Statistical inference with empty strata in judgment post stratified samples

Abstract

This article develops estimators for certain population characteristics using a judgment post stratified (JPS) sample. The paper first constructs a conditional JPS sample with a reduced set size K by conditioning on the ranks of the measured observations of the original JPS sample of set size $H \ge K$. The paper shows that the estimators of the population mean, median and distribution function based on this conditional JPS sample are consistent and have limiting normal distributions. It is shown that the proposed estimators, unlike the ratio and regression estimators, where they require a strong linearity assumption, only need a monotonic relationship between the response and auxiliary variable. For moderate sample sizes, the paper provides a bootstrap distribution to draw statistical inference. A small-scale simulation study shows that the proposed estimators based on a reduced set JPS sample perform better than the corresponding estimators based on a regular JPS sample.

Appendix

Proof of Theorem 1

For the proof of (i), from Eqs. (2) and (3) and from the law of large numbers, we write

$$\begin{aligned} \lim _{n\rightarrow \infty } \left\{ \frac{1}{n} \sum _{i=1}^na_{k:K|R_i} g(X_i-b)\right\} = \frac{1}{K} E\{g(X_{[k:K]}-b)\}+o_p(1) \end{aligned}$$

and

$$\begin{aligned} \lim _{n\rightarrow \infty } \left\{ \frac{1}{n} \sum _{i=1}^na_{k:K|R_i} \right\} = \frac{1}{K}+o_p(1). \end{aligned}$$

It is now easy to observe that

$$\begin{aligned} \lim _{n\rightarrow \infty } \hat{\theta }_{g,K}= & {} \sum _{k=1}^K \lim _{n\rightarrow \infty } \left( \frac{I_k}{d_n}\right) E\{g(X_{[k:K]}-b)\} +o_p(1) \nonumber \\= & {} \sum _{k=1}^K\frac{ E\{g(X_{[k:K]}-b)\}}{K}+o_p(1)= \theta _{g}+o_p(1). \end{aligned}$$

For the proof of (ii), without loss of generality assume that H is an even integer and $b=0$. We consider

$$\begin{aligned} E(\hat{\theta }_{g,K})= & {} E\left\{ E(\hat{\theta }_{g,K}|\varvec{R})\right\} =\sum _{h=1}^H E\left\{ \sum _{k=1}^K \frac{I_ka_{k:K|h}N_h}{d_n\sum _{h=1}^H a_{k:K|h}N_h}\right\} E(g(X_{[h:H]})) \\= & {} \sum _{h=1}^H E(w_h)E(g(X_{[h:H]}))=\sum _{h=1}^{H/2} \{E(w_h)E(g(X_{[h:H]}))\\&+\,E(w_{H+1-h})E(g(X_{[H+1-h:H]}))\}. \end{aligned}$$

In a JPS sample, since $E(N_{h})= E(N_{H+1-h})$, the expected weights also preserve the equality $E(w_h)=E(w_{H+1-h})$. By using the assumption of the theorem and the symmetry of the expected weights, we write

$$\begin{aligned} E(\hat{\theta }_{g,K})= & {} \sum _{h=1}^{H/2}E(w_h)\left\{ E(g(X_{[h:H]}-b))+E(g(X_{[H+1-h:H]}-b))\right\} \\= & {} \sum _{h=1}^{H/2}E(w_h)E(g(X-b))/2 = E(g(X-b))\sum _{h=1}^H E(w_h)=E(g(X-b)). \end{aligned}$$

This completes the proof. $\square$

Proof of Lemma (1)

$\bar{\varvec{T}}_{g}(b)$ is the mean of independent random vectors. Hence, central limit theorem gives the asymptotic normality. The expression $E(\varvec{\bar{T}}_g)$ follows from Eqs. (2) and (3). We now compute the variance and covariances. For $\sigma _{r,s}$, $1 \le r,s \le K$, we use the conditional covariance given $R_i$

$$\begin{aligned} \sigma _{r,s}= & {} \frac{1}{n}\sum _{h=1}^HE\left\{ \sum _{i=1}^nI(R_i=h)\mathrm{cov}(W_{r,R_i,g}(b),W_{s,R_i,g}(b))|R_i\right\} \\&+ \frac{1}{n}\mathrm{cov}\left\{ \sum _{i=1}^n I(R_i=h) E(W_{r,R_i,g}(b)|R_i),\sum _{i=1}^n I(R_i=h) E(W_{s,R_i,g}(b)|R_i)\right\} \\= & {} A+B. \end{aligned}$$

