Bias Reduction for Moment Estimates and MLEs

1. Introduction and Summary

Bias reduction for a functional $T\left(F\right)$ has a long history. While unbiased estimates do not exist for ratios of means, nor for standardised moments and cumulants, they do exist for $T\left(F\right)$ a product of moments. Fisher and others developed unbiased estimates for products of moments and cumulants. See Chapters 12 and 13 of [1]. These have the form

$$\begin{array}{cc}& {\widehat{T}}_{nk}=T_{nk}\left(F_n\right)={\displaystyle \sum _{i=0}^{k-1}}{S}_i\left(F_n\right)/{(n-1)}_i \mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e} S_0\left(F\right)=T\left(F\right),\hfill \\ & {(n-1)}_0=1,\mathrm{a}\mathrm{n}\mathrm{d} {(n-1)}_i=(n-1)(n-2)\dots (n-i)\mathrm{f}\mathrm{o}\mathrm{r} i\ge 1,\hfill \end{array}$$

(1)

and $F_n\left(x\right)$ is the empirical distribution of a sample of size n from a distribution $F\left(x\right)$. Unbiased estimates of cumulants are known as k-stastistics or poly-kays when multivariate. I call such an estimate a kth order Fisher estimate.

A 2nd analytic method was given in [2]. This was for general parametric problems with estimates of the form

$${\widehat{T}}_{nk}={\displaystyle \sum _{i=0}^{k-1}}{S}_i\left(F_n\right)/n^i.$$

It can in principle reduce bias to $O\left(n^{-k}\right)$ for any k, although the work increases rapidly with k and p, where p is the number of unknown parameters. I call such estimates a kth order Taylor series estimate. Examples in [2] include any function of a vector mean, and the reliability function.

Examples in [3] include any function of a vector mean, say $g\left(\mu \right)$,

$\left(E{a}^{\prime }X\right)/\left(E{b}^{\prime }X\right)$ for given $a,b\in R^s$, ${\left(E{a}^{\prime }X\right)}^p$ for given p, ${\Pi }_{j=2}^q{\mu }_j^{p_j},$${\mu }_r^p$ and ${\mu }_2^p$, $g\left({\mu }_2\right)$, ${\mu }_2^q$ and $\sigma$, $\mu /\sigma$, return periods, conditional means, and conditional exceedances, and the correlation of a bivariate distribution and its square. Its Tables 1–3 showed the superiority of ${\widehat{T}}_{n2}$ over ${\widehat{T}}_{n1}=T\left(F_n\right)$ for ${\mu }^{-1},{\mu }_4$ and $\sigma$ for various F. [4] gave estimates of low bias for inverse powers of a normal mean. [5] gave estimates of low bias for the multivariate normal. [6] gave bias-reduced estimates for skewness, kurtosis, L-skewness and L-kurtosis. [7] used the method to identify the bias of M-estimators in regression. See also [8].

rEf. [9] extended Fisher estimates, (1) to any $T\left(F\right)$. It showed that they consistently outperform bootstrapping, jackknifing and those due to [10,11,12]. It also provided computer programs in MAPLE for implementation of these estimates. For those seeking simulations, see [9]. Examples included $g\left(\mu \right)$ for a vector $\mu$ ${\mu }_1/{\mu }_2,$${\mu }_r$, $g(\mu ,{\mu }_2)$, $\mu /\sigma$, $g({\mu }_2,{\mu }_3)$, ${\mu }_3/{\sigma }^3$.

This Fisher method is the method I use in this paper. As a rule of thumb, ${\widehat{T}}_{n2}$ gives a quick improvement to $T\left(F_n\right)$, and ${\widehat{T}}_{n3}$ is a further improvement but requires more work, while ${\widehat{T}}_{n4}$ requires a lot more work.

Two popular methods for bias reduction are the jackknife and the bootstrap. For Quenouille’s jackknife, see Sections 5–8 of [13,14,15] for applications to maximum likelihood estimates (MLEs). See [16] for applications to variances. For the bootstrap method, see [17,18,19,20]. For a comparison of these methods, see [21]. As noted there, the bootstrap and the jackknife have bad computational efficiency compared to the analytic method in [2], and they only give second order bias reduction. That is, they only reduce the bias from $O\left(n^{-1}\right)$ to $O\left(n^{-2}\right)$, that is, from magnitude $n^{-1}$ to $n^{-2}$, where n is typically the sample size, or the minimum sample size in the case of more than one sample.

An early book on bias reduction is [22]. This includes applications to treatments, randomized and case-control studies, ANOVA, Logit analysis, log-linear analysis, and survival analysis. Other papers concerned with bias and bias reduction are [23,24,25,26,27,28,29,30,31,32,33,34,35].

I call any estimate with bias $O\left(n^{-k}\right)$ or $o\left(n^{-k+1}\right)$, a kth order estimate, where by $a_n=O\left(b_n\right)$ or $a_n\sim b_n$, I mean that $a_n/b_n$ is bounded, and by $a_n=o\left(b_n\right),$ I mean that $a_n/b_n\to 0$ as $n\to \infty .$ Here I give formulas for 4th order estimates of the Fisher form (1).

Ref. [9] gave a kth order estimate for $T\left(F\right)$ for $k\le 7$ in terms of its (functional) derivatives. Theorem 2.1 spells this out in detail for $k\le 4$. Let $g:R^q\to R$ be any function with finite (functional) derivatives at $U\left(F\right)\in R^q$. Theorem 2.2 gives a chain rule for $T\left(F\right)=g\left(U\right(F\left)\right)$. Theorem 2.3 applies this chain rule to give the terms needed for a kth order estimate of $g\left(U\right(F\left)\right)$ for $k\le 4$ in terms of the derivatives of $U\left(F\right)$. A k sample version of this was given without proof in (A.8)–(A.15) of [3].

Corollary 2.1 specializes to $q=1$, that is, $U\left(F\right)\in R$. Corollary 2.2 and Corollary 2.3 apply Theorem 2.3 to $U\left(F\right)$ a vector mean and a scalar mean. Example 2.2 is for ${\mu }_1/{\mu }_2$ for a bivariate distribution. Example 2.3 is for the mean of a gamma distribution using the MLE. Corollary 2.4 is for any function of ${\mu }_2$, and Example 2.4 is for $\sigma$. Corollary 2.5 is for any function of $(\mu ,{\mu }_2)$. Examples 2.5 and 2.6 are for ${\mu }_2/\mu$ and ${\mu }^2/{\mu }_2$, and cover moment estimates for the scaled Gamma distribution. Example 2.7 is for the moment estimate $\mu /{\mu }_2$. Example 2.8 is for the MLEs of the scaled Gamma distribution.

Section 3 gives a discussion and suggests future research directions. Theorems A.1 and A.2 extend the chain rule in Theorem 2.3 to $U\left(F\right)=h\left(V\right(F\left)\right)$. It can be applied to avoid the double use of chain rules in Example 2.7.

The method does not apply to $T\left(F\right)$ a quantile, such as the median of $F\left(x\right)$. To see the wide variety of possible expansions for moments of sample quantiles, see [36,37].

2. Bias Reduction for Moment Estimates

Let $F\left(x\right)$ be a distribution on $R^s$ where $s\ge 1$. Given a functional $T\left(F\right)$, its rth functional derivative, $T_{\cdot 1\dots r}=T_F(x_1,\dots ,x_r)$, was defined in [38]. The following theorem is given in [9].

Theorem 1. 

Let $F_n\left(x\right)$ be the empirical distribution of a random sample of size n from a distribution $F\left(x\right)$ on $R^s$. Let $T\left(F\right)$ be a functional with finite derivatives $T_{\cdot 1\dots r}.$ Then

$$for \widehat{T}=T\left(F_n\right),E\widehat{T}=T\left(F\right)+O\left(n^{-1}\right).$$

For $1\le k\le 7$, a kth order estimate of $T\left(F\right)$, that is, an estimate with bias $O\left(n^{-k}\right)$, is

$${\widehat{T}}_{nk}={\displaystyle \sum _{i=0}^{k-1}}{S}_i\left(F_n\right)/{(n-1)}_i where S_0\left(F\right)=T\left(F\right),$$

and $S_i\left(F\right)$ is given by (3.4)–(3.9) there for $i\le 6.$ So for $1\le k\le 4$, a kth order estimate of $T\left(F\right)$ is ${\widehat{T}}_{nk}$ where

$$\begin{array}{cc}& {\widehat{T}}_{n1}=\widehat{T},{\widehat{T}}_{n2}=\widehat{T}+S_1\left(F_n\right)/(n-1),\hfill \\ & {\widehat{T}}_{n3}=\widehat{T}+S_1\left(F_n\right)/(n-1)+S_2\left(F_n\right)/{(n-1)}_2, and\hfill \\ & {\widehat{T}}_{n4}=\widehat{T}+S_1\left(F_n\right)/(n-1)+S_2\left(F_n\right)/{(n-1)}_2+S_3\left(F_n\right)/{(n-1)}_3, where\hfill \end{array}$$

$$\begin{array}{cc}& S_1\left(F\right)=-T\left[2\right]/2,T\left[r\right]=\int T_{\cdot 1^r}d{F}_1,F_i=F\left(x_i\right),\hfill \end{array}$$

(1)

$$\begin{array}{cc}& S_2\left(F\right)=T\left[3\right]/3+T\left[2^2\right]/8,T\left[2^2\right]=\int \int T_{\cdot 1122}d{F}_{1}d{F}_2,\hfill \end{array}$$

(2)

$$\begin{array}{cc}& S_3\left(F\right)=-T\left[4\right]/4+3T\left[2^2\right]/8-T\left[32\right]/6-T\left[2^3\right]/48,\hfill \\ & T\left[32\right]=\int \int T_{\cdot 11122}d{F}_{1}d{F}_2,T\left[2^3\right]=\int \int \int T_{\cdot 112233}d{F}_{1}d{F}_{2}d{F}_3.\hfill \end{array}$$

(3)

(1) uses $1^r$ for a string of 1s. For example, $T_{\cdot 1^3}=T_{\cdot 111}.$ Similarly I write $T_{\cdot 1^3{2}^2}=T_{\cdot 11122}$, and $T_{\cdot 1^2{2}^2{3}^2}=T_{\cdot 112233}$. Readers may recall that $\int T_{\cdot 1}^{2}d{F}_1$ is the asymptotic variance of $n^{1/2}\left([T\left(F_n\right)-T\left(F\right)]\right).$

Note 2.1. 

For small n, the components of ${\widehat{T}}_{nk}$ of (1) may start to diverge. In that case, truncation is necessary. For example the 2nd term is less than half the first term if

$$n>n_0=1+|T\left[2\right]/T\left(F\right)|.$$

Example 1. 

Let $a\left(x\right):R^s:\to R$ be a given function, and $X\sim F\left(x\right)$ on $R^s$. Let $(\mu \left(F\right),{\mu }_2\left(F\right),{\mu }_3\left(F\right))=(\mu ,{\mu }_2,{\mu }_3)$ be the mean, variance and 3rd central moment of $a\left(X\right)$. By Appendix D of [3], their non-zero derivatives are

$$\begin{array}{cc}& {\mu }_{\cdot 1}=h_1,{\mu }_{2\cdot 1}=h_1^2-{\mu }_2,{\mu }_{2\cdot 12}=-2h_1{h}_2,{\mu }_{3\cdot 1}=h_1^3-{\mu }_3-3h_2{\mu }_2,\hfill \end{array}$$

$$\begin{array}{l}{\mu }_{3.12}=-3(h_1^2-{\mu }_2)h_2-3(h_2^2-{\mu }_2)h_1,{\mu }_{3.123}=12h_1{h}_2{h}_3,\hfill \\ \mathit{where} h_1=a\left(x_i\right)-\mu .\end{array}$$

(4)

So if $T\left(F\right)=\mu ,$ then $S_1\left(F\right)=0$ and ${\widehat{T}}_{n1}=\mu \left(F_n\right)$ is an unbiased estimate of μ If $T\left(F\right)={\mu }_2,$ then $T\left[2\right]=-2{\mu }_2$, $S_1\left(F\right)={\mu }_2,S_2\left(F\right)=0$, and

$${\widehat{T}}_{n2}={\mu }_2\left(F_n\right)[1+1/(n-1)]={\mu }_2\left(F_n\right)n/(n-1)$$

is an unbiased estimate of ${\mu }_2$. If $T\left(F\right)={\mu }_3,$ then $T\left[2\right]=-6{\mu }_3,T\left[3\right]=-12{\mu }_3$, $S_1\left(F\right)=3{\mu }_3,S_2\left(F\right)=4{\mu }_3,S_3\left(F\right)=0$, and

$${\widehat{T}}_{n3}={\mu }_3\left(F_n\right)[1+3/(n-1)+4/{(n-1)}_2]={\mu }_3\left(F_n\right)n^2/{(n-1)}_2$$

is an unbiased estimate of ${\mu }_3$. These unbiased estimates of ${\mu }_2$ and ${\mu }_3$ are Fisher’s $k_2$ and $k_3$. The derivatives of ${\mu }_r$ are given in Appendix D of [3] for $r\le 6,$ and for general r by (5.4) there.

Note 2.2. 

${\widehat{T}}_{nk}$ is an unbiased estimate of $T\left(F\right)$ if $T\left(F\right)$ is a polynomial of degree k or less in F, that is, a product of moments of total degree k or less. For example, ${\widehat{T}}_{nk}$ is an unbiased estimate of $T\left(F\right)={\mu }^k$ or ${\mu }_k$ or ${\kappa }_k$ where μ is the mean and ${\mu }_k,{\kappa }_k$ are the moments and cumulants of $a\left(X\right)$, where $a\left(X\right)$ is any real function of $X\sim F\left(x\right)$ on $R^s$.

To apply the theorem to a function of the moments of $F\left(x\right)$, for example to $T\left(F\right)=\mu /{\mu }_2^{1/2}$, we need a chain rule.

Chain rules for $T\left(x\right)=g\left(U\right(x\left)\right)$ and for $T\left(F\right)=g\left(U\right(F\left)\right)$.

(I) First consider $U\left(x\right),g\left(U\right):R\to R$. Set $T\left(x\right)=g\left(U\right(x\left)\right)$. Faa di Bruno’s chain rule for $T_{\cdot r}$, the rth derivative of $T\left(x\right)$, is given by [4c] of [39] as

$$\begin{array}{cc}& T_{\cdot r}={\displaystyle \sum _{k=1}^r}{g}_k{B}_{rk}(U_{\cdot 1},\dots ,U_{\cdot k}) \mathrm{f}\mathrm{o}\mathrm{r} r\ge 1,\hfill \end{array}$$

(5)

where $g_k$ is the kth derivative of $g\left(U\right)$, $U_{\cdot k}$ is the kth derivative of $U\left(x\right),$ and $B_{rk}=B_{rk}(U_{\cdot 1},\dots ,U_{\cdot k})$ is the partial exponential Bell polynomial tabled on p307 of [39]. So for $r\le 6$, the rth derivative, $T_{\cdot r}$, is given by

$$\begin{array}{cc}& B_{r1}=U_{\cdot r},B_{rr}=U_{\cdot 1}^r,B_{32}=3U_{\cdot 1}{U}_{\cdot 2},B_{42}=4U_{\cdot 1}{U}_{\cdot 3}+3U_{\cdot 2}^2,B_{43}=6U_{\cdot 1}^2{U}_{\cdot 2},\hfill \\ & B_{52}=5U_{\cdot 1}{U}_{\cdot 4}+10U_{\cdot 2}{U}_{\cdot 3},B_{53}=10U_{\cdot 1}^2{U}_{\cdot 3}+15U_{\cdot 1}{U}_{\cdot 2}^2,B_{54}=10U_{\cdot 1}^3{U}_{\cdot 2},\hfill \\ & B_{62}=6U_{\cdot 1}{U}_{\cdot 5}+15U_{\cdot 2}{U}_{\cdot 4}+10U_{\cdot 3}^2,B_{63}=15U_{\cdot 1}^2{U}_{\cdot 4}+60U_{\cdot 1}{U}_{\cdot 2}{U}_{\cdot 3}+15U_{\cdot 2}^3,\hfill \\ & B_{64}=20U_{\cdot 1}^3{U}_{\cdot 3}+45U_{\cdot 1}^2{U}_{\cdot 2}^2,B_{65}=15U_{\cdot 1}^4{U}_{\cdot 2}.\hfill \end{array}$$

For example, the 1st 3 derivatives of $T\left(x\right)=g\left(U\right(x\left)\right)$ are

$$\begin{array}{cc}& T_{\cdot 1}=g_1{U}_{\cdot 1},T_{\cdot 2}=g_1{U}_{\cdot 2}+g_2{U}_{\cdot 1}^2,T_{\cdot 3}=g_1{U}_{\cdot 3}+3g_2{U}_{\cdot 1}{U}_{\cdot 2}+g_3{U}_{\cdot 1}^3.\hfill \end{array}$$

(II) Now consider

$$\begin{array}{cc}& U\left(x\right):R^s\to R^q,g\left(U\right):R^q\to R. Set T\left(x\right)=g\left(U\left(x\right)\right).\hfill \end{array}$$

(6)

Let $U_{i\cdot 1\dots r}$ be the rth partial derivative of $U_i\left(x\right)$. Let $g_{1\dots k}$ be the kth partial derivative of $g\left(U\right)$ at $U=U\left(x\right)$

The rth partial derivative of $T\left(x\right)$ can be read off (5) by replacing $g_k$ by the partial derivative $g_{1\dots k}$, and the coefficients by a sum. That is,

Theorem 2. 

