Robust Bayesian Estimation

Abstract

Bayes methods in statistical inference are one of the important methods, and most of the research and messages tend to use the Bayes method in the estimation process. The regular Bayes method does not meet this problem, so in this thesis it is possible to verify the existence of prior data conflict by modeling the parameters of the prior distribution and then comparing the standard deviation of the prior distribution with the standard deviation of the posterior distribution, if the value of the standard deviation of the prior distribution is greater than the deviation. The standard distribution for the posterior distribution, it means that there is a problem of prior data conflict. Then we used an approach to solve this problem through a set of prior distributions called this approach by the robust Bayesian method, to identify the behavior of the estimators, two types of failure models were used, the first Weibull distribution to match it with continuous data. The second is a (Binomial) distribution to match the discrete data, the regular Bayes method is compared with the robust Bayesian method by using integrated mean square error (IMSE). In the Weibull distribution, the scale parameter (θ) and the survival function were estimated for two simulation experiments, the first was in the case of prior data unconflict the second was in the case of prior data conflict, so the simulation results showed that the robust Bayes method is the best by using the comparison criterion integrated mean square error (IMSE). On the practical side, real data were collected from Al-Manathira Hospital of the Najaf Health Department for the deaths of heart attack patients for 2018, the time of admission of the patient to the hospital until death was recorded, which is the time Exit where a sample of (15) patients was collected and the test of goodness of fit showed that the data follow a Weibull distribution with two parameters, the robust Bayes method was used to estimate the scale parameter and the survival function. As for the Binomial distribution, the parameter (P) and survival function were estimated for two experiments from the first simulation, which was in the case of prior data unconflict, as for the second experiment, it was in the case of prior data conflict. The simulation results showed that the robust Bayes method is the best by using the comparison criterion (IMSE). On the practical side, real data were collected from Yarmouk Teaching Hospital on breast cancer patients’ mortality from 2010 to 2017, and the test of goodness of fit showed that the data follow a Binomial distribution, the robust Bayes method was used to estimate the parameter (P) and survival function.

1. Introduction

Both the reliability function and the survival function have the same property, which is the measurement of the life span of a particular system or organism. In systems and equipment, it is called the reliability function, but for the organism, it is called the survival function.

Sometimes, especially in the analysis of survival functions, the failure events are few, so we need to include prior information, which can be used by the Bayes method. When merging the prior information with the observations to obtain the posterior distribution according to the Bayes rule, a problem may appear to us, which is the problem of prior data conflict with views, in regular Bayes method, this problem is not checked and is not addressed and thus unreal estimators are obtained, In this thesis, we will use an approach to address this problem, which is the prior data conflict problem, and thus this method is called the Robust Bayes Procedure.

What is meant by the prior data conflict is when the information of the prior distribution is combined with the distribution of observations, which may cause us this problem, meaning that the data under study are less homogeneous when the information of the prior distribution is combined with it, and thus we obtain unreal estimations without realizing.

In analyzing the problem, the researcher relied on two models of failure, the first model is the Weibull distribution with two parameters to match the continuous data, the second model is the Binomial distribution to match the discrete data to identify the behavior of the capabilities in these two types of data and the appropriateness of the robust methods to deal with the existence of the problem prior data conflict.

2. Robust Bayesian procedure

We also noted earlier that the concept of Bayes theory depends on prior information so that prior information is combined with the distribution of observations according to the Bayes rule, for the purpose of obtaining the posterior distribution, from here we may have a problem, which is a problem that prior data conflict. Whereas prior data they are the default values that are assumed for the parameters of the prior distribution, to find out this problem by updating the parameters of the prior distribution through two methods, namely Expected Conditional or Canonical Exponential Family and provided that the prior distribution is conjugate prior, after obtaining the prior distribution with the updated parameters, we extract the posterior distribution and then extract the standard deviation of the distribution if the value of the standard deviation of the prior distribution is greater than the standard deviation of the posterior distribution, then this means that there is a problem of prior data conflict, Thus, this problem can be addressed through the steps that we will explain later, This method is called the robust Bayesian method [1].

After the default values for the parameters of the prior distribution are chosen, the standard deviation of the prior distribution and the posterior distribution are extracted. If the value of the standard deviation of the prior distribution is greater than the standard deviation of the posterior distribution, this means that there is a problem of prior data conflict and provided that the posterior distribution is conjugate prior, this is the method that will be used in this chapter to verify the prior data conflict.

There are other ways to verify the prior data conflict that we did not used in this chapter, for example (Conflict checks based on relative belief, Connections between the relative belief and score checks and Other approaches to prior-data conflict checking) [2].

Then we move on to addressing the problem of prior data conflict through the proposal presented by (Walter and Augustin; 2009), this is for the purpose of generating a set of prior parameters, in short$\left({\prod }^0=\left[{\underset{¯}{n}}^0,{\overline{n}}^0\right]x\left[{\underset{¯}{y}}^0,{\overline{y}}^0\right]\right)$, So that this model that generates a set of prior parameters is called generalized iLuck-model and therefore we will get a set of posterior distributions, And then a Bayes estimator is obtained according to the type of loss function used, and thus this method is called the robust Bayesian method [3].

3. Weibull distribution

The Weibull distribution was used in 1951 by researcher Waldi Weibull in many experiments related to the reliability in the mechanical aspect and survival in the human aspect.

The emergence of this distribution, especially in the Second World War, and its wide applications in the field of reliability and life tests, was the focus of the attention of a number of researchers in this field, great in theory and practice.

The probability density function (pdf) for a two-parameter Weibull distribution is in the following form [4]:

$$\mathrm{f}\left(\mathrm{t}\right)=\frac{\mathrm{\beta }}{\mathrm{\theta }}{\mathrm{t}}^{\mathrm{\beta }-1}{\mathrm{e}}^{-\frac{{\mathrm{t}}^{\mathrm{\beta }}}{\mathrm{\theta }}};\mathrm{\theta }>0,t>0$$E1

Since:

$\mathrm{\beta }$: shape parameter.

$\mathrm{\theta }:$Scale parameter.

The formula for the (CDF) cumulative function is:

$$\mathrm{F}\left(\mathrm{t}\right)=1-{\mathrm{e}}^{-\frac{{\mathrm{t}}^{\mathrm{\beta }}}{\mathrm{\theta }}}$$E2

The formula for the survival function is:

$$\mathrm{S}\left(\mathrm{t}\right)={\mathrm{e}}^{-\frac{{\mathrm{t}}^{\mathrm{\beta }}}{\mathrm{\theta }}}$$E3

and the formula for moment $r^{\mathit{th}}$ is:

$${\mathrm{M}}_{\mathrm{r}}={\mathrm{\theta }}^{\frac{\mathrm{r}}{\mathrm{\beta }}}\mathrm{г}\left(1+\frac{\mathrm{r}}{\mathrm{\beta }}\right)$$E4

3.1 Bayesian estimation of scale parameter for Weibull distribution

Suppose we have a sample that follows the Weibull distribution shown in Eq. (1) and the appropriate prior distribution for the parameter (θ) is inverse gamma distribution according to the following formula [5]:

$$\mathrm{f}\left(\mathrm{\theta }/\mathrm{a},\mathrm{b}\right)=\frac{{\mathrm{b}}^{\mathrm{a}}}{\mathrm{г}\left(\mathrm{a}\right)}{\mathrm{\theta }}^{-\mathrm{a}-1}{\mathrm{e}}^{-\frac{\mathrm{b}}{\mathrm{\theta }}}$$E5

By using the Bayes rule, we get the posterior distribution, as shown below [6]:

$$\mathrm{f}\left(\mathrm{\theta }/\mathrm{t}\right)=\frac{\mathrm{f}\left({\mathrm{t}}_1,{\mathrm{t}}_2,\dots ,{\mathrm{t}}_{\mathrm{n}}/\mathrm{\theta },\mathrm{\beta }\right)\mathrm{f}\left(\mathrm{\theta }/\mathrm{a},\mathrm{b}\right)}{\underset{0}{\overset{\infty }{\int }}\mathrm{f}\left({\mathrm{t}}_1,{\mathrm{t}}_2,\dots ,{\mathrm{t}}_{\mathrm{n}}/\mathrm{\theta },\mathrm{\beta }\right)\mathrm{f}\left(\mathrm{\theta }/\mathrm{a},\mathrm{b}\right)\mathrm{d\theta }}$$E6

$$=\frac{{\left({\sum }_{\mathrm{i}=1}^n{\mathrm{t}}_{\mathrm{i}}^{\mathrm{\beta }}+\mathrm{b}\right)}^{\left(\mathrm{a}+\mathrm{n}\right)}}{\mathrm{г}\left(\mathrm{a}+\mathrm{n}\right)}{\mathrm{\theta }}^{-\left(\mathrm{a}+\mathrm{n}\right)-1}{\mathrm{e}}^{-\frac{\left({\sum }_{\mathrm{i}=1}^n{\mathrm{t}}_{\mathrm{i}}^{\mathrm{\beta }}+\mathrm{b}\right)}{\mathrm{\theta }}}$$

Since:

$$\sum _{\mathrm{i}=1}^n{\mathrm{t}}_{\mathrm{i}}^{\mathrm{\beta }}=\mathrm{\tau }\left(\mathrm{t}\right)$$

$$\mathrm{f}\left(\mathrm{\theta }/\mathrm{t}\right)\sim \text{Inverse Gamma}\left(\mathrm{a}+\mathrm{n},\mathrm{\tau }\left(\mathrm{t}\right)+\mathrm{b}\right)$$

Eq. (6) represents the posterior distribution for the parameter (θ), and according to the squared loss function, the Bayesian estimator for the parameter (θ) is the mean of the posterior distribution, as in the following steps [7]:

$$\mathrm{E}\left(\mathrm{\theta }/\mathrm{t}\right)=\widehat{\mathrm{\theta }}$$

$$\widehat{\mathrm{\theta }}=\underset{0}{\overset{\infty }{\int }}\mathrm{\theta f}\left(\mathrm{\theta }/\mathrm{t}\right)\mathrm{d\theta }$$

$$=\underset{0}{\overset{\infty }{\int }}\mathrm{\theta }\frac{{\left(\mathrm{\tau }\left(\mathrm{t}\right)+\mathrm{b}\right)}^{\left(\mathrm{a}+\mathrm{n}\right)}}{\mathrm{г}\left(\mathrm{a}+\mathrm{n}\right)}{\mathrm{\theta }}^{-\left(\mathrm{a}+\mathrm{n}\right)-1}{\mathrm{e}}^{-\frac{\left(\mathrm{\tau }\left(\mathrm{t}\right)+\mathrm{b}\right)}{\mathrm{\theta }}}\mathrm{d\theta }$$

$$=\underset{0}{\overset{\infty }{\int }}\frac{{\left(\mathrm{\tau }\left(\mathrm{t}\right)+\mathrm{b}\right)}^{\left(\mathrm{a}+\mathrm{n}\right)}}{\mathrm{г}\left(\mathrm{a}+\mathrm{n}\right)}{\mathrm{\theta }}^{-\left(\mathrm{a}+\mathrm{n}\right)-1+1}{\mathrm{e}}^{-\frac{\left(\mathrm{\tau }\left(\mathrm{t}\right)+\mathrm{b}\right)}{\mathrm{\theta }}}\mathrm{d\theta }$$

By using the transformation:

$$\mathrm{Let}\mathrm{y}=\frac{\left(\mathrm{\tau }\left(\mathrm{t}\right)+\mathrm{b}\right)}{\mathrm{\theta }}\Rightarrow \mathrm{\theta }=\frac{\left(\mathrm{\tau }\left(\mathrm{t}\right)+\mathrm{b}\right)}{\mathrm{y}},\mathrm{J}=\frac{\left(\mathrm{\tau }\left(\mathrm{t}\right)+\mathrm{b}\right)}{{\mathrm{y}}^2}$$

$$\widehat{\mathrm{\theta }}=\underset{0}{\overset{\infty }{\int }}\frac{{\left(\mathrm{\tau }\left(\mathrm{t}\right)+\mathrm{b}\right)}^{\left(\mathrm{a}+\mathrm{n}\right)}}{\mathrm{г}\left(\mathrm{a}+\mathrm{n}\right)}{\left(\frac{\mathrm{\tau }\left(\mathrm{t}\right)+\mathrm{b}}{\mathrm{y}}\right)}^{-\left(\mathrm{a}+\mathrm{n}\right)}{\mathrm{e}}^{-\mathrm{y}}\frac{\left(\mathrm{\tau }\left(\mathrm{t}\right)+\mathrm{b}\right)}{{\mathrm{y}}^2}\mathrm{dy}$$

$$\widehat{\mathrm{\theta }}=\frac{{\left(\mathrm{\tau }\left(\mathrm{t}\right)+\mathrm{b}\right)}^{\left(\mathrm{a}+\mathrm{n}-\mathrm{a}-\mathrm{n}+1\right)}}{\mathrm{г}\left(\mathrm{a}+\mathrm{n}\right)}\underset{0}{\overset{\infty }{\int }}{\left(\frac{1}{\mathrm{y}}\right)}^{-\left(\mathrm{a}+\mathrm{n}\right)+2}{\mathrm{e}}^{-\mathrm{y}}\mathrm{dy}$$

$$\widehat{\mathrm{\theta }}=\frac{\left(\mathrm{\tau }\left(\mathrm{t}\right)+\mathrm{b}\right)}{\mathrm{г}\left(\mathrm{a}+\mathrm{n}\right)}\underset{0}{\overset{\infty }{\int }}{\mathrm{y}}^{\mathrm{a}+\mathrm{n}-2}{\mathrm{e}}^{-\mathrm{y}}\mathrm{dy}$$

$$\widehat{\mathrm{\theta }}=\frac{\left(\mathrm{\tau }\left(\mathrm{t}\right)+\mathrm{b}\right)}{\mathrm{г}\left(\mathrm{a}+\mathrm{n}\right)}\mathrm{г}\left(\mathrm{a}+\mathrm{n}-1\right)$$

$$\widehat{\mathrm{\theta }}=\frac{\left(\mathrm{\tau }\left(\mathrm{t}\right)+\mathrm{b}\right)}{\mathrm{a}+\mathrm{n}-1}$$E7

