Poisson Counts, Square Root Transformation and Small Area Estimation

Abstract

The paper intends to serve two objectives. First, it revisits the celebrated Fay-Herriot model, but with homoscedastic known error variance. The motivation comes from an analysis of count data, in the present case, COVID-19 fatality for all counties in Florida. The Poisson model seems appropriate here, as is typical for rare events. An empirical Bayes (EB) approach is taken for estimation. However, unlike the conventional conjugate gamma or the log-normal prior for the Poisson mean, here we make a square root transformation of the original Poisson data, along with square root transformation of the corresponding mean. Proper back transformation is used to infer about the original Poisson means. The square root transformation makes the normal approximation of the transformed data more justifiable with added homoscedasticity. We obtain exact analytical formulas for the bias and mean squared error of the proposed EB estimators. In addition to illustrating our method with the COVID-19 example, we also evaluate performance of our procedure with simulated data as well.

Introduction

Small area estimation is now a topic of global importance. Methodologies abound, and many of these are finding real-life applications.

Normal theory small area estimation pervades the literature, and the pioneering (Fay and Herriot, 1979) method is most often used in real life applications. The Fay-Herriot model is a normal theory mixed effects area-level model. The model needs to assume known error variances in order to avoid non-identifiability, whereas in reality, these are sample estimates.

The present article deals with the Fay-Herriot model with known constant error variance. The motivation came from an analysis of count data, in particular, COVID-19 data, to find estimates of fatality for all counties in the state of Florida. The Poisson model is used, but we make a square transformation of the original data, and the corresponding mean parameters to attain a closer approximation to normality with added homoscedasticity. This is in contrast to the log-transformation, where also one typically assumes normality of the transformed data. However, transformation of the original data in the log-scale bears the potential hazard of leaving out zero counts, which on most occasions, can affect the conclusion significantly. Further, our approach allows one to develop Stein-type shrinkage estimators for small area means and study their properties analytically.

Variable transformation in the small area context has been addressed before. The logarithmic transformation with the assumption of log-normal distribution of the original data is most commonly used, for example, in modeling income distributions. See for example, Slud and Maiti (2006) and Ghosh et al. (2015). Recently, Hirose et al. (2021) considered an arc-sin transformation of binomial proportions for small area estimation. Sugasawa and Kubokawa (2017) suggested a non-explicit EB estimator and performed an analysis based on the dual power transformation similar to that of Hirose et al. (2021).

The remaining sections are as follows. In Section 2, we introduce the square root transformation, develop Stein-type shrinkage estimators for the transformed data motivated from an empirical Bayes point of view, and then back transform properly to estimate the original parameters of interest. In Sections 3 and 4, we obtain exact expressions for the bias and the mean squared error of our shrinkage estimator. Finally, Section 5 we also obtain an estimator of the mean squared error correct up to order $\mathcal{O}(m^{-1})$ where m is the number of small areas. Section 6 contains an illustration of the proposed method to estimate the number of deaths due to COVID-19 in each county. A simulation study is undertaken in Section 7. Some final remarks are made in Section 8.

Empirical Bayes Estimators

Suppose there are m areas with counts y_i for i-th area. We assume y_i are independently distributed from Poisson(λ_i) for i-th area. We transform $z_i=\sqrt{y_i}$ and with the usual variance stabilizing square root transformation so that V (z_i) is approximately 1/4. We begin with

$$z_i|{𝜃}_i\stackrel{ind}{\sim }N({𝜃}_i,1/4)\text{where}{𝜃}_i=\sqrt{{\lambda }_i}(i=1,\mathrm{...},m)\mathrm{.}$$

Following the customary approach, we consider independent $N(x_i^{\top }\beta ,A)$ priors for the 𝜃_i with p-dimensional auxiliary variables x_i and regression parameter $\beta \in {\mathbb{ℝ}}^p$ where m > p + 4. The posterior ${𝜃}_i|z_i\stackrel{ind}{\sim }N((1-B)z_i+B{x}_i^{\top }\beta ,(1-B)/4)$ where $B=\frac{1/4}{1/4+A}$. Thus, Bayes estimator of λ_i is

$${\widehat{\lambda }}_i^B=E({\lambda }_i|z_i)=E({𝜃}_i^2|z_i)=(1-B)/4+{\{(1-B)z_i+B{x}_i^{\top }\beta \}}^2$$ 1

We now turn towards empirical Bayes (EB) estimation of the λ_i. Writing X = (x₁,...,x_m)^⊤, Z = (z₁,...,z_m)^⊤ and $\widehat{\beta }={(X^{\top }X)}^{-1}{X}^{\top }Z$, it follows that marginally $||Z-X\widehat{\beta }|{|}^2\sim \frac{1}{4B}{\chi }_{m-p}^2$. Here X is a m × p matrix with rank p. Following Efron and Morris (1973), an EB estimator of B is $\widehat{B}=\frac{m-p-2}{4||Z-X\widehat{\beta }|{|}^2}$. Thus an EB estimator of λ_i is

$${\widehat{\lambda }}_i^{EB}=(1-\widehat{B})/4+{\{(1-\widehat{B})z_i+\widehat{B}{x}_i^{\top }\widehat{\beta }\}}^2$$ 2

For proving our technical results, we find it convenient also to define

$${\tilde{\lambda }}_i^{EB}=(1-B)/4+{\{(1-B)z_i+B{x}_i^{\top }\widehat{\beta }\}}^2,$$ 3

which is also an EB estimator of λ_i if the shrinkage factor B were known. Also, for notational simplicity we write m₀ = m − p hereafter.

Bias of ${\widehat{\lambda }}^{EB}$

For both bias and mean squared error (MSE) calculations for ${\widehat{\lambda }}_i^{EB}$ we need the following lemmas.

Lemma 1.

Let $\left(\begin{array}{c}X\\ Y\end{array}\right)\sim N\left[\left(\begin{array}{c}{\mu }_1\\ {\mu }_2\end{array}\right),\left(\begin{array}{cc}{\sigma }_1^2& {\sigma }_{12}\\ {\sigma }_{12}& {\sigma }_2^2\end{array}\right)\right]\mathrm{.}$ Then

$$\begin{array}{rcl}Var(X^2)=2{\sigma }_1^4+4{\mu }_1^2{\sigma }_1^2& & \\ Var(Y^2)=2{\sigma }_2^4+4{\mu }_2^2{\sigma }_2^2& & \\ Cov(X^2,Y^2)=2{\sigma }_{12}^2+4{\mu }_1{\mu }_2{\sigma }_{12}& & \\ & & \end{array}$$

Lemma 2.

Define $s_i=x_i^{\top }{(X^{\top }X)}^{-1}{x}_i$, $u_{i1}=(1-B)z_i+B{x}_i^{\top }\widehat{\beta }$ and $u_{i2}=(1-B)z_i+B{x}_i^{\top }\beta$ then,

$$\left[\begin{array}{c}{u}_{i1}\\ u_{i2}\end{array}\right]simN\left[\left(\begin{array}{c}{x}_i^{\top }\beta \\ x_i^{\top }\beta \end{array}\right),\left(\begin{array}{cc}A(1-B)+(2-B)s_i/4& (1-B)(A+s_i/4)\\ (1-B)(A+s_i/4)& A(1-B)\end{array}\right)\right]\mathrm{.}$$ 4

Proof 1.

The result follows from the independence of $\widehat{\beta }$ and $z_i-x_i\widehat{\beta }$ and noting that $\widehat{\beta }\sim N(\beta ,(1/4+A){(X^{\top }X)}^{-1})$ while $z_i-x_i^{\top }\widehat{\beta }\sim N(0,(1/4+A)(1-s_i))$. □

Lemma 3.