The rth derivative of $T\left(x\right)$ of (6) is

$$\begin{array}{cc}& T_{\cdot 1\dots r}={\displaystyle \sum _{k=1}^r}{g}_{i_1\dots i_k}{B}_k^r for B_k^r=B_{i_1\dots i_k}^{1\dots r}=B_{i_1\dots i_k}^{1\dots r}\left(U\right) as follows.\hfill \\ & B_1^r=U_{i_1\cdot 1\dots r},B_r^r=U_{i_1\cdot 1}\dots U_{i_r\cdot r},\hfill \end{array}$$

(7)

$$\begin{array}{cc}& B_2^3=\sum ^3{U}_{i_1\cdot 1}{U}_{i_2\cdot 23}=U_{i_1\cdot 1}{U}_{i_2\cdot 23}+U_{i_1\cdot 2}{U}_{i_2\cdot 13}+U_{i_1\cdot 3}{U}_{i_2\cdot 12}=S_3 say,\hfill \end{array}$$

(8)

$$\begin{array}{cc}& B_2^4=S_4+S_3^{\prime } where S_4=\sum ^4{U}_{i_1\cdot 1}{U}_{i_2\cdot 234},\hfill \end{array}$$

(9)

$$\begin{array}{cc}& S_3^{\prime }=\sum ^3{U}_{i_1\cdot 12}{U}_{i_2\cdot 34}=U_{i_1\cdot 12}{U}_{i_2\cdot 34}+U_{i_1\cdot 13}{U}_{i_2\cdot 24}+U_{i_1\cdot 14}{U}_{i_2\cdot 23},\hfill \end{array}$$

(10)

$$\begin{array}{cc}& B_3^4=\sum ^6{U}_{i_1\cdot 1}{U}_{i_2\cdot 2}{U}_{i_3\cdot 34}=S_6 say,\hfill \end{array}$$

(11)

$$\begin{array}{cc}& B_2^5=\sum ^5{U}_{i_1\cdot 1}{U}_{i_2\cdot 2345}+\sum ^{10}{U}_{i_1\cdot 12}{U}_{i_2\cdot 345}=S_5+S_{10} say,\hfill \end{array}$$

(12)

$$\begin{array}{cc}& B_3^5=\sum ^{10}{U}_{i_1\cdot 1}{U}_{i_2\cdot 2}{U}_{i_3\cdot 345}+\sum ^{15}{U}_{i_1\cdot 1}{U}_{i_2\cdot 23}{U}_{i_3\cdot 45}=S_{10}^{\prime }+S_{15} say,\hfill \end{array}$$

(13)

$$\begin{array}{cc}& B_4^5=\sum ^{10}{U}_{i_1\cdot 1}{U}_{i_2\cdot 2}{U}_{i_3\cdot 3}{U}_{i_4\cdot 45}=S_{10}^{\prime \prime } say,\hfill \\ & B_2^6=\sum ^6{U}_{i_1\cdot 1}{U}_{i_2\cdot 23456}+\sum ^{15}{U}_{i_1\cdot 12}{U}_{i_2\cdot 3456}+\sum ^{10}{U}_{i_1\cdot 123}{U}_{i_2\cdot 456}\hfill \end{array}$$

(14)

$$\begin{array}{cc}& =S_6^{\prime }+S_{15}^{\prime }+S_{10}^{\prime \prime \prime } say,\hfill \\ & B_3^6=\sum ^{15}{U}_{i_1\cdot 1}{U}_{i_2\cdot 2}{U}_{i_3\cdot 3456}+\sum ^{60}{U}_{i_1\cdot 1}{U}_{i_2\cdot 23}{U}_{i_3\cdot 456}+\sum ^{15}{U}_{i_1\cdot 12}{U}_{i_2\cdot 34}{U}_{i_3\cdot 56}\hfill \end{array}$$

(15)

$$\begin{array}{cc}& =S_{15}^{\prime \prime }+S_{60}+S_{15}^{\prime \prime \prime } say,\hfill \\ & B_4^6=\sum ^{20}{U}_{i_1\cdot 1}{U}_{i_2\cdot 2}{U}_{i_3\cdot 3}{U}_{i_4\cdot 456}+\sum ^{45}{U}_{i_1\cdot 1}{U}_{i_2\cdot 2}{U}_{i_3\cdot 34}{U}_{i_4\cdot 56}=S_{20}+S_{45}say,\hfill \end{array}$$

(16)

$$\begin{array}{cc}& B_5^6=\sum ^{15}{U}_{i_1\cdot 1}{U}_{i_2\cdot 2}{U}_{i_3\cdot 3}{U}_{i_4\cdot 4}{U}_{i_5\cdot 56}=S_{15}^{\prime \prime \prime \prime } say,\hfill \end{array}$$

(17)

${\sum }^N{f}_{1\dots r}$ sums over all N permutations of $1\dots r$ giving distinct terms, as in (8) and (10), and I use the tensor summation convention of summing repeated pairs $i_1,i_2,\dots$ over their range $1,2,\dots ,q$. In summary,

$$\begin{array}{cc}& T_{\cdot 1}=g_{i_1}{U}_{i_1\cdot 1}={\displaystyle \sum _{i_1=1}^q}{g}_{i_1}{U}_{i_1\cdot 1},T_{\cdot 12}=g_{i_1}{U}_{i_1\cdot 12}+g_{i_1{i}_2}{U}_{i_1\cdot 1}{U}_{i_2\cdot 2},\hfill \end{array}$$

(18)

$$\begin{array}{cc}& T_{\cdot 123}=g_{i_1}{U}_{i_1\cdot 123}+g_{i_1{i}_2}{S}_3+g_{123}{U}_{i_1\cdot 1}{U}_{i_2\cdot 2}{U}_{i_3\cdot 3},\hfill \end{array}$$

(19)

$$\begin{array}{cc}& T_{\cdot 1234}=g_{i_1}{U}_{i_1\cdot 1234}+g_{i_1{i}_2}(S_4+S_3^{\prime })+g_{123}{S}_6+g_{1234}{U}_{i_1\cdot 1}\dots U_{i_4\cdot 4},\hfill \end{array}$$

(20)

$$\begin{array}{cc}& T_{\cdot 1\dots 5}=g_{i_1}{U}_{i_1\cdot 1\dots 5}+g_{i_1{i}_2}(S_5+S_{10})+g_{123}(S_{10}^{\prime }+S_{15})+g_{1234}{S}_{10}^{\prime \prime }\hfill \end{array}$$

$$\begin{array}{cc}& +g_{1\dots 5}{U}_{i_1\cdot 1}\dots U_{i_5\cdot 5},\hfill \\ & T_{\cdot 1\dots 6}=g_{i_1}{U}_{i_1\cdot 1\dots 6}+g_{i_1{i}_2}(S_6^{\prime }+S_{15}^{\prime }+S_{10}^{\prime \prime \prime })+g_{123}(S_{15}^{\prime \prime }+S_{60}+S_{15}^{\prime \prime \prime })\hfill \end{array}$$

(21)

$$\begin{array}{cc}& +g_{1234}(S_{20}+S_{45})+g_{1\dots 5}{S}_{15}^{\prime \prime \prime \prime }+g_{1\dots 6}{U}_{i_1\cdot 1}\dots U_{i_6\cdot 6},\hfill \end{array}$$

(22)

for $S_3,\dots ,S_{15}^{\prime \prime \prime \prime }$ of (8)–(??).

These S will be used repeatedly. Note how in (8) and (10), our convention using ${\sum }^N$ is to keep $i_1,\dots ,i_k$ in that order, and only permute $1\dots r$.

So (7) gives $T_{\cdot 1\dots r}$ in terms of $t_r$ terms, where

$$\begin{array}{cc}& t_1=q,t_2=q+q^2,t_3=q+3q^2+q^3,t_4=q+6q^2+6q^3+q^4,\hfill \\ & t_5=q+15q^2+25q^3+10q^4+q^5,t_6=q+31q^2+90q^3+65q^4+15q^5+q^6.\hfill \end{array}$$

Most of the work in this paper was taken up obtaining formulas for $S_3\left(F\right)$ of (3), as it requires derivatives up to order 6, while $S_2\left(F\right)$ of (2) only requires derivatives up to order 4.

(III) Now suppose that $T\left(F\right)=g\left(U\right(F\left)\right)$ where $F\left(x\right)$ is a distribution function on $R^s$, $g:R^q\to R$ has partial derivatives $g_{i_1{i}_2\dots }$ at $U\left(F\right)\in R^q$, for $1\le i_1,i_2,\dots \le q$. Let $U_{i\cdot 1\dots r}$ be the rth functional derivative of $U_i\left(F\right)$. Then (7) holds with $T_{\cdot 1\dots r}$ the rth functional derivative of $T\left(F\right)$. This chain rule was given in (A.3)–(A.7) of [3]. For example, (4) gives the derivatives of $\mu ,{\mu }_2,{\mu }_3.$

I now apply it to obtain $T\left[2\right],\dots ,T\left[2^3\right]$ needed in Theorem 2.1 for a kth order estimate of $T\left(F\right)$ for $k\le 4$, using the notation

$$\begin{array}{cc}& \left[U_{i\cdot 11}\right]=\int U_{i\cdot 11}d{F}_1,\left[U_{i\cdot 1}{U}_{j\cdot 1}\right]=\int U_{i\cdot 1}{U}_{j\cdot 1}d{F}_1,\hfill \\ & \left[U_{i\cdot 12}{U}_{j\cdot 12}\right]=\int \int U_{i\cdot 12}{U}_{j\cdot 12}d{F}_{1}d{F}_2,\left[U_{i\cdot 1}{U}_{j\cdot 122}\right]=\int \int U_{i\cdot 1}{U}_{j\cdot 122}d{F}_{1}d{F}_2,\hfill \end{array}$$

and so on. (By construction, $\int U_{i\cdot 1\dots r}d{F}_k=0$ for $1\le k\le r$.) I give the following special labels to the terms that occur most often:

$$\begin{array}{cc}& \overline{i}=\left[U_{i\cdot 11}\right],(i_1\dots i_r)=[U_{i_1\cdot 1}\dots U_{i_r\cdot 1}],d_{i_1{i}_2}=\left[U_{i_1\cdot 1}{U}_{i_2\cdot 11}\right],\hfill \end{array}$$

(23)

$$\begin{array}{cc}& \left(i_1\right|i_2)=\left[U_{i_1\cdot 12}{U}_{i_2\cdot 12}\right],(i_1|i_2\left|i_3\right)=\left[U_{i_1\cdot 1}{U}_{i_2\cdot 12}{U}_{i_3\cdot 2}\right].\hfill \end{array}$$

(24)

$\overline{i}$ and $\left(i_1{i}_2\right)$ of (23) are used in $T\left[2\right]$ of (25) below and in $T\left[32\right]$ of (29) below. $d_{i_1{i}_2}$ of (23) and $\left(i_1{i}_2{i}_3\right)$ are used in $T\left[3\right]$ of (26) below. $\left(i_1{i}_2{i}_3\right)$ is also used in $T\left[32\right]$ of (29) below. $\left(i_1\right|i_2)$ and $\left(i_1\right|i_2\left|i_3\right)$ of (24) are used in $T\left[2^3\right]$ of (30) below.

Readers may recognize $\left(i_1{i}_2\right)$ and $(i_1\dots i_r)$ as the limits as $n\to \infty$, of $n\mu ({\widehat{\theta }}_i,{\widehat{\theta }}_j)$ and $n^{r-1}\mu ({\widehat{\theta }}_{i_1},\dots ,{\widehat{\theta }}_{i_r})$ where ${\widehat{\theta }}_i=U_i\left(F_n\right)$.

The next theorem gives the $T\left[\pi \right]$ needed by Theorem 2.1 for an estimate of $T\left(F\right)=g\left(U\right(F\left)\right)$ of bias $\sim n^{-k},k\le 4$, in terms of the derivatives of $U\left(F\right)$.

Theorem 3. 

Suppose that $T\left(F\right)=g\left(U\right(F\left)\right)$ where $g:R^q\to R$ has partial derivatives $g_{i_1{i}_2\dots }$ at $U\left(F\right)\in R^q$, for $1\le i_1,i_2,\dots \le q$.Then, using implicit summation of pairs of $i_1,\dots ,i_k$ over $1,\dots ,q$, the $T\left[\pi \right]$ needed for $S_1\left(F\right),S_2\left(F\right)$ and $S_3\left(F\right)$ of Theorem 2.1 are

$$\begin{array}{cc}& T\left[r\right]={\displaystyle \sum _{k=1}^r}{g}_{i_1\dots i_k}\left[B_{i_1\dots i_k}^{1^r}\right] where \left[B_{i_1\dots i_k}^{1^r}\right]=\int B_{i_1\dots i_k}^{1^r}d{F}_1,\hfill \\ & T\left[rs\right]={\displaystyle \sum _{k=1}^{r+s}}{g}_{i_1\dots i_k}\left[B_{i_1\dots i_k}^{1^r{2}^s}\right] where \left[B_{i_1\dots i_k}^{1^r{2}^s}\right]=\int \int B_{i_1\dots i_k}^{1^r{2}^s}d{F}_{1}d{F}_2,\hfill \\ & T\left[2^3\right]={\displaystyle \sum _{k=1}^6}{g}_{i_1\dots i_k}\left[B_{i_1\dots i_k}^{1^2{2}^2{3}^2}\right] where \left[B_{i_1\dots i_k}^{1^2{2}^2{3}^2}\right]=\int \int \int B_{i_1\dots i_k}^{1^2{2}^2{3}^2}d{F}_{1}d{F}_{2}d{F}_3.\hfill \end{array}$$

Suppressing the dependence of $\left[B_{Nk}\right]=\left[B_{i_1\dots i_k}^{1^{{\pi }_1}{2}^{{\pi }_2}\dots }\right]$ on $i_1,\dots ,i_k$, let us write these simply as

$$\begin{array}{cc}& T\left[\pi \right]={\displaystyle \sum _{k=1}^N}{g}_{i_1\dots i_k}\left[B_{Nk}\right] for \pi =({\pi }_1,{\pi }_2,\dots ) where N={\pi }_1+{\pi }_2+\dots .\hfill \end{array}$$

$$\begin{array}{cc}& Then{S}_1\left(F\right)=-T\left[2\right]/2whereT\left[2\right]=a_{12}+a_{22},\hfill \\ & a_{12}=g_{i_1}\left[B_{21}\right],\left[B_{21}\right]={\overline{i}}_{1}of\left(23\right),a_{22}=g_{i_1{i}_2}\left[B_{22}\right],\left[B_{22}\right]=\left(i_1{i}_2\right)of\left(23\right).\hfill \end{array}$$

(25)

$$\begin{array}{cc}& S_2\left(F\right)of\left(2\right)isgivenbyT\left[3\right]={\displaystyle \sum _{i=1}^3}{a}_{i3},where\hfill \\ & a_{13}=g_i\left[U_{i\cdot 111}\right],a_{23}/3=g_{i_1{i}_2}{d}_{i_1{i}_2},a_{33}=g_{i_1{i}_2{i}_3}\left(i_1{i}_2{i}_3\right)of\left(23\right),\hfill \end{array}$$

(26)

$$\begin{array}{cc}& and T\left[2^2\right]={\displaystyle \sum _{i=1}^4}{a}_{i4}, where a_{14}=g_i\left[B_{41}\right],\left[B_{41}\right]=\left[U_{i\cdot 1122}\right],\hfill \\ & a_{24}=g_{i_1{i}_2}\left[B_{42}\right],\left[B_{42}\right]={\overline{i}}_1{\overline{i}}_2+2\left(i_1\right|i_2)+4\left[U_{i_1\cdot 1}{U}_{i_2\cdot 122}\right],\hfill \\ & a_{34}=2g_{i_1{i}_2{i}_3}\{\left(i_1{i}_2\right){\overline{i}}_3+2(i_1|i_3|i_2\left)\right\},a_{44}=g_{i_1\dots i_4}\left(i_1{i}_2\right)\left(i_3{i}_4\right),of\left(24\right)\hfill \end{array}$$

(27)

$$\begin{array}{cc}& S_3\left(F\right) of \left(3\right) is given by T\left[4\right]={\displaystyle \sum _{k=1}^4} a_k where a_k=g_{i_1\dots i_k}\left[B_{4k}\right]\hfill \\ & \left[B_{41}\right]=\left[U_{i_1\cdot 1111}\right],\left[B_{42}\right]=4\left[U_{i_1\cdot 1}{U}_{i_2\cdot 111}\right]+3\left[U_{i_1\cdot 11}{U}_{i_2\cdot 11}\right],\hfill \\ & \left[B_{43}\right]=6\left[U_{i_1\cdot 1}{U}_{i_2\cdot 1}{U}_{i_3\cdot 11}\right],\left[B_{44}\right]=(i_1\dots i_4)of\left(23\right).\hfill \end{array}$$