In the above equation the first term reduces to

$$\begin{aligned} A= & {} \sum _{h=1}^H E\left( \frac{N_h}{n}\right) a_{r:K|h}a_{s:K|h} \mathrm{var}(g(X_{[h:H]}-b)) \\= & {} \frac{1}{H} \sum _{h=1} a_{r:K|h}a_{s:K|h} \mathrm{var}(g(X_{[h:H]}-b)). \end{aligned}$$

In a similar approach, we have

$$\begin{aligned} B= & {} \frac{1}{n} \mathrm{cov}\left\{ \sum _{h=1}^H N_h a_{r:K|h} E(g(X_{[h:H]}-b)),\sum _{h=1}^H N_ha_{s:K|h} E(g(X_{[h:H]}-b))\right\} \\= & {} \sum _{h=1}^H\sum _{h'=1}^H \frac{\mathrm{cov}(N_h,N_{h'})}{n} a_{r:K|h} E(g(X_{[h:H]}-b))a_{r:K|h'} E(g(X_{[h':H]}-b)) \\= & {} \sum _{h=1}^H\frac{\mathrm{Var}(N_h)}{n} a_{r:K|h}a_{s:K|h} (E(g(X_{[h:H]}-b)))^2 \\&+ \sum _{h=1}^H\sum _{h'\ne h}^H\frac{\mathrm{cov}(N_h,N_{h'})}{n} E(g(X_{[h:H]}-b))a_{r:K|h'} E(g(X_{[h':H]}-b)) \\= & {} \sum _{h=1}^H\frac{H-1}{H^2}a_{r:K|h}a_{s:K|h} (E(g(X_{[h:H]}-b)))^2 \\&+ \sum _{h=1}^H\sum _{h'\ne h}^H\frac{-1}{H^2} E(g(X_{[h:H]}-b))a_{r:K|h'} E(g(X_{[h':H]}-b)) \\= & {} \frac{1}{H} \sum _{h=1}^H a_{r:K|h}a_{s:K|h} (E(g(X_{[h:H]}-b)))^2 - \frac{1}{K^2} E(g(X_{[r:K]}-b)) E(g(X_{[s:K]}-b)). \end{aligned}$$

The second term in the right side of the above expression follows from Eq. (2). Combining expressions A and B, we obtain

$$\begin{aligned} \sigma _{r,s}=\frac{1}{H}\sum _{h=1}^H a_{r:K|h}a_{s:K|h} Eg^2(X_{[h:H]}-b) - \frac{1}{K^2} E(g(X_{[r:K]}-b)) E(g(X_{[s:K]}-b)). \end{aligned}$$

For $\tau _{r,s}$, $1 \le r \le k$ and $1\le s \le K-1$ we compute

$$\begin{aligned} \tau _{r,s}= & {} \frac{1}{n}\mathrm{cov}\left\{ \sum _{i=1}^nW_{r,R_i,g}(b),\sum _{i=1}^n a_{s:K|R_i}\right\} \\= & {} \frac{1}{n}E\left\{ \mathrm{cov} \left( \sum _{i=1}^nW_{r,R_i,g}(b), \sum _{i=1}^na_{s:K|R_i}\right) |\varvec{R}\right\} \nonumber \\&+\frac{1}{n}\mathrm{cov} \left\{ E\left( \sum _{i=1}^nW_{r,R_i,g}(b)|\varvec{R}\right) , E(\sum _{i=1}^na_{s:K|R_i})|\varvec{R}\right\} \\= & {} 0 +\frac{1}{n}\mathrm{cov} \left\{ \sum _{h=1}^H N_hW_{r,h,g}(b),\sum _{h=1}^H N_h a_{s:K|h}) \right\} \\= & {} \frac{1}{n} \sum _{h=1}^H \sum _{h'=1}^H \mathrm{cov}(N_h,N_{h'}) a_{r:K|h}a_{s:K|h'} Eg(X_{[h:H]}-b) \\= & {} \frac{1}{H}\sum _{h=1}^H a_{r:K|h}a_{s:K|h} Eg(X_{[h:H]}-b) -\frac{1}{K^2} Eg(X_{[r:K]}-b). \end{aligned}$$