• (i) $||Z-X\widehat{\beta }||$ is distributed independently of $\left(\frac{(z_1-x_1^{\top }\widehat{\beta })}{||Z-X\widehat{\beta }||},\dots ,\frac{(z_m-x_m^{\top }\widehat{\beta })}{||Z-X\widehat{\beta }||}\right)$.
• (ii) $E\left[\frac{{(z_i-x_i^{\top }\widehat{\beta })}^k}{||Z-X\widehat{\beta }|{|}^k}\right]=\frac{E{(z_i-x_i^{\top }\widehat{\beta })}^k}{E(||Z-X\widehat{\beta }|{|}^k)}$ for all positive integers k.
• (iii) $E\left[\frac{{(z_i-x_i^{\top }\widehat{\beta })}^k}{||Z-X\widehat{\beta }|{|}^l}\right]=0$ if k is an odd positive integer and 0 < l < k.

Proof 2.

Marginally, $Z\sim N(X\beta ,(1/4+A)I_m)$. Hence, $(\widehat{\beta },||Z-X\widehat{\beta }||)$ is complete sufficient for (β,A), while $\left(\frac{(z_1-x_1^{\top }\widehat{\beta })}{||Z-X\widehat{\beta }||},\dots ,\frac{(z_m-x_m^{\top }\widehat{\beta })}{||Z-X\widehat{\beta }||}\right)$ is ancillary. This proves (i) by an application of Basu’s Theorem (Basu, 1955).

Now for any positive integer k, and by part (i) in Lemma 3, we have,

$$\begin{array}{rcl}E{(z_i-x_i^{\top }\widehat{\beta })}^k& =& E\left[||Z-X\widehat{\beta }|{|}^k\frac{{(z_i-x_i^{\top }\widehat{\beta })}^k}{||Z-X\widehat{\beta }|{|}^k}\right]\\ & =& E\left[||Z-X\widehat{\beta }|{|}^k\right]E\left[\frac{{(z_i-x_i^{\top }\widehat{\beta })}^k}{||Z-X\widehat{\beta }|{|}^k}\right]\mathrm{.}\end{array}$$

This leads to $E\left[\frac{{(z_i-x_i^{\top }\widehat{\beta })}^k}{||Z-X\widehat{\beta }|{|}^k}\right]=\frac{E{(z_i-x_i^{\top }\widehat{\beta })}^k}{E(||Z-X\widehat{\beta }|{|}^k)}$.

Next noting that $((z_1-x_1^{\top }\widehat{\beta }),\dots ,(z_m-x_m^{\top }\widehat{\beta }))\stackrel{d}{=}-((z_1-x_1^{\top }\widehat{\beta }),\dots ,(z_m-x_m^{\top }\widehat{\beta }))$ it follows that

$$\begin{array}{rcl}\left(\frac{(z_1-x_1^{\top }\widehat{\beta })}{||Z-X\widehat{\beta }||},\dots ,\frac{(z_m-x_m^{\top }\widehat{\beta })}{||Z-X\widehat{\beta }||}\right)\stackrel{d}{=}\left(\frac{-(z_1-x_1^{\top }\widehat{\beta })}{||Z-X\widehat{\beta }||},\dots ,\frac{-(z_m-x_m^{\top }\widehat{\beta })}{||Z-X\widehat{\beta }||}\right)& & \end{array}$$

Therefore, $\frac{(z_i-x_i^{\top }\widehat{\beta })}{||Z-X\widehat{\beta }||}$ is symmetric random variable around 0 and its all odd moments are 0. Hence,

$$\begin{array}{rcl}E\left[\frac{{(z_i-x_i^{\top }\widehat{\beta })}^k}{||Z-X\widehat{\beta }|{|}^l}\right]& =& E\left[||Z-X\widehat{\beta }|{|}^{k-l}\frac{{(z_i-x_i^{\top }\widehat{\beta })}^k}{||Z-X\widehat{\beta }|{|}^k}\right]\\ & =& E(\left[||Z-X\widehat{\beta }|{|}^{k-l}\right]E\left[\frac{{(z_i-x_i^{\top }\widehat{\beta })}^k}{||Z-X\widehat{\beta }|{|}^k}\right]\end{array}$$

We get the second equality in the above equations using part (i) of this Lemma. Proof of (iii) is complete after observing k is an odd integer. □

Lemma 4.

• (i) $E(\widehat{B})=B$ if m₀ > 2.
• (ii) $E({\widehat{B}}^2)=\frac{m_0-2}{m_0-4}{B}^2$ if m₀ > 4.
• (iii) $E(1/\widehat{B})=\frac{m_0}{(m_0-2)B}$ if m₀ > 2.
• (iv) $E(1/{\widehat{B}}^2)=\frac{m_0(m_0+2)}{{(m_0-2)}^2{B}^2}$ if m₀ > 2.
• (v) $E\left[\frac{{(z_i-x_i^{\top }\widehat{\beta })}^2}{||Z-X\widehat{\beta }|{|}^2}\right]=\frac{(1-s_i)}{m_0}$.
• (vi) $E\left[\frac{{(z_i-x_i^{\top }\widehat{\beta })}^4}{||Z-X\widehat{\beta }|{|}^4}\right]=\frac{3{(1-s_i)}^2}{m_0(m_0+2)}$.

Proof 3.

The proof (i) to (iv) follows by noting that $\widehat{B}=\frac{m_0-2}{4||Z-X\widehat{\beta }|{|}^2}$ and $||Z-X\widehat{\beta }|{|}^2\sim \frac{1}{4B}{\chi }_{m_0}^2$. To prove (v) and (vi) we also need part (ii) of Lemma 3. □

Now we start with calculation of the bias $E({\widehat{\lambda }}_i^{EB}-{\lambda }_i)$. The following theorem is proved.

Theorem 1.

Suppose $z_i|{𝜃}_i\stackrel{ind}{\sim }N({𝜃}_i,1/4)$ with priors ${𝜃}_i\stackrel{ind}{\sim }N(x_i^{\top }\beta ,A)$ and m₀ > 2. Then bias of the EB estimator ${\widehat{\lambda }}_i^{EB}$ in Eq. 2 for ${\lambda }_i={𝜃}_i^2$ is given by

$$Bias({\widehat{\lambda }}_i^{EB})=E({\widehat{\lambda }}_i^{EB}-{\lambda }_i)=\frac{2-B}{4}\left(s_i+\frac{2(1-s_i)}{m_0}\right)\mathrm{.}$$ 5

Proof 4.

We begin with the partition

$$E({\widehat{\lambda }}_i^{EB}-{\lambda }_i)=E({\widehat{\lambda }}_i^B-{\lambda }_i)+E({\tilde{\lambda }}_i^{EB}-{\widehat{\lambda }}_i^B)+E({\widehat{\lambda }}_i^{EB}-{\tilde{\lambda }}_i^{EB})$$ 6

By Lemmas 1 and 2,

$$\begin{array}{rcl}E({\tilde{\lambda }}_i^{EB}-{\widehat{\lambda }}_i^B)& =& E(u_{i1}^2-u_{i2}^2)=Var(u_{i1})+{(x_i\beta )}^2-Var(u_{i2})-{(x_i\beta )}^2\\ & =& (2-B)s_i/4\end{array}$$ 7