3.2 Bayesian estimation of survival function for Weibull distribution

From Eq. (6), according to the squared loss function, the Bayesian estimator for the survival function is [8]:

$$\widehat{\mathrm{S}}\left(\mathrm{t}\right)=\underset{0}{\overset{\infty }{\int }}\mathrm{S}\left(\mathrm{t}\right)\mathrm{f}\left(\mathrm{\theta }/\mathrm{t}\right)\mathrm{d\theta }$$

$$\widehat{\mathrm{S}}\left(\mathrm{t}\right)={\left(\frac{\mathrm{b}+\mathrm{\tau }\left(\mathrm{t}\right)}{\mathrm{b}+\mathrm{\tau }\left(\mathrm{t}\right)+{\mathrm{t}}_i^{\mathrm{\beta }}}\right)}^{\mathrm{a}+\mathrm{n}}$$E8

3.3 Checking for prior data conflict for Weibull distribution

Suppose we have a sample that follows a Weibull distribution according to the Eq. (1) and the prior distribution suitable for the scale parameter (θ) is Inverse Gamma because it is a conjugate prior and as in the Eq. (5) [9].

Then we need to update the parameters of the prior distribution so that it is ${\mathrm{n}}^0>1,{\mathrm{y}}^0>0$ instead of the parameters (a, b), through two methods we get the prior distribution as shown in the following steps [10]:

The first method: It is the Expected Conditional method, as shown in the following steps:

$$\mathrm{E}\left(\mathrm{\theta }/\mathrm{a},\mathrm{b}\right)={\mathrm{y}}^0=\frac{\mathrm{b}}{\mathrm{a}-1}=\frac{\mathrm{b}}{{\mathrm{n}}^0}\Rightarrow \mathrm{b}={\mathrm{n}}^0{\mathrm{y}}^0,{\mathrm{n}}^0=\mathrm{a}-1\Rightarrow \mathrm{a}={\mathrm{n}}^0+1$$

Then the prior distribution with the updated parameters can be written in the following form:

$$\mathrm{f}\left(\mathrm{\theta }/{\mathrm{n}}^0{\mathrm{y}}^0\right)\propto \frac{{\left({\mathrm{n}}^0{\mathrm{y}}^0\right)}^{{\mathrm{n}}^0+1}}{\mathrm{г}\left({\mathrm{n}}^0+1\right)}{\mathrm{\theta }}^{-\left({\mathrm{n}}^0+1\right)-1}{\mathrm{e}}^{-\frac{{\mathrm{n}}^0{\mathrm{y}}^0}{\mathrm{\theta }}}$$E9

Since:

${\mathrm{y}}^0$: Pre-guessing the scale parameter.

${\mathrm{n}}^0$: Pre- guessing the sample size.

The second method: In this method, the prior distribution can be determined with parameters $\left({\mathrm{n}}^0,{\mathrm{y}}^0\right)$ through the following steps:

The first step: writing the model in the form of canonical exponential family and as shown below [10]:

$$\mathrm{f}\left(\mathrm{x}/\mathrm{\theta }\right)=\mathrm{a}\left(\mathrm{x}\right)exp(\mathrm{\psi }.\mathrm{\tau }\left(\mathrm{x}\right)-\mathrm{nb}\left(\mathrm{\psi }\right)$$E10

$$\mathrm{f}\left(\mathrm{t}/\mathrm{\theta },\mathrm{\beta }\right)={\mathrm{\beta }}^{\mathrm{n}}\prod _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{t}}_{\mathrm{i}}^{\mathrm{\beta }-1}{\mathrm{e}}^{-\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{t}}_{\mathrm{i}}^{\mathrm{\beta }}{\mathrm{\theta }}^{-1}-\mathrm{nln}\left(\mathrm{\theta }\right)}$$E11

$$\mathrm{a}\left(\mathrm{x}\right)={\mathrm{\beta }}^{\mathrm{n}}\prod _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{t}}_{\mathrm{i}}^{\mathrm{\beta }-1},\mathrm{\psi }=-\frac{1}{\mathrm{\theta }},\mathrm{\tau }\left(\mathrm{t}\right)=\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{t}}_{\mathrm{i}}^{\mathrm{\beta }},\mathrm{b}\left(\mathrm{\psi }\right)=ln\left(\mathrm{\theta }\right)$$

The second step: constructing the prior distribution with parameters $\left({\mathrm{n}}^0,{\mathrm{y}}^0\right)$ through the following form:

$$\mathrm{f}\left(\mathrm{\psi }/{\mathrm{n}}^0,{\mathrm{y}}^0\right)\mathrm{d\psi }\propto exp\left\{{\mathrm{n}}^0\left[{\mathrm{y}}^0.\mathrm{\psi }-\mathrm{b}\left(\mathrm{\psi }\right)\right]\right\}\mathrm{d\psi }$$E12

$$\mathrm{f}\left(\mathrm{\psi }/{\mathrm{n}}^0,{\mathrm{y}}^0\right)\mathrm{d\psi }\propto exp\left\{{\mathrm{n}}^0\left[{\mathrm{y}}^0\left(-\frac{1}{\mathrm{\theta }}\right)-ln\left(\mathrm{\theta }\right)\right]\right\}\mathrm{d\psi }$$

$$\left|\frac{\mathrm{d\psi }}{\mathrm{d\theta }}\right|=\frac{1}{{\mathrm{\theta }}^2}$$

$$\mathrm{f}\left(\mathrm{\theta }/{\mathrm{n}}^0,{\mathrm{y}}^0\right)\mathrm{d\theta }=\mathrm{f}\left(\mathrm{\psi }/{\mathrm{n}}^0,{\mathrm{y}}^0\right)\left|\frac{\mathrm{d\psi }}{\mathrm{d\theta }}\right|\mathrm{d\theta }\propto exp\left\{-\frac{{\mathrm{n}}^0{\mathrm{y}}^0}{\mathrm{\theta }}-{\mathrm{n}}^0\mathrm{ln\theta }\right\}\frac{1}{{\mathrm{\theta }}^2}$$

$$\mathrm{f}\left(\mathrm{\theta }/{\mathrm{n}}^0,{\mathrm{y}}^0\right)\propto {\mathrm{\theta }}^{-\left({\mathrm{n}}^0+1\right)-1}{\mathrm{e}}^{-\frac{{\mathrm{n}}^0{\mathrm{y}}^0}{\mathrm{\theta }}}$$E13

For the purpose of testing whether or not there is a problem of prior data conflict, we extract the standard deviation of the prior distribution and the standard deviation of the posterior distribution.

After we get the prior distribution with the updated parameters, we extract the standard deviation, as in the following steps:

$${\mathrm{M}}_{\mathrm{r}}={\int }_0^{\infty }{\mathrm{\theta }}^{\mathrm{r}}\mathrm{f}\left(\mathrm{\theta }/{\mathrm{n}}^0,{\mathrm{y}}^0\right)\mathrm{d\theta }$$

$${\mathrm{M}}_{\mathrm{r}}=\frac{{\left({\mathrm{n}}^0{\mathrm{y}}^0\right)}^{{\mathrm{n}}^0+1}}{\mathrm{г}\left({\mathrm{n}}^0+1\right)}\underset{0}{\overset{\infty }{\int }}{\mathrm{\theta }}^{-\left({\mathrm{n}}^0+1\right)-1+\mathrm{r}}{\mathrm{e}}^{-\frac{{\mathrm{n}}^0{\mathrm{y}}^0}{\mathrm{\theta }}}\mathrm{d\theta }$$

By using the transformation:

$$\mathrm{let}\mathrm{z}=\frac{{\mathrm{n}}^0{\mathrm{y}}^0}{\mathrm{\theta }}\Rightarrow \mathrm{\theta }=\frac{{\mathrm{n}}^0{\mathrm{y}}^0}{\mathrm{z}},\left|\mathrm{J}\right|=\frac{{\mathrm{n}}^0{\mathrm{y}}^0}{{\mathrm{z}}^2}$$

$${\mathrm{M}}_{\mathrm{r}}=\frac{{\left({\mathrm{n}}^0{\mathrm{y}}^0\right)}^{{\mathrm{n}}^0+1}}{\mathrm{г}\left({\mathrm{n}}^0+1\right)}\underset{0}{\overset{\infty }{\int }}{\left(\frac{{\mathrm{n}}^0{\mathrm{y}}^0}{\mathrm{z}}\right)}^{-{\mathrm{n}}^0-1-1+\mathrm{r}}{\mathrm{e}}^{-\mathrm{z}}\frac{{\mathrm{n}}^0{\mathrm{y}}^0}{{\mathrm{z}}^2}\mathrm{dz}$$

$${\mathrm{M}}_{\mathrm{r}}=\frac{{\left({\mathrm{n}}^0{\mathrm{y}}^0\right)}^{{\mathrm{n}}^0+1-{\mathrm{n}}^0-2+\mathrm{r}+1}}{\mathrm{г}\left({\mathrm{n}}^0+1\right)}\underset{0}{\overset{\infty }{\int }}{\mathrm{z}}^{{\mathrm{n}}^0+2-\mathrm{r}-2}{\mathrm{e}}^{-\mathrm{z}}\mathrm{dz}$$

$${\mathrm{M}}_{\mathrm{r}}=\frac{{\left({\mathrm{n}}^0{\mathrm{y}}^0\right)}^{\mathrm{r}}}{\mathrm{г}\left({\mathrm{n}}^0+1\right)}\mathrm{г}\left({\mathrm{n}}^0-\mathrm{r}+1\right)$$E14

After that, we extract the posterior distribution through the Bayes rule, as shown in the following steps:

$$\mathrm{f}\left(\mathrm{\theta }\backslash \mathrm{t}\right)=\frac{\frac{{\left({\mathrm{n}}^0{\mathrm{y}}^0\right)}^{{\mathrm{n}}^0+1}}{\mathrm{г}\left({\mathrm{n}}^0+1\right)}{\mathrm{\theta }}^{-\left({\mathrm{n}}^0+1\right)-1}{\mathrm{e}}^{-\frac{n^0{y}^0}{\mathrm{\theta }}}{\left(\frac{\mathrm{\beta }}{\mathrm{\theta }}\right)}^{\mathrm{n}}{\prod }_{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{t}}_{\mathrm{i}}^{\mathrm{\beta }-1}{\mathrm{e}}^{-\frac{\mathrm{\tau }\left(\mathrm{t}\right)}{\mathrm{\theta }}}}{\underset{0}{\overset{\infty }{\int }}\frac{{\left({\mathrm{n}}^0{\mathrm{y}}^0\right)}^{{\mathrm{n}}^0+1}}{\mathrm{г}\left({\mathrm{n}}^0+1\right)}{\mathrm{\theta }}^{-\left({\mathrm{n}}^0+1\right)-1}{\mathrm{e}}^{-\frac{n^0{y}^0}{\mathrm{\theta }}}{\left(\frac{\mathrm{\beta }}{\mathrm{\theta }}\right)}^{\mathrm{n}}{\prod }_{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{t}}_{\mathrm{i}}^{\mathrm{\beta }-1}{\mathrm{e}}^{-\frac{\mathrm{\tau }\left(\mathrm{t}\right)}{\mathrm{\theta }}}\mathrm{d\theta }}$$