(28)

$$\begin{array}{cc}& T\left[32\right]={\displaystyle \sum _{k=1}^5}{b}_k where b_k=g_{i_1\dots i_k}\left[B_{5k}\right],\hfill \\ & \left[B_{51}\right]=\left[U_{i_1\cdot 11122}\right],\hfill \\ & \left[B_{52}\right]=\left[S_5\right]+\left[S_{10}\right],\left[S_5\right]=3\left[U_{i_1\cdot 1}{U}_{i_2\cdot 1122}\right]+2\left[U_{i_1\cdot 2}{U}_{i_2\cdot 1112}\right],\hfill \\ & \left[S_{10}\right]=3\left[U_{i_1\cdot 11}{U}_{i_2\cdot 122}\right]+6\left[U_{i_1\cdot 12}{U}_{i_2\cdot 112}\right]+{\overline{i}}_1\left[U_{i_2\cdot 111}\right],\hfill \\ & \left[B_{53}\right]=\left[S_{10}^{\prime }\right]+\left[S_{15}\right],\left[S_{10}^{\prime }\right]=3\left[U_{i_1\cdot 1}{U}_{i_2\cdot 1}{U}_{i_3\cdot 122}\right]+6\left[U_{i_1\cdot 1}{U}_{i_2\cdot 2}{U}_{i_3\cdot 112}\right]+\left(i_1{i}_2\right)\left[U_{i_3\cdot 111}\right],\hfill \\ & \left[S_{15}\right]=3\left[U_{i_1\cdot 1}{U}_{i_2\cdot 11}\right]{\overline{i}}_3+6\left[U_{i_1\cdot 1}{U}_{i_2\cdot 12}{U}_{i_3\cdot 12}\right]+6\left[U_{i_1\cdot 2}{U}_{i_2\cdot 11}{U}_{i_3\cdot 12}\right],\hfill \\ & \left[B_{54}\right]=\left[S_{10}^{\prime \prime }\right]=\left(i_1{i}_2{i}_3\right){\overline{i}}_4+6\left[U_{i_1\cdot 1}{U}_{i_2\cdot 1}{U}_{i_3\cdot 2}{U}_{i_4\cdot 12}\right]+3\left(i_2{i}_3\right)d_{i_1{i}_4},\hfill \\ & \left[B_{55}\right]=\left(i_1{i}_2{i}_3\right)\left(i_4{i}_5\right)of\left(23\right),\hfill \end{array}$$

(29)

$$\begin{array}{cc}& and T\left[2^3\right]={\displaystyle \sum _{k=1}^6}{c}_k where c_k=g_{i_1\dots i_k}\left[B_{6k}\right],\left[B_{61}\right]=\left[U_{i_1\cdot 112233}\right],\hfill \end{array}$$

(30)

$$\begin{array}{cc}& for c_2 of \left(30\right):\left[B_{62}\right]=\left[S_6^{\prime }\right]+\left[S_{{15}^{\prime }}\right]+\left[S_{10}^{\prime \prime \prime }\right],\left[S_6^{\prime }\right]=6\left[U_{i_1\cdot 1}{U}_{i_2\cdot 12233}\right],\hfill \\ & \left[S_{15}^{\prime }\right]=3{\overline{i}}_1\left[U_{i_2\cdot 1122}\right]+12\left[U_{i_1\cdot 12}{U}_{i_2\cdot 1233}\right],\hfill \\ & \left[S_{10}^{\prime \prime \prime }\right]=6\left[U_{i_1\cdot 112}{U}_{i_2\cdot 233}\right]+4\left[U_{i_1\cdot 123}{U}_{i_2\cdot 123}\right],\hfill \\ & for c_3 of \left(30\right):\left[B_{63}\right]=\left[S_{15}^{\prime \prime }\right]+\left[S_{60}\right]+\left[S_{15}^{\prime \prime \prime }\right],\hfill \\ & \left[S_{15}^{\prime \prime }\right]=3\left(i_1{i}_2\right)\left[U_{i_3\cdot 1122}\right]+12\left[U_{i_1\cdot 1}{U}_{i_2\cdot 2}{U}_{i_3\cdot 1233}\right],\hfill \\ & \left[S_{60}\right]=24\left[U_{i_1\cdot 1}{U}_{i_2\cdot 12}{U}_{i_3\cdot 233}\right]+12\left[U_{i_1\cdot 1}{U}_{i_3\cdot 122}\right]{\overline{i}}_2+24\left[U_{i_1\cdot 1}{U}_{i_2\cdot 23}{U}_{i_3\cdot 123}\right],\hfill \\ & \left[S_{15}^{\prime \prime \prime }\right]=3{\overline{i}}_1{\overline{i}}_2{\overline{i}}_3+4{\displaystyle \sum _{23}^2}\left(i_1\right|i_2){\overline{i}}_3+4\left[U_{i_1\cdot 12}{U}_{i_2\cdot 13}{U}_{i_3\cdot 23}\right] where {\displaystyle \sum _{ij}^2}{f}_{ij}=f_{ij}+f_{ji},\hfill \end{array}$$

(31)

$$\begin{array}{cc}& for c_4 of \left(30\right):\left[B_{64}\right]=\left[S_{20}\right]+\left[S_{45}\right],\hfill \\ & \left[S_{20}\right]=12\left(i_1{i}_2\right)\left[U_{i_3\cdot 1}{U}_{i_4\cdot 122}\right]+8\left[U_{i_1\cdot 1}{U}_{i_2\cdot 2}{U}_{i_3\cdot 3}{U}_{i_4\cdot 123}\right],\hfill \\ & \left[S_{45}\right]=3\left(i_1{i}_2\right)\{{\overline{i}}_3{\overline{i}}_4+2\left(i_3\right|i_4)\}+12(i_1|i_3\left|i_2\right){\overline{i}}_4+24\left[U_{i_1\cdot 1}{U}_{i_2\cdot 2}{U}_{i_3\cdot 13}{U}_{i_4\cdot 23}\right],\hfill \end{array}$$

$$\begin{array}{cc}& for{c}_5 of \left(30\right):\left[B_{65}\right]=3\left(i_1{i}_2\right)\left(i_3{i}_4\right){\overline{i}}_5+12\left(i_1{i}_2\right)(i_3|i_5\left|i_4\right),\hfill \end{array}$$

(32)

$$\begin{array}{cc}& for{c}_6 of \left(30\right):\left[B_{66}\right]=\left(i_1{i}_2\right)\left(i_3{i}_4\right)\left(i_5{i}_6\right).\hfill \end{array}$$

(33)

Note 2.3. 

This gives 3 corrections to p274 of [3]. (i) $\left[S_{15}^{\prime \prime \prime }\right]$ is replaced by ${\overline{i}}_1{\overline{i}}_2{\overline{i}}_3+2{\sum }^3\left(i_1\right|i_2){\overline{i}}_3+8\left[U_{i_1\cdot 12}{U}_{i_2\cdot 13}{U}_{i_3\cdot 23}\right].$ (ii) The factor 12 was dropped in $\left[B_{65}\right]=B^{i_1\dots i_5}$ . (iii) The above expression for $c_5$ shows that a factor 4 should be added to the 2nd term in $B^{i_1,\dots ,i_5}$.

Note 2.4. 

We can write the components of these $T\left[\pi \right]$ symbolically as follows.

$T\left[2\right]$ and $S_1\left(F\right)$ need terms 1 and 2 $\sim 11,1.1,$ for $\overline{i}=\left[U_{i\cdot 11}\right]$ and $\left(i_1{i}_2\right)=\left[U_{i_1\cdot 1}{U}_{i_2\cdot 1}\right]$.

$T\left[3\right]$ needs terms 3,4,5 $\sim 111,1.11,1.1.1$ for $\left[U_{i\cdot 111}\right],d_{i_1{i}_2}=\left[U_{i_1\cdot 1}{U}_{i_2\cdot 11}\right],$ and $\left(i_1{i}_2{i}_3\right)=\left[U_{i_1\cdot 1}{U}_{i_2\cdot 1}{U}_{i_3\cdot 1}\right]$.

$T\left[2^2\right]$ needs terms $6-9\sim 1122,12.12,1.122,1.2.12$. These give $S_2\left(F\right)$.

$T\left[4\right]$ needs terms $10-14\sim 1111,1.111,11.11,1.1.11,1.1.1.1.$

$T\left[32\right]$ needs terms $15-24\sim 11122,1.1122,2.1112,11.122,12.112,1.1.122,1.2.112,1.12.12,2.11.12,1.1.2.12.$

$T\left[2^3\right]$ needs terms $25-35\sim 112233,1.12233,12.1233,112.233,123.123,1.2.1233,1.12.233,1.23.123,12.13.23,1.2.3.123,1.2.13.23.$ These give $S_3\left(F\right)$.

So $T\left[2\right]$ has 2 terms, $T\left[3\right]$ has 3 terms, $T\left[2^2\right]$ has 4 new terms, $T\left[4\right]$ has 5 new terms, $T\left[32\right]$ has 10 new terms, $T\left[2^3\right]$ has 11 new terms. This gives a total of 35 terms for $S_1\left(F\right)$, $S_2\left(F\right)$, and $S_3\left(F\right)$ needed for the 4th order estimate of $T\left(F\right),{\widehat{T}}_{n4}$. For the examples, many terms may be 0, as higher derivatives of $U\left(F\right)$ will be 0.

PROOF OF THEOREM 2.3. $g_{i_1\dots i_k}$ is symmetric in $i_1,\dots ,i_k$.

$$\begin{array}{cc}& \left(18\right)\Rightarrow T_{\cdot 11}=g_{i_1}{U}_{i_1\cdot 11}+g_{i_1{i}_2}{U}_{i_1\cdot 1}{U}_{i_2\cdot 1}\Rightarrow \left(25\right).\hfill \\ & \left(19\right)\Rightarrow T_{\cdot 111}=g_{i_1}{U}_{i_1\cdot 111}+3g_{i_1{i}_2}{U}_{i_1\cdot 1}{U}_{i_2\cdot 11}+g_{i_1{i}_2{i}_3}{U}_{i_1\cdot 1}{U}_{j\cdot 1}{U}_{i_2\cdot 1}\Rightarrow \left(26\right).\hfill \end{array}$$

For $T\left[2^2\right],$ replace $\left(1234\right)$ by $\left(1122\right)$ in (7) at $r=4$, and in (20), to get

$$\begin{array}{cc}& B_{41}=U_{i_1\cdot 1111},\hfill \\ & B_{42}=S_4+S_3^{\prime },S_4=2{\displaystyle \sum _{i_1{i}_2}^2}{U}_{i_1\cdot 1}{U}_{i_2\cdot 122},S_3^{\prime }=U_{i_1\cdot 11}{U}_{i_2\cdot 22}+2U_{i_1\cdot 12}{U}_{i_2\cdot 12},\hfill \\ & B_{43}={\displaystyle \sum _{12}^2}{U}_{i_1\cdot 1}{U}_{i_2\cdot 1}{U}_{i_3\cdot 22}+4U_{i_1\cdot 1}{U}_{i_2\cdot 2}{U}_{i_3\cdot 12},B_{44}=U_{i_1\cdot 1}{U}_{i_2\cdot 1}{U}_{i_3\cdot 2}{U}_{i_4\cdot 2}.\hfill \end{array}$$

For $T\left[32\right],$ replace $\left(12345\right)$ by $\left(11122\right)$ in (7) at $r=5$, and in (21), to get

$$\begin{array}{cc}& B_{51}=U_{i_1\cdot 11111},B_{52}=S_5+S_{10},S_5=3U_{i_1\cdot 1}{U}_{i_2\cdot 1122}+2U_{i_1\cdot 2}{U}_{i_2\cdot 1112},\hfill \\ & S_{10}=3U_{i_1\cdot 11}{U}_{i_2\cdot 122}+6U_{i_1\cdot 12}{U}_{i_2\cdot 112}+U_{i_1\cdot 22}{U}_{i_2\cdot 111},\hfill \\ & B_{53}=S_{10}^{\prime }+S_{15},S_{10}^{\prime }=3U_{i_1\cdot 1}{U}_{i_2\cdot 1}{U}_{i_3\cdot 122}+6U_{i_1\cdot 1}{U}_{i_2\cdot 2}{U}_{i_3\cdot 112}+U_{i_1\cdot 2}{U}_{i_2\cdot 2}{U}_{i_3\cdot 111},\hfill \\ & S_{15}=3U_{i_1\cdot 1}{U}_{i_2\cdot 11}{U}_{i_3\cdot 22}+6U_{i_1\cdot 1}{U}_{i_2\cdot 12}{U}_{i_3\cdot 12}+4U_{i_1\cdot 2}{U}_{i_2\cdot 11}{U}_{i_3\cdot 12}+2U_{i_1\cdot 2}{U}_{i_2\cdot 12}{U}_{i_3\cdot 11},\hfill \\ & B_{54}=S_{10}^{\prime \prime }=U_{i_1\cdot 1}{U}_{i_2\cdot 1}{U}_{i_3\cdot 1}{U}_{i_4\cdot 22}+6U_{i_1\cdot 1}{U}_{i_2\cdot 1}{U}_{i_3\cdot 2}{U}_{i_4\cdot 12}+3U_{i_1\cdot 1}{U}_{i_2\cdot 2}{U}_{i_3\cdot 2}{U}_{i_4\cdot 11},\hfill \\ & B_{55}=U_{i_1\cdot 1}{U}_{i_2\cdot 1}{U}_{i_3\cdot 1}{U}_{i_4\cdot 2}{U}_{i_5\cdot 2}.\hfill \end{array}$$

(29) follows. For $T\left[2^3\right]$, replace $\left(123456\right)$ by $\left(112233\right)$ in (7) at $r=6$, and in (22), to get

$$\begin{array}{cc}& B_{61}=U_{i_1\cdot 112233},B_{62}=S_6^{\prime }+S_{15}^{\prime }+S_{10}^{\prime \prime \prime },\hfill \\ & S_6^{\prime }=2U_{i_1\cdot 1}{U}_{i_2\cdot 12233}+2U_{i_1\cdot 2}{U}_{i_2\cdot 12233}+2U_{i_1\cdot 3}{U}_{i_2\cdot 11223},\hfill \\ & S_{15}^{\prime }=U_{i_1\cdot 11}{U}_{i_2\cdot 2233}+4U_{i_1\cdot 12}{U}_{i_2\cdot 1233}+4U_{i_1\cdot 13}{U}_{i_2\cdot 1223}+4U_{i_1\cdot 23}{U}_{i_2\cdot 1123}+U_{i_1\cdot 33}{U}_{i_2\cdot 1122},\hfill \\ & S_{10}^{\prime \prime \prime }=2U_{i_1\cdot 112}{U}_{i_2\cdot 233}+2U_{i_1\cdot 113}{U}_{i_2\cdot 223}+2U_{i_1\cdot 122}{U}_{i_2\cdot 133}+4U_{i_1\cdot 123}{U}_{i_2\cdot 123},\hfill \\ & B_{63}=S_{15}^{\prime \prime }+S_{60}+S_{15}^{\prime \prime \prime },S_{15}^{\prime \prime }=U_{i_1\cdot 1}{U}_{i_2\cdot 1}{U}_{i_3\cdot 2233}+4U_{i_1\cdot 1}{U}_{i_2\cdot 2}{U}_{i_3\cdot 1233}\hfill \\ & +4U_{i_1\cdot 1}{U}_{i_2\cdot 3}{U}_{i_3\cdot 1223}+U_{i_1\cdot 2}{U}_{i_2\cdot 2}{U}_{i_3\cdot 1133}+4U_{i_1\cdot 2}{U}_{i_2\cdot 3}{U}_{i_3\cdot 1123}+U_{i_1\cdot 3}{U}_{i_2\cdot 3}{U}_{i_3\cdot 1122},\hfill \\ & S_{60}=4U_{i_1\cdot 1}{U}_{i_2\cdot 12}{U}_{i_3\cdot 233}+4U_{i_1\cdot 1}{U}_{i_2\cdot 13}{U}_{i_3\cdot 223}+8U_{i_1\cdot 3}{U}_{i_2\cdot 12}{U}_{i_3\cdot 123}+2U_{i_1\cdot 3}{U}_{i_2\cdot 11}{U}_{i_3\cdot 223}\hfill \\ & +4U_{i_1\cdot 3}{U}_{i_2\cdot 13}{U}_{i_3\cdot 122}+2U_{i_1\cdot 2}{U}_{i_2\cdot 11}{U}_{i_3\cdot 233}+4U_{i_1\cdot 2}{U}_{i_2\cdot 12}{U}_{i_3\cdot 133}+8U_{i_1\cdot 2}{U}_{i_2\cdot 13}{U}_{i_3\cdot 123}\hfill \\ & +4U_{i_1\cdot 2}{U}_{i_2\cdot 23}{U}_{i_3\cdot 113}+2U_{i_1\cdot 2}{U}_{i_2\cdot 33}{U}_{i_3\cdot 112}+2U_{i_1\cdot 1}{U}_{i_2\cdot 22}{U}_{i_3\cdot 133}+8U_{i_1\cdot 1}{U}_{i_2\cdot 23}{U}_{i_3\cdot 123}\hfill \\ & +2U_{i_1\cdot 1}{U}_{i_2\cdot 33}{U}_{i_3\cdot 122}+2U_{i_1\cdot 3}{U}_{i_2\cdot 22}{U}_{i_3\cdot 113}+4U_{i_1\cdot 3}{U}_{i_2\cdot 23}{U}_{i_3\cdot 112},\hfill \\ & S_{15}^{\prime \prime \prime }=U_{i_1\cdot 11}{U}_{i_2\cdot 22}{U}_{i_3\cdot 33}+U_{i_1\cdot 22}{U}_{i_2\cdot 11}{U}_{i_3\cdot 33}+U_{i_1\cdot 33}{U}_{i_2\cdot 11}{U}_{i_3\cdot 22}\hfill \\ & +4U_{i1\cdot 12}{U}_{i_2\cdot 12}{U}_{i_3\cdot 33}+4U_{i1\cdot 13}{U}_{i_2\cdot 12}{U}_{i_3\cdot 23}+4U_{i1\cdot 23}{U}_{i_2\cdot 11}{U}_{i_3\cdot 23},\hfill \end{array}$$