Noticing $E({\lambda }_i)=E(E({\lambda }_i|z_i))=E({\widehat{\lambda }}_i^B)$, we have $E({\widehat{\lambda }}_i^B-{\lambda }_i)=0$. It is easy to see that

$$\begin{array}{rcl}{\widehat{\lambda }}_i^{EB}-{\tilde{\lambda }}_i^{EB}& =& (B-\widehat{B})/4+2x_i^{\top }\widehat{\beta }(B-\widehat{B})(z_i-x_i^{\top }\widehat{\beta })\\ & & +\{({\widehat{B}}^2-B^2)-2(\widehat{B}-B)\}{(z_i-x_i^{\top }\widehat{\beta })}^2\\ & =& A_1+A_2+A_3,\end{array}$$ 8

where $A_1=(B-\widehat{B})/4,A_2=2x_i^{\top }\widehat{\beta }(B-\widehat{B})$ and $A_3=\{({\widehat{B}}^2-B^2)-2(\widehat{B}-B)\}{(z_i-x_i^{\top }\widehat{\beta })}^2$. The expectation of A₁ is 0 since $\widehat{B}=\frac{m_0-2}{4||Z-X\widehat{\beta }|{|}^2}$ is unbiased estimator of B. Now,

$$\begin{array}{rcl}E(A_2)& =& E(x_i^{\top }\widehat{\beta })E\left[(B-\widehat{B})(z_i-x_i^{\top }\widehat{\beta })\right]\\ & =& E(x_i^{\top }\widehat{\beta })E(B(z_i-x_i^{\top }\widehat{\beta }))-E(x_i^{\top }\widehat{\beta })E\left[\frac{m_0-2}{4||Z-X\widehat{\beta }|{|}^2}(z_i-x_i^{\top }\widehat{\beta })\right]\\ & =& 0\end{array}$$ 9

The first equality in Eq. 9 is by independence of $\widehat{\beta }$ and $((z_1-x_1^{\top }\widehat{\beta }),\dots ,$ $(z_n-x_n^{\top }\widehat{\beta }))$. The third equality holds by part (iii) of Lemma 3 since $(z_i-x_i^{\top }\widehat{\beta })\sim N(0,(1/4+A)(1-s_i))$.

Finally, we simplify A₃. By (i) and (iii) of Lemma 3 and Lemma 4,

$$\begin{array}{rcl}E({\widehat{\lambda }}_i^{EB}-{\tilde{\lambda }}_i^{EB})& =& E[\{({\widehat{B}}^2-B^2)-2(\widehat{B}-B)\}{(z_i-x_i^{\top }\widehat{\beta })}^2]\\ & =& E\left[\frac{{(z_i-x_i^{\top }\widehat{\beta })}^2}{||Z-X\widehat{\beta }|{|}^2}\right]E\left[\{({\widehat{B}}^2-B^2)-2(\widehat{B}-B)\}\frac{m_0-2}{4\widehat{B}}\right]\\ & =& \frac{(1-s_i)}{m_0}\frac{(m_0-2)}{4}E[\widehat{B}-B^2/\widehat{B}-2+2B/\widehat{B}]\\ & =& \frac{(1-s_i)}{m_0}\frac{(m_0-2)}{4}\left[B-B\frac{m_0}{m_0-2}-2+2\frac{m_0}{m_0-2}\right]\\ & =& \frac{1-s_i}{2m_0}(2-B)\mathrm{.}\end{array}$$ 10

The proof of Eq. 5 follows now by combining (6) with Eqs. 7–10. □

Remark 1.

With the usual assumption, $s_i=x_i^{\top }{(X^{\top }X)}^{-1}{x}_i=\mathcal{O}(m^{-1})$ for large m, the bias of the EB estimator ${\widehat{\lambda }}_i^{EB}$, $E({\widehat{\lambda }}_i^{EB}-{\lambda }_i)=\mathcal{O}(m^{-1})$.

Remark 2.

We can estimate the bias in Eq. 5 by replacing the B by $\widehat{B}$, An unbiased estimator of the bias is

$$\widehat{bias}=\frac{2-\widehat{B}}{4}\left(s_i+\frac{2(1-s_i)}{m_0}\right)\mathrm{.}$$ 11

Thus, from Eq. 11, the EB estimator has positive bias, and the bias-corrected estimator of λ_i is

$${\widehat{\lambda }}_i^{CEB}={\widehat{\lambda }}^{EB}-\widehat{bias}\mathrm{.}$$

MSE of ${\widehat{\lambda }}^{EB}$

The following theorem provides an exact expression for the MSE of ${\widehat{\lambda }}^{EB}$.

Theorem 2.

Suppose $z_i|{𝜃}_i\stackrel{ind}{\sim }N({𝜃}_i,1/4)$ with priors ${𝜃}_i\stackrel{ind}{\sim }N(x_i^{\top }\beta ,A)$ and m₀ > 4. Then MSE of the EB estimator ${\widehat{\lambda }}_i^{EB}$ in Eq. 2 for ${\lambda }_i={𝜃}_i^2$ is given by

$$\begin{array}{rcl}MSE({\widehat{\lambda }}_i^{EB})& =& E{({\widehat{\lambda }}_i^{EB}-{\lambda }_i)}^2\\ & =& (1-B)\left[{(x_i^{\top }\beta )}^2+\frac{(1-B)(2-B)}{8B}\right]\\ & & +B{s}_i{(x_i^{\top }\beta )}^2+{(1-B)}^2{s}_i/4+s_i^2(1/2-B/4-B^2/16)\\ & & +\frac{B^2}{8(m_0-4)}+\frac{1-s_i}{2m_0}\left[4{(x_i^{\top }\beta )}^{2}B+s_i\right]\\ & & +3{(1-s_i)}^2\left[\frac{2m_0^2-9m_0+6}{4m_0(m_0+2)(m_0-4)}{B}^2-\frac{B}{m_0+2}+\frac{1}{2m_0}\right]\\ & & +\frac{(1-s_i)B}{2m_0}\left[1-\frac{B(m_0-3)}{m_0-4}\right]\\ & & +\frac{s_i(1-s_i)}{m_0}(2-2B+B^2/4)\mathrm{.}\end{array}$$

The proof of Theorem 2 is given in Appendix A.

Remark 3.

Theorem 2 shows that the MSE of EB estimator ${\widehat{\lambda }}_i^{EB}$, $E{({\widehat{\lambda }}_i^{EB}-E({\lambda }_i))}^2=\mathcal{O}(1)$ due to the first term in Eq. 12 for large m.

Estimation of the MSE of ${\widehat{\lambda }}^{EB}$

In this section we estimate the MSE of ${\widehat{\lambda }}_i^{EB}$ provided in Theorem 2 up to the order O(m^− 1) for large m. We now assume that $s_i=\mathcal{O}(m^{-1})$. Ignoring the O(m^− 2) terms, we rewrite

$$\begin{array}{rcl}MSE({\widehat{\lambda }}_i^{EB})& =& (1-B){(x_i^{\top }\beta )}^2+\frac{{(1-B)}^2(2-B)}{8B}+s_i\left[B{(x_i^{\top }\beta )}^2+\frac{{(1-B)}^2}{4}\right]\\ & & +\frac{B^2}{8m}+\frac{2B{(x_i^{\top }\beta )}^2}{m}+\frac{3{(1-B)}^2}{2m}+\frac{B(1-B)}{2m}+\mathcal{O}(m^{-2})\end{array}$$

It is easy to see that only first two terms in Eq. 12 do not depend on m and remaining terms are of $\mathcal{O}(m^{-1})$. Using Lemma 4 we get

$$\left(\begin{array}{c}E\left(\frac{1}{\widehat{B}}-\frac{2}{m\widehat{B}}\right)=\frac{1}{B}+\mathcal{O}(m^{-2})\\ E\left({\widehat{B}}^2-\frac{2}{m}{\widehat{B}}^2\right)=B^2+\mathcal{O}(m^{-2})\end{array}\right\}$$ 12

Using Eq. 12, we find

$$E\left[\frac{{(1-\widehat{B})}^2(2-\widehat{B})}{\widehat{B}}-\frac{4}{m\widehat{B}}+\frac{2{\widehat{B}}^2}{m}\right]=\left[\frac{{(1-B)}^2(2-B)}{B}\right]+\mathcal{O}(m^{-2})\mathrm{.}$$ 13

Now by $x_i^{\top }\widehat{\beta }\sim N(x_i^{\top }\beta ,\frac{s_i}{4B})$, Lemma 4 and the independence of $\widehat{\beta }$ and $\widehat{B}$, we also have

$$E[(1-\widehat{B})({(x_i^T\widehat{\beta })}^2-s_i/4\widehat{B})]=(1-B){(x_i^T\beta )}^2+O(m^{-2})\mathrm{.}$$ 14

Since we are ignoring $\mathcal{O}(m^{-2})$ terms in MSE estimation, we can estimate the $\mathcal{O}(m^{-1})$ terms in Eq. 12 simply by replacing B² by ${\widehat{B}}^2$ and ${(x_i^{\top }\beta )}^2$ by ${(x_i^{\top }\widehat{\beta })}^2$. By Eqs. 13 and 14, we derive estimator of the MSE of ${\widehat{\lambda }}_i^{EB}$ in Theorem 3.

