Exact distributions of tests of outliers for exponential samples

Abstract

In this paper, we propose an algorithm to derive the exact distributions of discordancy tests for exponential samples under the slippage alternative providing that their survival functions involve the linear combinations of independent and identically distributed exponential random variables with arbitrary real coefficients. In addition, we define the various performance measures in terms of conditional probabilities that the observed value of the test statistic exceeds the critical value given that the contaminants have the specific position numbers in the ordered sample. These make possible to calculate various performance measures of discordancy tests for the exponential samples to any desired degree of accuracy. For the purpose of illustration, we derive the distributions of the maximum likelihood ratio tests for testing single and multiple outliers in the exponential samples and then we calculate their performance measures accurately to six decimal places. Moreover, the definitions of the performance criteria are not restricted to the discordancy tests for exponential samples only, they are also equally applicable to the discordancy tests for samples from other distributions.

Appendices

Exact distribution of MLR test for testing a single outlier for the sample size $n=3$

Consider a single outlier problem with $n=3$. The probabilities $\Pr [T>d|B(r)]$ ($r=1,2,3)$ in (2) can be written as

$$\begin{aligned} \Pr [T>d|B(r)] =\Pr \left( \frac{1-3d}{u_1^r}Z_1+\frac{1-2d}{u_2^r}Z_2+\frac{1-d}{u_3^r}Z_3>0\right) . \end{aligned}$$

Using (1), we first calculate

$$\begin{aligned} \Pr [T>d|B(3)] =\Pr \left( \frac{1-3d}{b+2}Z_1+\frac{1-2d}{b+1}Z_2+\frac{1-d}{b}Z_3>0\right) = \Pr \,(\mathbf {AZ}>0), \end{aligned}$$

where $\mathbf {A}=((1-3d)/(b+2),(1-2d)/(b+1),(1-d)/(b))$. Clearly, $\Pr [T>d|B(3)]$ is 1 for $0<d\le 1/3$, and 0 for $d>1$. Letting $\mathbf {c}=(0,(b+1)(1-d)/[bd+(1-d)], -b(1-2d)/[bd+(1-d)])'$, we have $\mathbf {Ac}=0$. Now using recursion from (4) and the properties of deleting zero entries and renumbering the random variables again, we have

$$\begin{aligned}&\Pr [T>d|B(3)] =\frac{(b+1)(1-d)}{bd+(1-d)}\Pr \left( \frac{1-3d}{b+2}Z_1+\frac{1-d}{b}Z_2>0\right) \\&\quad +\frac{-b(1-2d)}{bd+(1-d)}\Pr \left( \frac{1-3d}{b+2}Z_1+\frac{1-2d}{b+1}Z_2>0\right) . \end{aligned}$$

The same process is applied to the obtained two probabilities with $\mathbf {A_1}=((1-3d)/(b+2),(1-d)/b)$ and $\mathbf {A_2}=((1-3d)/(b+2),(1-2d)/(b+1))$. Taking $\mathbf {c_1}=((b+2)(1-d)/[2(bd+(1-d))], -b(1-3d)/[2(bd+(1-d))])'$ to calculate $\Pr (\mathbf {A_1 Z}>0)$ and $\mathbf {c_2}=((b+2)(1-2d)/[(bd+(1-d))], -(b+1)(1-3d)/[(bd+(1-d))])'$ to calculate $\Pr (\mathbf {A_2 Z}>0)$, it follows that

$$\begin{aligned} \Pr [T>d|B(3)] ={\left\{ \begin{array}{ll} \frac{(b+1)(b+2)(1-d)^2}{2[bd+(1-d)]^2}-\frac{b(b+2)(1-2d)^2}{[bd+(1-d)]^2} &{} \frac{1}{3}<d\le \frac{1}{2},\\ {} \frac{(b+1)(b+2)(1-d)^2}{2[bd+(1-d)]^2} &{} \frac{1}{2}<d\le 1. \end{array}\right. } \end{aligned}$$

(31)