Finally, the $\gamma _{r,s}$, $1 \le r,s \le K-1$, follows from

$$\begin{aligned} \gamma _{r,s}= & {} \frac{1}{n}\mathrm{cov}\left\{ \sum _{i=1}^na_{r:K|R_i},\sum _{i=1}^na_{s:K|R_i} \right\} \\= & {} \frac{1}{n}E\left\{ \mathrm{cov} \left( \sum _{i=1}^n a_{r:K|R_i}, \sum _{i=1}^na_{s:K|R_i}\right) |\varvec{R}{}\right\} \nonumber \\&+\frac{1}{n}\mathrm{cov} \left\{ E\left( \sum _{i=1}^na_{r:K|R_i}\right) |\varvec{R}, E\left( \sum _{i=1}^na_{s:K|R_i}\right) |\varvec{R}\right\} \\= & {} 0+\frac{1}{H}\sum _{h=1}^H a_{r:K|h}a_{s:K|h} -\frac{1}{K^2}. \end{aligned}$$

The proof is completed. $\square$

Proof of Theorem 2

Let Q be a transformation from $\varvec{\bar{T}}_g(b)$ to $\hat{\theta }_g(b)$

$$\begin{aligned} \hat{\theta }_g(b)=\frac{1}{d_n} \sum _{k=1}^K \frac{I_k\bar{T}_{g,k}(b)}{\bar{T}_{g,K+k}(b)} \mathop {=}\limits ^{D}\frac{1}{K} \sum _{k=1}^K \frac{\bar{T}_{g,k}(b)}{\bar{T}_{g,K+k}(b)}=Q(\varvec{\bar{T}}_g(b)), \end{aligned}$$

where $\bar{T}_{g,k}(b)$ is the k th component of vector $\varvec{\bar{T}}_g(b)$ and $\bar{T}_{g,2K}(b)=1-\sum _{k=1}^{K-1} \bar{T}_{g,K+k}(b)$. The second equality in the above equation follows from the fact that $I_k/d_n$ converges in probability to 1 / K for $k=1,\ldots ,K$. It is clear that $Q(E(\varvec{\bar{T}}_g(b)))=E(g(X-b))$. For notational convenience, let $\varvec{\mu }_T=E(\varvec{\bar{T}}_g(b))$. By using a Taylor expansion of $Q(\varvec{\bar{T}}_g(b))$ around $\varvec{\mu }_T$ we write

$$\begin{aligned} \sqrt{n}\left( Q(\varvec{\bar{T}}_g(b))-E(g(X-b))\right) =\frac{1}{K}\varvec{L}_T^\top \sqrt{n}\left( \varvec{\bar{T}}_g(b) -\varvec{\mu }_T\right) +o_p(1), \end{aligned}$$

where $\varvec{L}_T$ is a $2K-1$ dimensional partial derivative vector of $Q(\varvec{\mu }_T)$,

$$\begin{aligned} L_r= & {} \frac{\mathrm{d}}{\mathrm{d}\mu _T(r)} Q(\varvec{\mu }_T)\\= & {} \left\{ \begin{array}{ll} K &{} \quad r=1,\ldots , K \\ K\left( Eg(X_{[K:K]}-b)-Eg(X_{[r-K:K]}-b)\right) &{}\quad r=K+1,\ldots , 2K-1, \end{array} \right. \end{aligned}$$

and $\mu _T(r)$ is the r th component of vector $\mu _T$. Let $\varvec{L}^\top =(\varvec{L}_1^\top ,\varvec{L}_2^\top )$ with $\varvec{L}_1^\top =(K,\ldots ,K)$ and $\varvec{L}_2=K(Eg(X_{[K:K]}-b)-Eg(X_{[1:K]}-b),\ldots , Eg(X_{[K:K]}-b)-Eg(X_{[K-1:K]}-b))$. It is then easy to see that $\sqrt{n}\left( Q(\varvec{\bar{T}}_g(b))-E(g(X-b))\right)$ converges to a normal distribution with mean zero and variance