Theorem 3.

Assume conditions of Theorem 2. Then $MSE({\widehat{\lambda }}_i^{EB})$ as given in (13) is estimated by

$$\begin{array}{rcl}\widehat{MSE({\widehat{\lambda }}_i^{EB})}& =& (1-\widehat{B}){(x_i^{\top }\widehat{\beta })}^2+\left[\frac{{(1-\widehat{B})}^2(2-\widehat{B})}{8\widehat{B}}\right]\\ & & +s_i\left[\widehat{B}{(x_i^{\top }\widehat{\beta })}^2+\frac{{(1-\widehat{B})}^2}{4}-\frac{(1-\widehat{B})}{4\widehat{B}}\right]+\frac{3{\widehat{B}}^2}{8m}\\ & & +\frac{2\widehat{B}}{m}{(x_i^{\top }\widehat{\beta })}^2+\frac{3}{2m}{(1-\widehat{B})}^2+\frac{\widehat{B}(1-\widehat{B})}{2m}-\frac{1}{2m\widehat{B}}+{\mathcal{O}}_p(m^{-2})\mathrm{.}\end{array}$$

The next theorem shows that MSE of the bias-corrected estimator ${\widehat{\lambda }}_i^{CEB}$ equals $MSE({\widehat{\lambda }}_i^{EB})+\mathcal{O}(m^{-2})$. In other words, bias correction does not lead to any significant improvement over $MSE({\widehat{\lambda }}_i^{EB})$, at least, when calculated up to $\mathcal{O}(m^{-1})$.

Theorem 4.

Assume conditions of Theorem 2. Then

$$\begin{array}{rcl}E[{({\widehat{\lambda }}_i^{CEB}-{\lambda }_i)}^2]=E[{({\widehat{\lambda }}_i^{EB}-{\lambda }_i)}^2]+\mathcal{O}(m^{-2})\mathrm{.}& & \end{array}$$

Proof 5.

Write $\widehat{bias}=\frac{(2-\widehat{B})}{4}{d}_i$, where $d_i=s_i+2\frac{1-s_i}{m_0}=\mathcal{O}(m^{-1})$. We begin with

$$\begin{array}{rcl}E[{({\widehat{\lambda }}_i^{CEB}-{\lambda }_i)}^2]& =& V[{\widehat{\lambda }}_i^{CEB}-{\lambda }_i],\\ & =& V[{\widehat{\lambda }}_i^{EB}-{\lambda }_i]+V[\widehat{bias}]-2Cov({\widehat{\lambda }}_i^{EB}-{\lambda }_i,\widehat{bias})\mathrm{.}\end{array}$$ 15

It is immediate that $V[{\widehat{\lambda }}_i^{EB}-{\lambda }_i]=E[{({\widehat{\lambda }}_i^{EB}-{\lambda }_i)}^2]-{(bias)}^2$, where ${(bias)}^2=\frac{1}{16}{(2-B)}^2{d}_i^2=\mathcal{O}(m^{-2})$. Next

$$\begin{array}{rcl}V[\widehat{bias}]& =& E[{\widehat{B}}^2]-B^2=E\left[\frac{{(m_0-2)}^2}{16||Z-X\widehat{\beta }|{|}^4}\right]-B^2\\ & =& \frac{{(m_0-2)}^2}{16}\frac{16B^2}{(m_0-2)(m_0-4)}-B^2=\frac{2B^2}{m_0-4}\mathrm{.}\end{array}$$

Hence,

$$V[\widehat{bias}]=\frac{d_i^2}{8}\frac{B^2}{m_0-4}=\mathcal{O}(m^{-3})\mathrm{.}$$ 16

Finally, $Cov({\widehat{\lambda }}_i^{EB}-{\lambda }_i,\widehat{bias})=Cov({\widehat{\lambda }}_i^{EB},\widehat{bias})-Cov({\lambda }_i,\widehat{bias})$, But

$$\begin{array}{rcl}Cov({\lambda }_i,\widehat{bias})& =& E[{\lambda }_i\widehat{bias}]-E[{\lambda }_i]E[\widehat{bias}],\\ & =& Cov(\widehat{{\lambda }_i^B},\widehat{bias}),\\ & =& Cov({\{(1-B)z_i+B{x}_i^{\top }\beta \}}^2,\widehat{bias}),\\ & =& \frac{(1-s_i)}{8m_0}{(1-B)}^2{d}_i=\mathcal{O}(m^{-2})\mathrm{.}\end{array}$$ 17

Thus, $Cov({\widehat{\lambda }}_i^{EB}-{\lambda }_i,\widehat{bias})=Cov({\widehat{\lambda }}_i^{EB},\widehat{bias})+\mathcal{O}(m^{-2})$. Now,

$$\begin{array}{rcl}Cov({\widehat{\lambda }}_i^{EB},\widehat{bias})& =& Cov\left(\frac{1-\widehat{B}}{4}+{\{(1-\widehat{B})z_i+\widehat{B}{x}_i^{\top }\widehat{\beta }\}}^2,\frac{2-\widehat{B}}{4}{d}_i\right),\\ & =& \frac{d_i}{16}V[\widehat{B}]-\frac{d_i}{4}Cov\left({\{(1-\widehat{B})z_i+\widehat{B}{x}_i^{\top }\widehat{\beta }\}}^2,\widehat{B}\right),\\ & =& \frac{B^2}{8(m_0-4)}{d}_i-\frac{d_i}{4}Cov\left(\right){\{x_i^{\top }\widehat{\beta }\}}^2+2(1-\widehat{B})(z_i-x_i^{\top }\widehat{\beta })x_i^{\top }\widehat{\beta }\\ & & +{(1-\widehat{B})}^2{(z_i-x_i^{\top }\widehat{\beta })}^2,\widehat{B}\left)\right)\mathrm{.}\end{array}$$ 18

Due to the independence of $\widehat{\beta }$ and $Z-X\widehat{\beta }$, $Cov({\{x_i^{\top }\widehat{\beta }\}}^2,\widehat{B})=0$. Next, again invoking the symmetry of Z − Xβ around 0, and $E(z_i-x_i^{\top }\widehat{\beta })=0$,

$$\begin{array}{rcl}Cov((1-\widehat{B})(z_i-x_i^{\top }\widehat{\beta })x_i^{\top }\widehat{\beta },\widehat{B})& =& E[\{(1-\widehat{B})(z_i-x_i^{\top }\widehat{\beta })x_i^{\top }\widehat{\beta }\widehat{B}]\\ & & -E[\{(1-\widehat{B})(z_i-x_i^{\top }\widehat{\beta })x_i^{\top }\widehat{\beta }]E[\widehat{B}],\\ & =& 0.\end{array}$$

Hence Eq. 18 reduces to

$$\begin{array}{rcl}Cov({\widehat{\lambda }}_i^{EB},\widehat{bias})& =& (-d_i/4)Cov((1-2\widehat{B}+{\widehat{B}}^2){(z_i-x_i^{\top }\widehat{\beta })}^2,\widehat{B})\\ & & +\mathcal{O}(m^{-2}),\end{array}$$ 19

$$\begin{array}{rcl}Cov({(z_i-x_i^{\top }\widehat{\beta })}^2,\widehat{B})& =& E\left[\frac{(m_0-2)}{4}\frac{{(z_i-x_i^{\top }\widehat{\beta })}^2}{||Z-X\widehat{\beta }|{|}^2}\right]-E[{(z_i-x_i^{\top }\widehat{\beta })}^2]B\mathrm{.}\\ & =& \frac{1-s_i}{2m_0}\mathrm{.}\end{array}$$ 20