In a similar manner, we can obtain

$$\begin{aligned} \Pr [T>d|B(2)] ={\left\{ \begin{array}{ll} \frac{(b+1)(b+2)(1-d)^2}{[d+b(1-d)][1+d+b(1-d)]}-\frac{(b+2)(1-2d)^2}{[d+b(1-d)][bd+(1-d)]} &{} \frac{1}{3}<d\le \frac{1}{2},\\ {} \frac{(b+1)(b+2)(1-d)^2}{[d+b(1-d)][1+d+b(1-d)]} &{} \frac{1}{2}<d\le 1, \end{array}\right. } \end{aligned}$$

(32)

and

$$\begin{aligned} \Pr [T>d|B(1)] ={\left\{ \begin{array}{ll} \frac{2(b+2)(1-d)^2}{[1+d+b(1-d)]}-\frac{(b+2)(1-2d)^2}{[2d+b(1-2d)]} &{} \frac{1}{3}<d\le \frac{1}{2},\\ {} \frac{2(b+2)(1-d)^2}{[1+d+b(1-d)]} &{} \frac{1}{2}<d\le 1. \end{array}\right. } \end{aligned}$$

(33)

From (6), we have

$$\begin{aligned} \Pr [B(1)] = \frac{b}{b+2}, \quad \Pr [B(2)] = \frac{2b}{(b+1)(b+2)}, \quad \text {and}\quad \Pr [B(3)] = \frac{2}{(b+1)(b+2)}. \end{aligned}$$

(34)

Now using (31–34), we get the distribution function of T under H for $n=3$ which is

$$\begin{aligned} \Pr [T<d|H] = 1-\sum _{r=1}^{3}{\Pr [T>d|B(r)] \Pr [B(r)]}. \end{aligned}$$

(35)

The density function of T for $n=3$ can be obtained by differentiating (35) with respect to d. Also, the density function under the labelled slippage $H_n$ can be obtained by differentiating $1-\Pr [T>d|B(3)]$ with respect to d and is equivalent to the density obtained by Chikkagoudar and Kunchur (1983) for $n=3$.

The critical value of test T for a single outlier testing can be obtained from (35) by substituting $b=1$ for the significance level $\alpha =0.05$ which yields to be $d=0.8709$. Now, using (18–19), we can calculate the various performance measures. For example, for the size of the shift $b=0.5$, the P, $\textit{NSP}$, $\textit{SP}$, $\textit{NSE}$ and $\textit{SE}$ are 0.0701, 0.0523, 0.0178, 0.4810 and 0.4488 respectively.

Exact distribution of MLR test for testing two outliers for the sample size $n=3$

Consider two outliers problem with $n=3$. The probabilities $\Pr [T_2>d|B(r,s)]$$((r,s)\in S^{(2)}=\{(1,2),(1,3),(2,3)\})$ (14) can be written as

$$\begin{aligned} \Pr [T_2>d|B(r,s)] =\Pr \left( \frac{2-3d}{u_1^{r,s}}Z_1+\frac{2(1-d)}{u_2^{r,s}}Z_2+\frac{1-d}{u_3^{r,s}}Z_3 >0\right) . \end{aligned}$$

Using (10), we first calculate

$$\begin{aligned} \Pr [T_2>d|B(1,2)]{=} \Pr \left( \frac{2-3d}{2b+1}Z_1+\frac{2(1-d)}{2b}Z_2{+}\frac{1-d}{b}Z_3>0 \right) {=}\Pr (\mathbf {AZ}>0), \end{aligned}$$

where $\mathbf {A}=(((2-3d)/(2b+1),2(1-d)/(2b),(1-d)/b)$. Clearly, $\Pr [T>d|B(1,2)]$ is 1 for $0<d\le 2/3$ and 0 for $d>1$. Letting $\mathbf {c}=(-2(2b+1)(d-1)/(2b+d-bd), (3d-2)(b+1)/(2b+d-b d),0)'$, we have $\mathbf {AC}=0$. Now using recursion from (4) and the properties of deleting zero entries and renumbering the random variables again, we have

$$\begin{aligned} \Pr [T>d|B(1,2)]&=\frac{-2(2b+1)(d-1)}{2b+d-bd}\Pr \left( \frac{2(1-d)}{2b}Z_1+\frac{1-d}{b}Z_2>0\right) \\&\quad +\frac{(3d-2)(b+1)}{2b+d-b d}\Pr \left( \frac{2-3d}{2b+1}Z_1+\frac{1-d}{b}Z_2>0\right) . \end{aligned}$$