$$\begin{array}{cc}& B_{64}=S_{20}+S_{45},S_{20}=2U_{i_1\cdot 112}{U}_{i_2\cdot 2}{U}_{i_3\cdot 3}{U}_{i_4\cdot 3}+2U_{i_1\cdot 113}{U}_{i_2\cdot 2}{U}_{i_3\cdot 2}{U}_{i_4\cdot 3}\hfill \\ & +2U_{i_1\cdot 122}{U}_{i_2\cdot 1}{U}_{i_3\cdot 3}{U}_{i_4\cdot 3}+8U_{i_1\cdot 123}{U}_{i_2\cdot 1}{U}_{i_3\cdot 2}{U}_{i_4\cdot 3}+2U_{i_1\cdot 133}{U}_{i_2\cdot 1}{U}_{i_3\cdot 2}{U}_{i_4\cdot 2}\hfill \\ & +2U_{i_1\cdot 223}{U}_{i_2\cdot 1}{U}_{i_3\cdot 1}{U}_{i_4\cdot 3}+2U_{i_1\cdot 233}{U}_{i_2\cdot 1}{U}_{i_3\cdot 1}{U}_{i_4\cdot 2},\hfill \\ & S_{45}=U_{i_1\cdot 1}{U}_{i_2\cdot 1}(U_{i_3\cdot 22}{U}_{i_4\cdot 33}+2U_{i_3\cdot 23}{U}_{i_4\cdot 23})+4U_{i_1\cdot 1}{U}_{i_2\cdot 2}(U_{i_3\cdot 12}{U}_{i_4\cdot 33}+2U_{i_3\cdot 13}{U}_{i_4\cdot 23})\hfill \\ & +4U_{i_1\cdot 1}{U}_{i_2\cdot 3}(2U_{i_3\cdot 12}{U}_{i_4\cdot 23}+U_{i_3\cdot 13}{U}_{i_4\cdot 22})+U_{i_1\cdot 2}{U}_{i_2\cdot 2}(U_{i_3\cdot 11}{U}_{i_4\cdot 33}+2U_{i_3\cdot 13}{U}_{i_4\cdot 13})\hfill \\ & +4U_{i_1\cdot 2}{U}_{i_2\cdot 3}(U_{i_3\cdot 11}{U}_{i_4\cdot 23}+U_{i_3\cdot 12}{U}_{i_4\cdot 13}+U_{i_3\cdot 13}{U}_{i_4\cdot 12})\hfill \\ & +U_{i_1\cdot 3}{U}_{i_2\cdot 3}(U_{i_3\cdot 11}{U}_{i_4\cdot 22}+2U_{i_3\cdot 12}{U}_{i_4\cdot 12}),\hfill \\ & B_{65}=S_{15}^{\prime \prime \prime \prime }=U_{i_1\cdot 11}{U}_{i_2\cdot 2}{U}_{i_3\cdot 2}{U}_{i_4\cdot 3}{U}_{i_5\cdot 3}+U_{i_1\cdot 22}{U}_{i_2\cdot 1}{U}_{i_3\cdot 1}{U}_{i_4\cdot 3}{U}_{i_5\cdot 3}\hfill \\ & +U_{i_1\cdot 33}{U}_{i_2\cdot 1}{U}_{i_3\cdot 1}{U}_{i_4\cdot 2}{U}_{i_5\cdot 2}+4U_{i_1\cdot 12}{U}_{i_2\cdot 1}{U}_{i_3\cdot 2}{U}_{i_4\cdot 3}{U}_{i_5\cdot 3}\hfill \\ & +4U_{i_1\cdot 13}{U}_{i_2\cdot 1}{U}_{i_3\cdot 3}{U}_{i_4\cdot 2}{U}_{i_5\cdot 3}+4U_{i_1\cdot 23}{U}_{i_2\cdot 1}{U}_{i_3\cdot 1}{U}_{i_4\cdot 2}{U}_{i_5\cdot 3},\hfill \\ & B_{66}=U_{i_1\cdot 1}{U}_{i_2\cdot 1}{U}_{i_3\cdot 2}{U}_{i_4\cdot 2}{U}_{i_5\cdot 3}{U}_{i_6\cdot 3}.\hfill \end{array}$$

$\left[B_{61}\right],\dots ,\left[B_{66}\right]$ of (30) follow. □

Note 2.5. 

Theorem 2.3 allows us to write $S_j\left(F\right)$ of (1)–(3) in the form

$$\begin{array}{cc}& S_j\left(F\right)={\displaystyle \sum _{k=1}^{2i}}{g}_{i_1\dots i_k}{S}_j^{i_1\dots i_k}.\hfill \\ & For example, S_2^i=\left[U_{i\cdot 111}\right]/3+\left[U_{i\cdot 1122}\right]/8,\hfill \\ & S_2^{ij}=d_{ij}/3+\{\overline{i}\overline{j}+2\left(i\right|j)+4\left[U_{i\cdot 1}{U}_{j\cdot 122}\right]\}/8,\hfill \\ & S_3^{ijk}=\left(ijk\right)/3+\{\left(ij\right)\overline{k}+2\left(ikj\right)\}/4,S_4^{ijkl}=\left(ij\right)\left(kl\right).\hfill \end{array}$$

For moment estimates, one can apply Theorem 2.3 with $U\left(F\right)=(\mu ,{\mu }_2,{\mu }_3,\dots ,{\mu }_q)$.

Corollary 1. 

Suppose that $T\left(F\right)=g\left(U\right(F)$ with $U\left(F\right)\in R$. Set $g_k={(d/du)}^{k}g\left(u\right)$ at $u=U\left(F\right)$. So, the expressions in (23)–(24) are

$$\begin{array}{cc}& \overline{1}=\left[U_{\cdot 11}\right],\left(1^r\right)=\left[U_{\cdot 1}^r\right],d_{11}=\left[U_{\cdot 1}{U}_{\cdot 11}\right],\hfill \\ & \left(1\right|1)=\left[U_{\cdot 12}^2\right],\left(1\right|1\left|1\right)=\left[U_{\cdot 1}{U}_{\cdot 12}{U}_{\cdot 2}\right].\hfill \\ & Also,T\left[2\right]=g_1\overline{1}+g_2\left(1^2\right),\hfill \\ & T\left[3\right]=g_1\left[U_{\cdot 111}\right]+3g_2{d}_{11}+g_3\left(1^3\right),\hfill \\ & T\left[2^2\right]=g_1\left[U_{\cdot 1122}\right]+g_2\{{\overline{1}}^2+2\left(1\right|1)+4\left[U_{\cdot 1}{U}_{\cdot 122}\right]\}+2g_3\{\left(1^2\right)\overline{1}+2\left(1\right|1\left|1\right)\}\hfill \\ & +g_4{\left(1^2\right)}^2,\hfill \end{array}$$

$$\begin{array}{cc}& T\left[4\right]=g_1\left[U_{\cdot 1111}\right]+g_2(4\left[U_{\cdot 1}{U}_{\cdot 111}\right]+3\left[U_{\cdot 11}{U}_{\cdot 11}\right])+6g_3\left[U_{\cdot 1}^2{U}_{i_3\cdot 11}\right]+g_4\left(1^4\right),\hfill \\ & T\left[32\right]=g_1\left[U_{\cdot 11122}\right]+g_2(3\left[U_{\cdot 1}{U}_{\cdot 1122}\right]+2\left[U_{\cdot 2}{U}_{\cdot 1112}\right]+3\left[U_{\cdot 11}{U}_{\cdot 122}\right]\hfill \\ & +6\left[U_{\cdot 12}{U}_{\cdot 112}\right]+\overline{1}\left[U_{\cdot 111}\right])+g_3(3\left[U_{\cdot 1}^2{U}_{\cdot 122}\right]+6\left[U_{\cdot 1}{U}_{\cdot 2}{U}_{\cdot 112}\right]\hfill \\ & +\left(1^2\right)\left[U_{\cdot 111}\right]+3\left[U_{\cdot 1}{U}_{\cdot 11}\right]\overline{1}+6\left[U_{\cdot 1}{U}_{\cdot 12}^2\right]+4\left[U_{\cdot 2}{U}_{\cdot 11}{U}_{\cdot 12}\right]+2\left[U_{\cdot 2}{U}_{\cdot 12}{U}_{\cdot 11}\right])\hfill \\ & +g_4\{\left(1^3\right)\overline{1}+9\left(1^2\right)d_{11}\}+g_5\left(1^2\right)\left(1^3\right),\hfill \\ & T\left[2^3\right]={\displaystyle \sum _{k=1}^6}{g}_k{a}_k where a_1=\left[U_{\cdot 112233}\right],\hfill \\ & a_2=6\left[U_{\cdot 1}{U}_{\cdot 12233}\right]+3\overline{1}\left[U_{\cdot 1122}\right]+12\left[U_{\cdot 12}{U}_{\cdot 1233}\right]+6\left[U_{\cdot 112}{U}_{\cdot 233}\right]+4\left[U_{\cdot 123}^2\right],\hfill \\ & a_3=3\left(1^2\right)\left[U_{\cdot 1122}\right]+12\left[U_{\cdot 1}{U}_{\cdot 2}{U}_{\cdot 1233}\right]+24\left[U_{\cdot 1}{U}_{\cdot 12}{U}_{\cdot 233}\right]+12\left[U_{\cdot 1}{U}_{\cdot 122}\right]\overline{1}\hfill \\ & +24\left[U_{\cdot 1}{U}_{\cdot 23}{U}_{\cdot 123}\right]+3{\overline{1}}^3+8\left(1\right|1)\overline{1}+4\left[U_{\cdot 12}{U}_{\cdot 13}{U}_{\cdot 23}\right],\hfill \\ & a_4=12\left(1^2\right)\left[U_{\cdot 1}{U}_{\cdot 122}\right]+8\left[U_{\cdot 1}{U}_{\cdot 2}{U}_{\cdot 3}{U}_{\cdot 123}\right]+3\left(1^2\right)\{{\overline{1}}^2+2\left(1\right|1)\}+12\left(1\right|1\left|1\right)\overline{1}\hfill \\ & +24\left[U_{\cdot 1}{U}_{\cdot 2}{U}_{\cdot 13}{U}_{\cdot 23}\right],\hfill \\ & a_5=3{\left(1^2\right)}^2\overline{1}+12\left(1^2\right)\left(1\right|1\left|1\right), and a_6={\left(1^2\right)}^3.\hfill \end{array}$$

Note 2.6. 

$T\left[2\right],T\left[3\right]$ and $T\left[2^2\right]$ for $q=1$ were given in Example 6.4 of [9].

Corollary 2. 

Let $a\left(x\right):R^s:\to R^q$ be a given function, and $X\sim F\left(x\right)$ on $R^s$ with mean $\mu =U\left(F\right)=\int a\left(x\right)dF\left(x\right)\in R^q$. Set $T\left(F\right)=g\left(\mu \right)$ for g as in Theorem 2.2, with partial derivatives $g_{i_1{i}_2\dots }$ at $\mu \in R^q$. So $\overline{i}=\left[U_{i\cdot 111}\right]=d_{ij}=\left(i\right|j)=\left(i\right|j\left|k\right)=0$ for the terms in (23)–(24) , and the non-zero terms in Theorem 2.2 needed for a kth order estimate of $T\left(F\right)=g\left(U\right)$ for $k\le 4$, are

$$\begin{array}{cc}& U_{\cdot i}=h_i where h_i=a\left(x_i\right)-{\mu }_i\in R,\hfill \end{array}$$

$$\begin{array}{cc}& (i_1\dots i_r)=\mu (a_{i_1}\left(X\right),\dots ,a_{i_r}\left(X\right)),\hfill \end{array}$$

(34)

$$\begin{array}{cc}& T\left[2\right]=a_{22}=g_{i_1{i}_2}\left(i_1{i}_2\right),T\left[3\right]=a_{33}=g_{i_1{i}_2{i}_3}\left(i_1{i}_2{i}_3\right),\hfill \end{array}$$

(35)

$$\begin{array}{cc}& T\left[2^2\right]=a_{44}=\left(i_1{i}_2\right)g_{i_1\dots i_4}\left(i_3{i}_4\right),\hfill \end{array}$$

(36)

$$\begin{array}{cc}& T\left[4\right]=g_{i_1\dots i_4}(i_1\dots i_4),T\left[32\right]=g_{i_1\dots i_5}\left(i_1{i}_2{i}_3\right)\left(i_4{i}_5\right),\hfill \end{array}$$

(37)

$$\begin{array}{cc}& T\left[2^3\right]=g_{i_1\dots i_6}\left(i_1{i}_2\right)\left(i_3{i}_4\right)\left(i_5{i}_6\right).\hfill \end{array}$$

(38)

So for $q=2$ and ${\sum }_{12}^{2}f(1,2)=f(1,2)+f(2,1)$ for any $f(1,2)$,

$$\begin{array}{cc}& T\left[2\right]=g_{11}\left(11\right)+2g_{12}\left(12\right)+g_{22}\left(22\right),\hfill \\ & T\left[3\right]=g_{111}\left(111\right)+3g_{112}\left(112\right)+3g_{122}\left(122\right)+g_{222}\left(222\right),\hfill \\ & T\left[2^2\right]={\displaystyle \sum _{12}^2}\{g_{1^4}{\left(11\right)}^2+4g_{1^{3}2}\left(11\right)\left(12\right)\}+2g_{1^2{2}^2}\left(1^2{2}^2\right)\{\left(11\right)\left(22\right)+2{\left(12\right)}^2\},\hfill \\ & T\left[4\right]={\displaystyle \sum _{12}^2}\{g_{1^4}\left(1^4\right)+4g_{1^{3}2}\left(1^{3}2\right)\}+6g_{1^2{2}^2}\left(1^2{2}^2\right),\hfill \\ & T\left[32\right]={\displaystyle \sum _{12}^2}\{g_{1^5}\left(1^2\right)\left(1^3\right)+4g_{1^{4}2}[3\left(11\right)\left(112\right)+2\left(12\right)\left(111\right)]\hfill \\ & +g_{1^3{2}^2}[6\left(12\right)\left(112\right)+\left(22\right)\left(111\right)]+3g_{1^2{2}^3}\left(22\right)\left(112\right)\left]\right\},\hfill \\ & T\left[2^3\right]={\displaystyle \sum _{12}^2}\{g_{1^6}{\left(11\right)}^3+6g_{1^{5}2}{\left(11\right)}^2\left(12\right)+3g_{1^4{2}^2}[{\left(11\right)}^2\left(22\right)+4\left(11\right){\left(22\right)}^2]\}\hfill \\ & +4g_{1^3{2}^3}[3\left(11\right)\left(12\right)\left(22\right)+2{\left(12\right)}^3].\hfill \end{array}$$

PROOF (34)–(38) follow from Theorem 2.3 as terms with higher derivatives of $U\left(F\right)$ are zero. The result for $q=2$ follows after some work. □ (34)–(38) were given in more obscure notation in Example 5.1 of [3].

Example 2. 

Take $q=2,g\left(\mu \right)={\mu }_1/{\mu }_2.$ Then

$$\begin{array}{cc}& T\left[2\right]=2g_{12}\left(12\right)+g_{22}\left(22\right),\hfill \\ & T\left[3\right]=3g_{122}\left(122\right)+g_{222}\left(222\right),\hfill \\ & T\left[2^2\right]=g_{2^4}{\left(22\right)}^2+4g_{12^3}\left(22\right)\left(12\right),\hfill \\ & T\left[4\right]=g_{2^4}\left(2^4\right)+4g_{12^3}\left(12^3\right),\hfill \\ & T\left[32\right]=g_{2^5}\left(2^2\right)\left(2^3\right)+4g_{12^4}[3\left(22\right)\left(122\right)+2\left(12\right)\left(222\right)],\hfill \\ & T\left[2^3\right]=g_{2^6}{\left(22\right)}^3+6g_{12^5}\left(12\right){\left(22\right)}^2,\hfill \\ & where g_{12^r}={(-1)}^{r-1}(r-1)!{\mu }_1/{\mu }_2^r,g_{2^r}={(-1)}^{r-1}(r-1)!/{\mu }_2^r:\hfill \\ & g_{12}=-{\mu }_2^{-2},g_{22}=2{\mu }_1{\mu }_2^{-3},g_{122}=2{\mu }_2^{-3},g_{222}=-6{\mu }_1{\mu }_2^{-4},\hfill \\ & g_{12^3}=-6{\mu }_2^{-4},g_{2^4}=24{\mu }_1{\mu }_2^{-5},g_{12^4}=24{\mu }_2^{-5},g_{2^5}=-120{\mu }_1{\mu }_2^{-6},\hfill \\ & g_{12^5}=-120{\mu }_2^{-6},g_{2^5}=720{\mu }_1{\mu }_2^{-7}.\hfill \end{array}$$

Corollary 3. 