Next

$$\begin{array}{rcl}Cov(\widehat{B}{(z_i-x_i^{\top }\widehat{\beta })}^2,\widehat{B})=Cov\left(\frac{(m_0-2)}{4}\frac{{(z_i-x_i^{\top }\widehat{\beta })}^2}{||Z-X\widehat{\beta }|{|}^2},\frac{(m_0-2)}{4||Z-X\widehat{\beta }|{|}^2}\right)=0& & \end{array}$$

Since $\frac{{(z_i-x_i^{\top }\widehat{\beta })}^2}{||Z-X\widehat{\beta }|{|}^2}$ is ancillary, and $||Z-X\widehat{\beta }|{|}^2$ is a function of the complete sufficient statistic $(\widehat{\beta },Z-X\widehat{\beta })$, by Basu’s Theorem,

$$\begin{array}{rcl}Cov({\widehat{B}}^2{(z_i-x_i^{\top }\widehat{\beta })}^2,\widehat{B})& =& E[{\widehat{B}}^3{(z_i-x_i^{\top }\widehat{\beta })}^2]-E[{\widehat{B}}^2{(z_i-x_i^{\top }\widehat{\beta })}^2]B\\ & =& \frac{{(m_0-2)}^3}{64}E\left[\frac{{(z_i-x_i^{\top }\widehat{\beta })}^2}{||Z-X\widehat{\beta }|{|}^2}\cdot \frac{1}{||Z-X\widehat{\beta }|{|}^4}\right]\\ & & -\frac{{(m_0-2)}^2}{16}E\left[\frac{{(z_i-x_i^{\top }\widehat{\beta })}^2}{||Z-X\widehat{\beta }|{|}^2}\cdot \frac{1}{||Z-X\widehat{\beta }|{|}^2}\right],\\ & =& \frac{(m_0-2)(1-s_i)B^2}{2m_0(m_0-4)}\end{array}$$ 21

Hence, from Eqs. 18–21, $Cov({\widehat{\lambda }}_i^{EB},\widehat{bias})=\mathcal{O}(m^{-2})$. This along with Eqs. 15–17 proves the theorem. □

Data Analysis

In this section, we now deploy our approach on the 2020 COVID-19 pandemic dataset, which is available at https://usafacts.org/visualizations/coronavirus-covid-19-spread-map/usafacts.org. This example is used mainly for illustration. We are using the figures provided as the sampled estimates. Our study shows that the coefficient of determination (R²) does not increase much if we include other demographic variables such as the population size, number of people over age 60, and income in the linear model for the number of deaths regressing on number of confirmed cases. It suggests that the number of confirmed cases is the most crucial variable in estimation of the number of deaths by Coronavirus than the aforementioned demographic variables. We have also studied a few more county-level data sources1 and we found out that adjusted gross income (AGI)2 of the year 2017 is really relevant for estimating the number of deaths. In our model, we have transformed the number of confirmed cases and adjusted gross income (AGI) by taking the square root. All data are aggregated at the county level. We are interested in estimating the counts of death due to Coronavirus for all counties in Florida. Here m = 57 since Florida has 57 counties. From Section 2 we know that $\sqrt{y_i}\stackrel{ind}{\sim }N(x_i^{\top }\beta ,A+0.25)$ and we estimate β by ordinary least square method. Based on our analysis, we get $\widehat{\beta }={(-0.2786,0.0917,0.0003)}^{\top }$, the respective estimates for the intercept, number of confirmed cases and AGI. We have summarized our results based on our model in Table 1 and the shrinkage factor $\widehat{B}=0.3777$. It seems that our model-based approach seems to pull the direct estimates towards some grand average, as one anticipates in a typical EB analysis. Figure 1 shows that the estimates are higher in south east Florida than the rest of the state.

Table 1.

COVID19in Florida as of December 28, 2020

County name	Confirm	AGI1	Deaths2	EB3	BIAS	Corrected BIAS EB4	RMSE5	Death rate6	Esti-mated rate7
Alachua	15473	7.294066	128	128.55	0.023	128.527	12.82	4.76	4.78
Baker	2414	0.554345	36	36.68	0.025	36.655	9.37	12.32	12.56
Bay	11403	4.571565	203	201.86	0.024	201.836	11.48	11.62	11.55
Bradford	2118	0.478624	23	23.75	0.025	23.725	9.32	8.16	8.42
Brevard	20355	17.959964	513	509.65	0.027	509.623	17.42	8.52	8.47
Broward	133480	68.207212	1828	1832.21	0.092	1832.118	45.89	9.36	9.38
Calhoun	1169	0.207429	28	28.70	0.025	28.675	9.16	19.85	20.35
Charlotte	7128	5.233347	231	229.35	0.023	229.327	11.46	12.23	12.14
Citrus	6467	3.447436	260	257.45	0.024	257.426	10.73	17.37	17.20
Clay	10829	6.127044	183	182.44	0.023	182.417	12.05	8.35	8.32
Collier	22004	27.616972	328	329.86	0.048	329.812	21.46	8.52	8.57
Columbia	5837	1.304030	117	116.71	0.025	116.685	9.86	16.32	16.28
DeSoto	2843	0.527165	52	52.52	0.025	52.495	9.38	13.68	13.82
Dixie	1088	0.540883	11	11.73	0.025	11.705	9.28	6.54	6.97
Duval	58491	28.523854	722	720.17	0.030	720.140	24.32	7.54	7.52
Escambia	21507	8.068327	360	357.02	0.024	356.996	13.52	11.31	11.22
Flagler	3565	3.411798	48	48.81	0.024	48.786	10.53	4.17	4.24
Franklin	878	0.283821	4	4.61	0.025	4.585	9.17	3.30	3.80
Gadsden	3836	0.824111	59	59.48	0.025	59.455	9.56	12.92	13.03
Gilchrist	984	0.328444	23	23.73	0.025	23.705	9.19	12.38	12.77
Glades	762	0.232870	11	11.72	0.025	11.695	9.14	7.96	8.49
Gulf	1273	0.321757	27	27.71	0.025	27.685	9.21	19.80	20.32
Hamilton	1224	0.190548	12	12.73	0.025	12.705	9.15	8.32	8.82
Hardee	2057	0.417308	20	20.75	0.025	20.725	9.29	7.42	7.70
Hendry	3235	0.709752	49	49.57	0.025	49.545	9.48	11.66	11.80
Hernando	7074	3.958244	264	261.49	0.024	261.466	10.96	13.61	13.48
Highlands	4822	1.882105	199	197.30	0.025	197.275	10.02	18.73	18.57
Hillsborough	74788	46.284944	1064	1064.35	0.053	1064.297	32.76	7.23	7.23
Holmes	1603	0.280332	24	24.73	0.025	24.705	9.21	12.23	12.61
Indian	6553	7.607307	156	155.99	0.024	155.966	12.36	9.75	9.75
Jackson	4502	0.819253	115	114.65	0.025	114.625	9.60	24.78	24.70
Jefferson	957	0.309690	16	16.74	0.025	16.715	9.18	11.23	11.75
Lafayette	1414	0.106457	21	21.73	0.025	21.705	9.13	24.93	25.80
Lake	14841	9.534133	293	291.32	0.023	291.297	13.67	7.98	7.94
Lee	39332	27.089733	652	649.84	0.031	649.809	22.42	8.46	8.43
Leon	18742	7.918907	173	173.09	0.023	173.067	13.28	5.89	5.90
Levy	1798	0.721669	19	19.77	0.025	19.745	9.39	4.58	4.76
Liberty	716	0.113382	14	14.73	0.025	14.705	9.09	16.76	17.63
Madison	1516	0.324513	33	33.68	0.025	33.655	9.22	17.84	18.21
Manatee	21539	13.804729	412	409.42	0.023	409.397	15.82	10.22	10.15
Marion	17125	7.971880	456	450.97	0.023	450.947	13.20	12.47	12.34
Martin	7656	9.243197	207	206.49	0.024	206.466	13.08	12.86	12.83
Miami-Dade	290363	92.544722	4155	4160.13	0.427	4159.703	67.08	15.29	15.31
Monroe	4168	5.227991	35	36.04	0.024	36.016	11.27	4.72	4.86
Nassau	4521	3.389901	66	66.65	0.024	66.626	10.58	7.45	7.52
Okaloosa	12407	6.812245	224	222.92	0.023	222.897	12.43	10.63	10.58
Okeechobee	2347	0.720267	51	51.54	0.025	51.515	9.42	12.09	12.22
Orange	73691	41.729653	732	735.09	0.044	735.046	30.80	5.25	5.28
Osceola	24512	7.340821	286	284.32	0.026	284.294	13.43	7.61	7.57
Palm	80865	84.441622	1866	1873.59	0.279	1873.311	49.08	12.47	12.52
Pasco	21222	13.333365	360	358.20	0.023	358.177	15.61	6.50	6.47
Pinellas	43480	35.959433	1035	1029.81	0.049	1029.761	26.35	10.62	10.56
Polk	35942	15.491829	767	758.92	0.024	758.896	17.47	10.58	10.47
Putnam	3819	1.217151	71	71.36	0.025	71.335	9.70	9.53	9.58
St.	12481	12.966328	112	113.41	0.025	113.385	14.88	4.23	4.28
St.	13864	7.683491	393	389.01	0.023	388.987	12.87	11.97	11.85
Santa	10575	5.192317	122	122.24	0.023	122.217	11.67	6.62	6.63
Sarasota	17916	20.149336	502	499.32	0.033	499.287	18.14	11.57	11.51
Seminole	17445	16.199289	311	310.49	0.026	310.464	16.51	6.59	6.58
Sumter	4819	4.709867	118	118.11	0.024	118.086	11.11	8.91	8.92
Suwannee	3913	0.701966	94	93.97	0.025	93.945	9.51	21.16	21.16
Taylor	1914	0.362501	25	25.73	0.025	25.705	9.26	11.59	11.93
Union	1416	0.217377	65	65.30	0.025	65.275	9.18	42.66	42.86
Volusia	21365	14.096377	430	427.17	0.023	427.147	15.93	7.77	7.72
Wakulla	2032	0.710564	21	21.77	0.025	21.745	9.40	6.22	6.45
Walton	4707	3.317614	43	43.86	0.024	43.836	10.57	5.81	5.92
Washington	1907	0.397700	30	30.71	0.025	30.685	9.27	11.78	12.06

