=q_{1}+q_{2},$ (241) $\displaystyle
d_{330}$ $\displaystyle=q_{3}+p_{1}^{2}q_{2},$ (242)

which yields

$\displaystyle E(f_{2})$ $\displaystyle=\frac{3}{4p_{1}^{2}d_{10}^{3}}$
$\displaystyle\cdot\left\\{-2p_{1}^{3}(q_{1}+q_{2})^{2}+4p_{1}(q_{3}+p_{1}^{2}q_{2})(q_{1}+q_{2})+d_{340}d_{10}\right\\}$
$\displaystyle=\frac{3}{4p_{1}^{2}d_{10}^{3}}\left\\{2p_{1}^{3}(q_{2}^{2}-q_{1}^{2})+4p_{1}q_{3}(q_{1}+q_{2})+d_{340}d_{10}\right\\}.$
(243)

Finally, we have from (35) and (162) that

$\frac{1}{2}E(Q_{n})=\frac{1}{2}+n^{-1}R+O(n^{-3/2}),$ (244)

where with $\rho_{1}=n^{1/2}E(f_{1})$ as in (228) and $\rho_{2}=E(f_{2})$ as
in (243), we have

$R=\rho_{0}+\rho_{1}+\rho_{2},$ (245)

where $\rho_{0}$ is as in (196).

This expression may be simplified into

$R=-\frac{K}{24d_{10}^{3}p_{1}},$

where

$\displaystyle K$
$\displaystyle=6d_{10}^{2}p_{1}\\{-2(p_{11}+6p_{11}s_{0;11}+2s_{1})+p_{1}(2+12s_{1}+5s_{1}^{2}-17s_{2})\\}$
$\displaystyle+3\Big{(}4p_{131}^{2}-8p_{1}p_{131}\\{p_{21}+p_{11}(1+s_{0;11}+s_{1;11}^{(12)})\\}$
$\displaystyle+p_{1}^{2}[8p_{131}+p_{21}^{2}+24p_{131}s_{1}+8p_{131}s_{1}^{2}+4p_{11}p_{21}(5+5s_{0;11}-9s_{1}+5s_{1;11}^{(12)})$
$\displaystyle-4p_{11}^{2}\\{15s_{1}^{2}-18s_{1}(1+s_{1;11}^{(12)})+2(1+s_{1;11}^{(12)})^{2}+s_{0;11}(4-6s_{1}+4s_{1;11}^{(12)})\\}-8p_{131}s_{2}]$
$\displaystyle+p_{1}^{4}\\{1+40s_{1}^{3}+9s_{1}^{4}+10s_{2}-15s_{2}^{2}+8s_{1}(3+s_{2})+s_{1}^{2}(58+6s_{2})\\}$
$\displaystyle+2p_{1}^{3}[-p_{21}(7+12s_{1}+s_{1}^{2}-13s_{2})+2p_{11}\\{-3s_{1}^{3}-s_{1}^{2}(17+5s_{1;11}^{(12)})$
$\displaystyle-3s_{1}(5+2s_{1;11}^{(12)}-5s_{2})+s_{0;11}(1+2s_{1}+3s_{1}^{2}-3s_{2})-(1+s_{1;11}^{(12)})(-1+7s_{2})\\}]\Big{)}$
$\displaystyle+d_{10}\Big{(}-8p_{131}-6p_{1761}+3p_{1}(4p_{11}^{2}+4p_{21}-p_{241}+4p_{451}+4p_{131}s_{0;131}^{(11)}+8p_{11}s_{1})$
$\displaystyle+4p_{1}^{2}[-1-3p_{21}-6s_{1}+27p_{21}s_{1}-3s_{1}^{2}+6p_{12}s_{1}s_{0;12}-6p_{12}s_{1;12}$
$\displaystyle-3p_{21}s_{1;21}^{(43)}-s_{2}-6p_{21}s_{0;21}+3p_{11}\\{-1+2s_{0;11}+18s_{1}^{2}+2s_{1;11}+2s_{1;11}^{(12)}+(s_{1;11}^{(12)})^{2}$
$\displaystyle-
s_{1}(19+s_{1;11}+16s_{1;11}^{(12)})+3s_{2}+s_{2;11}^{(151)}-s_{2;121}^{(144)}\\}]$
$\displaystyle+3p_{1}^{3}\\{3-4(-9+m)s_{1}^{2}+4s_{1}^{3}+4(-4+m)s_{2}-4s_{1}(-7+4s_{2}+s_{2}^{(11)})-6s_{3}+4s_{3}^{(11)}\\}\Big{)}.$