Given $a\left(x\right):R^s\to R$, and $X\sim F\left(x\right),$ set

$$\begin{array}{cc}& A=a\left(X\right),\mu =EA=\int a\left(x\right)dF\left(x\right), and {\mu }_r={\mu }_r\left(A\right).\hfill \end{array}$$

(39)

Given $g:R\to R$, set $g_k={(d/d\mu )}^{k}g\left(\mu \right)$. Then for $T\left(F\right)=g\left(\mu \right),$

$$\begin{array}{cc}& T\left[2\right]={\mu }_2{g}_2,T\left[3\right]={\mu }_3{g}_3,T\left[2^2\right]={\mu }_2^2{g}_4,T\left[4\right]={\mu }_4{g}_4,\hfill \\ & T\left[32\right]={\mu }_2{\mu }_3{g}_5,T\left[2^3\right]={\mu }_2^3{g}_6.\hfill \end{array}$$

Example 3. 

Suppose that we have a random sample from the gamma distribution, say $X=G\sim F\left(x\right)$ where $G\sim Gamma\left(\gamma \right)$. That is, G has density $x^{\gamma -1}{e}^{-x}/\Gamma \left(\gamma \right)$ on $(0,\infty )$. Then by (2.2) of [40], γ has MLE $\widehat{\gamma }={\psi }^{-1}\left(\overline{Y}\right)$ where $Y=lnX$ and $\psi \left(\gamma \right)=(d/d\gamma )ln\Gamma \left(\gamma \right)$. So for $k\le 4,k$th order estimates of γ is given by Corollary 2.3 in terms of the 1st 4 central moments of $Y=lnX$. These central moment are given in terms of the non-central moments,

$$E{Y}^k={\tau }_{\cdot k}\left(0\right)where\tau \left(t\right)=E{G}^t=\Gamma (\gamma +t)/\Gamma \left(\gamma \right),$$

in the usual way, by binomial expansions. To obtain the derivatives of $g\left(\mu \right)={\psi }^{-1}\left(\mu \right)$, apply [41]. This gives them in terms of the derivatives of $\psi \left(\gamma \right)$. These are asily obtained from

$$\psi (1+t)=-{\gamma }_0+\zeta \left(2\right)t-\zeta \left(3\right)t^2+\dots where\zeta \left(s\right)={\displaystyle \sum _{n=1}^{\infty }}{n}^{-s},$$

the zeta function, and ${\gamma }_0$ is Euler’s constant. (Differentiate the expression on p175 of [39].)

In what follows, it is convenient to set

$$\begin{array}{cc}& M_4={\mu }_4-{\mu }_2^2,\hfill \end{array}$$

the asymptotic variance of $n^{1/2}({\mu }_2\left(F_n\right)-{\mu }_2\left(F\right))$.

Corollary 4. 

Take $a\left(x\right),A,\mu ,{\mu }_r$ of (39), and $T\left(F\right)=g\left({\mu }_2\right).$ Then

$$\begin{array}{cc}& \overline{1}=-2{\mu }_2,\left(1^r\right)=E{({(A-\mu )}^2-{\mu }_2)}^r,\left(1^2\right)=M_4,\hfill \\ & \left(1^3\right)={\mu }_6-3{\mu }_4{\mu }_2+2{\mu }_2^3,\left(1^4\right)={\mu }_8-4{\mu }_6{\mu }_2+6{\mu }_4{\mu }_2^2-3{\mu }_2^4,\hfill \\ & d_{11}=-2M_4,\left(1\right|1)=4{\mu }_2^2,\left(1\right|1\left|1\right)=-2{\mu }_3^2.\hfill \\ & Also,T\left[2\right]=g_1\overline{1}+g_2\left(1^2\right)=-2{\mu }_2{g}_1+M_4{g}_2,\hfill \\ & T\left[3\right]=3g_2{d}_{11}+g_3\left(1^3\right)=-6m_4{g}_2+({\mu }_6-3{\mu }_4{\mu }_2+2{\mu }_2^3)g_3,\hfill \\ & T\left[2^2\right]=g_2\{{\overline{1}}^2+2\left(1\right|1)\}+2g_3\{\left(1^2\right)\overline{1}+2\left(1\right|1\left|1\right)\}+g_4{\left(1^2\right)}^2\hfill \\ & =12{\mu }_2^2{g}_2-4(2{\mu }_3^2+{\mu }_2{M}_4)g_3+M_4^2{g}_4,\hfill \\ & T\left[4\right]=3g_2\left[U_{\cdot 11}^2\right]+6g_3\left[U_{\cdot 1}^2{U}_{\cdot 11}\right]+g_4\left(1^4\right)\hfill \\ & =12{\mu }_4{g}_2-12({\mu }_6-2{\mu }_2{\mu }_4+{\mu }_2^3)g_3+({\mu }_8-4{\mu }_6{\mu }_2+6{\mu }_4{\mu }_2^2-3{\mu }_2^4)g_4,\hfill \\ & T\left[32\right]=g_3(3d_{11}\overline{1}+6\left[U_{\cdot 1}{U}_{\cdot 12}^2\right]+4\left[U_{\cdot 2}{U}_{\cdot 11}{U}_{\cdot 12}\right]+2\left[U_{\cdot 2}{U}_{\cdot 12}{U}_{\cdot 11}\right])\hfill \\ & +g_4\{\left(1^3\right)\overline{1}+9\left(1^2\right)d_{11}\}+g_5\left(1^2\right)\left(1^3\right)\hfill \\ & =(36{\mu }_2{M}_4+24{\mu }_3^2)g_3-2(11{\mu }_2^4-21{\mu }_2^2{\mu }_4+9{\mu }_4^2+{\mu }_2{\mu }_6)g_4\hfill \\ & +M_4({\mu }_6-3{\mu }_4{\mu }_2+2{\mu }_2^3)g_5\hfill \\ & T\left[2^3\right]={\displaystyle \sum _{k=3}^6}{g}_k{a}_k where a_3=3{\overline{1}}^3+8\left(1\right|1)\overline{1}+4\left[U_{\cdot 12}{U}_{\cdot 13}{U}_{\cdot 23}\right]=-120{\mu }_2^3,\hfill \\ & a_4=3\left(1^2\right)\{{\overline{1}}^2+2\left(1\right|1)\}+12\left(1\right|1\left|1\right)\overline{1}+24\left[U_{\cdot 1}{U}_{\cdot 2}{U}_{\cdot 13}{U}_{\cdot 23}\right]=36{\mu }_2({\mu }_2{M}_4+4{\mu }_3^2),\hfill \\ & a_5=3{\left(1^2\right)}^2\overline{1}+12\left(1^2\right)\left(1\right|1\left|1\right)=-6{\mu }_2{M}_4^2-24{\mu }_3^2{M}_4,\hfill \\ & a_6={\left(1^2\right)}^3=M_4^3.\hfill \end{array}$$

(40)

Example 4. 

Suppose that $g\left(u\right)=u^{1/2}$, so that $T\left(F\right)=\sigma ={\mu }_2^{1/2}$, the standard deviation of $A=a\left(X\right)$. Set $u_r={\mu }_r/{\sigma }^r$, the standardized rth central moment. Then,

$$\begin{array}{cc}& T\left[2\right]/\sigma =-(u_4+3)/4,S_1\left(F\right)=\sigma (u_4+3)/8,\hfill \\ & T\left[3\right]/\sigma =3(u_6+u_4-2)/8,T\left[2^2\right]/\sigma =-3(5u_4^2+16u_3^2-2u_4+13)/16,\hfill \\ & S_2\left(F\right)=-\sigma (14u_4^2+16u_6+48u_3^2+10u_4+7)/128,\hfill \\ & T\left[4\right]/\sigma =-3(5u_8+4u_6-2u_4+9)/16,\hfill \\ & T\left[32\right]=3\sigma (35u_4{u}_6+75u_4^2-15u_6+96u_3^2-101u_4-70)/32,\hfill \end{array}$$

$$\begin{array}{cc}& T\left[2^3\right]=45\sigma (-112u_4{u}_3^2-49u_4^2-192u_3^2+50u_4+47)/64,\hfill \\ & S_3\left(F\right)=\sigma (5040u_3^2{u}_4-8u_4{u}_6-245u_8+1425u_4^2+7296u_3^2+24u_6-1322u_4\hfill \\ & -2239)/512.\hfill \end{array}$$

This corrects $S_2\left(F\right)$ on p215 of [9] and gives for the 1st time $S_3\left(F\right)$ for σ.

I now apply Theorem 2.3 with $q=2,U\left(F\right)=(\mu ,{\mu }_2)$.

Corollary 5. 

Estimates of $T\left(F\right)=g(\mu ,{\mu }_2)$.

Let $a\left(x\right):R^s:\to R$ be a given function, and $X\sim F\left(x\right)$ on $R^s$. Let μ and ${\mu }_2$ be the mean and variance of $a\left(X\right)$. Their non-zero derivatives are given by (4). Take $q=2$. Let $g_{1\dots k}$ be the kth partial derivative of $g\left(U\right)$ at $U=(\mu ,{\mu }_2)$. By (1), the terms in (23)–(24) are

$$\begin{array}{cc}& \overline{1}=0,\overline{2}=\left[{\mu }_{2\cdot 11}\right]=-2{\mu }_2,\hfill \\ & \left(11\right)=\left[{\mu }_{\cdot 1}^2\right]={\mu }_2,\left(12\right)=\left[{\mu }_{\cdot 1}{\mu }_{2\cdot 1}\right]={\mu }_3,\left(22\right)=\left[{\mu }_{2\cdot 1}^2\right]={\mu }_4-{\mu }_2^2,\hfill \\ & d_{11}=0,d_{12}=-2{\mu }_3,d_{22}=-2M_4,\left(111\right)={\mu }_3,\left(112\right)={\mu }_4-{\mu }_2^2,\hfill \\ & \left(122\right)={\mu }_5-2{\mu }_2{\mu }_3,\left(222\right)={\mu }_6-3{\mu }_2{\mu }_4+2{\mu }_2^3,\hfill \\ & \left(i\right|j)=4{\mu }_2^{2}I(i=j=2),\hfill \end{array}$$

$$\begin{array}{cc}& \left(i\right|1\left|k\right)=0,\left(1\right|2\left|1\right)=-2{\mu }_2^2,\left(1\right|2\left|2\right)=-2{\mu }_2{\mu }_3,\left(2\right|2\left|2\right)=-2{\mu }_3^2.\hfill \end{array}$$

(41)

For $2\le k\le 4$, the kth order estimate of $g(\mu ,{\mu }_2)$, ${\widehat{T}}_{nk}$ of Theorem 2.1, is given in terms of the following.

$$\begin{array}{cc}& For k=2:T\left[2\right]=a_{12}+a_{22} where\hfill \\ & a_{12}=-2g_2{\mu }_2,a_{22}=g_{11}{\mu }_2+2g_{12}{\mu }_3+g_{22}{M}_4.\hfill \end{array}$$

(42)

$$\begin{array}{c}\hfill For k=3:T\left[3\right]=a_{23}+a_{33} where a_{23}/3=-4g_{12}{\mu }_3-2g_{22}{M}_4,\end{array}$$

(43)

$$\begin{array}{cc}& a_{33}=g_{111}{\mu }_3+3g_{112}{M}_4+3g_{122}({\mu }_5-2{\mu }_2{\mu }_3)+g_{222}({\mu }_6-3{\mu }_2{\mu }_4+2{\mu }_2^3),\hfill \end{array}$$

(44)

$$\begin{array}{cc}& T\left[2^2\right]={\displaystyle \sum _{i=2}^4}{a}_{i4} where a_{24}=12g_{22}{\mu }_2^2,a_{34}/4=c_1{g}_{112}+c_2{g}_{122}+c_3{g}_{222},\hfill \\ & c_1=-3{\mu }_2^2,c_2=-6{\mu }_2{\mu }_3,c_3=-{\mu }_2{M}_4-2{\mu }_3^2.\hfill \\ & a_{44}=g_{1^4}{\mu }_2^2+4g_{1^{3}2}{\mu }_2{\mu }_3+g_{1122}(2{\mu }_2{M}_4+4{\mu }_3^2)+4g_{12^3}{\mu }_3{M}_4+g_{2^4}{M}_4^2.\hfill \end{array}$$

(45)

$$\begin{array}{cc}& For k=4:T\left[4\right]={\displaystyle \sum _{k=2}^4}{a}_k where a_2=3g_{22}\left[{\mu }_{2\cdot 11}^2\right]=12g_{22}{\mu }_4,\hfill \\ & a_3=-12{\displaystyle \sum _{i=1}^3}{a}_{3i}^{\prime }{g}_{1^{3-i}{2}^i},a_{31}^{\prime }={\mu }_4,a_{32}^{\prime }=2({\mu }_5-{\mu }_2{\mu }_3),\hfill \end{array}$$

(46)

$$\begin{array}{cc}& a_{33}^{\prime }={\mu }_6-2{\mu }_2{\mu }_4+{\mu }_2^3,\hfill \\ & a_4={\displaystyle \sum _{i=0}^4}{a}_{4i}^{\prime }{g}_{1^{4-i}{2}^i},a_{40}^{\prime }={\mu }_4,a_{41}^{\prime }=4({\mu }_5-{\mu }_2{\mu }_3),a_{42}^{\prime }=6({\mu }_6-2{\mu }_2{\mu }_4+{\mu }_2^3),\hfill \end{array}$$

(47)

$$\begin{array}{cc}& a_{43}^{\prime }=4({\mu }_7-3{\mu }_2{\mu }_5+3{\mu }_2^2{\mu }_3),a_{44}^{\prime }=({\mu }_8-4{\mu }_2{\mu }_6+6{\mu }_2^2{\mu }_4-3{\mu }_2^4).\hfill \end{array}$$

(48)

$$\begin{array}{cc}& T\left[32\right]={\displaystyle \sum _{k=3}^5}{b}_k where b_3=60g_{122}{\mu }_2{\mu }_3+12g_{222}(3{\mu }_2{M}_4+2{\mu }_3^2).\hfill \end{array}$$

(49)

$$\begin{array}{cc}& b_4=-2{\displaystyle \sum _{i=1}^4}{a}_i{g}_{1^{4-i}{2}^i} with\hfill \\ & a_1=10{\mu }_2{\mu }_3,a_2={\mu }_2{M}_4+9{\mu }_3^2,a_3={\mu }_2{\mu }_5+9{\mu }_3{\mu }_4-10{\mu }_3{\mu }_2^2,\hfill \end{array}$$

(50)

$$\begin{array}{cc}& a_4={\mu }_2{\mu }_6+6{\mu }_3({\mu }_5-{\mu }_2{\mu }_3)+3{\mu }_4^2-9{\mu }_2^2{\mu }_4+5{\mu }_2^4.\hfill \end{array}$$

(51)

$$\begin{array}{cc}& b_5={\displaystyle \sum _{i=0}^5}{a}_i{g}_{1^{5-i}{2}^i}, with with a_0={\mu }_2{\mu }_3,a_1=3{\mu }_2{M}_4+2{\mu }_3^2,\hfill \\ & a_2=7{\mu }_3{\mu }_4+3{\mu }_2{\mu }_5-10{\mu }_2^2{\mu }_3,a_3=6{\mu }_3({\mu }_5-{\mu }_2{\mu }_3)+3{\mu }_4^2-9{\mu }_2^2{\mu }_4+5{\mu }_2^4,\hfill \\ & a_4=3M_4{\mu }_5-9{\mu }_2{\mu }_3{\mu }_4+7{\mu }_2^3{\mu }_3+2{\mu }_3{\mu }_6,\hfill \end{array}$$

(52)

$$\begin{array}{cc}& a_5=M_4({\mu }_6-3{\mu }_2{\mu }_4+2{\mu }_2^3).\hfill \end{array}$$

(53)

$$\begin{array}{cc}& T\left[2^3\right]={\displaystyle \sum _{k=3}^6}{c}_k where c_3=-120{\mu }_2^3{g}_{2^3},\hfill \end{array}$$

(54)

$$\begin{array}{cc}& c_4/36=a_2{g}_{1^2{2}^2}+a_3{g}_{12^3}+a_4{g}_{2^4} with\hfill \end{array}$$

(55)

$$\begin{array}{cc}& a_2=5{\mu }_2^3,a_3=10{\mu }_2^2{\mu }_3,a_4={\mu }_2^2{M}_4+4{\mu }_2{\mu }_3^2,\hfill \end{array}$$

(56)

$$\begin{array}{cc}& c_5=-6{\mu }_2{\displaystyle \sum _{i=1}^5}{a}_i{g}_{1^{5-i}{2}^i} with\hfill \end{array}$$

(57)

$$\begin{array}{cc}& a_1={\mu }_2^2,a_2=4{\mu }_2{\mu }_3,a_3=2{\mu }_2{M}_4+4{\mu }_3^2,a_4=4{\mu }_3{M}_4,a_5=M_4^2,\hfill \end{array}$$

(58)

$$\begin{array}{cc}& and c_6={\displaystyle \sum _{i=0}^6}{a}_i{g}_{1^{6-i}{2}^i} with\hfill \\ & a_0={\mu }_2^3,a_1=4{\mu }_2^2{\mu }_3,a_2=3{\mu }_2^2{M}_4+12{\mu }_2{\mu }_3^2,a_3=8{\mu }_3({\mu }_2{M}_4+{\mu }_3^2),\hfill \end{array}$$

(59)

$$\begin{array}{cc}& a_4=3M_4({\mu }_2{M}_4+4{\mu }_3^2),a_5=4{\mu }_3{M}_4^2,a_6=M_4^3.\hfill \end{array}$$

(60)

Note 2.7. 