¹ Calculated in millions

² Number of deaths is our direct estimates

³ Empirical Bayes estimator based on our methodology

⁴ Corrected BIAS EB= EB- BIAS

⁵ RMSE= $\sqrt{MSE}$

⁶ Death rate= Deaths/(Population Size)

⁷ E7stimated rate= EB/(Population Size)

This rate is calculated for every 10,000 population

Figure 1.

The top two map compares the EB and BIAS corrected EB for the number of deaths due to COVID-19 in each county of Florida, and the bottom two maps show the BIAS and RMSE of the EB

Simulation

In this section, we will measure the performance of our model via a simulation study. The choice of the auxiliary parameters is guided by the case study of the previous section. For illustration purposes, we have considered only one covariate- the number of confirmed cases to estimate the number of deaths due to COVID 19. This data is available for 3,142 counties of the United States. For simulation purposes, we have taken a random sample from this data without any replacement for each choice of m. The number of small areas (counties), m, is set to be 25, 50, 100, 200, 500, or 1000. For each choice of m, we generated data from the model :

$$\begin{array}{rcl}{z}_i|{𝜃}_i\stackrel{ind}{\sim }N({𝜃}_i,1/4),{𝜃}_i=\sqrt{{\lambda }_i}\stackrel{ind}{\sim }N(x_i^{\top }\beta ,A)& & \end{array}$$

The design matrix X includes a column of ones and one explanatory variable. To set the value of the parameter for β and A, we first create a linear regression model for the number of deaths on the number of confirmed cases using entire data for 3,142 counties. The estimated value for regression coefficient vector β is (5.281570,0.000272)^⊤ and mean square residuals is 22.75. For simulation we set β = (5.281570,0.000272)^⊤ and A = 22.75 − 0.25 = 22.50, hence shrinkage factor B = 0.011. Due to this variance stabilizing transformation, the shrinkage factor does not change between counties. Now using Eq. 22 we generate λ_i and z_i for all i = (1,…,m). The explanatory variable is again number of confirmed cases which is simulated randomly without replacement from the entire populations of 3,142 counties in the United States.

Here we will compare the true RMSE and estimated RMSE of ${\widehat{\lambda }}_i^{EB}$. We examine our findings in Theorems 2, 3 and 4 based on six different settings for m. Here we will vary the m and only one dataset is generated for each m, the latter taking values 25,50,100,200,500,1000. We have estimated the true RMSE of ${\widehat{\lambda }}_i^{CEB}$ based on 1,000 simulated samples since we do not have exact expression for RMSE of ${\widehat{\lambda }}_i^{CEB}$.

Figure 2 substantiates that the approximations given in Theorems 3 and 4 are fairly close to the true RMSE. In addition, they also point out one particular small area where the MSE is significantly higher than rest of the small areas.

Figure 2.

figure compares True RMSE: root of Eq. 12, estimated RMSE: root of Eq. 15 and simulated RMSE of bias corrected estimators. The RMSE of bias corrected EB estimators (${\widehat{\lambda }}_i^{CEB}$) is based on 1,000 simulations provided in Fig. 2 and it also verifies the result in Theorem 4

Conclusion

The paper introduces square root transformation of Poisson count data, and attains approximately both normality of the transformed data as well as variance stabilization. In this way, we obtain explicit estimates of bias and MSE for Poisson means. Based on the simulation, it seems that our estimates closely resemble the truth. Data analysis part tells us that estimates are higher on south-east Florida when the model appropriate.

There are many potential extensions. One that immediately comes to mind is consideration of unit level models with corresponding square root transformation. Gonçalves and Ghosh (2021) have addressed this problem using a pure hierarchical Bayesian framework, but an empirical Bayes approach with all its theoretical properties should also be a topic of future investigation. Even under the present framework, one may add a spatial component and using something like a CAR model (see for example, Ghosh et al., 1999). A final interesting problem is to consider an overdispersed Poisson model, i.e. a negative binomial model for count data with variable transformation as in Yu (2009) which also leads to homoscedasticity.

Proof of Theorem 2

Proof 6.