$$=\frac{\frac{{\left({\mathrm{n}}^0{\mathrm{y}}^0\right)}^{{\mathrm{n}}^0+1}}{\mathrm{г}\left({\mathrm{n}}^0+1\right)}{\mathrm{\beta }}^{\mathrm{n}}{\prod }_{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{t}}_{\mathrm{i}}^{\mathrm{\beta }-1}{\mathrm{\theta }}^{-\left({\mathrm{n}}^0+\mathrm{n}+1\right)-1}{\mathrm{e}}^{-\frac{\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}{\mathrm{\theta }}}}{\frac{{\left({\mathrm{n}}^0{\mathrm{y}}^0\right)}^{{\mathrm{n}}^0+1}}{\mathrm{г}\left({\mathrm{n}}^0+1\right)}{\mathrm{\beta }}^{\mathrm{n}}{\prod }_{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{t}}_{\mathrm{i}}^{\mathrm{\beta }-1}\underset{0}{\overset{\infty }{\int }}{\mathrm{\theta }}^{-\left({\mathrm{n}}^0+\mathrm{n}+1\right)-1}{\mathrm{e}}^{-\frac{\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}{\mathrm{\theta }}}\mathrm{d\theta }}$$

$$=\frac{{\mathrm{\theta }}^{-\left({\mathrm{n}}^0+\mathrm{n}+1\right)-1}{\mathrm{e}}^{-\frac{\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}{\mathrm{\theta }}}}{\underset{0}{\overset{\infty }{\int }}{\mathrm{\theta }}^{-\left({\mathrm{n}}^0+\mathrm{n}+1\right)-1}{\mathrm{e}}^{-\frac{\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}{\mathrm{\theta }}}\mathrm{d\theta }}=\frac{{\mathrm{\theta }}^{-\left({\mathrm{n}}^0+\mathrm{n}+1\right)-1}{\mathrm{e}}^{-\frac{\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}{\mathrm{\theta }}}}{\frac{\mathrm{г}\left({\mathrm{n}}^0+\mathrm{n}+1\right)}{{\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}^{{\mathrm{n}}^0+\mathrm{n}+1}}\underset{0}{\overset{\infty }{\int }}\frac{{\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}^{{\mathrm{n}}^0+\mathrm{n}+1}}{\mathrm{г}\left({\mathrm{n}}^0+\mathrm{n}+1\right)}{\mathrm{\theta }}^{-\left({\mathrm{n}}^0+\mathrm{n}+1\right)-1}{\mathrm{e}}^{-\frac{\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}{\mathrm{\theta }}}\mathrm{d\theta }}$$

$$=\frac{{\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}^{{\mathrm{n}}^0+\mathrm{n}+1}}{\mathrm{г}\left({\mathrm{n}}^0+\mathrm{n}+1\right)}{\mathrm{\theta }}^{-\left({\mathrm{n}}^0+\mathrm{n}+1\right)-1}{\mathrm{e}}^{-\frac{\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}{\mathrm{\theta }}}$$E15

$$\mathrm{f}\left(\mathrm{\theta }\backslash \mathrm{t}\right)\sim \mathrm{IG}\left({\mathrm{n}}^0+\mathrm{n}+1,{\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)$$

After we get the posterior distribution and according to the above equation, we extract the standard deviation, as in the following steps:

$${\mathrm{M}}_{\mathrm{r}}=\frac{{\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}^{{\mathrm{n}}^0+\mathrm{n}+1}}{\mathrm{г}\left({\mathrm{n}}^0+\mathrm{n}+1\right)}\underset{0}{\overset{\infty }{\int }}{\mathrm{\theta }}^{-\left({\mathrm{n}}^0+\mathrm{n}+1\right)-1+\mathrm{r}}{\mathrm{e}}^{-\frac{\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}{\mathrm{\theta }}}\mathrm{d\theta }$$

By using the transformation:

$$\mathrm{let}\mathrm{z}=\frac{\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}{\mathrm{\theta }}\Rightarrow \mathrm{\theta }=\frac{\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}{\mathrm{z}},\left|\mathrm{J}\right|=\frac{\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}{{\mathrm{z}}^2}$$

$${\mathrm{M}}_{\mathrm{r}}=\frac{{\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}^{{\mathrm{n}}^0+\mathrm{n}+1}}{\mathrm{г}\left({\mathrm{n}}^0+\mathrm{n}+1\right)}\underset{0}{\overset{\infty }{\int }}{\left(\frac{{\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{\tau }\left(\mathrm{t}\right)}{\mathrm{z}}\right)}^{-{\mathrm{n}}^0-\mathrm{n}-2+\mathrm{r}}{\mathrm{e}}^{-\mathrm{z}}\frac{\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}{{\mathrm{z}}^2}\mathrm{dz}$$

$${\mathrm{M}}_{\mathrm{r}}=\frac{{\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}^{{\mathrm{n}}^0+\mathrm{n}+1-{\mathrm{n}}^0-\mathrm{n}-2+\mathrm{r}+1}}{\mathrm{г}\left({\mathrm{n}}^0+\mathrm{n}+1\right)}\underset{0}{\overset{\infty }{\int }}{\left(\mathrm{z}\right)}^{{\mathrm{n}}^0+\mathrm{n}-\mathrm{r}}{\mathrm{e}}^{-\mathrm{z}}\mathrm{dz}$$

$${\mathrm{M}}_{\mathrm{r}}=\frac{{\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}^{\mathrm{r}}}{\mathrm{г}\left({\mathrm{n}}^0+\mathrm{n}+1\right)}\mathrm{г}\left({\mathrm{n}}^0+\mathrm{n}-\mathrm{r}+1\right)$$E16

$$\mathrm{s}.\mathrm{d}\text{posterior}=\sqrt{\frac{{\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}^2}{{\left({\mathrm{n}}^0+\mathrm{n}\right)}^2\left({\mathrm{n}}^0+\mathrm{n}-1\right)}}$$E17

The above equation represents the standard deviation of the posterior distribution, after that the comparison is made between the value of the standard deviation of the prior distribution with the standard deviation of the posterior distribution. If the value of the standard deviation of the prior distribution is greater than the value of the standard deviation of the posterior distribution, this means that there is a problem of prior data conflict.

A second way to get the standard deviation of the posterior distribution is through the following steps:

Through the following form which represents the posterior distribution:

$$\mathrm{f}\left(\mathrm{\theta }/\mathrm{t}\right)=\frac{{\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}^{{\mathrm{n}}^0+\mathrm{n}+1}}{\mathrm{г}\left({\mathrm{n}}^0+\mathrm{n}+1\right)}{\mathrm{\theta }}^{-\left({\mathrm{n}}^0+\mathrm{n}+1\right)-1}{\mathrm{e}}^{-\frac{\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}{\mathrm{\theta }}}$$

Compensation for:

$${\mathrm{y}}^{\mathrm{n}}=\frac{{\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{\tau }\left(\mathrm{t}\right)}{{\mathrm{n}}^0+\mathrm{n}},{\mathrm{n}}^{\mathrm{n}}={\mathrm{n}}^0+\mathrm{n}$$

The posterior distribution becomes as follows:

$$\mathrm{f}\left(\mathrm{\theta }/{\mathrm{n}}^{\mathrm{n}}{\mathrm{y}}^{\mathrm{n}}\right)=\frac{{\left({\mathrm{n}}^{\mathrm{n}}{\mathrm{y}}^{\mathrm{n}}\right)}^{{\mathrm{n}}^{\mathrm{n}}+1}}{\mathrm{г}\left({\mathrm{n}}^{\mathrm{n}}+1\right)}{\mathrm{\theta }}^{-\left({\mathrm{n}}^{\mathrm{n}}+1\right)-1}{\mathrm{e}}^{-\frac{{\mathrm{n}}^{\mathrm{n}}{\mathrm{y}}^{\mathrm{n}}}{\mathrm{\theta }}}$$E18

From the above, we conclude that $\mathrm{f}\left(\mathrm{\theta }/\mathrm{t}\right)=\mathrm{f}\left(\mathrm{\theta }/{\mathrm{n}}^{\mathrm{n}}{\mathrm{y}}^{\mathrm{n}}\right)$ this means that the standard deviation of the prior distribution and the standard deviation of the posterior distribution will be according to the following formula:

$$\mathrm{s}.\mathrm{d}\text{prior}=\sqrt{\frac{{\left({\mathrm{y}}^0\right)}^2}{{\mathrm{n}}^0-1}}$$E19

$$\mathrm{s}.\mathrm{d}\text{posterior}=\sqrt{\frac{{\left({\mathrm{y}}^{\mathrm{n}}\right)}^2}{{\mathrm{n}}^{\mathrm{n}}-1}}$$E20

3.4 Standard error for mean

The indicator was employed for the purpose of testing the problem of prior data conflict. If the value of the standard error of the mean of the prior distribution is greater than the standard error of the mean of the posterior distribution, then this means that there is a problem of prior data conflict, and its formula is in the following form:

$${\mathrm{SE}}_{\overline{\mathrm{x}}}.\text{prior}=\frac{\mathrm{s}.\mathrm{d}\text{prior}}{\sqrt{\mathrm{n}}}$$E21

$${\mathrm{SE}}_{\overline{\mathrm{x}}}.\text{posterior}=\frac{\mathrm{s}.\mathrm{d}\text{posterior}}{\sqrt{\mathrm{n}}}$$E22

3.5 Address the problem of prior data conflict for Weibull distribution

Although this problem is represented by the prior data conflict problem, we can use a model to address the prior data conflict problem, This is done through the use of a set of prior parameters and according to the proposal presented by (Quaeghebeur and Cooman; 2005) [11], In short ${\prod }^0=n^{0}x\left[{\underset{¯}{y}}^0,{\overline{y}}^0\right]$, Another proposal was submitted (Walter and Augustin; 2009) In order to obtain a set of prior parameters, in brief ${\prod }^0=\left[{\underset{¯}{n}}^0,{\overline{n}}^0\right]x\left[{\underset{¯}{y}}^0,{\overline{y}}^0\right]$, In general, the model presented to obtain a set of prior parameters is called (Generalized iLuck-Model), after that we get a set of posterior distributions, as shown below [3]:

$${\mathrm{f}}_1\left(\mathrm{\theta }/{\underset{¯}{\mathrm{n}}}^0,{\underset{¯}{\mathrm{y}}}^0\right)=\frac{{\left({\underset{¯}{\mathrm{n}}}^0{\underset{¯}{\mathrm{y}}}^0\right)}^{{\underset{¯}{\mathrm{n}}}^0+1}}{\mathrm{г}\left({\underset{¯}{\mathrm{n}}}^0+1\right)}{\mathrm{\theta }}^{-\left({\underset{¯}{\mathrm{n}}}^0+1\right)-1}{\mathrm{e}}^{-\frac{{\underset{¯}{\mathrm{n}}}^0{\underset{¯}{\mathrm{y}}}^0}{\mathrm{\theta }}}$$E23

$${\mathrm{f}}_2\left(\mathrm{\theta }/{\underset{¯}{\mathrm{n}}}^0,{\overline{\mathrm{y}}}^0\right)=\frac{{\left({\underset{¯}{\mathrm{n}}}^0{\overline{\mathrm{y}}}^0\right)}^{{\mathrm{n}}^0+1}}{\mathrm{г}\left({\underset{¯}{\mathrm{n}}}^0+1\right)}{\mathrm{\theta }}^{-\left({\underset{¯}{\mathrm{n}}}^0+1\right)-1}{\mathrm{e}}^{-\frac{{\underset{¯}{\mathrm{n}}}^0{\overline{\mathrm{y}}}^0}{\mathrm{\theta }}}$$E24

$${\mathrm{f}}_3\left(\mathrm{\theta }/{\overline{\mathrm{n}}}^0,{\underset{¯}{\mathrm{y}}}^0\right)=\frac{{\left({\overline{\mathrm{n}}}^0{\underset{¯}{\mathrm{y}}}^0\right)}^{{\overline{\mathrm{n}}}^0+1}}{\mathrm{г}\left({\overline{\mathrm{n}}}^0+1\right)}{\mathrm{\theta }}^{-\left({\overline{\mathrm{n}}}^0+1\right)-1}{\mathrm{e}}^{-\frac{{\overline{\mathrm{n}}}^0{\underset{¯}{\mathrm{y}}}^0}{\mathrm{\theta }}}$$E25

$${\mathrm{f}}_4\left(\mathrm{\theta }/,{\overline{\mathrm{n}}}^0,{\overline{\mathrm{y}}}^0\right)=\frac{{\left({\overline{\mathrm{n}}}^0{\overline{\mathrm{y}}}^0\right)}^{{\overline{\mathrm{n}}}^0+1}}{\mathrm{г}\left({\overline{\mathrm{n}}}^0+1\right)}{\mathrm{\theta }}^{-\left({\overline{\mathrm{n}}}^0+1\right)-1}{\mathrm{e}}^{-\frac{{\overline{\mathrm{n}}}^0{\overline{\mathrm{y}}}^0}{\mathrm{\theta }}}$$E26

Since:

${\underset{¯}{\mathrm{n}}}^0$:Minimum.