The same process is applied to calculate the two probabilities with $\mathbf {A_1}=(2(1-d)/(2b),(1-d)/b)$ and $\mathbf {A_2}=((2-3d)/(2b+1),(1-d)/b)$. Note that $\Pr (\mathbf {A_1 Z}>0)$ is 1 for $0<d\le 1$. Taking $\mathbf {c_2}=((2b+1)(1-d)/[(2b+2d-2bd-1)], (3d-2)/[(2b+2d-2bd-1)])'$ to calculate $\Pr (\mathbf {A_2 Z}>0)$, it follows that

$$\begin{aligned} \Pr [T>d|B(1,2)]= & {} \frac{-2(2b + 1)(d - 1)}{2b + d - bd}\nonumber \\&-\frac{(2b + 1)(3d - 2)(b + 1)(d - 1)}{(2b + d - bd)(2b + 2d - 2bd - 1)},\quad \frac{2}{3}<d\le 1.\nonumber \\ \end{aligned}$$

(36)

In similar manner, we can obtain

$$\begin{aligned} \Pr [T>d|B(1,3)]= & {} \frac{-2(2b + 1)(d - 1)}{2b + d - bd}\nonumber \\&-\frac{(2b + 1)(3d - 2)(b + 1)(d - 1)}{(bd - d + 1)(2b + d - bd)},\quad \frac{2}{3}<d\le 1. \end{aligned}$$

(37)

$$\begin{aligned} \Pr [T>d|B(2,3)]= & {} \frac{-(2b + 1)(d - 1))}{bd - d + 1}\nonumber \\&\quad -\frac{b(2b + 1)(3d - 2)(d - 1)}{(bd - d + 1)^2},\quad \frac{2}{3}<d\le 1.\qquad \end{aligned}$$

(38)

From (11), we have

$$\begin{aligned} \Pr [B(1,2)] =&\frac{2b^2}{(2b + 1)(b + 1)}, \quad \Pr [B(1,3)] = \frac{2b}{(2b + 1)(b + 1)}, \nonumber \\&\quad \text {and}\quad \Pr [B(2,3)] = \frac{1}{2b + 1}. \end{aligned}$$

(39)

Combining the probabilities in (36–38) and using (39), we obtain the distribution function of test statistic $T_2$ as follows.

$$\begin{aligned} \Pr [T_2<d]&= 1-\Pr [T>d|B(1,2)]\Pr [B(1,2)]-\Pr [T>d|B(1,3)]\Pr [B(1,3)]\nonumber \\&\quad -\Pr [T>d|B(2,3)]\Pr [B(2,3)]. \end{aligned}$$

(40)

The null distribution of $T_2$ can be obtained from (40) by letting $b=1$. For the significance level $\alpha =0.05$, the critical value for the test $T_2$ can be calculated from the null distribution of $T_2$ which is equal to $d=0.991559$.

Now, using (22–24) and plugging (36–39) into them, we can calculate the various performance measures for the test. For example, for the size of the shift $b=0.5$, the P, $\textit{NSP}$, $\textit{SP}$, $\textit{SW}$, $\textit{NSE}$, $\textit{SE}$ and $\textit{PSE}$ are 0.0579, 0.0329, 0, 0.0250, 0.4671, 0, 0.4750 and 0.0658 respectively.