Here, via (95) and (166),

$d_{10}=p_{11}-p_{1}(1+s_{1}).$

In particular, for $m=2$, we get

$R=\frac{1}{12\mu_{1}\mu_{2}}(1+9\mu_{1}+9\mu_{2}+15\mu_{1}\mu_{2}).$

##  4 Bartlett correction

In proposition 1 and corollary 2, asymptotically the expectation of
$Q_{n}|\hat{\lambda}>0$ approaches 1, and we write
$E(Q_{\infty}|\hat{\lambda}>0)=1$.

Now, consider a situation where no conditioning is involved. If, for the
unconditional distribution, a result such as proposition 1 is present, the
principle of Bartlett correction is to adjust the asymptotic statistic
$Q_{\infty}$ according to

$Q_{n}\approx\frac{E(Q_{n})}{E(Q_{\infty})}Q_{\infty}\approx(1+n^{-1}R)Q_{\infty}.$
(246)

It is clear from (246) that the first moment of
$\\{E(Q_{n})/E(Q_{\infty})\\}Q_{\infty}$ equals the first moment of $Q_{n}$.
In fact, Lawley (1956) shows that under very general conditions, this is in
fact true even for all higher moments.

Now, define the $\alpha$ percentile of $Q_{n}$, $q_{n}(\alpha)$, through
$\alpha=P(Q_{n}>q_{n}(\alpha))$. Then, from (246),

$\alpha\approx
P\left\\{(1+n^{-1}R)Q_{\infty}>q_{n}(\alpha)\right\\}=P\left\\{Q_{\infty}>\frac{q_{n}(\alpha)}{1+n^{-1}R}\right\\}.$

Now, in the standard situation where $Q_{\infty}$ is $\chi^{2}_{1}$
distributed, we have
$\alpha=P\\{Q_{\infty}>\chi^{2}_{1}(\alpha)\\}$, i.e.

$q_{n}(\alpha)\approx(1+n^{-1}R)\chi^{2}_{1}(\alpha).$ (247)

However, in our case we have that the distribution of $Q_{\infty}$ conditional
on $\hat{\lambda}>0$ is $\chi^{2}_{1}$. With $\pi_{n}=P(\hat{\lambda}>0)$, we
have

$\alpha=P\\{Q_{n}>q_{n}(\alpha)\\}=\pi_{n}P\\{Q_{n}>q_{n}(\alpha)|\hat{\lambda}>0\\},$

i.e. $\pi_{n}^{-1}\alpha=P\\{Q_{n}>q_{n}(\alpha)|\hat{\lambda}>0\\}$. Then,
analogous to (247), we have

$q_{n}(\alpha)\approx(1+n^{-1}R)\chi^{2}_{1}(\pi_{n}^{-1}\alpha).$ (248)

Furthermore, we have that as $n\to\infty$, $\pi_{n}\to 1/2$. In a practical
situation, we do not know $\pi_{n}$. Neither do we know $R$, but we can
replace it by an estimate, $\hat{R}$, where we insert $\bar{y}_{j}$ for
$\mu_{j}$, for all $j=1,2,...,m$. In particular, if $m=2$, corollary 2 gives
us

$\hat{R}=\frac{1}{6\bar{y}_{1}\bar{y}_{2}}(1+9\bar{y}_{1}+9\bar{y}_{2}+15\bar{y}_{1}\bar{y}_{2}).$
(249)

Hence, putting $\pi_{n}=1/2$, the practical analogue of (248) is

$\hat{q}_{n}(\alpha)=(1+n^{-1}\hat{R})\chi^{2}_{1}(2\alpha).$ (250)

##  5 Simulations

To evaluate the Bartlett correction in a simulation study, we may compare
estimates of

  1. 1.