${\overline{\mathrm{n}}}^0$: Maximum.

${\underset{¯}{\mathrm{y}}}^0$: Minimum.

${\overline{\mathrm{y}}}^0$: Maximum.

The above equations represent a set of the prior distributions obtained through the iLuck-Model, after that we extract the posterior set of distributions according to the following steps:

The first posterior distribution: From Eq. (23) and by using the Bayes rule, we get the first posterior distribution, as in the following equation:

$${\mathrm{f}}_1\left(\mathrm{\theta }\backslash \mathrm{t}\right)=\frac{{\left({\underset{¯}{\mathrm{n}}}^0{\underset{¯}{\mathrm{y}}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}^{{\underset{¯}{\mathrm{n}}}^0+\mathrm{n}+1}}{\mathrm{г}\left({\underset{¯}{\mathrm{n}}}^0+\mathrm{n}+1\right)}{\mathrm{\theta }}^{-\left({\underset{¯}{\mathrm{n}}}^0+\mathrm{n}+1\right)-1}{\mathrm{e}}^{-\frac{\left({\underset{¯}{\mathrm{n}}}^0{\underset{¯}{\mathrm{y}}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}{\mathrm{\theta }}}$$E27

The above equation represents the first posterior distribution which is the Inverse Gamma distribution and by taking advantage of the properties of the Inverse Gamma distribution we get the central moments as in the following equation:

$${\mathrm{M}}_{\mathrm{r}}=\frac{{\left({\underset{¯}{\mathrm{n}}}^0{\underset{¯}{\mathrm{y}}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}^{\mathrm{r}}}{\mathrm{г}\left({\underset{¯}{\mathrm{n}}}^0+\mathrm{n}+1\right)}\mathrm{г}\left({\underset{¯}{\mathrm{n}}}^0+\mathrm{n}-\mathrm{r}+1\right)$$E28

The second posterior distribution: From Eq. (24) and by using the Bayes rule we get the second posterior distribution as in the following equation:

$${\mathrm{f}}_2\left(\mathrm{\theta }\backslash \mathrm{t}\right)=\frac{{\left({\underset{¯}{\mathrm{n}}}^0{\overline{\mathrm{y}}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}^{{\underset{¯}{\mathrm{n}}}^0+\mathrm{n}+1}}{\mathrm{г}\left({\underset{¯}{\mathrm{n}}}^0+\mathrm{n}+1\right)}{\mathrm{\theta }}^{-\left({\underset{¯}{\mathrm{n}}}^0+\mathrm{n}+1\right)-1}{\mathrm{e}}^{-\frac{\left({\underset{¯}{\mathrm{n}}}^0{\overline{\mathrm{y}}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}{\mathrm{\theta }}}$$E29

In short:

$${\mathrm{f}}_2\left(\mathrm{\theta }\backslash \mathrm{t}\right)\sim \mathrm{IG}\left({\underset{¯}{\mathrm{n}}}^0+\mathrm{n}+1,{\underset{¯}{\mathrm{n}}}^0{\overline{\mathrm{y}}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)$$

The above equation represents the second posterior distribution, which is the Inverse Gamma distribution, and by taking advantage of the properties of the Inverse Gamma distribution we get the central moments as in the following equation:

$${\mathrm{M}}_{\mathrm{r}}=\frac{{\left({\underset{¯}{\mathrm{n}}}^0{\overline{\mathrm{y}}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}^{\mathrm{r}}}{\mathrm{г}\left({\underset{¯}{\mathrm{n}}}^0+\mathrm{n}+1\right)}\mathrm{г}\left({\underset{¯}{\mathrm{n}}}^0+\mathrm{n}-\mathrm{r}+1\right)$$E30

The third posterior distribution: From Eq. (25) and by using the Bayes rule, we get the third posterior distribution as in the following equation:

$${\mathrm{f}}_3\left(\mathrm{\theta }\backslash \mathrm{t}\right)=\frac{{\left({\overline{\mathrm{n}}}^0{\underset{¯}{\mathrm{y}}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}^{{\overline{\mathrm{n}}}^0+\mathrm{n}+1}}{\mathrm{г}\left({\overline{\mathrm{n}}}^0+\mathrm{n}+1\right)}{\mathrm{\theta }}^{-\left({\underset{¯}{\mathrm{n}}}^0+\mathrm{n}+1\right)-1}{\mathrm{e}}^{-\frac{\left({\overline{\mathrm{n}}}^0{\underset{¯}{\mathrm{y}}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}{\mathrm{\theta }}}$$E31

$${\mathrm{f}}_3\left(\mathrm{\theta }\backslash \mathrm{t}\right)\sim \mathrm{IG}\left({\overline{\mathrm{n}}}^0+\mathrm{n}+1,{\overline{\mathrm{n}}}^0{\underset{¯}{\mathrm{y}}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)$$

The above equation represents the third posterior distribution which is the Inverse Gamma distribution and by taking advantage of the properties of the Inverse Gamma distribution we get the central moments as in the following equation:

$${\mathrm{M}}_{\mathrm{r}}=\frac{{\left({\overline{\mathrm{n}}}^0{\underset{¯}{\mathrm{y}}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}^{\mathrm{r}}}{\mathrm{г}\left({\overline{\mathrm{n}}}^0+\mathrm{n}+1\right)}\mathrm{г}\left({\overline{\mathrm{n}}}^0+\mathrm{n}-\mathrm{r}+1\right)$$E32

The fourth posterior distribution: From Eq. (26) and by using the Bayes rule we get the fourth posterior distribution as in the following equation:

$${\mathrm{f}}_4\left(\mathrm{\theta }\backslash \mathrm{t}\right)=\frac{{\left({\overline{\mathrm{n}}}^0{\overline{\mathrm{y}}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}^{{\overline{\mathrm{n}}}^0+\mathrm{n}+1}}{\mathrm{г}\left({\overline{\mathrm{n}}}^0+\mathrm{n}+1\right)}{\mathrm{\theta }}^{-\left({\underset{¯}{\mathrm{n}}}^0+\mathrm{n}+1\right)-1}{\mathrm{e}}^{-\frac{\left({\overline{\mathrm{n}}}^0{\overline{\mathrm{y}}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}{\mathrm{\theta }}}$$E33

$${\mathrm{f}}_4\left(\mathrm{\theta }\backslash \mathrm{t}\right)\sim \mathrm{IG}\left({\overline{\mathrm{n}}}^0+\mathrm{n}+1,{\overline{\mathrm{n}}}^0{\overline{\mathrm{y}}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)$$

The above equation represents the fourth posterior distribution which is the Inverse Gamma distribution and by taking advantage of the properties of the Inverse Gamma distribution we get the central moments as in the following equation:

$${\mathrm{M}}_{\mathrm{r}}=\frac{{\left({\overline{\mathrm{n}}}^0{\overline{\mathrm{y}}}^0+\mathrm{\tau }\left(\mathrm{t}\right)\right)}^{\mathrm{r}}}{\mathrm{г}\left({\overline{\mathrm{n}}}^0+\mathrm{n}+1\right)}\mathrm{г}\left({\overline{\mathrm{n}}}^0+\mathrm{n}-\mathrm{r}+1\right)$$E34

After taking the arithmetic mean of the posterior distributions we get the iLuck-Model [10]:

$${\underset{¯}{\mathrm{y}}}^{\mathrm{n}}=\text{lower}\left({\mathrm{y}}^{\mathrm{n}}\right)=\left\{\begin{array}{c}\frac{{\overline{\mathrm{n}}}^0{\underset{¯}{\mathrm{y}}}^0+\mathrm{\tau }\left(\mathrm{t}\right)}{{\overline{\mathrm{n}}}^0+\mathrm{n}}\text{if}\overline{\mathrm{\tau }}\left(\mathrm{t}\right)\ge {\underset{¯}{\mathrm{y}}}^0\\ \frac{{\underset{¯}{\mathrm{n}}}^0{\underset{¯}{\mathrm{y}}}^0+\mathrm{\tau }\left(\mathrm{t}\right)}{{\underset{¯}{\mathrm{n}}}^0+\mathrm{n}}\text{if}\overline{\mathrm{\tau }}\left(\mathrm{t}\right)<{\underset{¯}{\mathrm{y}}}^0\end{array}\right.$$E35

$${\overline{\mathrm{y}}}^{\mathrm{n}}=\text{uper}\left({\mathrm{y}}^{\mathrm{n}}\right)=\left\{\begin{array}{c}\frac{{\overline{\mathrm{n}}}^0{\overline{\mathrm{y}}}^0+\mathrm{\tau }\left(\mathrm{t}\right)}{{\overline{\mathrm{n}}}^0+\mathrm{n}}\text{if}\overline{\mathrm{\tau }}\left(\mathrm{t}\right)\le {\overline{\mathrm{y}}}^0\\ \frac{{\underset{¯}{\mathrm{n}}}^0{\overline{\mathrm{y}}}^0+\mathrm{\tau }\left(\mathrm{t}\right)}{{\underset{¯}{\mathrm{n}}}^0+\mathrm{n}}\text{if}\overline{\mathrm{\tau }}\left(\mathrm{t}\right)>{\overline{\mathrm{y}}}^0\end{array}\right.$$E36

Eqs. (35), (36) represent a generalized iLuck-Model, a model that represents the lower bound and the model that represents the upper bound is chosen based on the value of $\overline{\mathrm{\tau }}\left(\mathrm{t}\right)$ the estimator we obtain will be in the form of an interval. Therefore we will take the average for that period and from the above the posterior distribution will be in the following form:

$$\mathrm{f}\left(\mathrm{\theta }/{\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}}\right)=\frac{{\left({\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}}\right)}^{{\mathrm{n}}^{\mathrm{m}}+1}}{\mathrm{г}\left({\mathrm{n}}^{\mathrm{m}}+1\right)}{\mathrm{\theta }}^{-\left({\mathrm{n}}^{\mathrm{m}}+1\right)-1}{\mathrm{e}}^{-\frac{{\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}}}{\mathrm{\theta }}}$$E37

$${\mathrm{n}}^{\mathrm{m}}=\frac{\text{lower}\left({\mathrm{n}}^{\mathrm{n}}\right)+\text{uper}\left({\mathrm{n}}^{\mathrm{n}}\right)}{2},{\mathrm{y}}^{\mathrm{m}}=\frac{\text{lower}\left({\mathrm{y}}^{\mathrm{n}}\right)+\text{uper}\left({\mathrm{y}}^{\mathrm{n}}\right)}{2}$$

3.6 Robust Bayesian estimation for scale parameter for Weibull distribution

From Eq. (37) and by using the squared loss function, we get a Bayes estimator for the scale parameter as follows [12]:

$$\mathrm{E}\left(\mathrm{\theta }/{\mathrm{n}}^{\mathrm{m}},{\mathrm{y}}^{\mathrm{m}}\right)={\widehat{\mathrm{\theta }}}_{\mathrm{Rob}}$$

$${\widehat{\mathrm{\theta }}}_{\mathrm{Rob}}={\mathrm{y}}^{\mathrm{m}}$$E38

3.7 Robust Bayesian estimation for survival function for Weibull distribution

From Eq. (37) and by using the squared loss function we get a Bayes estimator for the survival function as follows [12]:

$$\widehat{\mathrm{S}}\left(\mathrm{t}\right)=\underset{0}{\overset{\infty }{\int }}S\left(\mathrm{t}\right)\mathrm{f}\left(\mathrm{\theta }/\mathrm{t}\right)\mathrm{d\theta }$$

$$\widehat{\mathrm{S}}\left(\mathrm{t}\right)=\frac{{\left({\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}}\right)}^{{\mathrm{n}}^{\mathrm{m}}+1}}{\mathrm{г}\left({\mathrm{n}}^{\mathrm{m}}+1\right)}\underset{0}{\overset{\infty }{\int }}{\mathrm{e}}^{-\frac{{\mathrm{t}}^{\mathrm{\beta }}}{\mathrm{\theta }}}{\mathrm{\theta }}^{-\left({\mathrm{n}}^{\mathrm{m}}+1\right)-1}{\mathrm{e}}^{-\frac{{\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}}}{\mathrm{\theta }}}\mathrm{d\theta }$$

$$\widehat{\mathrm{S}}\left(\mathrm{t}\right)=\frac{{\left({\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}}\right)}^{{\mathrm{n}}^{\mathrm{m}}+1}}{\mathrm{г}\left({\mathrm{n}}^{\mathrm{m}}+1\right)}\underset{0}{\overset{\infty }{\int }}{\mathrm{\theta }}^{-\left({\mathrm{n}}^{\mathrm{m}}+1\right)-1}{\mathrm{e}}^{-\frac{\left({\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}}+{\mathrm{t}}^{\mathrm{\beta }}\right)}{\mathrm{\theta }}}\mathrm{d\theta }$$