Exact distribution of MLR sequential test for testing two outliers for the sample size $n=3$

To calculate the joint probability expression $\Pr [U_1<d_1, U_2<d_2|B(2,3)]$, we first need to calculate the probabilities $\Pr [U_1<d_1|B(2,3)]$ and $\Pr [U_2<d_2|B(2,3)]$. Thus

$$\begin{aligned} \Pr [U_1>d_1|B(2,3)]&= P\left[ \sum _{j=1}^{n}[1-(n-j+1)d_1]\frac{Z_j}{u^{2,3}_j}>0\right] , \end{aligned}$$

and

$$\begin{aligned} \Pr [U_2>d_2|B(2,3)]&= P\left[ \sum _{j=1}^{n-1}[1-(n-j)d_2]\frac{Z_j}{u^{2,3}_j}>0\right] , \end{aligned}$$

where $u^{2,3}_1 =2 b+1$, $u^{2,3}_2 = 2 b$ and $u^{2,3}_3 = b$ have been defined previously in (10).

Following the lines of arguments in calculating the probability in appendix A, we have

$$\begin{aligned} \Pr [U_1>d_1|B(2,3)] = {\left\{ \begin{array}{ll} - \frac{(2b + 1)(2d_1 - 1)}{2bd_1 - 2d_1 + 1} - \frac{2b(2b + 1)(3d_1 - 1)(d_1 - 1)}{(2bd_1 - 2d_1 + 1)(b - d_1 + bd_1 + 1)} &{} 1/3<d_1\le 1/2,\\ {} \frac{(2b + 1)(2d_1 - 1)(2d_1 - 2)}{2bd_1 - 2d_1 + 1} - \frac{2b(2b + 1)(3d_1 - 1)(d_1 - 1)}{(2bd_1 - 2d_1 + 1)(b - d_1 + bd_1 + 1)} &{} 1/2<d_1\le 1, \end{array}\right. } \end{aligned}$$

(41)

and

$$\begin{aligned} \Pr [U_2>d_2|B(2,3)] = -\frac{(2b + 1)(d_2 - 1)}{2bd_2 - d_2 + 1} \quad 1/2<d_2\le 1. \end{aligned}$$

(42)

For $0<d_1\le 1, 0<d_2\le 1$, the joint probability $\Pr [U_1>d_1, U_2>d_2|B(2,3)]$ can be written as $P(\mathbf {AZ}>0|B(2,3))$ using (10) and (14) where

$$\begin{aligned} \mathbf {A} =\begin{pmatrix} \frac{1-3 d_1}{u^{2,3}_1} &{} \frac{1-2 d_1}{u^{2,3}_2} &{} \frac{1- d_1}{u^{2,3}_3}\\ \frac{1-2 d_2}{u^{2,3}_1} &{} \frac{1- d_2}{u^{2,3}_2} &{} 0 \end{pmatrix} \end{aligned}$$

and $u^{2,3}_1 =2 b+1$, $u^{2,3}_2 = 2 b$ and $u^{2,3}_3 = b$.

Let $\mathbf {c}=(-((2b + 1)(d_2 - 1))/(2bd_2 - d_2 + 1),(2b(2d_2 - 1))/(2bd_2 - d_2 + 1),0)'$, then $\mathbf {AC}=((d_2 - d_1 + d_1 d_2)/(2b d_2 - d_2 + 1),0)'$ and using recursion (4), we get

$$\begin{aligned} \Pr [\mathbf {AZ}>0|B(2,3)]= & {} -\frac{(2b + 1)(d_2 - 1)}{2bd_2 - d_2 + 1}\Pr [\mathbf {A_1 Z}>0|B(2,3)]\\&+ \frac{2b(2d_2 - 1)}{2bd_2 - d_2 + 1}\Pr [\mathbf {A_2 Z}>0|B(2,3)], \end{aligned}$$