In this section we will do calculate the MSE of the EB estimator ${\widehat{\lambda }}_i^{EB}$. We observe that $E(({\widehat{\lambda }}_i^B-{\lambda }_i)({\tilde{\lambda }}_i^{EB}-{\widehat{\lambda }}_i^B))=0$ and $E(({\widehat{\lambda }}_i^B-{\lambda }_i)({\widehat{\lambda }}_i^{EB}-{\tilde{\lambda }}_i^{EB}))=0$ since $E(({\widehat{\lambda }}_i^B-{\lambda }_i)|z_i)=0$. Thus we have

$$\begin{array}{rcl}E{({\widehat{\lambda }}_i^{EB}-{\lambda }_i)}^2& =& E{({\widehat{\lambda }}_i^B-{\lambda }_i)}^2+E{({\tilde{\lambda }}_i^{EB}-{\widehat{\lambda }}_i^B)}^2+E{({\widehat{\lambda }}_i^{EB}-{\tilde{\lambda }}_i^{EB})}^2\\ & & +2E(({\widehat{\lambda }}_i^{EB}-{\tilde{\lambda }}_i^{EB})({\tilde{\lambda }}_i^{EB}-{\widehat{\lambda }}_i^B))\end{array}$$ 5.11

Each term in the right side of Eq. 5.11 will be computed separately. Since ${𝜃}_i|z_i\stackrel{ind}{\sim }N((1-B)z_i+B{x}_i^{\top }\beta ,(1-B)/4)$, by Lemma 1,

$$\begin{array}{rcl}E{({\widehat{\lambda }}_i^B-{\lambda }_i)}^2& =& E{(E({\lambda }_i|z_i)-{\lambda }_i)}^2=E[E\{{(E({\lambda }_i)-{\lambda }_i)}^2|z_i\}]=E[Var({\lambda }_i|z_i)]\\ & =& E[{(1-B)}^2/8+u_{i2}^2(1-B)]\\ & =& (1-B)[(1-B)/8+(A(1-B)+{(x_i^{\top }\beta )}^2)]\\ & =& (1-B)\left[{(x_i^{\top }\beta )}^2+\frac{(1-B)(2-B)}{8B}\right]\mathrm{.}\end{array}$$

Next we compute the second term in the right side of Eq. 5.11. By Lemmas 2 and Eq. 7,

$$\begin{array}{rcl}E{({\tilde{\lambda }}_i^{EB}-{\widehat{\lambda }}_i^B)}^2& =& Var({\widehat{\lambda }}_i^B)+Var({\tilde{\lambda }}_i^{EB})-2Cov({\widehat{\lambda }}_i^B,{\tilde{\lambda }}_i^{EB})+{(E({\tilde{\lambda }}_i^{EB}-{\widehat{\lambda }}_i^B))}^2\\ & =& Var(u_{i1}^2)+Var(u_{i2}^2)-2Cov(u_{i1}^2,u_{i2}^2)+{(E(u_{i1}^2-u_{i2}^2))}^2\mathrm{.}\end{array}$$ 5.12

Also,

$$Var(u_{i1}^2)=2{(A(1-B)+(2-B)s_i/4)}^2+(4A(1-B)+(2-B)s_i){(x_i^{\top }\beta )}^2$$ 5.13

$$Var(u_{i2}^2)=2{(A(1-B))}^2+4A(1-B){(x_i^{\top }\beta )}^2$$ 5.14

$$Cov(u_{i1}^2,u_{i2}^2)=2{(1-B)}^2{(A+s_i/4)}^2+4{(x_i^{\top }\beta )}^2(1-B)(A+s_i/4)$$ 5.15

$${[E(u_{i1}^2-u_{i2}^2)]}^2={(2-B)}^2{s}_i^2/16$$ 5.16

By Eqs. 5.13–5.16, 5.12 simplifies to

$$E{({\tilde{\lambda }}_i^{EB}-{\widehat{\lambda }}_i^B)}^2=B{s}_i{(x_i^{\top }\beta )}^2+{(1-B)}^2{s}_i/4+s_i^2(1/2-B/4-B^2/16)$$ 5.17

Now we evaluate the last expression in right side of Eq. 5.11.

$$\begin{array}{rcl}{\tilde{\lambda }}_i^{EB}-{\widehat{\lambda }}_i^B& =& {((1-B)z_i+B{x}_i^{\top }\widehat{\beta })}^2-{((1-B)z_i+B{x}_i^{\top }\beta )}^2\\ & =& B^2[{(x_i^{\top }\widehat{\beta }-x_i^{\top }\beta )}^2+2x_i^{\top }\beta (x_i^{\top }\widehat{\beta }-x_i^{\top }\beta )]\\ & & +2B(1-B)(x_i^{\top }\widehat{\beta }-x_i^{\top }\beta )x_i^{\top }\widehat{\beta }\\ & +& 2B(1-B)(x_i^{\top }\widehat{\beta }-x_i^{\top }\beta )(z_i-x_i^{\top }\widehat{\beta })\\ & =& B_1+B_2+B_3\end{array}$$ 5.18

We define $B_1=B^2[{(x_i^{\top }\widehat{\beta }-x_i^{\top }\beta )}^2+2x_i^{\top }\beta (x_i^{\top }\widehat{\beta }-x_i^{\top }\beta )],B_2=2B(1-B)(x_i^{\top }\widehat{\beta }-x_i^{\top }\beta )x_i^{\top }\widehat{\beta },B_3=2B(1-B)(x_i^{\top }\widehat{\beta }-x_i^{\top }\beta )(z_i-x_i^{\top }\widehat{\beta })$. The expressions A₁,A₂ and A₃ in Eq. 8 are functions of residuals $((z_i-x_i^{\top }\widehat{\beta })\mathrm{...},\dots ,(z_i-x_i^{\top }\widehat{\beta }))$ and the expressions B₁ and B₂ in Eq. 5.18 are functions of $\widehat{\beta }$. Therefore, (A₁,A₂,A₃) is independent of (B₁,B₂) and their covariances are 0. The expectation of B₃ is 0 since $(x_i^{\top }\widehat{\beta }-x_i^{\top }\beta )\text{is independent of}(z_i-x_i^{\top }\widehat{\beta })$. Also,

$$\begin{array}{rcl}Cov(A_1,B_3)& =& \frac{2B(1-B)}{4}E((B-\widehat{B})(z_i-x_i^{\top }\widehat{\beta }))E(x_i^{\top }\widehat{\beta }-x_i^{\top }\beta )=0.\\ Cov(A_3,B_3)& =& 2B(1-B)E\left[({\widehat{B}}^2-B^2-2(\widehat{B}-B)){(z_i-x_i^{\top }\widehat{\beta })}^3\right]\\ & & \times E(x_i^{\top }\widehat{\beta }-x_i^{\top }\beta )\\ & =& 0.\end{array}$$

Hence, again using part (v) of Lemma 4, Eqs. 8 and 5.18, we have

$$\begin{array}{rcl}Cov({\widehat{\lambda }}_i^{EB}-{\tilde{\lambda }}_i^{EB},{\tilde{\lambda }}_i^{EB}-{\widehat{\lambda }}_i^B)& =& Cov(A_2,B_3)\\ & =& E(2x_i^{\top }\widehat{\beta }(B-\widehat{B})(z_n-x_n^{\top }\widehat{\beta })2B(1-B)\\ & & \times (x_i^{\top }\widehat{\beta }-x_i^{\top }\beta )(z_i-x_i^{\top }\widehat{\beta }))\\ & =& 4B(1-B)E((B-\widehat{B}){(z_i-x_i^{\top }\widehat{\beta })}^2)\\ & & \times E(x_i^{\top }\widehat{\beta }(x_i^{\top }\widehat{\beta }-x_i^{\top }\beta ))\\ & =& 4B(1-B)\frac{(1-s_i)}{2m_0}Var(x_i^{\top }\widehat{\beta })\\ & =& 4B(1-B)\frac{(1-s_i)}{2m_0}(1/4+A)s_i\\ & =& (1-B)\frac{s_i(1-s_i)}{2m_0}\mathrm{.}\end{array}$$

Hence, from Eqs. 7 and 10,

$$\begin{array}{rcl}E(({\widehat{\lambda }}_i^{EB}-{\tilde{\lambda }}_i^{EB})({\tilde{\lambda }}_i^{EB}-{\widehat{\lambda }}_i^B))& =& (1-B)\frac{s_i(1-s_i)}{2m_0}\\ & & +\frac{2-B}{2}\frac{(1-s_i)}{m_0}\frac{2-B}{4}{s}_i\\ & =& \frac{s_i(1-s_i)}{2m_0}(2-2B+B^2/4)\mathrm{.}\end{array}$$

Next we compute the remaining third term in the right side of Eq. 5.11.