This was given without proof in Example 6.5 of [9]. This had an incorrect value for $U_{112}$ in $T\left[4\right]$, so that $S_2\left(F\right)$ there is correct but not $S_3\left(F\right)$. $S_2\left(F\right)$, but not $S_3\left(F\right)$, was also covered by Example 5.7 of [3]. Example 6.5.1 p217 of [9] applies this to $\mu /\sigma$.

PROOF (41) follows from $\left(i\right|2\left|k\right)=-2\left[U_{i\cdot 1}{h}_1\right]\left[U_{k\cdot 1}{h}_1\right]$. By (25), $a_{12}=-2g_2{\mu }_2$ as $\overline{1}=0$.

$$a_{22}=g_{11}\left(11\right)+2g_{12}\left(12\right)+g_{22}\left(22\right).$$

As $U_{i\cdot 111}=0,a_{13}$ of (26) is 0, (43) and (44) follow from (26) by substituting into $a_{23}/3={\sum }_{j=1}^2{g}_{j2}{d}_{j2}=g_{11}{d}_{11}+2g_{12}{d}_{12}+g_{22}{d}_{22},$ and

$$\begin{array}{cc}& a_{33}=g_{111}\left(111\right)+3g_{112}\left(112\right)+3g_{122}\left(122\right)+g_{222}\left(222\right),\hfill \end{array}$$

$a_{14}=0$ as $U_{i\cdot 1122}=0.$ (45) follows from

$$\begin{array}{cc}& a_{24}=g_{22}(u_1+2u_2),u_1={\left[{\mu }_{2\cdot 11}\right]}^2=4{\mu }_2^2,u_2=\left[{\mu }_{2\cdot 12}^2\right]=4{\mu }_2^2,\hfill \\ & a_{34}=2g_{ij2}(v_{1ij}+2v_{2ij})=2v_1+4v_2 where,\hfill \\ & v_1=g_{ij2}{v}_{1ij},v_{1ij}=\left(ij\right)\overline{2},v_2=g_{ij2}{v}_{2ij},v_{2ij}=\left(i\right|2\left|j\right)=-2\left(1i\right)\left(1j\right),.\hfill \\ & a_{44}=g_{1^4}{\left(11\right)}^2+4g_{1^{3}2}\left(11\right)\left(12\right)+6g_{1122}\left(11\right)\left(22\right)+4g_{12^3}\left(12\right)\left(22\right)+g_{2^4}{\left(22\right)}^2.\hfill \end{array}$$

Now substitute $\left(11\right),\dots ,\left(222\right).$ $T\left[4\right]$ is given by (28) with $\left[B_{ij}^4\right]=0$ unless $i=j=2$, and $\left[B_{ijk}^4\right]=0$ unless $k=2$. Also, for $h_1\left(x_1\right)=x_1-\mu$,

$$\begin{array}{cc}& \left[B_{22}^4\right]=-\left[B_{112}^4\right]=12{\mu }_4,\left[B_{122}^4\right]=-12({\mu }_5-{\mu }_2{\mu }_3),\left[B_{222}^4\right]=-12({\mu }_6-2{\mu }_2{\mu }_4+{\mu }_2^3);\hfill \\ & \mathrm{So}, \mathrm{for} a_k \mathrm{of} \left(28\right),a_2=3g_{22}\left[{\mu }_{2\cdot 11}^2\right]=12{\mu }_4{g}_{22},\hfill \\ & a_3/(-12)=g_{i_1{i}_{2}2}\left[U_{i_1\cdot 11}{U}_{i_2\cdot 1}{h}_1^2\right]=g_{112}\left[h_1^4\right]+2g_{122}\left[h_1^3(h_1^2-{\mu }_2)\right]+g_{222}\left[h_1^2{(h_1^2-{\mu }_2)}^2\right],\hfill \end{array}$$

and (46) holds. $T\left[32\right]$ is given by (29) with $b_1=b_2=0$ as $\left[B_{51}\right]=\left[B_{52}\right]=0,$ $\left[B_{i_1{i}_2{i}_3}^5\right]=\left[B_{53}\right]$ is 0 if $i_2=1$ or $i_3=1$.

$$\begin{array}{cc}& \left[B_{i22}^5\right]=-6\left[U_{i\cdot 1}{h}_1^2\right]\overline{2}+12\left[U_{i\cdot 1}{h}_1^2\right]{\mu }_2+24\left[U_{i\cdot 2}{h}_2\right]{\mu }_3:\hfill \\ & \left[B_{122}^5\right]=60{\mu }_2{\mu }_3,\left[B_{222}^5\right]=36{\mu }_2{M}_4+24{\mu }_3^2;\hfill \\ & b_3=3g_{i_{1}22}(\left[U_{i_1\cdot 1}{\mu }_{2\cdot 11}\right]\overline{2}+2\left[U_{i_1\cdot 1}{\mu }_{2\cdot 12}^2\right]+2\left[U_{i_1\cdot 2}{\mu }_{2\cdot 12}{\mu }_{2\cdot 11}\right]).\hfill \end{array}$$

So (49) holds. Given two functions $a_{i_1\dots i_k}$ and $b_{i_1\dots i_k}$, set

$$\begin{array}{cc}& {\left(ab\right)}_{i_1\dots i_k}=a_{i_1\dots i_k}{b}_{i_1\dots i_k},{\left[ab\right]}_{i_1\dots i_k}={\displaystyle \sum _{i_1,\dots ,i_k=1}^q}{\left(ab\right)}_{i_1\dots i_k}.\hfill \end{array}$$

(61)

$$\begin{array}{cc}& \mathrm{Set} G_{i_1{i}_2{i}_3}=g_{i_1{i}_2{i}_{3}2},D_{i_1{i}_2{i}_3}={\mu }_2\left(i_1{i}_2{i}_3\right)+6\left(i_{3}1\right)\left(i_1{i}_{2}1\right)+3\left(i_2{i}_3\right)\left(i_{1}11\right).\hfill \\ & \mathrm{Then} b_4=g_{i_1{i}_2{i}_{3}2}\{\left(i_1{i}_2{i}_3\right)\overline{2}+6\left[U_{i_1\cdot 1}{U}_{i_2\cdot 1}{U}_{i_3\cdot 2}{\mu }_{2\cdot 12}\right]+3\left[U_{i_1\cdot 1}{U}_{i_2\cdot 2}{U}_{i_3\cdot 2}{\mu }_{2\cdot 11}\right]\}\hfill \\ & =-2{\left[GD\right]}_{i_1{i}_2{i}_3}=-2[{\left(GD\right)}_{1^3}+3{\left(GD\right)}_{112}+3{\left(GD\right)}_{122}+{\left(GD\right)}_{2^3}]\mathrm{by} \left(61\right),\hfill \\ & where D_{1^3}=10{\mu }_2{\mu }_3,D_{112}={\mu }_2\left(112\right)+9\left(21\right)\left(111\right),\hfill \\ & D_{122}={\mu }_2\left(122\right)+6\left(21\right)\left(112\right)+3\left(22\right)\left(111\right),D_{2^3}={\mu }_2\left(222\right)+6\left(21\right)\left(221\right)\hfill \\ & +3\left(22\right)\left(211\right).\hfill \end{array}$$

So (50) holds with $a_i=D_{1^{4-i}{2}^{i-1}}$.

$$\begin{array}{cc}& Set H_{i_1{i}_2{i}_3}=g_{i_1{i}_2{i}_{3}11}\left(11\right)+2g_{i_1{i}_2{i}_{3}12}\left(12\right)+g_{i_1{i}_2{i}_{3}22}\left(22\right).\mathrm{Then} \mathrm{by}\left(61\right),\hfill \\ & b_5=\left(i_1{i}_2{i}_3\right)H_{i_1{i}_2{i}_3}={\left[dH\right]}_{i_1{i}_2{i}_3}={\left(dH\right)}_{1^3}+3{\left(dH\right)}_{112}+3{\left(dH\right)}_{122}+{\left(dH\right)}_{2^3},\hfill \end{array}$$

where $d_{i_1{i}_2{i}_3}=\left(i_1{i}_2{i}_3\right).$ (52) follows. $T\left[2^3\right]$ needs $c_k$ of (30). $c_1=c_2=0$ as $\left[B_{61}\right]=\left[B_{62}\right]=0$. For $\left[B_{63}\right],\left[S_{15}^{\prime \prime }\right]=\left[S_{60}\right]=0,\left[S_{15}^{\prime \prime \prime }\right]=0$ unless $i_1=i_2=i_3=2$, in which case, since $\left(2\right|2)=4{\mu }_2^2,\left[S_{15}^{\prime \prime \prime }\right]=-96{\mu }_2^3$. So, $\left[B_{63}\right]=\left[S_{15}^{\prime \prime \prime }\right]I(i_1=i_2=i_3=2)$ and $c_3$ is given by (54). For $\left[B_{64}\right],\left[S_{20}\right]=0$; and $\left[S_{45}\right]=0$ unless $i_3=i_4=2$, in which case,

$$\begin{array}{cc}& \left[S_{45}\right]/12=3{\mu }_2^2\left(i_1{i}_2\right)-2{\mu }_2\left(i_{1}2i_2\right)+8{\mu }_2\left(1i_1\right)\left(1i_2\right)\hfill \\ & =5{\mu }_2^{3}if{i}_1=i_2=1,\hfill \\ & =10{\mu }_2^2{\mu }_{3}if{i}_1=1,i_2=2,\hfill \\ & ={\mu }_2^2{M}_4+4{\mu }_2{\mu }_3^{2}if{i}_1=i_2=2,by\left(41\right).\hfill \end{array}$$

So $c_4$ is given by (55).

$$\begin{array}{cc}& c_5=3(\overline{1}{J}_1+\overline{2}{J}_2)=-6{\mu }_2{J}_2 where\hfill \\ & J_j=\left(i_1{i}_2\right)\left(i_3{i}_4\right)g_{i_1\dots i_{4}j}=\left(i_1{i}_2\right)J_j^{i_1{i}_2},\hfill \\ & J_j^{i_1{i}_2}=\left(i_3{i}_4\right)g_{i_1\dots i_{4}j}=\left(11\right)g_{i_1{i}_{2}11j}+2\left(12\right)g_{i_1{i}_{2}12j}+\left(22\right)g_{i_1{i}_{2}22j}.\hfill \end{array}$$

$$\begin{array}{cc}& So,J_2=\left(11\right)J_2^{11}+2\left(12\right)J_2^{12}+\left(22\right)J_2^{22} where\hfill \\ & J_2^{11}=\left(11\right)g_{1^{4}2}+2\left(12\right)g_{1^3{2}^2}+\left(22\right)g_{1^2{2}^3},\hfill \\ & J_2^{12}=\left(11\right)g_{1^3{2}^2}+2\left(12\right)g_{1^2{2}^3}+\left(22\right)g_{12^4},\hfill \\ & J_2^{22}=\left(11\right)g_{1^2{2}^3}+2\left(12\right)g_{12^4}+\left(22\right)g_{2^5}.\hfill \\ & So,J_2=g_{1^{4}2}{\left(11\right)}^2+4g_{1^3{2}^2}\left(11\right)\left(12\right)+2g_{1^2{2}^3}[\left(11\right)\left(22\right)+2{\left(12\right)}^2]+4g_{12^4}\left(12\right)\left(22\right)\hfill \\ & +g_{2^5}{\left(22\right)}^2.\hfill \end{array}$$

So, (57) holds with

$$\begin{array}{cc}& a_0=0,a_1={\left(11\right)}^2,a_2=4\left(11\right)\left(12\right),a_3=2\left(11\right)\left(22\right)+4{\left(12\right)}^2,\hfill \\ & a_4=4\left(12\right)\left(22\right),a_5={\left(22\right)}^2.\hfill \\ & Also,c_6=\left(i_1{i}_2\right)K^{i_1{i}_2},for{K}^{i_1{i}_2}=\left(i_3{i}_4\right)K^{i_1\dots i_4},\hfill \\ & K^{i_1\dots i_4}=\left(i_5{i}_6\right)g_{i_1\dots i_6}=\left(11\right)g_{i_1\dots i_{4}11}+2\left(12\right)g_{i_1\dots i_{4}12}+\left(22\right)g_{i_1\dots i_{4}22},\hfill \\ & K^{i_1{i}_2}=\left(11\right)K^{i_1{i}_{2}11}+2\left(12\right)K^{i_1{i}_{2}12}+\left(22\right)K^{i_1{i}_{2}22},\hfill \\ & K^{i_1{i}_{2}11}=\left(11\right)g_{i_1{i}_2{1}^4}+2\left(12\right)g_{i_1{i}_2{1}^{3}2}+\left(22\right)g_{i_1{i}_{2}1122},\hfill \\ & K^{i_1{i}_{2}12}=\left(11\right)g_{i_1{i}_2{1}^{3}2}+2\left(12\right)g_{i_1{i}_{2}1122}+\left(22\right)g_{i_1{i}_{2}12^3},\hfill \\ & K^{i_1{i}_{2}22}=\left(11\right)g_{i_1{i}_{2}1122}+2\left(12\right)g_{i_1{i}_{2}12^3}+\left(22\right)g_{i_1{i}_2{2}^4}.\hfill \\ & \mathrm{So} c_6=\left(11\right)K^{11}+2\left(12\right)K^{12}+\left(22\right)K^{22},\hfill \\ & K^{11}=\left(11\right)K^{1^4}+2\left(12\right)K^{1^{3}2}+\left(22\right)K^{1122},K^{1^4}=\left(11\right)g_{1^6}+2\left(12\right)g_{1^{5}2}+\left(22\right)g_{1^4{2}^2},\hfill \\ & K^{1^{3}2}=\left(11\right)g_{1^{5}2}+2\left(12\right)g_{1^4{2}^2}+\left(22\right)g_{1^3{2}^3},K^{1122}=\left(11\right)g_{1^4{2}^2}+2\left(12\right)g_{1^3{2}^3}+\left(22\right)g_{1^2{2}^4}.\hfill \\ & K^{12}=\left(11\right)K^{1^{3}2}+2\left(12\right)K^{1^2{2}^2}+\left(22\right)K^{12^3},K^{1^{3}2}=\left(11\right)g_{1^{5}2}+2\left(12\right)g_{1^4{2}^2}+\left(22\right)g_{1^3{2}^3},\hfill \\ & K^{1^2{2}^2}=\left(11\right)g_{1^4{2}^2}+2\left(12\right)g_{1^3{2}^3}+\left(22\right)g_{1^2{2}^4},K^{12^3}=\left(11\right)g_{1^3{2}^3}+2\left(12\right)g_{1^2{2}^4}+\left(22\right)g_{12^5}.\hfill \\ & K^{22}=\left(11\right)K^{1^2{2}^2}+2\left(12\right)K^{12^3}+\left(22\right)K^{2^4},K^{1^2{2}^2}=\left(11\right)g_{1^4{2}^2}+2\left(12\right)g_{1^3{2}^3}+\left(22\right)g_{1^2{2}^4},\hfill \\ & K^{12^3}=\left(11\right)g_{1^3{2}^3}+2\left(12\right)g_{1^2{2}^4}+\left(22\right)g_{12^5},K^{2^4}=\left(11\right)g_{1^2{2}^4}+2\left(12\right)g_{12^5}+\left(22\right)g_{2^6}.\hfill \\ & \mathrm{So}, c_6\mathrm{is} \mathrm{given} \mathrm{by} \left(59\right)\mathrm{with} a_0={\left(11\right)}^3,a_1=4{\left(11\right)}^2\left(12\right),\hfill \\ & a_2=3{\left(11\right)}^2\left(22\right)+12\left(11\right){\left(12\right)}^2,a_3=8\left(12\right)\{\left(11\right)\left(22\right)+{\left(12\right)}^2\},\hfill \\ & a_4=3\left(11\right){\left(22\right)}^2+12{\left(12\right)}^2\left(22\right)=3\left(22\right)[\left(11\right)\left(22\right)+4{\left(12\right)}^2],\hfill \\ & a_5=4\left(12\right){\left(22\right)}^2,a_6={\left(22\right)}^3.\square \hfill \end{array}$$

Example 5. 