where

$$\begin{aligned} \mathbf {A_1} =\begin{pmatrix} \frac{d_2 - d_1 + d_1 d_2}{2b d_2 - d_2 + 1} &{} \frac{1-2 d_1}{u^{2,3}_2} &{} \frac{1- d_1}{u^{2,3}_3}\\ 0 &{} \frac{1- d_2}{u^{2,3}_2} &{} 0 \end{pmatrix} \end{aligned}$$

and

$$\begin{aligned} \mathbf {A_2} =\begin{pmatrix} \frac{1-3 d_1}{u^{2,3}_1} &{} -\frac{d_1 - d_2 - d_1 d_2}{2b d_2 - d_2 + 1} &{} \frac{1- d_1}{u^{2,3}_3}\\ \frac{1-2 d_2}{u^{2,3}_1} &{} 0 &{} 0 \end{pmatrix}. \end{aligned}$$

Since $1/2 < d_2 \le 1$, all the entries in second row of $\mathbf {A_2}$ are less than or equal to 0 which implies that $\Pr [\mathbf {A_2 Z} > 0|B(2,3)] = 0$. Moreover, all the entries in second row of $\mathbf {A_1}$ are greater than or equal to 0 which implies that second row of $\mathbf {A_1}$ can be deleted.

Thus, we have

$$\begin{aligned} \Pr [\mathbf {A_1 Z}>0|B(2,3)]= \Pr \left[ \frac{d_2 - d_1 - d_1 d_2}{2b d_2 - d_2 + 1}Z_1+\frac{1-2 d_1}{u^{2,3}_2} Z_2+\frac{1- d_1}{u^{2,3}_3} Z_3 >0\right] . \end{aligned}$$

Before proceeding further, we first check the value of $(d_2 - d_1 - d_1 d_2)/(2b d_2 - d_2 + 1)$. It can be easily shown that

$$\begin{aligned} \frac{d_2 - d_1 - d_1 d_2}{2b d_2 - d_2 + 1}&={\left\{ \begin{array}{ll} -\frac{\beta + 2\sqrt{1 - \beta } - 1}{6b + 2\beta - 2b\beta } &{} \quad \text {for}\; 1/3<d_1\le 1/2, 1/2<d_2\le 1, \\ {} -\frac{\beta + 2\sqrt{3 \beta } + 3}{6b + 2\beta - 2b\beta } &{} \quad \text {for}\; 1/2<d_1 \le 1, 1/2<d_2\le 1, \end{array}\right. }\\&<0, \end{aligned}$$

where

$$\begin{aligned} d_1&= {\left\{ \begin{array}{ll} \frac{1+\sqrt{1 - \beta }}{3} &{} \quad \text {for}\; 1/3<d_1\le 1/2, \\ 1 - \frac{\sqrt{3 \beta }}{3} &{} \quad \text {for}\; 1/2<d_1 \le 1, \end{array}\right. } \end{aligned}$$

(43)

$$\begin{aligned} d_2&=\frac{3-\beta }{\beta + 3} \quad \text {for}\; 1/2 <d_2\le 1. \end{aligned}$$

(44)

The values of $d_1$ and $d_2$ are obtained such that under the null hypothesis $H_0$, $\Pr [U_1>d_1]=\Pr [U_2>d_2]=\beta \$ to satisfy the condition $\Pr [U_1<d_1,U_2<d_2]=1-\Pr [U_1>d_1]-\Pr [U_2>d_2]+\Pr [U_1>d_1,U_2>d_2]=1-\alpha$ where $\alpha$ is the significance level. Note that under the null hypothesis the $\Pr [U_1>d_1]$ and $\Pr [U_2>d_2]$ are equivalent to the $\Pr [U_1>d_1|B(2,3)]$ and $\Pr [U_2>d_2|B(2,3)]$ respectively.