$$\begin{aligned} \sigma ^2_{\hat{\theta }_g}= \frac{1}{K^2}\varvec{L}_1^\top \varvec{\Sigma }_{1,1}\varvec{L}_1+\frac{1}{K^2} 2\varvec{L}_1^\top \varvec{\Sigma }_{1,2}\varvec{L}_2+\frac{1}{K^2}\varvec{L}_2^\top \varvec{\Sigma }_{2,2}\varvec{L}_2. \end{aligned}$$

By using Lemma 1, we write

$$\begin{aligned} \frac{1}{K^2}\varvec{L}_1^\top \varvec{\Sigma }_{1,1}\varvec{L}_1= & {} \sum _{r=1}^K\sum _{s=1}^K\left\{ \frac{1}{H} \sum _{h=1}^H a_{r:K|h}a_{s:K|h}E\left\{ g^2(X_{[h:H]}-b)\right\} \right. \\&- \left. \frac{E\left\{ g(X_{[r:K]}-b)\right\} E\left\{ g(X_{[s:K]}-b)\right\} }{K^2}\right\} \\= & {} \frac{1}{H} \sum _{h=1}^H E\left\{ g^2(X_{[h:H]}-b)\right\} -\left\{ Eg(X-b)\right\} ^2=\sigma ^2_g. \end{aligned}$$

Let $d_s=Eg(X_{[K:K]}-b)-Eg(X_{[s:K]}-b)$, $s=1,\ldots , K-1$.

$$\begin{aligned} \frac{1}{K^2}\varvec{L}_1^\top \varvec{\Sigma }_{1,2}\varvec{L}_2= & {} \sum _{r=1}^{K}\sum _{s=1}^{K-1} d_s\tau _{r,s} \\= & {} \sum _{r=1}^{K}\sum _{s=1}^{K-1} d_s \left[ \left\{ \frac{1}{H}\sum _{h=1}^H \alpha _{r:K|h}\alpha _{s:K|h}Eg(X_{[h:H]}-b)\right\} -\frac{Eg(X_{[r:K]}-b)}{K^2}\right] \\= & {} \sum _{s=1}^{K-1}d_s\left\{ \frac{Eg(X_{[s:K]}-b)}{K}-\frac{ Eg(X-b)}{K}\right\} \\= & {} \sum _{s=1}^{K-1} \left\{ Eg(X_{[K:K]}-b)- Eg(X_{[s:K]}-b)\right\} \left\{ \frac{Eg(X_{[s:K]}-b)}{K}-\frac{ Eg(X-b)}{K}\right\} \\= & {} -\sum _{s=1}^{K}\left\{ Eg(X_{[s:K]}-b)-Eg(X-b)\right\} ^2/K. \end{aligned}$$

For the last expression, we consider

$$\begin{aligned} \frac{1}{K^2}\varvec{L}_2^\top \varvec{\Sigma }_{2,2}\varvec{L}_2= & {} \sum _{r=1}^{K-1}\sum _{s=1}^{K-1} d_rd_s \gamma _{r,s} \\= & {} \sum _{r=1}^{K-1}\sum _{s=1}^{K-1} \left\{ Eg(X_{[K:K]}-b)- Eg(X_{[r:K]}-b)\right\} \\&\times \left\{ Eg(X_{[K:K]}-b)- Eg(X_{[s:K]}-b)\right\} \gamma _{r,s} \\= & {} \sum _{r=1}^{K}\sum _{s=1}^{K} \left\{ Eg(X_{[K:K]}-b)- Eg(X_{[r:K]}-b)\right\} \\&\times \left\{ Eg(X_{[K:K]}-b)- Eg(X_{[s:K]}-b)\right\} \gamma _{r,s} \\= & {} \sum _{r=1}^{K}\sum _{s=1}^{K}\frac{1}{H}\sum _{h=1}^H \alpha _{r:K|h}\alpha _{s:K|h}Eg(X_{[r:K]}-b) Eg(X_{[s:K]}-b)\\&-\left\{ Eg(x-b)\right\} ^2. \end{aligned}$$