Take $T\left(F\right)=g(\mu ,{\mu }_2)={\mu }_2/\mu$. The non-zero derivatives of g to order 6 are

$$\begin{array}{cc}& g_1=-{\mu }^{-2}{\mu }_2,g_2={\mu }^{-1},g_{11}=2{\mu }^{-3}{\mu }_2,g_{12}=-{\mu }^{-2},g_{111}=-6{\mu }^{-4}{\mu }_2,\hfill \\ & g_{112}=2{\mu }^{-3},g_{1^4}=24{\mu }^{-5}{\mu }_2,g_{1^{3}2}=-6{\mu }^{-4},\hfill \end{array}$$

$$\begin{array}{cc}& g_{1^5}=-120{\mu }^{-6}{\mu }_2,g_{1^{4}2}=24{\mu }^{-5},g_{1^6}=720{\mu }^{-7}{\mu }_2,g_{1^{5}2}=-120{\mu }^{-6}.\hfill \end{array}$$

(62)

Set $w_r={\mu }^{-r}{\mu }_r$. For $k\le 4$, a kth order estimate of $T\left(F\right)$ is given by Theorem 2.1 with

$$\begin{array}{cc}& T\left[2\right]=2\mu (-w_2+w_2^2-w_3),sothat{S}_1\left(F\right)=\mu (w_2-w_2^2+w_3),\hfill \\ & T\left[3\right]=6\mu (w_3-w_2{w}_3+w_4-w_2^2),\hfill \\ & T\left[2^2\right]=24\mu (-w_2^2+w_2^3-w_2{w}_3).\hfill \\ & T\left[4\right]=6\mu [w_3-4w_4+4w_2{w}_4-4(w_5-w_2{w}_3)],\hfill \\ & T\left[32\right]=24\mu [5w_2(1-w_2)w_3+3w_2(w_4-w_2^2)+2w_3^2].\hfill \\ & T\left[2^3\right]=c_5+c_6 where c_5=-144{\mu }^{-5}{\mu }_2^3,\hfill \\ & c_6=720{\mu }^{-7}{\mu }_2^4-480{\mu }^{-6}{\mu }_2^2{\mu }_3=240{\mu }^{-1}{w}_2^2(3w_2-2w_3).\hfill \\ & So, S_2\left(F\right)/\mu =2w_3+w_2{w}_3+2w_4-5w_2^2+w_2{w}_3+3w_2^3,\hfill \\ & and S_3\left(F\right)/\mu =-3w_3/2+6w_4+6w_5-17w_2{w}_3-9w_2^2-18w_2{w}_4-8w_3^2+21w_2^3\hfill \\ & +14w_2^2{w}_3-15w_2^4.\hfill \end{array}$$

For example, by (42)–(58),

$$\begin{array}{cc}& T\left[2\right]=a_{12}+a_{22} where a_{12}=-2\mu w_2,a_{22}=2\mu (w_2^2-w_3),\hfill \\ & T\left[3\right]=a_{23}/3+a_{33} where a_{23}/3=2\mu w_3,\hfill \\ & a_{33}=6\mu (-w_2{w}_3+w_4-w_2^2),\hfill \\ & T\left[2^2\right]={\displaystyle \sum _{i=3}^4}{a}_{i4} where a_{34}=-24\mu w_2^2,a_{44}=24\mu (w_2^3-w_2{w}_3),\hfill \end{array}$$

and $T\left[2^3\right]$ is given by (54)–(60) with$\left\{g_{1^r{2}^s}\right\}$ of (62).

Example 6. 

Take $T\left(F\right)=g(\mu ,{\mu }_2)={\mu }^2/{\mu }_2$. We need the partial derivatives

$$\begin{array}{cc}& g_1=2\mu /{\mu }_2,g_2=-{\mu }^2/{\mu }_2^2,g_{11}=2/{\mu }_2,g_{12}=-2\mu /{\mu }_2^2,g_{22}=2{\mu }^2/{\mu }_2^3,\hfill \\ & g_{112}=-2/{\mu }_2^2,g_{122}=4\mu /{\mu }_2^3,g_{2^3}=-6{\mu }^2/{\mu }_2^4,\hfill \\ & g_{1^2{2}^2}=4/{\mu }_2^3,g_{{12}^3}=-12\mu /{\mu }_2^4,g_{2^4}=24{\mu }^2/{\mu }_2^5,\hfill \\ & g_{1^2{2}^3}=-12/{\mu }_2^4,g_{{12}^4}=48\mu /{\mu }_2^5,g_{2^5}=-120{\mu }^2/{\mu }_2^6,\hfill \\ & g_{1^2{2}^4}=48/{\mu }_2^5,g_{12^5}=-240\mu /{\mu }_2^6,g_{2^6}=720{\mu }^2/{\mu }_2^7.\hfill \end{array}$$

Substituting into Corollary 2.5, and simplifying gives

$$\begin{array}{cc}& T\left[2\right]=2u_1^2{u}_4-4u_1{u}_3+2,\hfill \\ & T\left[3\right]/6=(1+2u_1^2)(1-u_4)+2u_1{u}_5-u_1^2(u_6-3u_4+2),\hfill \\ & T\left[2^2\right]/24=u_1^2(2u_4^2+2u_3^3-8u_4+3)-2u_1{u}_3(u_4+1)+u_4,\hfill \\ & T\left[4\right]/24=u_1^2(u_8-u_6+u_4)-2u_1(u_7-5u_5+u_3)+u_6-u_4+1,\hfill \\ & T\left[32\right]/12={\displaystyle \sum _{i=0}^2}{a}_i{u}_1^i where a_0=-6u_3{u}_5-3u_4^2+(25u_4-13)/3,\hfill \\ & a_1/2=4u_3{u}_6+6u_4{u}_5-9u_3{u}_4-5u_5+14u_3,\hfill \\ & -a_2/2=5u_4{u}_6+24u_3{u}_5-9u_4^2-18u_3^2-3u_6+16u_4-9,\hfill \\ & T\left[2^3\right]/48={\displaystyle \sum _{i=0}^2}{a}_i{u}_1^i where a_0=6u_3^2(2u_4-1)+3u_4^2-3u_4+15,\hfill \\ & a_1=-20u_3{u}_4^2+34u_3{u}_4-59u_3,a_2=15u_4^3-30u_4^2+72u_3^2+231u_4-211.\hfill \end{array}$$

Example 7. 

Take $T\left(F\right)=\mu /{\mu }_2$. Set

$$\sigma ={\mu }_2^{1/2},u_1=\mu /\sigma ,and{u}_r={\mu }_r/{\sigma }^{r}forr\ge 2.$$

For $k\le 4$, a kth order estimate of $T\left(F\right)$ is given by Theorem 2.1 in terms of

$$\begin{array}{cc}& S_1\left(F\right)=-T\left[2\right]/2={\sigma }^{-1}\{u_3+u_1(-2u_4+1)\},\hfill \\ & T\left[3\right]/3=2{\sigma }^{-1}{L}_1 where L_1=u_5-2u_1(u_4-1),\hfill \\ & T\left[2^2\right]/8=3{\sigma }^{-1}{L}_2 where L_2=2u_1{u}_3^2-u_3(u_4+1)+u_1(u_4^2-u_4+1),\hfill \end{array}$$

$$\begin{array}{cc}& S_2\left(F\right)={\sigma }^{-1}(2L_1+3L_2).Also by \left(46\right),T\left[4\right]={\displaystyle \sum _{k=2}^4}{a}_k where\hfill \\ & a_2=-6\{{\mu }_2^{-2}{\mu }_3+2\mu {\mu }_2^{-3}{M}_4\}=6{\sigma }^{-1}{M}_1 for M_1=u_3-2u_1{u}_2^{-3}(u_4-u_2^2),\hfill \\ & a_3={\displaystyle \sum _{i=0}^3}{a}_{3i}^{\prime }{g}_{1^{3-i}{2}^i}=2{\mu }_2^{-3}{a}_{32}^{\prime }-6\mu {\mu }_2^{-4}{a}_{33}^{\prime } for a_{3i}^{\prime }of\left(47\right)\hfill \\ & =6{\sigma }^{-1}{M}_2, for M_2=8u_2^{-3}(u_5-u_2{u}_3)-u_1{u}_2^{-4}(u_6-2u_2{u}_4+u_2^3),\hfill \\ & a_4=-6{\mu }_2^{-4}{a}_{43}^{\prime }+24\mu {\mu }_2^{-3}{a}_{44}^{\prime } for a_{4i}^{\prime }of\left(48\right).\hfill \end{array}$$

$$\begin{array}{cc}& Also, by \left(29\right) and \left(52\right),T\left[32\right]={\displaystyle \sum _{k=3}^5}{b}_k where\hfill \\ & b_3/24=8{\mu }_2^{-3}{\mu }_3-3\mu {\mu }_2^{-4}(3{\mu }_2{M}_4+2{\mu }_3^2),\hfill \\ & b_4/12={\mu }_2^{-4}{a}_3-4\mu {\mu }_2^{-5}{a}_4 for a_3,a_{4}of\left(51\right),\hfill \\ & b_5/24={\mu }_2^{-5}{a}_4-5\mu {\mu }_2^{-6}{a}_5 for a_4,a_{5}of\left(53\right).\hfill \\ & By \left(54\right)--\left(59\right),T\left[2^3\right]={\sigma }^{-1}{\displaystyle \sum _{k=3}^6}{c}_k where c_3=720\mu /\sigma =720u_1,\hfill \\ & c_4=432\{-5u_3+2u_1(4u_3^2+u_4-1)\},\hfill \\ & c_5=144{\mu }_3(-M_4{\mu }_2^3+5\mu M_4^2/{\mu }_2^5),\hfill \\ & c_6=240(-2{\mu }_3{M}_4^2/{\mu }_2^6+3{\mu }_4^3{\mu }_2^7).\hfill \end{array}$$

For, the non-zero derivatives of g up to order 6 are

$$\begin{array}{cc}& g_1={\mu }_2^{-1},g_2=-\mu {\mu }_2^{-2},g_{12}=-{\mu }_2^{-2},g_{22}=2\mu {\mu }_2^{-3},\hfill \\ & g_{122}=2{\mu }_2^{-3},g_{2^3}=-6\mu {\mu }_2^{-4},g_{{12}^3}=-6{\mu }_2^{-4},g_{2^4}=24\mu {\mu }_2^{-5},\hfill \\ & g_{{12}^4}=24{\mu }_2^{-4},g_{2^5}=-120\mu {\mu }_2^{-6},g_{12^5}=-120{\mu }_2^{-6},g_{2^6}=720\mu {\mu }_2^{-7}.\hfill \end{array}$$

So by Corollary 2.5,

$$\begin{array}{cc}& for a_{i2}of\left(42\right),\sigma a_{12}=2u_1,\sigma a_{22}/2=-u_3+2u_1(u_4-1),\hfill \\ & for a_{i3}of\left(43\right),\sigma a_{23}/12=u_3-u_1(u_4-1),\hfill \\ & \sigma a_{33}/6=u_5-2u_3-u_1(u_6-3u_4+2);\hfill \\ & for a_{i4}of\left(45\right),\sigma a_{24}=24u_1,\sigma a_{34}/24=-2u_3+u_1(u_4-1+2u_3^2),\hfill \\ & \sigma a_{44}/24=-u_3(u_4-1)+u_1{(u_4-1)}^2.\hfill \end{array}$$

Example 8. 

Suppose that we have a random sample from the scaled gamma distribution,

$$\begin{array}{cc}& X={\theta }_{1}G\sim F\left(x\right) where G\sim Gamma\left({\theta }_2\right).\hfill \end{array}$$

(63)

Then by Section 3 of [40], moment estimates of ${\theta }_1$ and ${\theta }_2$ are

$$\begin{array}{cc}& \widehat{{\theta }_1}=\widehat{{\mu }_2}/\widehat{\mu } and \widehat{{\theta }_2}={\widehat{\mu }}^2/\widehat{{\mu }_2}, where\hfill \\ & \mu =EX,{\mu }_2=var\left(X\right),\widehat{\mu }=\overline{X},{\widehat{\mu }}_2=\overline{X^2}-{\overline{X}}^2.\hfill \end{array}$$

the sample mean and the empirical variance. So for $k\le 4,k$th order estimates of these are given by Examples 2.5 and 2.6.

I now give the MLEs. The calculations needed are much more lengthy than those for the moment estimates above. Set

$$\begin{array}{cc}& a_1\left(x\right)=x,a_2\left(x\right)=lnx,\hfill \end{array}$$

$$\begin{array}{cc}& U_i=U_i\left(F\right)=\int a_i\left(x\right)dF\left(x\right),{\widehat{U}}_i=\int a_i\left(x\right)d{F}_n\left(x\right)fori=1,2.\hfill \end{array}$$

(64)

By (3.2) of [40] or c(2.51) of [42], the MLEs are

$$\begin{array}{cc}& \widetilde{{\theta }_1}={\widehat{U}}_1/\widetilde{{\theta }_2}=g_1\left(\widehat{U}\right),say,\hfill \end{array}$$

(65)

$$\begin{array}{cc}& \widetilde{{\theta }_2}=ϵ(ln{\widehat{U}}_1-\widehat{U_2})=g_2\left(\widehat{U}\right),say,\hfill \end{array}$$

(66)

$$\begin{array}{cc}& whereϵ\left(x\right)istheinverseof\hfill \end{array}$$

(67)

$$\begin{array}{cc}& \lambda \left(\rho \right)=ln\rho -\psi \left(\rho \right),and\psi \left(\rho \right)=(d/d\rho )ln\Gamma \left(\rho \right).\hfill \end{array}$$

(68)

These both have the form $g\left(U\left(F_n\right)\right)$ where $U\left(F\right)\in R^2$.

The kth order estimate of ${\theta }_2$ using its MLE (66) for $k\le 4$.

$$\begin{array}{cc}& {\theta }_2=ϵ\left(\nu \right) where \nu =\nu \left(F\right)=ln{U}_1-U_2=f\left(U\right)say,\hfill \end{array}$$

(69)

with rth order functional derivative

$$\begin{array}{cc}& {\nu }_{\cdot 1\dots r}={(-1)}^{r-1}{U}_1^{-r}{U}_{1\cdot 1}\dots U_{1\cdot r}-U_{2\cdot 1}I(r=1), where U_{i\cdot j}=a_i\left(x_j\right)-U_i.\hfill \end{array}$$

Now apply Corollary 2.1 with

$$Set{ϵ}_r=g_{1^r}={(d/d\nu )}^rϵ\left(\nu \right)at\nu =\nu \left(F\right).$$

${ϵ}_r$ can be written in terms of the derivatives of $\lambda \left(\rho \right)$ of (68) using [41]. The terms in (23)–(24) are

$$\begin{array}{cc}& {\overline{1}}^{\nu }=\left[{\nu }_{11}\right]=-U_1^{-2}\left(1^2\right),{\left(1^2\right)}^{\nu }=[\cdot {\nu }_1^2]=-U_1^{-2}\left(1^2\right),\hfill \\ & {\left(1^3\right)}^{\nu }=\left[{\nu }_{\cdot 1}^3\right]=U_1^{-3}\left(1^3\right)-3U_1^{-2}\left(1^{2}2\right)+3U_1^{-1}\left(12^2\right)-\left(2^3\right),\hfill \\ & {\left(1^4\right)}^{\nu }=\left[{\nu }_{\cdot 1}^4\right]=U_1^{-4}\left(1^4\right)-4U_1^{-3}\left(1^{3}2\right)+6U_1^{-2}\left(1^2{2}^2\right)-4U_1^{-1}\left(12^3\right)+\left(2^4\right),\hfill \\ & d_{11}^{\nu }=\left[{\nu }_{\cdot 1}{\nu }_{\cdot 11}\right]=-U_1^{-3}\left(1^3\right)+U_1^{-2}\left(1^{2}2\right)],{\left(1\right|1)}^{\nu }=\left[{\nu }_{\cdot 12}^2\right]=U_1^{-4}{\left(1^2\right)}^2,\hfill \\ & {\left(1\right|1\left|1\right)}^{\nu }=\left[{\nu }_{\cdot 1}{\nu }_{\cdot 12}{\nu }_{\cdot 2}\right]=-2U_1^{-3}{\{U_1^{-1}\left(1^2\right)-\left(12\right)\}}^2.\hfill \end{array}$$

To apply Corollary 2.1, write ${\theta }_2$ as

$${\theta }_2=ϵ\left(\nu \right)=g\left(U\left(F\right)\right),\nu =f\left(U\right),g\left(U\right)=g_2\left(U\right)=ϵ\left(f\left(U\right)\right).$$

As $lnX=ln{\theta }_1+lnG,$

$$\left(1^{r}2\right)=\mu (X,\dots ,X,lnX)={\theta }_1^r\mu (G,\dots ,G,lnG),$$

and the $(i_1\dots i_r)$ of (34) needed for $T\left[2\right],\dots ,T\left[2^3\right]$ of (35)–(38), are

$$\begin{array}{cc}& \left(1^r{2}^s\right)=\mu (X,\dots ,X,lnX,\dots ,lnX)={\theta }_1^r\mu (G,\dots ,G,lnG,\dots lnG).\hfill \end{array}$$

(70)