$P\\{Q_{n}>\chi^{2}_{1}(2\alpha)\\}$

  2. 2.

$P\\{Q_{n}>(1+n^{-1}\hat{R})\chi^{2}_{1}(2\alpha)\\}$

  3. 3.

$P\\{Q_{n}>(1+n^{-1}R)\chi^{2}_{1}(2\alpha)\\}$

  4. 4.

$P\\{Q_{n}>(1+n^{-1}R)\chi^{2}_{1}(\hat{\pi}_{n}^{-1}\alpha)\\}$

These should go from worse to better. Note that the first one is the
asymptotic approximation, the second one is the workable version of the
Bartlett approximation while the other two are not useful in practice, but
interesting by means of comparison. In the last one, $\hat{\pi}_{n}$ simply
denotes the empirical proportion of positive $\hat{\lambda}$ in the
simulation.

To check the validity and usefulness of the Bartlett correction, we have
performed a small simulation study. Matlab R2109a was used. In each case, 1
000 000 replications were run. Optimization was performed using the procedure
fmincon, and program codes are available upon request.

Below, in each case studied we depict expected simulated statistics together
with the Bartlett approximation from proposition 1 (corollary 2 for $m=2$).
Observe that asymptotically, $E(Q_{n})$ tends to $1/2$, and in the figures we
give $2E(Q_{n})$ as the expectation. This is motivated by the fact that
$E(Q_{n})=\pi_{n}*E(Q_{n}|\hat{\lambda}>0)$, i.e. $2E(Q_{n})\approx
E(Q_{n}|\hat{\lambda}>0)$ when $\pi_{n}\approx 1/2$. Because of this, we give
the fractions of positive $\hat{\lambda}$ for different models in a separate
graph at the end of the section. We also separately graph the simulated
variances of $Q_{n}$ for the different models.

Casewise, we also graph the empirical rejection probabilities (erp) under the
null when using the asymptotic distribution, the Bartlett correction, the
simulated Bartlett correction or (in some cases) the latter which, in
addition, takes the simulated proportion of positive $\hat{\lambda}$ into
account. In all cases, the nominal test sizes are 5%.

We start by looking at dimension $m=2$. The first case studied is when
$\mu_{1}=\mu_{2}=1$. The sample size goes from $n=20$ in steps of 20 up to
$n=140$. We give the expected statistic in figure 1. We find that for small
$n$, the Bartlett approximation overestimates the simulated expectation, but
the accuracy improves as $n$ grows. Part of the explanation of the deviation
is that the fraction of positive $\hat{\lambda}$, shown in figure 9, is lower
than 0.5 for small samples, which means that the up weighting of the simulated
$E(Q_{n})$ by a factor 2 is too little in these cases.

As for the empirical test sizes (in per cent) in figure 2, we can see that the
asymptotic erp values are too high for small $n$. This is no surprise, since
the asymptotic expectation is too large, and also the simulated variances (see
figure 10) are large. In view of figure 1, it is also to be expected that the
Bartlett approximation over corrects, producing sizes a bit below the nominal
5%. The simulated Bartlett correction does not suffer from this problem, and
it also produces fairly accurate erp values.

It may perhaps be a bit surprising to see that, when taking positive
$\hat{\lambda}$ proportions into account, we get slight over estimations. But
this is probably due to the fact that the simulated variances are a bit large
for small $n$, see figure 10. A large variance shifts the right tail of the
distribution to the right, whereas a small fraction of positive
$\hat{\lambda}$ shifts it to the left. In this case, when it comes to the
simulated Bartlett correction with no correction for $\hat{\lambda}$, it seems
that these two effects compensate each other. Imposing the $\hat{\lambda}$
correction, only the variance effect remains, and we get over estimation.