By using the transformation:

$$\mathrm{let}\mathrm{z}=\frac{\left({\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}}+{\mathrm{t}}^{\mathrm{\beta }}\right)}{\mathrm{\theta }}⟹\mathrm{\theta }=\frac{\left({\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}}+{\mathrm{t}}^{\mathrm{\beta }}\right)}{\mathrm{z}};\left|\mathrm{J}\right|=\mid -\frac{\left({\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}}+{\mathrm{t}}^{\mathrm{\beta }}\right)}{{\mathrm{z}}^2}\mid$$

$$\widehat{\mathrm{S}}\left(\mathrm{t}\right)=\frac{{\left({\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}}\right)}^{{\mathrm{n}}^{\mathrm{m}}+1}}{\mathrm{г}\left({\mathrm{n}}^{\mathrm{m}}+1\right)}\underset{0}{\overset{\infty }{\int }}{\left(\frac{\left({\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}}+{\mathrm{t}}^{\mathrm{\beta }}\right)}{\mathrm{z}}\right)}^{-\left({\mathrm{n}}^{\mathrm{m}}+1\right)-1}{\mathrm{e}}^{-\mathrm{z}}\frac{\left({\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}}+{\mathrm{t}}^{\mathrm{\beta }}\right)}{{\mathrm{z}}^2}\mathrm{dz}$$

$$\widehat{\mathrm{S}}\left(\mathrm{t}\right)=\frac{{\left({\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}}\right)}^{{\mathrm{n}}^{\mathrm{m}}+1}}{\mathrm{г}\left({\mathrm{n}}^{\mathrm{m}}+1\right)}{\left({\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}}+{\mathrm{t}}^{\mathrm{\beta }}\right)}^{-\left({\mathrm{n}}^{\mathrm{m}}+1\right)}\underset{0}{\overset{\infty }{\int }}{\left(\mathrm{z}\right)}^{{\mathrm{n}}^{\mathrm{m}}}{\mathrm{e}}^{-\mathrm{z}}\mathrm{dz}$$

$${\widehat{\mathrm{S}}}_{\mathrm{Rob}}\left(\mathrm{t}\right)={\left(\frac{{\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}}}{{\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}}+{\mathrm{t}}^{\mathrm{\beta }}}\right)}^{{\mathrm{n}}^{\mathrm{m}}+1}$$E39

4. The experimental side of the Weibull distribution

In this section, Weibull distribution data will be generated by using the R program to estimate the scale parameter and the survival function in the prior data conflict, as shown in the following Tables 1–4:

Model	β = 2				${\mathrm{n}}^0$		Best
	${\mathrm{y}}^0$		Lower		Upper
	Lower	Upper	2		5
			$\mathrm{\theta }$	n	$\widehat{\mathrm{pdf}}$	${\widehat{\mathrm{pdf}}}_{\mathrm{rob}}$
1	1.5	2	1.5	10	0.008738	0.004018	Robust Bayesian
				20	0.004564	0.002699
				40	0.002184	0.001634
2	2	3	2	10	0.006128	0.002333
				20	0.003421	0.001845
				40	0.001539	0.001094
3	2.5	4	2.5	10	0.004801	0.001671
				20	0.002811	0.001428
				40	0.001295	0.000895

Table 1.

Integrated mean square error (IMSE) of the probability density function (pdf) for the Weibull distribution in the case of prior data conflict.

Model	$\mathrm{\beta }=2$			${\mathrm{n}}^0$
				Lower		upper
				2		5
	${\mathrm{y}}^0$		$\mathrm{\theta }$	n	$\widehat{\mathrm{S}}\left(\mathrm{t}\right)$	${\widehat{\mathbf{S}}}_{\mathbf{rob}}\left(\mathbf{t}\right)$	Best
	Lower	Upper
1	1.5	2	1.5	10	0.005332	0.002705	Robust Bayesian
				20	0.002894	0.001948
				40	0.001570	0.001263
2	2	3	2	10	0.004891	0.002070
				20	0.002916	0.001729
				40	0.001462	0.001120
3	2.5	4	2.5	10	0.005310	0.002075
				20	0.002802	0.001591
				40	0.001412	0.001050

Table 2.

Integrated mean square error (IMSE) of the survival function for the Weibull distribution in the case of prior data conflict.

B = 2		${\mathrm{n}}^0$
${\mathbf{y}}^{\mathbf{0}}$		Lower	Upper
		6	8
lower	upper	${\widehat{\mathbf{\theta }}}_{\mathbf{rob}}$	Standard deviation prior	Standard deviation posterior
16	20	16.10352	280.4339	242.4159

Table 3.

Estimation of the scale parameter of the Weibull distribution in the case of prior data conflict.

B = 2		${\mathrm{n}}^0$
${\mathbf{y}}^{\mathbf{0}}$		Lower	Upper
		6	8
lower	upper	t	${\widehat{\mathbf{S}}}_{\mathbf{rob}}\left(\mathbf{t}\right)$
16	20	1	0.937227322
		2	0.772418759
		3	0.561586411
		7	0.050817125
		10	0.003284835

Table 4.

Estimation of the survival function of the Weibull distribution in the case of prior data conflict.

Through Tables 1 and 2 the simulation results showed that the robust Bayesian estimator is the best through the comparison standard (IMSE) in the case of prior data conflict. From the above, the robust Bayesian estimator will be applied to the real data to estimate the scale parameter and the survival function.

5. Real data description for Weibull distribution

Real data of a size of (15) for heart attack patients were collected from Al-Manathira General Hospital of the Najaf Health Department for the year 2018, as the time of admission of the patient to the hospital until discharge was recorded and that all of them were in a state of death upon discharge. This data is complete data, ti = (2,1,1,1,1,2,1,3,7,1,2,10,7,1,1), as these times are in days.

6. Binomial distribution

It is one of the discrete distributions in which the experiment can be repeated for (n) times so that the probability of success is (p) and the probability of failure is (1-p), so that the probability density function has the following formula [13]:

$$\mathrm{f}\left(\mathrm{x}\right)=\left(\genfrac{}{}{0pt}{}{\mathrm{n}}{\mathrm{x}}\right){\mathrm{p}}^{\mathrm{x}}{\left(1-\mathrm{p}\right)}^{\mathrm{n}-\mathrm{x}},\mathrm{x}=0,1,\dots ,\mathrm{n}$$E40

and the average is:

$$\mathrm{E}\left(\mathrm{x}\right)=\mathrm{np}$$E41

The variance is:

$${\mathrm{\sigma }}^2=\mathrm{np}\left(1-\mathrm{p}\right)$$E42

6.1 Bayesian estimation for parameter p for binomial distribution

Suppose we have a sample that follows the Binomial distribution shown in the Eq. (40) and the appropriate prior distribution is the Beta distribution according to the following formula [13]:

$$\mathrm{f}\left(\mathrm{p}/\mathrm{a},\mathrm{b}\right)=\frac{1}{\mathrm{\beta }\left(\mathrm{\alpha },\mathrm{\beta }\right)}{\mathrm{p}}^{\mathrm{\alpha }-1}{\left(1-\mathrm{p}\right)}^{\mathrm{\beta }-1};0<p<1$$E43

To obtain the posterior distribution according to the Bayes rule as follows:

$$\mathrm{f}\left(\mathrm{p}/\mathrm{x}\right)=\frac{\mathrm{f}\left({\mathrm{x}}_1,{\mathrm{x}}_2,\dots ,{\mathrm{x}}_{\mathrm{n}}/\mathrm{p}\right)\mathrm{f}\left(\mathrm{p}/\mathrm{\alpha },\mathrm{\beta }\right)}{\underset{0}{\overset{1}{\int }}\mathrm{f}\left({\mathrm{x}}_1,{\mathrm{x}}_2,\dots ,{\mathrm{x}}_{\mathrm{n}}/\mathrm{p}\right)\mathrm{f}\left(\mathrm{p}/\mathrm{\alpha },\mathrm{\beta }\right)\mathrm{dp}}$$

$$=\frac{\left(\genfrac{}{}{0pt}{}{\mathrm{n}}{{\sum }_{\mathrm{i}=1}^{\mathrm{n}}\mathrm{x}}\right){\mathrm{p}}^{{\sum }_{\mathrm{i}=1}^{\mathrm{n}}\mathrm{x}}{\left(1-\mathrm{p}\right)}^{\mathrm{n}-{\sum }_{\mathrm{i}=1}^{\mathrm{n}}\mathrm{x}}\frac{1}{\mathrm{\beta }\left(\mathrm{\alpha },\mathrm{\beta }\right)}{\mathrm{p}}^{\mathrm{\alpha }-1}{\left(1-\mathrm{p}\right)}^{\mathrm{\beta }-1}}{{\int }_0^1\left(\genfrac{}{}{0pt}{}{\mathrm{n}}{{\sum }_{\mathrm{i}=1}^{\mathrm{n}}\mathrm{x}}\right){\mathrm{p}}^{{\sum }_{\mathrm{i}=1}^{\mathrm{n}}\mathrm{x}}{\left(1-\mathrm{p}\right)}^{\mathrm{n}-{\sum }_{\mathrm{i}=1}^{\mathrm{n}}\mathrm{x}}\frac{1}{\mathrm{\beta }\left(\mathrm{\alpha },\mathrm{\beta }\right)}{\mathrm{p}}^{\mathrm{\alpha }-1}{\left(1-\mathrm{p}\right)}^{\mathrm{\beta }-1}\mathrm{dp}}$$

$$=\frac{{\mathrm{p}}^{{\sum }_{\mathrm{i}=1}^{\mathrm{n}}\mathrm{x}+\mathrm{\alpha }-1}{\left(1-\mathrm{p}\right)}^{\mathrm{n}-{\sum }_{\mathrm{i}=1}^{\mathrm{n}}\mathrm{x}+\mathrm{\beta }-1}}{{\int }_0^1{\mathrm{p}}^{{\sum }_{\mathrm{i}=1}^{\mathrm{n}}\mathrm{x}+\mathrm{\alpha }-1}{\left(1-\mathrm{p}\right)}^{\mathrm{n}-{\sum }_{\mathrm{i}=1}^{\mathrm{n}}\mathrm{x}+\mathrm{\beta }-1}\mathrm{dp}}$$

Since:

$$\sum _{\mathrm{i}=1}^{\mathrm{n}}\mathrm{x}=s$$

$$=\frac{1}{\mathrm{\beta }\left(\mathrm{\alpha }+\mathrm{s},\mathrm{n}+\mathrm{\beta }-\mathrm{s}\right)}{\mathrm{p}}^{\mathrm{\alpha }+\mathrm{s}-1}{\left(1-\mathrm{p}\right)}^{\mathrm{n}+\mathrm{\beta }-\mathrm{s}-1}$$E44

From Eq. (44) and by using the squared loss function we get a Bayes estimator for parameter (p) as follows:

$$\mathrm{E}\left(\mathrm{p}/\mathrm{s}\right)=\widehat{\mathrm{p}}$$

$$\widehat{\mathrm{p}}=\frac{\mathrm{\alpha }+\mathrm{s}}{\mathrm{n}+\mathrm{\alpha }+\mathrm{\beta }}$$E45

6.2 Bayesian estimation for survival function for binomial distribution

Through Eq. (44) and by using the squared loss function we get a Bayes estimator for the survival function as follows [11] (Figures 1 and 2):

Figure 1.

It shows the behavior of the survival function by using the robust Bayes estimator, which is decreasing as the value of (t) increases, and this is consistent with the statistical theory [11].

Figure 2.