This completes the proof. $\square$

Proof of Corollary 1

We first rewrite $\sigma ^2_{\theta :K}$ as

$$\begin{aligned} \sigma ^2_{\hat{\theta }_{g:K}}= & {} \sigma ^2_g-\sum _{r=1}^K (Eg(X_{[r:K]}-b)-Eg(X-b))^2/K -\left\{ \sum _{r=1}^K (Eg(X_{[r:K]}-b))^2/K\right. \\&\left. -\sum _{r=1}^K \sum _{s=1}^K\sum _{h=1}^H\frac{ a_{r:K|h}a_{s:K|h}}{H}Eg(X_{[r:K]}-b)Eg(X_{[s:K]}-b)\right\} \\= & {} \sigma ^2_{\hat{\theta }_{g:RSS(K)}}- \left\{ \sum _{r=1}^K \frac{(Eg(X_{[r:K]}-b))^2}{K}\right. \\&\left. -\sum _{r=1}^K \sum _{s=1}^K\sum _{h=1}^H\frac{ a_{r:K|h}a_{s:K|h}}{H}Eg(X_{[r:K]}-b)Eg(X_{[s:K]}-b) \right\} \\= & {} \sigma ^2_{\hat{\theta }_{g:RSS(K)}}-A_d. \end{aligned}$$

We need to show $A_d$ is non-negative. We first consider

$$\begin{aligned} B_d= \sum _{r=1}^K \sum _{s=1}^K \frac{1}{H} \sum _{h=1}^H a_{r:K|h}a_{s:K|h}\left\{ E\left( g(X_{[r:K]}-b)\right) -E\left( g(X_{[r:K]}-b)\right) \right\} ^2 \ge 0. \end{aligned}$$

We re-write expression $B_d$ as

$$\begin{aligned} B_d= & {} 2\sum _{r=1}^K\sum _{s=1}^K \frac{1}{H}\sum _{h=1}^Ha_{r:K|h}a_{s:K|h}\left\{ E\left( g(X_{[r:K]}-b)\right) \right\} ^2 \\&-2\sum _{r=1}^K\sum _{s=1}^K \frac{1}{H}\sum _{h=1}^Ha_{r:K|h}a_{s:K|h}\left\{ E\left( g(X_{[r:K]}-b)\right) \right\} \left\{ E\left( g(X_{[r:K]}-b)\right) \right\} . \end{aligned}$$

Use of equalities $\sum _{s=1}^K a_{s:K|h}=1$ and $\sum _{h=1}^H a_{s:K|h}=H/K$ in the first term of the above equation reduces $B_d$ to $2A_d$

$$\begin{aligned} B_d= & {} \frac{2}{K}\sum _{r=1}^K\left\{ E\left( g(X_{[r:K]}-b)\right) \right\} ^2 \\&-2\sum _{r=1}^K\sum _{s=1}^K \frac{1}{H}\sum _{h=1}^Ha_{r:K|h}a_{s:K|h}\left\{ E\left( g(X_{[r:K]}-b)\right) \right\} \left\{ E\left( g(X_{[r:K]}-b)\right) \right\} \\= & {} 2A_d \ge 0. \end{aligned}$$

This completes the proof. $\square$

Proof of Theorem 3

Let $U_n(b/\sqrt{n})=\left\{ S_n(\eta _p+b/\sqrt{n})-S_n(\eta _p))/(b/\sqrt{n}\right\}$. By using conditional expectation given rank vector $\varvec{R}$, we obtain

$$\begin{aligned} E(U_n(b/\sqrt{n})= \sum _{h=1}^H E \left\{ \sum _{k=1}^K\frac{I_kN_h a_{k:K|h}}{d_n \sum _{t=1}^H N_t a_{k:K|t}}\right\} \frac{F_{[h]}(\eta _p)-F_{[h]}(\eta _p+b/\sqrt{n})}{b/\sqrt{n}}. \end{aligned}$$