The cases $\mu_{1}=\mu_{2}=5$ and $\mu_{1}=\mu_{2}=20$ are plotted in figures
3-4 and 5-6, respectively. Here, the Bartlett corrections turn out to work
much better. Observe that $\hat{\lambda}$ corrected values are not given in
these figures, the reason being that $\hat{\lambda}$ is so close to 0.5 (see
figure 9) so that these are virtually the same as the values obtained by the
simulated Bartlett correction.

Regrettably, for dimensions $m>2$ the Bartlett correction does not seem so
useful in practice. One example is the case with $m=3$ and
$\mu_{1}=\mu_{2}=\mu_{3}=1$, that is depicted in figures 7 and 8. As is seen
from figure 7, the main problem is that, for the studied sample sizes $n$, the
simulated expectation does not tend to one in a linear way as $n$ increases.
(Supposedly, it does so for large enough $n$, but then the asymptotic
distribution probably works so well that Bartlett correction is not relevant.)

Still, it seems from figure 8 that Bartlett correction works quite well for
$n$ values of 100 and higher. But this is merely a stroke of luck, because as
is seen from figures 9 and 10, the proportions of positive $\hat{\lambda}$ as
well as the variances of $Q_{n}$ are low here, and this compensates for the
fact that the Bartlett approximation over estimates the simulated expectation.
This is also why the simulated Bartlett correction over estimates the nominal
test size, even more so when imposing the $\hat{\lambda}$ correction.

Many more cases were simulated by the author, and further results may be
obtained at request.

Finally, figure 9 gives the estimated proportions of positive $\hat{\lambda}$
for the models studied, and here we see that these are closer to 0.5 in cases
with larger $\mu_{j}$ and further away in the dimension 3 case. In figure 10,
we find that the variances of $Q_{n}$ are very similar in the dimension 2
cases, and a little smaller in the dimension 3 case. Observe that, as is
easily seen, the asymptotic variance of $Q_{n}$ is $5/4=1.25$.

##  6 Conclusion

In the present paper, we have presented Bartlett corrections for the maximum
log likelihood test of independence in the multivariate Poisson model proposed
by e.g. Karlis (2003). This topic may be considered of interest, if not only
by itself, so for giving an example of the Bartlett correction in the non
standard type of situation where the parameter of the null hypothesis lies on
the border of the parameter space.

In general, the resulting Bartlett correction factor is given by a very
complicated formula, although explicit. But in the special case of dimension
two (which is the case to be applied in Larsson, 2020), it simplifies into a
nice and short expression. Our simulation studies also indicate that it is in
dimension two that the correction is useful in practice. In higher dimensions,
it seems that for sample sizes of practical interest, the expected statistic
is far from a linear function of the sample size.

One possible improvement of the correction would be to incorporate the
probability of a positive parameter estimate in some way. Another improvement
could be to make the correction non linear. Computer intensive methods could
also come into play. These would be interesting scopes for future research on
the subject.

## References

Bartlett, M.S. (1937) Properties of sufficiency and statistical tests.
Proceedings of the Royal Statistical Society, Series A, 160, 268-282.
Hayakawa, T. (1977) The likelihood ratio criterion and the asymptotic
expansion of its distribution. Annals of the Institute of Statistical
Mathematics 29, 359-378. Karlis, D. (2003) An EM algorithm for multivariate
Poisson distribution and related models. Journal of Applied Statistics 30,
63-77. Larsson, R. (2020) Discrete factor analysis using a dependent Poisson
model. Computational Statistics 35, 1133-1152. Lawley, D.N. (1956) A general
method for approximating to the the distribution of the likelihood ratio
criteria. Biometrika 43, 295-303. Self, S.G., Liang, K.Y. (1987) Asymptotic
properties of maximum likelihood estimators and likelihood ratio tests under
nonstandard conditions. Journal of the American Statistical Association 82,
605-610.