Using the recursion (4), we obtain

$$\begin{aligned}&\Pr [U_1>d_1,U_2>d_2|B(2,3)] \nonumber \\&\quad ={\left\{ \begin{array}{ll} -\frac{(2b + 1)(3 d_1 + d_2 - 3b d_1 + b d_2 - 3 d_1 d_2 + 4b d_1^2 + 2d_1^2 d_2 - 2d_1^2 - b d_1 d_2 - 1)}{(2b d_1 - 2d_1 {+} 1)(b d_1 - d_2 - d_1 + b d_2 + d_1 d_2 - b d_1 d_2 + 1)} &{} 1/3<d_1 {\le } 1/2, 1/2 {<}d_2\le 1,\\ {} -\frac{2 (2b + 1)(d_1 - 1)^2 (d_2 - 1)}{b d_1 - d_2 - d_1 + b d_2 + d_1 d_2 - b d_1 d_2 + 1} &{} 1/2<d_1 \le 1, 1/2 <d_2\le 1. \end{array}\right. } \end{aligned}$$

(45)

Combining the probabilities obtained in (41), (42) and (45), we can obtain the required probability using

$$\begin{aligned}&\Pr [U_1<d_1,U_2<d_2|B(2,3)]=1-\Pr [U_1>d_1|B(2,3)]-\Pr [U_2>d_2|B(2,3)]\\&\quad +\Pr [U_1>d_1,U_2>d_2|B(2,3)]. \end{aligned}$$

It is also worthwhile to mention that the range of $\beta$ can be determined using Bonferroni’s inequality (Lin and Balakrishnan 2009). Since under $H_0$,

$$\begin{aligned} \Pr \left[ \bigcup \limits _{i=1}^{k}(U_i>d_i)|B(2,3)\right] \le \sum \limits _{i=1}^{k}\Pr [U_i>d_i|B(2,3)] \end{aligned}$$

which gives $\beta \ge \alpha /k$. In particular, when $k=2$, we have $\Pr [U_1>d_1,U_2>d_2|B(2,3)]=2\beta -\alpha \le 1$ which leads to have the range of $\beta$ for testing two upper outliers as $\alpha /k \le \beta \le (1+\alpha )/2$.

In similar way, in order to calculate the probability $\Pr [U_1<d_1,U_2<d_2|B(1,2)]$, we can obtain

$$\begin{aligned} \Pr [U_1>d_1|B(1,2)] = {\left\{ \begin{array}{ll} - \frac{(2b + 1)(2 d_1 - 1)}{b + d_1 - b d_1} - \frac{(2 b + 1)(3 d_1 - 1)(b + 1)(d_1 - 1)}{2(b + d_1 - b d_1)^2} &{} 1/3<d_1\le 1/2,\\ {} \frac{(2b + 1)(2 d_1 - 1)(b + 1)(d1 - 1)}{(b + d_1 - b d_1)^2} - \frac{(2b + 1)(3 d_1 - 1)(b + 1)(d_1 - 1)}{2(b + d_1 - b d_1)^2} &{} 1/2<d_1 \le 1. \end{array}\right. } \end{aligned}$$

(46)

$$\begin{aligned} \Pr [U_2>d_2|B(1,2)] = \frac{(2b + 1)(1-d_2)}{b + d_2} \quad 1/2<d_2 \le 1. \end{aligned}$$

(47)

$$\begin{aligned}&\Pr [U_1>d_1,U_2>d_2|B(1,2)] \nonumber \\&\quad ={\left\{ \begin{array}{ll} -\frac{(2b + 1)(d_2 - 2 d_1 - b + 2b d_1 + b d_2 - 2d_1 d_2 - b d_1^2 + d_1^2 d_2 + 3 d_1^2 - 2 b d_1 d_2 + b d_1^2 d_2)}{(b + d_1 - b d_1)^2} &{} 1/3{<}d_1 \le 1/2, 1/2 {<}d_2\le 1,\\ {} \frac{(2b {+} 1)(b {+} 1)(1{-}d_1)^2 (1-d_2)}{(b + d_1 - b d_1)^2} &{} 1/2<d_1 \le 1, 1/2 <d_2\le 1. \end{array}\right. } \end{aligned}$$

(48)