It shows the behavior of the survival function of a binomial distribution which is decreasing and this is consistent with the statistical theory [13].

$$\widehat{\mathrm{S}}\left(\mathrm{t}\right)=\int \mathrm{S}\left(\mathrm{t}\right)\mathrm{f}\left(\mathrm{p}/\mathrm{s}\right)\mathrm{dp}$$

$$\mathrm{f}\left(\mathrm{p}/\mathrm{s}\right)=\frac{1}{\mathrm{\beta }\left(\mathrm{\alpha }+\mathrm{s},\mathrm{n}+\mathrm{\beta }-\mathrm{s}\right)}{\mathrm{p}}^{\mathrm{\alpha }+\mathrm{s}-1}{\left(1-\mathrm{p}\right)}^{\mathrm{n}+\mathrm{\beta }-\mathrm{s}-1}$$

$$\mathrm{S}\left(\mathrm{t}\right)=\sum _{\mathrm{j}=\mathrm{t}}^{\mathrm{n}}\left(\genfrac{}{}{0pt}{}{n}{\mathrm{j}}\right){\mathrm{p}}^{\mathrm{j}}{\left(1-\mathrm{p}\right)}^{\mathrm{n}-\mathrm{j}}$$

$$\widehat{\mathrm{S}}\left(\mathrm{t}\right)=\frac{1}{\mathrm{\beta }\left(\mathrm{\alpha }+\mathrm{s},\mathrm{n}+\mathrm{\beta }-\mathrm{s}\right)}\sum _{\mathrm{j}=\mathrm{t}}^{\mathrm{n}}\frac{\mathrm{n}!}{\mathrm{j}!\left(\mathrm{n}-\mathrm{j}\right)!}\underset{0}{\overset{1}{\int }}{\mathrm{p}}^{\mathrm{\alpha }+\mathrm{s}+\mathrm{j}-1}{\left(1-\mathrm{p}\right)}^{\mathrm{n}+\mathrm{\beta }-\mathrm{s}-1+\mathrm{n}-\mathrm{j}}\mathrm{dp}$$

$$\widehat{\mathrm{S}}\left(\mathrm{t}\right)=\frac{1}{\mathrm{\beta }\left(\mathrm{\alpha }+\mathrm{s},\mathrm{n}+\mathrm{\beta }-\mathrm{s}\right)}\sum _{\mathrm{j}=\mathrm{t}}^{\mathrm{n}}\frac{\mathrm{n}!}{\mathrm{j}!\left(\mathrm{n}-\mathrm{j}\right)!}\underset{0}{\overset{1}{\int }}{\mathrm{p}}^{\mathrm{\alpha }+\mathrm{s}+\mathrm{j}-1}{\left(1-\mathrm{p}\right)}^{2\mathrm{n}+\mathrm{\beta }-\mathrm{s}-\mathrm{j}-1}\mathrm{dp}$$

$$\widehat{\mathrm{S}}\left(\mathrm{t}\right)=\frac{1}{\mathrm{\beta }\left(\mathrm{\alpha }+\mathrm{s},\mathrm{n}+\mathrm{\beta }-\mathrm{s}\right)}\sum _{\mathrm{j}=\mathrm{t}}^{\mathrm{n}}\frac{\mathrm{n}!}{\mathrm{j}!\left(\mathrm{n}-\mathrm{j}\right)!}\underset{0}{\overset{1}{\int }}{\mathrm{p}}^{\mathrm{\alpha }+\mathrm{s}+\mathrm{j}-1}{\left(1-\mathrm{p}\right)}^{2\mathrm{n}+\mathrm{\beta }-\mathrm{s}-\mathrm{j}-1}\mathrm{dp}$$

Multiply and divide by:

$$\mathrm{\beta }\left(\mathrm{\alpha }+\mathrm{s}+\mathrm{j},2\mathrm{n}+\mathrm{\beta }-\mathrm{s}-\mathrm{j}\right)$$

$$\hat{\mathrm{S}}\left(\mathrm{t}\right)=\begin{array}{l}\frac{1}{\mathrm{\beta }\left(\mathrm{\alpha }+\mathrm{s},\mathrm{n}+\mathrm{\beta }-\mathrm{s}\right)}\sum _{\mathrm{j}=\mathrm{t}}^{\mathrm{n}}\frac{\mathrm{n}!}{\mathrm{j}!\left(\mathrm{n}-\mathrm{j}\right)!}\frac{\mathrm{\beta }\left(\mathrm{\alpha }+\mathrm{s}+\mathrm{j},2\mathrm{n}+\mathrm{\beta }-\mathrm{s}-\mathrm{j}\right)}{\mathrm{\beta }\left(\mathrm{\alpha }+\mathrm{s}+\mathrm{j},2\mathrm{n}+\mathrm{\beta }-\mathrm{s}-\mathrm{j}\right)}\underset{0}{\overset{1}{\int }}{\mathrm{p}}^{\mathrm{\alpha }+\mathrm{s}+\mathrm{j}-1}(1\\ {-\mathrm{p})}^{2\mathrm{n}+\mathrm{\beta }-\mathrm{s}-\mathrm{j}-1}\mathrm{dp}\end{array}$$

$$\widehat{\mathrm{S}}\left(\mathrm{t}\right)=\frac{1}{\mathrm{\beta }\left(\mathrm{\alpha }+\mathrm{s},\mathrm{n}+\mathrm{\beta }-\mathrm{s}\right)}\sum _{\mathrm{j}=\mathrm{t}}^{\mathrm{n}}\frac{\mathrm{n}!}{\mathrm{j}!\left(\mathrm{n}-\mathrm{j}\right)!}\mathrm{\beta }\left(\mathrm{\alpha }+\mathrm{s}+\mathrm{j},2\mathrm{n}+\mathrm{\beta }-\mathrm{s}-\mathrm{j}\right)$$E46

6.3 Checking of prior data conflict for binomial distribution

Suppose that we have a sample that follows the binomial distribution and in short x ∼ bin (n,p) and as shown in the Eq. (40) then we need to determine the prior distribution through two methods [14]:

The first method: It is the Expected Conditional method we have previously explained this method so we go through the following steps [9]:

$$\mathrm{E}\left(\mathrm{p}/\mathrm{\alpha },\mathrm{\beta }\right)={\mathrm{y}}^0=\frac{\mathrm{\alpha }}{\mathrm{\alpha }+\mathrm{\beta }}$$

$${\mathrm{y}}^0=\frac{\mathrm{\alpha }}{{\mathrm{n}}^0},\text{where}{\mathrm{n}}^0=\mathrm{\alpha }+\mathrm{\beta }$$

$${\mathrm{y}}^0=\frac{\mathrm{\alpha }}{{\mathrm{n}}^0}\Rightarrow \mathrm{\alpha }={\mathrm{n}}^0{\mathrm{y}}^0,{\mathrm{n}}^0={\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{\beta }\Rightarrow \mathrm{\beta }={\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)$$

Then we substitute the parameters $\mathrm{\alpha }={\mathrm{n}}^0{\mathrm{y}}^0$, $\mathrm{\beta }={\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)$ with the prior distribution to get the prior distribution with the updated parameters:

$$\mathrm{f}\left(\mathrm{p}/{\mathrm{n}}^0,{\mathrm{y}}^0\right)=\frac{1}{\mathrm{\beta }\left({\mathrm{n}}^0{\mathrm{y}}^0,{\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)\right)}{\mathrm{p}}^{{\mathrm{n}}^0{\mathrm{y}}^0-1}{\left(1-\mathrm{p}\right)}^{{\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)-1}$$E47

The second method: In this method the prior distribution can be determined with the updated parameters through two steps, which are as follows [10]:

The first step: If the model can be written in the form of canonical exponential family, as shown below:

$$\begin{array}{l}\mathrm{f}\left(\mathrm{x}/\mathrm{p}\right)=\mathrm{a}\left(\mathrm{x}\right)exp(\mathrm{\psi }.\mathrm{\tau }\left(\mathrm{x}\right)-\mathrm{nb}\left(\mathrm{\psi }\right))\\ \mathrm{f}\left(\mathrm{s}/\mathrm{p}\right)=\left(\genfrac{}{}{0pt}{}{\mathrm{n}}{\mathrm{s}}\right){\mathrm{p}}^{\mathrm{s}}{\left(1-\mathrm{p}\right)}^{\mathrm{n}-\mathrm{s}}\\ =\left(\genfrac{}{}{0pt}{}{\mathrm{n}}{\mathrm{s}}\right)exp\left\{ln\left(\frac{\mathrm{p}}{1-\mathrm{p}}\right)\mathrm{s}-\mathrm{n}\left(-ln\left(1-\mathrm{p}\right)\right)\right\}\\ \mathrm{a}\left(\mathrm{x}\right)=\left(\genfrac{}{}{0pt}{}{\mathrm{n}}{\mathrm{s}}\right),\mathrm{\psi }=ln\left(\frac{\mathrm{p}}{1-\mathrm{p}}\right),\mathrm{\tau }\left(\mathrm{x}\right)=\mathrm{s},\mathrm{b}\left(\mathrm{\psi }\right)=-ln\left(1-\mathrm{p}\right)\end{array}$$E48

The second step: the prior distribution can be built by using the following model:

$$\mathrm{f}\left(\mathrm{\psi }/{\mathrm{n}}^0,{\mathrm{y}}^0\right)\mathrm{d\psi }\propto exp\left\{{\mathrm{n}}^0\left[{\mathrm{y}}^0.\mathrm{\psi }-\mathrm{b}\left(\mathrm{\psi }\right)\right]\right\}\mathrm{d\psi }$$

$$\mathrm{f}\left(\mathrm{\psi }/{\mathrm{n}}^0,{\mathrm{y}}^0\right)\mathrm{d\psi }\propto exp\left\{{\mathrm{n}}^0\left[{\mathrm{y}}^{0}ln\left(\frac{\mathrm{p}}{1-\mathrm{p}}\right)+ln\left(1-\mathrm{p}\right)\right]\right\}\mathrm{d\psi }$$

$$\left|\frac{\mathrm{d\psi }}{\mathrm{dp}}\right|=\frac{1}{\mathrm{p}\left(1-\mathrm{p}\right)}$$

$$\begin{array}{rcl}\mathrm{f}\left(\mathrm{p}/{\mathrm{n}}^0,{\mathrm{y}}^0\right)\mathrm{dp}& =& \mathrm{f}\left(\mathrm{\psi }/{\mathrm{n}}^0,{\mathrm{y}}^0\right)\left|\frac{\mathrm{d\psi }}{\mathrm{dp}}\right|\mathrm{dp}\\ & \propto & exp\left\{{\mathrm{n}}^0{\mathrm{y}}^{0}ln\left(\mathrm{p}\right)+\left({\mathrm{n}}^0-{\mathrm{n}}^0{\mathrm{y}}^0)ln(1-\mathrm{p}\right)\right\}\frac{1}{\mathrm{p}\left(1-\mathrm{p}\right)}\mathrm{dp}\end{array}$$

Then the prior distribution with the updated parameters is:

$$\mathrm{f}\left(\mathrm{p}/{\mathrm{n}}^0,{\mathrm{y}}^0\right)=\frac{1}{\mathrm{\beta }\left({\mathrm{n}}^0{\mathrm{y}}^0,{\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)\right)}{\mathrm{p}}^{{\mathrm{n}}^0{\mathrm{y}}^0-1}{\left(1-\mathrm{p}\right)}^{{\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)-1}$$E49

From the above equation, we extract the standard deviation of the prior distribution, as follows:

$${\mathrm{M}}_{\mathrm{r}}=\frac{1}{\mathrm{\beta }\left({\mathrm{n}}^0{\mathrm{y}}^0,{\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)\right)}\underset{0}{\overset{1}{\int }}{\mathrm{p}}^{{\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{r}-1}{\left(1-\mathrm{p}\right)}^{{\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)-1}\mathrm{dp}$$

$$=\frac{1}{\mathrm{\beta }\left({\mathrm{n}}^0{\mathrm{y}}^0,{\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)\right)}\mathrm{\beta }\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{r},{\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)\right)$$

$$=\frac{\mathrm{г}\left({\mathrm{n}}^0{\mathrm{y}}^0+{\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)\right)}{\mathrm{г}\left({\mathrm{n}}^0{\mathrm{y}}^0\right)\mathrm{г}\left({\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)\right)}\ast \frac{\mathrm{г}\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{r}\right)\mathrm{г}\left({\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)\right)}{\mathrm{г}\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{r}+{\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)\right)}$$

$$=\frac{\mathrm{г}\left({\mathrm{n}}^0\right)\mathrm{г}\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{r}\right)}{\mathrm{г}\left({\mathrm{n}}^0{\mathrm{y}}^0\right)\mathrm{г}\left({\mathrm{n}}^0+\mathrm{r}\right)}$$E50

$$\mathrm{s}.\mathrm{d}\text{prior}=\sqrt{\frac{{\mathrm{y}}^0\left(1-{\mathrm{y}}^0\right)}{{\mathrm{n}}^0+1}}$$E51

From Eq. (47) and by using the Bayes rule, we extract the posterior distribution as shown in the following steps:

$$\mathrm{f}\left(\mathrm{p}/\mathrm{s}\right)=\frac{\left(\genfrac{}{}{0pt}{}{\mathrm{n}}{\mathrm{s}}\right){\mathrm{p}}^{\mathrm{s}}{\left(1-\mathrm{p}\right)}^{\mathrm{n}-\mathrm{s}}\frac{1}{\mathrm{\beta }\left({\mathrm{n}}^0{\mathrm{y}}^0,{\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)\right)}{\mathrm{p}}^{{\mathrm{n}}^0{\mathrm{y}}^0-1}{\left(1-\mathrm{p}\right)}^{{\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)-1}}{\underset{0}{\overset{1}{\int }}\left(\genfrac{}{}{0pt}{}{\mathrm{n}}{\mathrm{s}}\right){\mathrm{p}}^{\mathrm{s}}{\left(1-\mathrm{p}\right)}^{\mathrm{n}-\mathrm{s}}\frac{1}{\mathrm{\beta }\left({\mathrm{n}}^0{\mathrm{y}}^0,{\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)\right)}{\mathrm{p}}^{{\mathrm{n}}^0{\mathrm{y}}^0-1}{\left(1-\mathrm{p}\right)}^{{\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)-1}\mathrm{dp}}$$

$$=\frac{{\mathrm{p}}^{{\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{s}-1}{\left(1-\mathrm{p}\right)}^{{\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)+\mathrm{n}-\mathrm{s}-1}}{\underset{0}{\overset{1}{\int }}{\mathrm{p}}^{{\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{s}-1}{\left(1-\mathrm{p}\right)}^{{\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)+\mathrm{n}-\mathrm{s}-1}\mathrm{dp}}$$

$$=\frac{1}{\mathrm{\beta }\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{s},{\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)+\mathrm{n}-\mathrm{s}\right)}{\mathrm{p}}^{{\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{s}-1}{\left(1-\mathrm{p}\right)}^{{\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)+\mathrm{n}-\mathrm{s}-1}$$E52