For large n, we can write

$$\begin{aligned} \lim _{n \rightarrow \infty }E\left\{ \frac{I_kN_h a_{k:K|h}}{d_n \sum _{t=1}^H N_t a_{k:K|t}}\right\}= & {} E\left\{ \lim _{n\rightarrow \infty }\frac{I_k}{d_n} \lim _{n\rightarrow \infty }\frac{N_h a_{k:K|h}}{d_n \sum _{t=1}^H N_t a_{k:K|t}}\right\} \\= & {} \frac{1}{K} \frac{a_{k:K|h}/H}{\sum _{t=1}^Ha_{k:K|t}/H} = a_{k:K|h}/H. \end{aligned}$$

It is now easy to observe that $E(U_n(b/\sqrt{n})$ has a limit at $-bf(\eta _p)$

$$\begin{aligned} \lim _{n\rightarrow \infty } E(U_n(b/\sqrt{n})= & {} \sum _{h=1}^H \sum _{k=1}^K \frac{a_{k:K|h}}{H} \lim _{n \rightarrow \infty } \frac{F_{[h]}(\eta _p)-F_{[h]}(\eta _p+b/\sqrt{n})}{b/\sqrt{n}}\\= & {} -b \sum _{k=1}^K \sum _{h=1}^H \frac{a_{k:K|h}}{H}f_{[h]}(\eta _p)=-b \sum _{k=1}^K \frac{H}{KH}f(\eta _p)=- bf(\eta _p). \end{aligned}$$

We now show that the variance of $U_n(b/\sqrt{n})$ converges to zero as n goes to infinity. Without loss of generality, assume that $b>0$ and $\eta _p=0$. In this case, $U_n(b/\sqrt{n})$ can be written as

$$\begin{aligned} U_n(b/\sqrt{n})= \sum _{i=1}^n \sum _{k=1}^K \frac{I_k a_{k:K|R_i}}{ d_n\sum _{j=1}^n a_{k:K|R_j}} I( 0 \le X_i \le b/\sqrt{n}). \end{aligned}$$

The variance of $U_n(b/\sqrt{n})$ can be written as

$$\begin{aligned} \mathrm{Var}(U_n(b/\sqrt{n})= \mathrm{Var}\left( E\left\{ U_n(b/\sqrt{n})|\varvec{R}\right\} \right) +E\left( \mathrm{Var}\left\{ U_n(b/\sqrt{n})|\varvec{R}\right\} \right) =D_n+G_n. \end{aligned}$$

The expression $D_n$ is given by

$$\begin{aligned} D_n= & {} \frac{1}{b^2}\sum _{h=1}^H \sum _{h'=1}^H \left( F_{[h]}(b/\sqrt{n})-F_{[h]}(0)\right) \left( F_{[h']}(b/\sqrt{n})-F_{[h']}(0)\right) \\&\times \sum _{k=1}^K \sum _{k'=1}^K \mathrm{Cov}\left( \frac{\sqrt{n}I_kN_h a_{k:K|h}}{d_n \sum _{t=1}^H N_t a_{k:K|t}}, \frac{\sqrt{n}I_{k'} N_{h'} a_{k':K|h'}}{d_n \sum _{t=1}^H N_t a_{k':K|t}}\right) . \end{aligned}$$

In the above expression, the covariances are all finite and the double sum in the first term converges to zero as n gets large. Hence, $D_n$ has a limit at 0. In a similar fashion, $G_n$ can be written as

$$\begin{aligned} G_n= & {} \sum _{k=1}^K \sum _{h=1}^H E \left\{ \frac{I_k^2N_h a^2_{k:K|h}}{d^2_n (\sum _{t=1}^H N_t a_{k:K|t})^2}\right\} \left( F_{[h']}(b/\sqrt{n})-F_{[h']}(0)\right) \\&\times \left\{ 1- \left( F_{[h']}(b/\sqrt{n})+F_{[h']}(0)\right) \right\} . \end{aligned}$$

Since the expectation in the above expression has a finite limit, $G_n$ converges to zero as n goes to infinity. We then conclude that variance $U_n(b/\sqrt{n})$ goes to zero as n approaches to infinity. This establishes the point-wise convergence

$$\begin{aligned} \sqrt{n}S_n(\eta _p+b/\sqrt{n})= \sqrt{n}S_n(\eta _p)-b f(\eta _p)+o_p(1). \end{aligned}$$