To calculate, $\Pr [U_1<d_1,U_2<d_2|B(1,3)]$, we get

$$\begin{aligned} \Pr [U_1>d_1|B(1,3)] = {\left\{ \begin{array}{ll} \frac{(2b + 1)(1-2 d_1)}{b + d_1 - b d_1} - \frac{(2b + 1)(1-3 d_1)(b + 1)(1-d_1)}{(b + d_1 - b d_1) (b - d_1 + b d_1 + 1)} &{} 1/3<d_1 \le 1/2,\\ {} \frac{(2b + 1)(1-2 d_1)(b + 1)(1-d_1)}{(b d_1 - d_1 + 1)(b + d_1 - b d_1)} - \frac{(2b + 1)(1-3 d_1)(b + 1)(1-d_1)}{(b + d_1 - b d_1)(b - d_1 + b d_1 + 1)} &{} 1/2<d_1 \le 1. \end{array}\right. } \end{aligned}$$

(49)

$$\begin{aligned} \Pr [U_2>d_2|B(1,3)] = \frac{(2b + 1)(1-d_2)}{b + d_2} \quad 1/2<d_2 \le 1. \end{aligned}$$

(50)

$$\begin{aligned} Z&\Pr [U_1>d_1,U_2>d_2|B(1,3)] \nonumber \\ \quad&={\left\{ \begin{array}{ll} -\frac{(2b + 1)(b d_1 - d_1 - b + 2b d_2 + d_1 d_2 + b d_1^2 - d_1^2 d_2 + d_1^2 - 5b d_1 d_2 + 3b d_1^2 d_2)}{(b + d_1 - b d_1)(b + d_2 - b d_2 - d_1 d_2 + b d_1 d_2)} &{} 1/3{<}d_1\le 1/2, 1/2 {<}d_2\le 1,\\ {} \frac{(2b + 1)(b + 1)(1-d_1)^2(1-d_2)}{(b d_1 - d_1 + 1)(b + d_2 - b d_2 - d_1 d_2 + b d_1 d_2)} &{} 1/2<d_1\le 1, 1/2 <d_2\le 1. \end{array}\right. } \end{aligned}$$

(51)

Finally, we can obtain the joint distribution of $(U_1,U_2)$ by plugging the probabilities in (41–42), (45–51) and (39) as follows

$$\begin{aligned} \Pr [U_1<d_1,U_2<d_2|\bar{H}]&=1-\sum _{(r,s)\in S^{(2)}}{\Pr [U_1>d_1|B(r,s)]\Pr [B(r,s)]}\\&\quad - \sum _{(r,s)\in S^{(2)}}{\Pr [U_2>d_2|B(r,s)]\Pr [B(r,s)]} \\&\quad +\sum _{(r,s)\in S^{(2)}}{\Pr [U_1>d_1,U_2>d_2|B(r,s)]\Pr [B(r,s)]}, \end{aligned}$$

where $S^{(2)}=\{(1,2), (1,3), (2,3)\}$.

Note that the exact critical values of the sequential tests for testing $k=2$ upper outliers can be obtained by plugging the expressions of $d_1$ in (43) and $d_2$ in (44) into

$$\begin{aligned} \Pr [U_1<d_1,U_2<d_2|\bar{H}] =1-2 \beta + \Pr [U_1>d_1,U_2>d_2|\bar{H}] =1-\alpha , \end{aligned}$$

where $\alpha$ is the significance level.

Therefore, for $\alpha =0.05$, we obtain the value of $\beta =0.025427$ which yields the critical values to be $d_1= 0.907936$ and $d_2=0.983191$ using (43) and (44).

Once we calculate the critical values, we can compute the various performance measures discussed in Sect. 6 by plugging the probabilities in (41–51) and (39) for different size of shifts $b<1$. For example, for $b=0.5$, the P, $\textit{NSP}$, $\textit{SP}$, $\textit{NSE}$, ${ PL}_1$, ${ PL}_2$ and ${ PL}_3$ are 0.00113, 0.00057, 0, 0.48379, 0.00008, 0.00048 and 0.01564 respectively.