Eq. (52) represents the posterior distribution, after that we extract the standard deviation of the posterior distribution as follows:

$${\mathrm{M}}_{\mathrm{r}}=\frac{1}{\mathrm{\beta }\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{s},{\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)+\mathrm{n}-\mathrm{s}\right)}\underset{0}{\overset{1}{\int }}{\mathrm{p}}^{{\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{s}+\mathrm{r}-1}{\left(1-\mathrm{p}\right)}^{{\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)+\mathrm{n}-\mathrm{s}-1}\mathrm{dp}$$

$$=\frac{1}{\mathrm{\beta }\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{s},{\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)+\mathrm{n}-\mathrm{s}\right)}\mathrm{\beta }\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{s}+\mathrm{r},{\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)+\mathrm{n}-\mathrm{s}\right)$$

$$=\frac{\mathrm{г}\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{s}+{\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)+\mathrm{n}-\mathrm{s}\right)}{\mathrm{г}\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{s}\right)\mathrm{г}\left({\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)+\mathrm{n}-\mathrm{s}\right)}\ast \frac{\mathrm{г}\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{s}+\mathrm{r}\right)\mathrm{г}\left({\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)+\mathrm{n}-\mathrm{s}\right)}{\mathrm{г}\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{s}+\mathrm{r}+{\mathrm{n}}^0\left(1-{\mathrm{y}}^0\right)+\mathrm{n}-\mathrm{s}\right)}$$

$$=\frac{\mathrm{г}\left({\mathrm{n}}^0+\mathrm{n}\right)\mathrm{г}\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{s}+\mathrm{r}\right)}{\mathrm{г}\left({\mathrm{n}}^0+\mathrm{n}+\mathrm{r}\right)\mathrm{г}\left({\mathrm{n}}^0{\mathrm{y}}^0+\mathrm{s}\right)}$$E53

Since

$$\mathrm{f}\left(\mathrm{p}/\mathrm{s}\right)=\mathrm{f}\left(\mathrm{p}/{\mathrm{n}}^{\mathrm{n}},{\mathrm{y}}^{\mathrm{n}}\right)$$

Then the standard deviation of the posterior distribution is according to the following formula:

$$\mathrm{s}.\mathrm{d}\text{posterior}=\sqrt{\frac{{\mathrm{y}}^{\mathrm{n}}\left(1-{\mathrm{y}}^{\mathrm{n}}\right)}{{\mathrm{n}}^{\mathrm{n}}+1}}$$E54

6.4 Address the problem of prior data conflict for binomial distribution

In the part on the distribution of Weibull, how to solve this problem was explained, so we will enter the following steps [3]:

$${\mathrm{f}}_1\left(\mathrm{p}/\mathrm{s}\right)=\frac{1}{\mathrm{\beta }\left({\underset{¯}{\mathrm{n}}}^0{\underset{¯}{\mathrm{y}}}^0+\mathrm{s},{\underset{¯}{\mathrm{n}}}^0\left(1-{\underset{¯}{\mathrm{y}}}^0\right)+\mathrm{n}-\mathrm{s}\right)}{\mathrm{p}}^{{\underset{¯}{\mathrm{n}}}^0{\underset{¯}{\mathrm{y}}}^0+\mathrm{s}-1}{\left(1-\mathrm{p}\right)}^{{\underset{¯}{\mathrm{n}}}^0\left(1-{\underset{¯}{\mathrm{y}}}^0\right)+\mathrm{n}-\mathrm{s}-1}$$E55

$${\mathrm{f}}_2\left(\mathrm{p}/\mathrm{s}\right)=\frac{1}{\mathrm{\beta }\left({\underset{¯}{\mathrm{n}}}^0{\overline{\mathrm{y}}}^0+\mathrm{s},{\underset{¯}{\mathrm{n}}}^0\left(1-{\overline{\mathrm{y}}}^0\right)+\mathrm{n}-\mathrm{s}\right)}{\mathrm{p}}^{{\underset{¯}{\mathrm{n}}}^0{\overline{\mathrm{y}}}^0+\mathrm{s}-1}{\left(1-\mathrm{p}\right)}^{{\underset{¯}{\mathrm{n}}}^0\left(1-{\overline{\mathrm{y}}}^0\right)+\mathrm{n}-\mathrm{s}-1}$$E56

$${\mathrm{f}}_3\left(\mathrm{p}/\mathrm{s}\right)=\frac{1}{\mathrm{\beta }\left({\overline{\mathrm{n}}}^0{\underset{¯}{\mathrm{y}}}^0+\mathrm{s},{\overline{\mathrm{n}}}^0\left(1-{\underset{¯}{\mathrm{y}}}^0\right)+\mathrm{n}-\mathrm{s}\right)}{\mathrm{p}}^{{\overline{\mathrm{n}}}^0{\underset{¯}{\mathrm{y}}}^0+\mathrm{s}-1}{\left(1-\mathrm{p}\right)}^{{\overline{\mathrm{n}}}^0\left(1-{\underset{¯}{\mathrm{y}}}^0\right)+\mathrm{n}-\mathrm{s}-1}$$E57

$${\mathrm{f}}_4\left(\mathrm{p}/\mathrm{s}\right)=\frac{1}{\mathrm{\beta }\left({\overline{\mathrm{n}}}^0{\overline{\mathrm{y}}}^0+\mathrm{s},{\overline{\mathrm{n}}}^0\left(1-{\overline{\mathrm{y}}}^0\right)+\mathrm{n}-\mathrm{s}\right)}{\mathrm{p}}^{{\overline{\mathrm{n}}}^0{\overline{\mathrm{y}}}^0+\mathrm{s}-1}{\left(1-\mathrm{p}\right)}^{{\overline{\mathrm{n}}}^0\left(1-{\overline{\mathrm{y}}}^0\right)+\mathrm{n}-\mathrm{s}-1}$$E58

The first posterior distribution: From Eq. (55) and by using the Bayes rule, we get the posterior distribution, as in the following equation:

$${\mathrm{f}}_1\left(\mathrm{p}/\mathrm{s}\right)=\frac{1}{\mathrm{\beta }\left({\underset{¯}{\mathrm{n}}}^0{\underset{¯}{\mathrm{y}}}^0+\mathrm{s},{\underset{¯}{\mathrm{n}}}^0\left(1-{\underset{¯}{\mathrm{y}}}^0\right)+\mathrm{n}-\mathrm{s}\right)}{\mathrm{p}}^{{\underset{¯}{\mathrm{n}}}^0{\underset{¯}{\mathrm{y}}}^0+\mathrm{s}-1}{\left(1-\mathrm{p}\right)}^{{\underset{¯}{\mathrm{n}}}^0\left(1-{\underset{¯}{\mathrm{y}}}^0\right)+\mathrm{n}-\mathrm{s}-1}$$E59

The above equation represents the first posterior distribution which is the Beta distribution and by using the properties of the Beta distribution we get the central moments, as in the following equation:

$${\mathrm{M}}_r=\frac{\mathrm{г}\left({\underset{¯}{\mathrm{n}}}^0+\mathrm{n}\right)\mathrm{г}\left({\underset{¯}{\mathrm{n}}}^0{\underset{¯}{\mathrm{y}}}^0+\mathrm{s}+\mathrm{r}\right)}{\mathrm{г}\left({\underset{¯}{\mathrm{n}}}^0+\mathrm{n}+\mathrm{r}\right)\mathrm{г}\left({\underset{¯}{\mathrm{n}}}^0{\underset{¯}{\mathrm{y}}}^0+\mathrm{s}\right)}$$E60

The second posterior distribution: From Eq. (56) and by using the Bayes rule, we get the second posterior distribution, as in the following equation:

$${\mathrm{f}}_2\left(\mathrm{p}/\mathrm{s}\right)=\frac{1}{\mathrm{\beta }\left({\underset{¯}{\mathrm{n}}}^0{\overline{\mathrm{y}}}^0+\mathrm{s},{\underset{¯}{\mathrm{n}}}^0\left(1-{\overline{\mathrm{y}}}^0\right)+\mathrm{n}-\mathrm{s}\right)}{\mathrm{p}}^{{\underset{¯}{\mathrm{n}}}^0{\overline{\mathrm{y}}}^0+\mathrm{s}-1}{\left(1-\mathrm{p}\right)}^{{\underset{¯}{\mathrm{n}}}^0\left(1-{\overline{\mathrm{y}}}^0\right)+\mathrm{n}-\mathrm{s}-1}$$E61

The above equation represents the second posterior distribution, which is the Beta distribution, and by using the properties of the Beta distribution, we get the central moments, as in the following equation:

$${\mathrm{M}}_{\mathrm{r}}=\frac{\mathrm{г}\left({\underset{¯}{\mathrm{n}}}^0+\mathrm{n}\right)\mathrm{г}\left({\underset{¯}{\mathrm{n}}}^0{\overline{\mathrm{y}}}^0+\mathrm{s}+\mathrm{r}\right)}{\mathrm{г}\left({\underset{¯}{\mathrm{n}}}^0+\mathrm{n}+\mathrm{r}\right)\mathrm{г}\left({\underset{¯}{\mathrm{n}}}^0{\overline{\mathrm{y}}}^0+\mathrm{s}\right)}$$E62

The third posterior distribution: From Eq. (57) and by using the Bayes rule, we get the third posterior distribution, as in the following equation:

$${\mathrm{f}}_3\left(\mathrm{p}/\mathrm{s}\right)=\frac{1}{\mathrm{\beta }\left({\overline{\mathrm{n}}}^0{\underset{¯}{\mathrm{y}}}^0+\mathrm{s},{\overline{\mathrm{n}}}^0\left(1-{\underset{¯}{\mathrm{y}}}^0\right)+\mathrm{n}-\mathrm{s}\right)}{\mathrm{p}}^{{\overline{\mathrm{n}}}^0{\underset{¯}{\mathrm{y}}}^0+\mathrm{s}-1}{\left(1-\mathrm{p}\right)}^{{\overline{\mathrm{n}}}^0\left(1-{\underset{¯}{\mathrm{y}}}^0\right)+\mathrm{n}-\mathrm{s}-1}$$E63

The above equation represents the third posterior distribution, which is the Beta distribution, and by using the properties of the Beta distribution, we get the central moments, as in the following equation:

$${\mathrm{M}}_{\mathrm{r}}=\frac{\mathrm{г}\left({\overline{\mathrm{n}}}^0+\mathrm{n}\right)\mathrm{г}\left({\overline{\mathrm{n}}}^0{\underset{¯}{\mathrm{y}}}^0+\mathrm{s}+\mathrm{r}\right)}{\mathrm{г}\left({\overline{\mathrm{n}}}^0+\mathrm{n}+\mathrm{r}\right)\mathrm{г}\left({\overline{\mathrm{n}}}^0{\underset{¯}{\mathrm{y}}}^0+\mathrm{s}\right)}$$E64

The fourth posterior distribution: From Eq. (58) and by using the Bayes rule, we get the fourth posterior distribution, as in the following equation:

$${\mathrm{f}}_4\left(\mathrm{p}/\mathrm{s}\right)=\frac{1}{\mathrm{\beta }\left({\overline{\mathrm{n}}}^0{\overline{\mathrm{y}}}^0+\mathrm{s},{\overline{\mathrm{n}}}^0\left(1-{\overline{\mathrm{y}}}^0\right)+\mathrm{n}-\mathrm{s}\right)}{\mathrm{p}}^{{\overline{\mathrm{n}}}^0{\overline{\mathrm{y}}}^0+\mathrm{s}-1}{\left(1-\mathrm{p}\right)}^{{\overline{\mathrm{n}}}^0\left(1-{\overline{\mathrm{y}}}^0\right)+\mathrm{n}-\mathrm{s}-1}$$E65