The uniform convergence follows from the monotonicity of $S_n(b)$. $\square$

Proof of Theorem 4

We first observe that sample size vector $\varvec{N}^{\top }= (N_1,\ldots , N_H)$ has a multinomial distribution with parameters n and $(1/H,\ldots , 1/H)$. The probability that $W_n(q)$ equals K is

$$\begin{aligned} P(W_n(q)=K)= \left( \begin{array}{l} {H} \\ {K}\end{array} \right) P( N_1> q, \ldots , N_K> q, N_{K+1} \le q, \ldots , N_{H} \le q). \end{aligned}$$

Let

$$\begin{aligned} A_{K,q}=\{ N_1> q, \ldots , N_K> q\} \text{ and } B_{H-K,q}=\{ N_{K+1} \le q, \ldots , N_{H} \le q\}. \end{aligned}$$

For a fixed j, we also define $B_{j,H-K,q}$ to be the event that each one of the j judgment classes in set $B_{H-K,q}$ has exactly q observations. Note that by the definition of the set $B_{H-K,q}$, $B_{j,H-K,q}= B_{H-K,H-K,q}$ when $q=0$. With this new notation we write

$$\begin{aligned} P((W_n(q)=K)=\left( \begin{array}{l} {H} \\ {K} \end{array}\right) \sum _{j=0}^{q(H-K)} \left( \begin{array}{c} {q(H-K)} \\ {j}\end{array}\right) P( A_{K,q}| B_{j,H-K,q}) P(B_{j,H-K,q}). \end{aligned}$$

It is clear that

$$\begin{aligned} P(B_{j,H-K,q})= & {} \frac{n!}{(n-qj)!}\left( \frac{K}{H}\right) ^{n-qj}\left( \frac{1}{H}\right) ^{qj} \end{aligned}$$

and

$$\begin{aligned} P( A_{K,q}| B_{j,H-K,q})=\sum _{n_1>q, \ldots , n_K>q} \left( \begin{array}{c} {n-qj}\\ {n_1,\ldots , n_K} \end{array}\right) (1/K)^{n-qj}. \end{aligned}$$

Let

$$\begin{aligned} A_{K,q}=\{A_{1:K,q} \cap \cdots \cap A_{K:K,q}\}, \end{aligned}$$

where $A_{h:K,q}$ is the event that $N_{h} > q$ for $h \le K$. By using DeMorgan’s law, we write

$$\begin{aligned} P( A_{K,q}| B_{j,H-K,q})= & {} 1- P\left( A^C_{K,q}| B_{j,H-K,q}\right) = 1- P\left( \cup _{h=1}^K A^C_{h:K,q} | B_{j,H-K,q}\right) \nonumber \\= & {} 1- \sum _{i=1}^K \left( \begin{array}{l} K \\ i \end{array}\right) (-1)^{i-1}P\left( A^C_{1:K,q}\cap \cdots \cap A^C_{i:K,q}|B_{j,H-K,q}\right) .\nonumber \\ \end{aligned}$$

(11)

We now evaluate the following conditional probability

$$\begin{aligned} P\left( A^C_{1:K,q}\cap \cdots \cap A^C_{i:K,q}|B_{j,H-K,q}\right)= & {} \sum _{r_1=0}^q \cdots \sum _{r_i=0}^q \left( \begin{array}{c} {n-qj}\\ {r_1,\ldots , r_i, n-qj-T_i} \end{array}\right) \nonumber \\&\times (1/K)^{T_i}\{1-i/K\}^{(n-qj-T_i)}, \end{aligned}$$

where $T_i= \sum _{y=1}^i r_y$. With some algebra, the above equation simplifies to

$$\begin{aligned} P\left( A^C_{1:K,q}\cap \cdots \cap A^C_{i:K,q}|B_{j,H-K,q}\right) =\sum _{s=0}^{iq} \left( \begin{array}{l} {qi} \\ s \end{array}\right) \left( \begin{array}{l} {n-qj} \\ {s} \end{array}\right) \frac{s! (K-i)^{n-qj-s}}{K^{n-qj}}. \end{aligned}$$

We complete the proof by putting this expression in (11). $\square$