These can be found from the non-central moments

$$E{G}^t{(lnG)}^s={\tau }_{\cdot s}\left(t\right)where\tau \left(t\right)=E{G}^t=\Gamma (t+{\theta }_2)/\Gamma \left({\theta }_2\right).$$

To obtain the $g_{j_1\dots j_r}$ for $r\le 6$ needed for (34)–(38) and Corollary 2.2, apply (II) of 24 and (6) with $s=2,q=1$ and $(x,U,g,T)$ replaced by $(U,f,ϵ,g)$. Replace $T_{\cdot 1\dots r}$ by $g_{j_1\dots j_r}$, to get

$$\begin{array}{cc}& g_{j_1\dots j_r}={\displaystyle \sum _{k=1}^r}{B}_k^r{ϵ}_k:B_1^r=f_{j_1\dots j_r},B_r^r=f_{j_1}\dots f_{j_r},\hfill \end{array}$$

(71)

and so on. The non-zero derivatives of $f\left(U\right)$ of (69) are

$$\begin{array}{cc}& f_{1^r}={(-1)}^{r-1}(r-1)!U_1^{-r},f_2=-1.\hfill \end{array}$$

(72)

$$\begin{array}{cc}& So, T\left[2\right]=g_{j_1{j}_2}\left(j_1{j}_2\right)=g_{11}\left(11\right)+2g_{12}\left(12\right)+g_{22}\left(22\right)\hfill \\ & where g_{11}=U_1^{-2}(-{ϵ}_1+{ϵ}_2),g_{12}=-U_1^{-1}{ϵ}_2,g_{22}={ϵ}_2.\hfill \\ & g_{j_1{j}_2{j}_3}={\displaystyle \sum _{k=1}^r}{B}_k^3{ϵ}_k=f_{j_1{j}_2{j}_3}{ϵ}_1+S_3{ϵ}_2+f_{j_1}\dots f_{j_3}{ϵ}_3\hfill \\ & where B_2^3=S_3=\sum ^3{f}_{j_1}{f}_{j_2{j}_3}.\hfill \\ & By \left(35\right),T\left[3\right]=g_{j_1{j}_2{j}_3}\left(j_1{j}_2{j}_3\right)\hfill \end{array}$$

$$\begin{array}{cc}& =g_{111}\left(111\right)+3g_{112}\left(112\right)+3g_{122}\left(122\right)+g_{222}\left(222\right) where\hfill \\ & g_{111}=2U_1^{-3}{ϵ}_1+S_3{ϵ}_2+U_1^{-3}{ϵ}_3=U_1^{-3}(2{ϵ}_1-3{ϵ}_2+{ϵ}_3)\hfill \\ & as{S}_3=3f_1{f}_{11}=-3U_1^{-3},\hfill \end{array}$$

(73)

$$\begin{array}{cc}& g_{112}=U_1^{-2}({ϵ}_2-{ϵ}_3)as{S}_3=3f_2{f}_{11}=U_1^{-2},\hfill \\ & g_{122}=f_1{f}_2^2{ϵ}_3=U_1^{-1}{ϵ}_{3}as{S}_3=0,g_{222}=f_2^3{ϵ}_3=-{ϵ}_3,as{S}_3=0.\hfill \\ & By \left(36\right),T\left[2^2\right]=\left(j_1{j}_2\right)\left(j_3{j}_4\right)g_{j_1\dots j_4}={\displaystyle \sum _{i=0}^4}{a}_i{g}_{1^{4-i}{2}^i} where a_0={\left(11\right)}^2,\hfill \end{array}$$

(74)

$$\begin{array}{cc}& a_1=4\left(11\right)\left(12\right),a_2=2\left(11\right)\left(2^2\right)+4{\left(12\right)}^2,a_3=4\left(12\right)\left(22\right),a_4={\left(22\right)}^2.\hfill \\ & g_{j_1\dots j_4}={\displaystyle \sum _{k=1}^r}{B}_k^4{ϵ}_k=f_{j_1\dots j_4}{ϵ}_1+B_2^4{ϵ}_2+B_3^4{ϵ}_3+f_{j_1}\dots f_{j_4}{ϵ}_4 where\hfill \\ & B_2^4=S_4+S_3^{\prime },S_4=\sum ^4{f}_{j_1}{f}_{j_2{j}_3{j}_4},S_3^{\prime }=\sum ^3{f}_{j_1{j}_2}{f}_{j_3{j}_4},\hfill \\ & B_3^4=S_6=\sum ^6{f}_{j_1}{f}_{j_2}{f}_{j_3{j}_4},B_4^4=f_{j_1}\dots f_{j_4}.\hfill \\ & By \left(37\right),T\left[4\right]=g_{j_1\dots j_4}(j_1\dots j_4)\hfill \\ & =g_{1^4}\left(1^4\right)+4g_{1^{3}2}\left(1^{3}2\right)+6g_{1^2{2}^2}\left(1^2{2}^2\right)+4g_{12^3}\left(12^3\right)+g_{2^4}\left(2^4\right)\hfill \end{array}$$

(75)

$$\begin{array}{cc}& where g_{1^4}=f_{1^4}{ϵ}_1+(S_4+S_3^{\prime }){ϵ}_2+S_6{ϵ}_3+f_1^4{ϵ}_4,\hfill \\ & S_4=4f_1{f}_{11}=8U_1^{-4},S_3^{\prime }=3f_{11}^2=3U_1^{-4},S_6=6f_1^2{f}_{11}=-6U_1^{-4}.\hfill \\ & So, g_{1^4}=U_1^{-4}(-6{ϵ}_1+2{ϵ}_2-6{ϵ}_3+{ϵ}_4).\hfill \\ & g_{1^{3}2}=(S_4+S_3^{\prime }){ϵ}_2+S_6{ϵ}_3+f_1^3{f}_2{ϵ}_4\hfill \\ & where S_4=f_2{f}_{111}=-2U_1^{-3},S_3^{\prime }=0,S_6=3f_1{f}_2{f}_{11}=-3U_1^{-3}.\hfill \end{array}$$

(76)

$$\begin{array}{cc}& So, g_{1^{3}2}=-U_1^{-3}(2{ϵ}_2+3{ϵ}_3+{ϵ}_4);\hfill \\ & g_{1^2{2}^2}=(S_4+S_3^{\prime }){ϵ}_2+S_6{ϵ}_3+f_1^2{f}_2^2{ϵ}_4 where\hfill \\ & S_4=S_3^{\prime }=0,S_6=f_2^2{f}_{11}=-U_1^{-2}.\hfill \end{array}$$

(77)

$$\begin{array}{cc}& So, g_{1^2{2}^2}=U_1^{-2}(-{ϵ}_3+{ϵ}_4);\hfill \\ & g_{12^3}=(S_4+S_3^{\prime }){ϵ}_2+S_6{ϵ}_3+f_1{f}_2^3{ϵ}_4 where S_4=S_3^{\prime }=S_6=0.\hfill \\ & So, g_{12^3}=-U_1^{-1}{ϵ}_4;g_{2^4}=f_2^4{ϵ}_4={ϵ}_4.\hfill \\ & By \left(37\right),T\left[32\right]={\displaystyle \sum _{i=0}^5}{a}_i{g}_{1^{5-i}{2}^i} where\hfill \\ & a_0=\left(111\right)\left(11\right),a_1=3\left(112\right)\left(11\right)+2\left(111\right)\left(12\right),\hfill \\ & a_2=3\left(122\right)\left(11\right)+6\left(112\right)\left(12\right)+\left(111\right)\left(22\right),\hfill \\ & a_3=\left(2^3\right)\left(1^2\right)+6\left(122\right)\left(12\right)+3\left(112\right)\left(22\right),\hfill \\ & a_4=2\left(2^3\right)\left(12\right)+3\left(112\right)\left(2^2\right),a_5=\left(2^3\right)\left(2^2\right).\hfill \end{array}$$

(78)

$$\begin{array}{cc}& By \left(38\right),T\left[2^3\right]={\displaystyle \sum _{i=0}^6}{a}_i{g}_{1^{6-i}{2}^i} where\hfill \end{array}$$

(79)

$$\begin{array}{cc}& a_0={\left(11\right)}^3,a_1=6{\left(11\right)}^2\left(12\right),a_2=3{\left(11\right)}^2\left(22\right)+12\left(11\right){\left(12\right)}^2,\hfill \end{array}$$

(80)

$$\begin{array}{cc}& a_3=12\left(11\right)\left(12\right)\left(22\right)+8({12}^3,a_4=3\left(11\right){\left(22\right)}^2+12{\left(12\right)}^2\left(22\right),\hfill \end{array}$$

(81)

$$\begin{array}{cc}& a_5=6\left(12\right){\left(22\right)}^2,a_6={\left(2^2\right)}^3.\hfill \end{array}$$

(82)

This completes the kth order estimate of ${\theta }_2$, given a random sample distributed as (63), using the MLE (66).

The kth order estimate of ${\theta }_1$ for $k\le 4$ using its MLE (65).

$$\begin{array}{cc}& {\theta }_1=g_1\left(U\right)=U_1/f\left(U\right)=h_1{h}_2^{-1}=G\left(h\left(U\right)\right)say,\hfill \\ & where h_2=f\left(U\right)={\theta }_2=g_2\left(U\right)={\theta }_{2}ln{U}_1-U_2,h_1=U_1.\hfill \end{array}$$

for $U=(U_1,U_2)$ of (64.

The derivatives of $g_2\left(U\right)$ are given by (71). Denote the partial derivatives of $h_i$ by $h_{j_1\dots j_r}^i$. The derivatives of $g\left(U\right)=g_1\left(U\right)=G\left(h\left(U\right)\right)$ are given by Theorem 2.2 with $q=s=2$ and $(x,U,g,T)$ replaced by $(U,h,G,g)$:

$$\begin{array}{cc}& g_{j_1\dots j_r}={\displaystyle \sum _{k=1}^r}{B}_k^r{G}_{i_1\dots i_k} where B_1^r=h_{j_1\dots j_r}^{i_1},B_r^r=h_{j_1}^{i_1}\dots h_{j_r}^{i_r},\hfill \end{array}$$

and so on. The non-zero derivatives of G and h are

$$\begin{array}{cc}& G_{12^r}={(-1)}^{r}r!h_2^{-r-1},G_{2^r}={(-1)}^{r}r!h_1{h}_2^{-r-1},h_1^1=1,h_{j_1\dots j_r}^2=f_{j_1\dots j_r}.\hfill \\ & So, T\left[2\right]=g_{j_1{j}_2}\left(j_1{j}_2\right)=g_{11}\left(11\right)+2g_{12}\left(12\right)+g_{22}\left(22\right) where\hfill \\ & g_{11}=2G_{12}{f}_1,g_{12}=G_{11}{f}_1+G_{12}{f}_2+G_{22}{f}_1{f}_2,g_{22}=G_2{f}_{22}+G_{22}{f}_2^2.\hfill \end{array}$$

(83)

$T\left[3\right]$ is given by (73) with 3rd order derivatives

$$\begin{array}{cc}& g_{j_1{j}_2{j}_3}={\displaystyle \sum _{k=1}^3}{a}_k, where a_k=G_{i_1\dots i_k}{B}_k^3,\hfill \\ & B_1^3=h_{i_1\cdot j_1\dots j_3},B_2^3=S_3=\sum ^3{h}_{i_1\cdot j_1}{h}_{i_2\cdot j_2{j}_3},B_3^3=h_{i_1\cdot j_1}\dots h_{i_3\cdot j_3}.\hfill \\ & So for g_{1^3},a_1=G_2{f}_{111},a_2/3=G_{12}{f}_{11}+G_{22}{f}_1{f}_{11},\hfill \\ & a_3=G_{111}+3G_{112}{f}_2+3G_{122}{f}_2^2+G_{222}{f}_2^3,\hfill \\ & for g_{1^{2}2},a_1=G_2{f}_{112},a_2=2G_{12}{f}_{12}+G_{22}(2f_1{f}_{12}+f_2{f}_{11}),\hfill \\ & a_3=2G_{112}{f}_1{f}_2+G_{122}{f}_1^2{f}_2,\hfill \\ & for g_{12^2},a_1=G_2{f}_{122},a_2=G_{22}{f}_{12}+G_{22}(2f_2{f}_{12}+f_1{f}_{22}),a_3=G_{222}{f}_2^3.\hfill \end{array}$$

$T\left[2^2\right]$ is given by (75) with 4th order derivatives

$$\begin{array}{cc}& g_{j_1\dots j_4}={\displaystyle \sum _{k=1}^4}{c}_k, where c_1=G_i{h}_{i\cdot j_1\dots j_4},\hfill \\ & c_2=G_{i_1{i}_2}(S_4+S_3^{\prime }),S_4=\sum ^4{h}_{i_1\cdot j_1}{h}_{i_2\cdot j_2{j}_3{j}_4},S_3^{\prime }=\sum ^3{h}_{i_1\cdot j_1{j}_2}{h}_{i_2\cdot j_3{j}_4},\hfill \\ & c_3=G_{i_1{i}_2{i}_3}{S}_6,S_6=\sum ^6{h}_{i_1\cdot j_1}{h}_{i_2\cdot j_2}{h}_{i_3\cdot j_3{j}_4},c_4=G_{i_1\dots i_4}{h}_{i_1\cdot j_1}\dots h_{i_4\cdot j_4}.\hfill \end{array}$$

This completes the terms needed for $S_2\left(F\right)$ and ${\widehat{T}}_{n3}$ of (1), for the 3rd order estimate of ${\theta }_1$ using its MLE. The 4th order estimate can be found similarly using the 5th and 6th order derivatives of g.

To compute $\psi \left(\gamma \right)$, see p2665 of [40]. A more direct approach to obtain this kth order estimate of ${\theta }_1$ based on the MLE (65), is to apply Corollary A.1 with $(g,h,V)$ of (A1) replaced by $(G,h,U)$. See Example A.1 below.

For confidence intervals for ${\theta }_1$ and ${\theta }_2$, see Section 5 of [40].

3. Discussion

As noted, [9] showed that Fisher estimates, (1), outperform bootstrapping, jackknifing and those due to [10,11,12]. It also provided computer programs in MAPLE for implementation of these estimates. Yet this method, and the earlier method of analytic bias reduction in [2], have largely been overlooked by many potential users. For example, https://zbmath.org/?q=analytic+bias+reduction only lists 26 analytic bias reduction papers. These include [29,30]. These only give at best 2nd order estimates.

Yet expansions for the bias of functions of estimates suitable for Theorems 3.1 and 5.1 go back to [42] for MLEs. While these expansions for MLEs can be a lot more complicated than for a moment estimate, they are often worth it, as MLEs are more efficient. On the other hand, estimates can often be framed as functions of mean-type estimates, and so dealt with using Corollaries 2.2 and 2.3.

Future directions.

1. [3,9] extended the theory of kth order bias reduction, to samples from more than 1 distribution, say $F=(F_1,\dots ,F_k)$ with means $\mu =({\mu }_1,\dots ,{\mu }_k)$. Examples include $T\left(F\right)=g\left(\mu \right)$, see Example 6.2 in [9] and Example 5.4 of [3], in particular in [9], $T\left(F\right)={\left({\alpha }^{\prime }\mu \right)}^q$ for given $q\in R,\alpha \in R^k$, and $T\left(F\right)={\mu }_1{\mu }_2/({\mu }_1+{\mu }_2)$, and Example 5.5 of [3], $T\left(F\right)=\left(E{a}^{\prime }X\right)/\left(E{b}^{\prime }X\right)$ for given $a,b\in R^k$,

2. There are a huge number of papers on moment estimates, cumulant estimates, MLEs, M-estimates, and exponential families. The results here, or their k sample extension in [9], can be applied to many of these. (Moment and cumulant estimates generally provide much more tractable results than using MLEs, but are less efficient.)

3. The results here can be applied to obtain bias reduction for an extreme estimate, that is, a standard estimate with asymptotic variance $O\left(n^{-1}\right)$. See [37,43,44,45,46,47]. Applications can also be made to the extremes of ARMA processes. See for example [48].

4. Except for [49], the results above are for real functions of real parameters. For applications to control theory and other branches of electrical engineering, parameters often lie in C. This is the case for example for the cell-phone problem considered in [50], and also for the spectral density estimate: see [51]. In order to extend these result to analytic functions $t\left(\theta \right):C^p\to C$, the concepts in [52] of a mean-type estimate, and a standard estimate need to be extended to $\widehat{\theta }\in C^p$. For a way to do this, see [53,54].

5. Another important advance would be to sampling from a finite population. The groundwork has already been laid in the following remarkable inversion principle for obtaining unbiased estimates given in [55].

Let $F=F_N$ be the distribution of a finite real population of size N. Let $F_n$ be the empirical distribution of a sample of size $n\le N$ drawn from the population without replacement. $T\left(F_N\right)$ be any product of the moments or cumulants of $F_N$.

$$Let{T}_{nN}\left(F_N\right)=ET\left(F_n\right).ThenE{T}_{Nn}\left(F_n\right)=T\left(F_N\right).$$

It also gave an explicit expression for $T_{nN}\left(F_N\right)$ for all products of order up to 6. This has the potential to provide an estimate of bias $O\left(n^{-7}\right)$ for $T\left(F_N\right)$ any smooth functional of the population distribution.