The above equation represents the fourth posterior distribution, which is the Beta distribution, and by using the properties of the Beta distribution, we get the central moments, as in the following equation:

$${\mathrm{M}}_{\mathrm{r}}=\frac{\mathrm{г}\left({\overline{\mathrm{n}}}^0+\mathrm{n}\right)\mathrm{г}\left({\overline{\mathrm{n}}}^0{\overline{\mathrm{y}}}^0+\mathrm{s}+\mathrm{r}\right)}{\mathrm{г}\left({\overline{\mathrm{n}}}^0+\mathrm{n}+\mathrm{r}\right)\mathrm{г}\left({\overline{\mathrm{n}}}^0{\overline{\mathrm{y}}}^0+\mathrm{s}\right)}$$E66

After taking the average of the posterior distributions, we get the iLuck-Model [10]:

$${\underset{¯}{\mathrm{y}}}^{\mathrm{n}}=\text{lower}\left({\mathrm{y}}^{\mathrm{n}}\right)=\left\{\begin{array}{c}\frac{{\overline{\mathrm{n}}}^0{\underset{¯}{\mathrm{y}}}^0+\mathrm{\tau }\left(\mathrm{x}\right)}{{\overline{\mathrm{n}}}^0+\mathrm{n}}\text{if}\overline{\mathrm{\tau }}\left(\mathrm{x}\right)\ge {\underset{¯}{\mathrm{y}}}^0\\ \frac{{\underset{¯}{\mathrm{n}}}^0{\underset{¯}{\mathrm{y}}}^0+\mathrm{\tau }\left(\mathrm{x}\right)}{{\underset{¯}{\mathrm{n}}}^0+\mathrm{n}}\text{if}\overline{\mathrm{\tau }}\left(\mathrm{x}\right)<{\underset{¯}{\mathrm{y}}}^0\end{array}\right.$$E67

$${\overline{\mathrm{y}}}^{\mathrm{n}}=\text{uper}\left({\mathrm{y}}^{\mathrm{n}}\right)=\left\{\begin{array}{c}\frac{{\overline{\mathrm{n}}}^0{\overline{\mathrm{y}}}^0+\mathrm{\tau }\left(\mathrm{x}\right)}{{\overline{\mathrm{n}}}^0+\mathrm{n}}\text{if}\overline{\mathrm{\tau }}\left(\mathrm{x}\right)\le {\overline{\mathrm{y}}}^0\\ \frac{{\underset{¯}{\mathrm{n}}}^0{\overline{\mathrm{y}}}^0+\mathrm{\tau }\left(\mathrm{x}\right)}{{\underset{¯}{\mathrm{n}}}^0+\mathrm{n}}\text{if}\overline{\mathrm{\tau }}\left(\mathrm{x}\right)>{\overline{\mathrm{y}}}^0\end{array}\right.$$E68

Eqs. (67), (68) represent a generalized iLuck-Model, a model that represents the lower bound and the model that represents the upper bound is chosen based on the value of $\overline{\mathrm{\tau }}\left(x\right)$, the estimator we obtain will be in the form of an interval Therefore, we will take the average for that period, and from the above, the posterior distribution model will be in the final form and as in the following equation:

$$\mathrm{f}\left(\mathrm{p}/{\mathrm{n}}^{\mathrm{m}},{\mathrm{y}}^{\mathrm{m}}\right)=\frac{1}{\mathrm{\beta }\left({\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}},{\mathrm{n}}^{\mathrm{m}}\left(1-{\mathrm{y}}^{\mathrm{m}}\right)\right)}{\mathrm{p}}^{{\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}}-1}{\left(1-\mathrm{p}\right)}^{{\mathrm{n}}^{\mathrm{m}}\left(1-{\mathrm{y}}^{\mathrm{m}}\right)-1}$$E69

$${\mathrm{n}}^{\mathrm{m}}=\frac{\text{lower}\left({\mathrm{n}}^{\mathrm{n}}\right)+\text{uper}\left({\mathrm{n}}^{\mathrm{n}}\right)}{2},{\mathrm{y}}^{\mathrm{m}}=\frac{\text{lower}\left({\mathrm{y}}^{\mathrm{n}}\right)+\text{uper}\left({\mathrm{y}}^{\mathrm{n}}\right)}{2}$$

6.5 Robust Bayesian estimation for parameter for binomial distribution

From Eq. (69) and by using the squared loss function, we get the robust Bayesian estimator for parameter (P) as follows [15]:

$$\mathrm{E}\left(\mathrm{p}/{\mathrm{n}}^{\mathrm{m}},{\mathrm{y}}^{\mathrm{m}}\right)={\widehat{\mathrm{p}}}_{\mathrm{Rob}}$$

$${\widehat{\mathrm{p}}}_{\mathrm{Rob}}=\frac{{\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}}}{{\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}}+{\mathrm{n}}^{\mathrm{m}}\left(1-{\mathrm{y}}^{\mathrm{m}}\right)}$$E70

6.6 Robust Bayesian estimation for survival function for binomial distribution

From Eq. (69) and by using the squared loss function, we get a Bayesian estimator for the survival function, as follows [15]:

$$\widehat{\mathrm{S}}\left(\mathrm{x}\right)=\underset{0}{\overset{1}{\int }}\mathrm{S}\left(\mathrm{x}\right)\mathrm{f}\left(\mathrm{p}/\mathrm{s}\right)\mathrm{dp}$$

$${\widehat{\mathrm{S}}}_{\mathrm{rob}}\left(\mathrm{x}\right)=\frac{1}{\mathrm{\beta }({\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}},{\mathrm{n}}^{\mathrm{m}}\left(1-{\mathrm{y}}^{\mathrm{m}}\right)}\sum _{\mathrm{j}=\mathrm{x}}^{\mathrm{n}}\frac{\mathrm{n}!}{\mathrm{j}!\left(\mathrm{n}-\mathrm{j}\right)!}\underset{0}{\overset{1}{\int }}{\mathrm{p}}^{\mathrm{j}}{\left(1-\mathrm{p}\right)}^{\mathrm{n}-\mathrm{j}}{\mathrm{p}}^{{\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}}-1}{\left(1-\mathrm{p}\right)}^{{\mathrm{n}}^{\mathrm{m}}\left(1-{\mathrm{y}}^{\mathrm{m}}\right)-1}\mathrm{dp}$$

$${\widehat{\mathrm{S}}}_{\mathrm{rob}}\left(\mathrm{x}\right)=\frac{1}{\mathrm{\beta }({\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}},{\mathrm{n}}^{\mathrm{m}}\left(1-{\mathrm{y}}^{\mathrm{m}}\right)}\sum _{\mathrm{j}=\mathrm{x}}^{\mathrm{n}}\frac{\mathrm{n}!}{\mathrm{j}!\left(\mathrm{n}-\mathrm{j}\right)!}\underset{0}{\overset{1}{\int }}{\mathrm{p}}^{{\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}}+\mathrm{j}-1}{\left(1-\mathrm{p}\right)}^{{\mathrm{n}}^{\mathrm{m}}\left(1-{\mathrm{y}}^{\mathrm{m}}\right)+\mathrm{n}-\mathrm{j}-1}\mathrm{dp}$$

Multiply and divide the equation by:

$$\mathrm{\beta }\left({\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}}+\mathrm{j},{\mathrm{n}}^{\mathrm{m}}\left(1-{\mathrm{y}}^{\mathrm{m}}\right)+\mathrm{n}-\mathrm{j}\right)$$

$${\hat{\mathrm{S}}}_{\mathrm{rob}}\left(\mathrm{x}\right)=\frac{1}{\mathrm{\beta }({\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}},{\mathrm{n}}^{\mathrm{m}}\left(1-{\mathrm{y}}^{\mathrm{m}}\right)}\sum _{\mathrm{j}=\mathrm{x}}^{\mathrm{n}}\frac{\mathrm{n}!}{\mathrm{j}!\left(\mathrm{n}-\mathrm{j}\right)!}\frac{\mathrm{\beta }\left({\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}}+\mathrm{j},{\mathrm{n}}^{\mathrm{m}}\left(1-{\mathrm{y}}^{\mathrm{m}}\right)+\mathrm{n}-\mathrm{j}\right)}{\mathrm{\beta }\left({\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}}+\mathrm{j},{\mathrm{n}}^{\mathrm{m}}\left(1-{\mathrm{y}}^{\mathrm{m}}\right)+\mathrm{n}-\mathrm{j}\right)}\underset{0}{\overset{1}{\int }}{\mathrm{p}}^{{\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}}+\mathrm{j}-1}{\left(1-\mathrm{p}\right)}^{{\mathrm{n}}^{\mathrm{m}}\left(1-{\mathrm{y}}^{\mathrm{m}}\right)+\mathrm{n}-\mathrm{j}-1}\mathrm{dp}$$

$${\widehat{\mathrm{S}}}_{\mathrm{Rob}}\left(\mathrm{x}\right)=\frac{1}{\mathrm{\beta }\left({\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}},{\mathrm{n}}^{\mathrm{m}}\left(1-{\mathrm{y}}^{\mathrm{m}}\right)\right)}\sum _{\mathrm{j}=\mathrm{x}}^{\mathrm{n}}\frac{\mathrm{n}!}{\mathrm{j}!\left(\mathrm{n}-\mathrm{j}\right)!}\mathrm{\beta }\left({\mathrm{n}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{m}}+\mathrm{j},{\mathrm{n}}^{\mathrm{m}}\left(1-{\mathrm{y}}^{\mathrm{m}}\right)+\mathrm{n}-\mathrm{j}\right)$$E71

7. The experimental side of the binomial distribution

In this section, Binomial distribution data will be generated by using the R program to estimate the (P) parameter and the survival function in the prior data conflict, as shown in the following tables:

Through Tables 5 and 6 the simulation results showed that the robust Bayesian estimator is the best through the comparison standard (IMSE) in the case of prior data conflict. From the above, the robust Bayesian estimator will be applied to the real data to estimate the (P) parameter and the survival function for the Binomial distribution.

Model	${\mathbf{n}}^{\mathbf{0}}$
	lower					upper
	2					4
	${\mathrm{y}}^0$		P	K	n	$\widehat{\mathrm{pmf}}$	${\widehat{\mathrm{pmf}}}_{\mathrm{rob}}$	Best
	Lower	upper
1	0.2	0.4	0.4	5	10	0.006340	0.005353	Robust Bayesian Estimator
				10	20	0.002334	0.002143
				20	40	0.000830	0.000799
2	0.3	0.5	0.5	5	10	0.006121	0.005225
				10	20	0.002207	0.002041
				20	40	0.000832	0.000802
3	0.4	0.6	0.6	5	10	0.005728	0.004898
				10	20	0.002369	0.002206
				20	40	0.000770	0.000743

Table 5.

The integrated mean square error (IMSE) of the probability mass function (pmf) for the binomial distribution in the case of prior data conflict.

Model	${\mathrm{n}}^0$
	Lower					upper
	2					4
	${\mathrm{y}}^0$		$\mathrm{p}$	k	n	$\widehat{\mathrm{S}}\left(\mathrm{x}\right)$	${\widehat{\mathrm{S}}}_{\mathrm{rob}}\left(\mathrm{x}\right)$	Best
	Lower	Upper
1	0.2	0.4	0.4	5	10	0.034387	0.027765	Robust Bayesian Estimator
				10	20	0.024367	0.022128
				20	40	0.018268	0.017509
2	0.3	0.5	0.5	5	10	0.034540	0.028961
				10	20	0.024535	0.022408
				20	40	0.016549	0.015884
3	0.4	0.6	0.6	5	10	0.032064	0.026952
				10	20	0.022817	0.021020
				20	40	0.016745	0.016075

Table 6.

The integrated mean square error (IMSE) of the survival function for the binomial distribution in the case of prior data conflict.

8. Real data description for the binomial distribution

Mortality data for patients with breast cancer were collected from Yarmouk Teaching Hospital for the period from 2010 to 2017, and the data collected are as follows (Tables 7–9):

Year	2010	2011	2012	2013	2014	2015	2016	2017
Xi	3	4	2	3	2	4	1	0

Table 7.

The real data for the binomial distribution.

${\mathrm{y}}^0$		${\mathrm{n}}^0$
		Lower	upper
		4	6
		${\widehat{\mathrm{p}}}_{\mathrm{rob}}$	Standard deviation prior	Standard deviation posterior
Lower	Upper
0.3	0.6	0.603595	0.042	0.006646

Table 8.

Robust Bayesian estimator for parameter (P) of the binomial distribution.

${\mathrm{y}}^0$		${\mathrm{n}}^0$
		Lower	upper
		4	6
		x	${\widehat{\mathrm{S}}}_{\mathrm{rob}}\left(\mathrm{x}\right)$
Lower	Upper
0.3	0.6	0	1
		1	0.999
		2	0.996
		3	0.994
		4	0.986

Table 9.

Robust Bayesian estimator for survival function of the binomial distribution.