$$\begin{array}{rcl}E{({\widehat{\lambda }}_i^{EB}-{\tilde{\lambda }}_i^{EB})}^2& =& E{(A_1+A_2+A_3)}^2\\ & =& E{(B-\widehat{B})}^2/16+4E{(x_i^{\top }\widehat{\beta }(B-\widehat{B})(z_i-x_i^{\top }\widehat{\beta }))}^2\\ & & +E[{\{({\widehat{B}}^2-B^2)-2(\widehat{B}-B)\}}^2{(z_i-x_i^{\top }\widehat{\beta })}^4]\\ & & +[E(B-\widehat{B})\{({\widehat{B}}^2-B^2)-2(\widehat{B}-B)\}{(z_i-x_i^{\top }\widehat{\beta })}^2]/2.\end{array}$$ 5.19

In the above calculation, the cross terms E(A₁A₂) and E(A₂A₃) in Eq. 5.19 vanish by part (iii) of Lemma 3. Again, by part (ii) of Lemma 4, we have,

$$E{(B-\widehat{B})}^2=B^2-2BE(\widehat{B})+E({\widehat{B}}^2)=-B^2+\frac{m_0-2}{m_0-4}{B}^2=\frac{2B^2}{(m_0-4)}\mathrm{.}$$ 5.20

Again, applying Lemmas 3 and 4,

$$\begin{array}{rcl}E{[x_i^{\top }\widehat{\beta }(B-\widehat{B})(z_i-x_i^{\top }\widehat{\beta })]}^2& =& E[{(x_i^{\top }\widehat{\beta })}^2]E[{(B-\widehat{B})}^2{(z_i-x_i^{\top }\widehat{\beta })}^2]\\ & =& \left[{(x_i^{\top }\beta )}^2+\frac{s_i}{4B}\right]\\ & & \times E\left[\frac{{(z_i-x_i^{\top }\widehat{\beta })}^2}{||Z-X\widehat{\beta }|{|}^2}{(B-\widehat{B})}^2\frac{m_0-2}{4\widehat{B}}\right]\\ & =& \left[{(x_i^{\top }\beta )}^2+\frac{s_i}{4B}\right]\frac{(1-s_i)(m_0-2)}{4m_0}\\ & & \times E[B^2/\widehat{B}-2B+\widehat{B}]\\ & =& \left[{(x_i^{\top }\beta )}^2+\frac{s_i}{4B}\right]\frac{(1-s_i)(m_0-2)}{4m_0}\frac{2B}{m_0-2}\\ & =& \frac{1-s_i}{2m_0}\left[{(x_i^{\top }\beta )}^{2}B+\frac{s_i}{4}\right]\mathrm{.}\end{array}$$

Now we calculate the third term, $1/2E[{\{({\widehat{B}}^2-B^2)-2(\widehat{B}-B)\}}^2{(z_i-x_i^{\top }\widehat{\beta })}^4]$ in Eq. 5.19. By Lemma 3 and 4 and recalling $||Z-X\widehat{\beta }|{|}^2=\frac{m_0-2}{4\widehat{B}}$, we obtain,

$$\begin{array}{rcl}& & E[{\{({\widehat{B}}^2-B^2)-2(\widehat{B}-B)\}}^2{(z_i-x_i^{\top }\widehat{\beta })}^4]\\ & =& E\left[\frac{{(z_i-x_i^{\top }\widehat{\beta })}^4}{||Z-X\widehat{\beta }|{|}^4}\right]E\left[{\{({\widehat{B}}^2-B^2)-2(\widehat{B}-B)\}}^2\}\frac{{(m_0-2)}^2}{16{\widehat{B}}^2}\right]\\ & =& \frac{3{(1-s_i)}^2{(m_0-2)}^2}{16m_0(m_0+2)}E[({\widehat{B}}^2-2B^2+B^4/{\widehat{B}}^2)-4(\widehat{B}-B^2/\widehat{B}-B+B^3/{\widehat{B}}^2)\\ & & +4(1-2B/\widehat{B}+B^2/{\widehat{B}}^2)]\\ & =& \frac{3{(1-s_i)}^2{(m_0-2)}^2}{16m_0(m_0+2)}\left[\right)B^2\left(\frac{m_0-2}{m_0-4}-2+\frac{m_0(m_0+2)}{{(m_0-2)}^2}\right)\\ & & +4B\left(\frac{m_0}{m_0-2}-\frac{m_0(m_0+2)}{{(m_0-2)}^2}\right)+4\left(1-2\frac{m_0}{m_0-2}+\frac{m_0(m_0+2)}{{(m_0-2)}^2}\right)\left]\right)\\ & =& \frac{3{(1-s_i)}^2{(m_0-2)}^2}{16m_0(m_0+2)}\left[\right)B^2\frac{4(2m_0^2-9m_0+6)}{(m_0-4){(m_0-2)}^2}\\ & & +4B\frac{-4m_0}{{(m_0-2)}^2}+4\frac{2(m_0+2)}{{(m_0-2)}^2})\left]\right)\\ & =& 3{(1-s_i)}^2\left[\frac{2m_0^2-9m_0+6}{4m_0(m_0+2)(m_0-4)}{B}^2-\frac{B}{m_0+2}+\frac{1}{2m_0}\right]\mathrm{.}\end{array}$$ 5.21

Finally, once again by Lemmas 3 and 4, and recalling $||Z-X\widehat{\beta }|{|}^2=\frac{m_0-2}{4\widehat{B}}$, we get (5.19),

$$\begin{array}{rcl}& & E[(B-\widehat{B})\{({\widehat{B}}^2-B^2)-2(\widehat{B}-B)\}{(z_i-x_i^{\top }\widehat{\beta })}^2]\\ & & =E\left[\frac{{(z_i-x_i^{\top }\widehat{\beta })}^2}{||Z-X\widehat{\beta }|{|}^2}\right]\left[E(B-\widehat{B})\{({\widehat{B}}^2-B^2)-2(\widehat{B}-B)\}\frac{m_0-2}{4\widehat{B}}\right]\\ & & =\frac{(1-s_i)(m_0-2)}{4m_0}E\left[B\widehat{B}-{\widehat{B}}^2-B^3/\widehat{B}+B^2-2(B-\widehat{B})-2B+\frac{2B^2}{\widehat{B}}\right]\\ & & =\frac{(1-s_i)(m_0-2)}{4m_0}\left[B^2-B^2\frac{m_0-2}{m_0-4}-\frac{m_0{B}^2}{m_0-2}+B^2-2B+\frac{m_{0}2B}{m_0-2}\right]\\ & & =\frac{(1-s_i)(m_0-2)}{4m_0}\left[\frac{4B}{m_0-2}-\frac{4B^2(m_0-3)}{(m_0-2)(m_0-4)}\right]\\ & & =\frac{(1-s_i)B}{m_0}\left[1-\frac{B(m_0-3)}{m_0-4}\right]\mathrm{.}\end{array}$$ 5.22

From Eqs. 5.19–5.22, we obtain,

$$\begin{array}{rcl}E{({\widehat{\lambda }}_i^{EB}-{\tilde{\lambda }}_i^{EB})}^2& =& \frac{B^2}{8(m_0-4)}+\frac{1-s_i}{2m_0}\left[4{(x_i^{\top }\beta )}^{2}B+s_i\right]\\ & & +3{(1-s_i)}^2\left[\frac{2m_0^2-9m_0+6}{4m_0(m_0+2)(m_0-4)}{B}^2-\frac{B}{m_0+2}+\frac{1}{2m_0}\right]\\ & & +\frac{(1-s_i)B}{2m_0}\left[1-\frac{B(m_0-3)}{m_0-4}\right]\mathrm{.}\end{array}$$

Theorem 2 follows from Eqs. 5.11, 5.12, 5.17, 5.19 and 5.23. □

Funding

The third author’s research was partially supported by JSPS KAKENHI grant number 18K12758.

Compliance with Ethical Standards

The Author(s) declare(s) that there is no conflict of interest that are relevant to the content of this article.