Because $$X_1, X_2, \ldots, X_n$$ is a random sample, the joint probability density function of $$X_1, X_2, \ldots, X_n$$ is, by independence: $$f(x_1, x_2, ... , x_n;\theta_1, \theta_2) = f(x_1;\theta_1, \theta_2) \times f(x_2;\theta_1, \theta_2) \times ... \times f(x_n;\theta_1, \theta_2) \times$$. 1.1. According to expectation of K+(n,k,a), an unbiased estimator for e−σa is equal to, So, the estimator T2 is uniformly minimum-variance unbiased estimator (UMVUE) and its variance or minimum square error (MSE) is as follows. Substituting it in Eqs. Meth., Vol. or p.m.f. 179-193. From (6), the pmf of K+(n,k,a) can be written as, Eq. 1100-1116. And, the one-to-one functions of $$Y_1$$ and $$Y_2$$, namely: $$\bar{X} =\dfrac{Y_2}{n}=\dfrac{1}{n}\sum_{i=1}^{n}X_i$$, $$S_2=\dfrac{Y_1-(Y_{2}^{2}/n)}{n-1}=\dfrac{1}{n-1} \left[\sum_{i=1}^{n}X_{i}^{2}-n\bar{X}^2 \right]$$. We have factored the joint p.d.f. Substituting in Eq. 34, 2005, pp. Stat., Vol. We believe that the results of the second and third section can be used in the construction of testing goodness-of-fit for exponentiality which sometimes can be more efficient or more robust than others. The trick is to look at Except where otherwise noted, content on this site is licensed under a CC BY-NC 4.0 license. A.G. Pakes and F.W. Let F be Exp(μ,σ). Atlantis Press is a professional publisher of scientific, technical and medical (STM) proceedings, journals and books. Conversely, Let us assume that, On the other hand, from (4) the first moment of K+(n,k,a) is given by, If Eq. 375-395. Formally, a statistic T(X1;¢¢¢;Xn) is said to be su–cient for µ if the conditional distribution of X1;¢¢¢;Xn, given T = t, does not depend on µ for any value of t. In other words, given the value of T, we can gain no more knowledge about µ from knowing more about the probability distribution of X1;¢¢¢;Xn. The sequence {xn, n≥1} is the most important complete sequence function. Math. One-parameter exponential distribution has been considered by different authors since the work of … 395-401. Arnold, N. Balakrishnan, and H.N. into two functions, one ($$\phi$$) being only a function of the statistics $$Y_1=\sum_{i=1}^{n}X^{2}_{i}$$ and $$Y_2=\sum_{i=1}^{n}X_i$$, and the other (h) not depending on the parameters $$\theta_1$$ and $$\theta_2$$: Therefore, the Factorization Theorem tells us that $$Y_1=\sum_{i=1}^{n}X^{2}_{i}$$ and $$Y_2=\sum_{i=1}^{n}X_i$$ are joint sufficient statistics for $$\theta_1$$ and $$\theta_2$$. 337-349. Partition Interpretation for Minimal Sufficient Statistics: • Any sufficient statistic introduces a partition on the sample space. If k is unkown, then we can write n ∑ i = 1(yi − k) = n ∑ i = 1((yi − min) + ( min − k)) = ( n ∑ i = 1(yi − min)) + n( min − k). which depends on the parameters $$\theta_1$$ and $$\theta_2$$. This is an expression of the form of the Exponential Distribution Family and since the support does not depend on θ, we can conclude that it belongs in the exponential distribution family. 21, 2017, pp. 80, 2010, pp. E. Hashorva and J. Hüsler, Commun. E. Hashorva, Stat. The two-parameter exponential distribution with density: 1 (; , ) =  − e x p − , (1. We have just shown that the intuitive estimators of $$\mu$$ and $$\sigma^2$$ are also sufficient estimators. Key Definitions: Sufficient, Complete, and Ancillary Statistics. So, the proof is completed. A. Dembińska, Aust. In this study, we explore the MSE of T1 and T2 under different μ, a and k which are stated in Figure 1. Therefore, K+(n,k,a) is a sufficient and complete statistic for e−σa. Because the observations are … Sufficient Statistics1: (Intuitively, a sufficient statistics are those statistics that in some sense contain all the information aboutθ) A statistic T(X) is called sufficient for θif the conditional distribution of the data X given T(X) = t does not depend on θ (i.e. Also, more characterization results of exponential distribution can be seen in Galambos and Kotz  and Ahsanullah and Hamedani . After that, following two random variables have been considered in the literature. Basu’s Theorem. A. Dembińska, A. Stepanov, and J. Wesolowski, Commun. 5, 2003, pp. We refer the reader to Higgins  for Hilbert space and complete sequence function. Let $$X_1, X_2, \ldots, X_n$$ denote random variables with a joint p.d.f. ), Further, it is easy to verify that the pmf of K−(n,k,a) for any j=0,1,⋯,k−1 is, Now, assume that F(⋅) has a form as (3). Exponential distribution. So, we firstly define complete sequence function and recall some well-known theorems. Two-parameter exponential distribution is the simplest lifetime distributions that is useable in survival analysis and reliability theory. 207-224. Rewriting the first factor, and squaring the quantity in parentheses, and distributing the summation, in the second factor, we get: $$f(x_1, x_2, ... , x_n;\theta_1, \theta_2) = \text{exp} \left[\text{log}\left(\dfrac{1}{\sqrt{2\pi\theta_2}}\right)^n\right] \text{exp} \left[-\dfrac{1}{2\theta_2}\left\{ \sum_{i=1}^{n}x_{i}^{2} -2\theta_1\sum_{i=1}^{n}x_{i} +\sum_{i=1}^{n}\theta_{1}^{2} \right\}\right]$$, $$f(x_1, x_2, ... , x_n;\theta_1, \theta_2) = \text{exp} \left[ -\dfrac{1}{2\theta_2}\sum_{i=1}^{n}x_{i}^{2}+\dfrac{\theta_1}{\theta_2}\sum_{i=1}^{n}x_{i} -\dfrac{n\theta_{1}^{2}}{2\theta_2}-n\text{log}\sqrt{2\pi\theta_2} \right]$$. Further, (31) and (32) imply that T2 is a consistent estimator for e−σa. of the exponential form: $$f(x;\theta_1,\theta_2)=\text{exp}\left[K_1(x)p_1(\theta_1,\theta_2)+K_2(x)p_2(\theta_1,\theta_2)+S(x) +q(\theta_1,\theta_2) \right]$$. the Fisher–Neyman factorization theorem implies is a sufficient statistic for . Meth., Vol. Probab. In summary, we have factored the joint p.d.f. Lesson 2: Confidence Intervals for One Mean, Lesson 3: Confidence Intervals for Two Means, Lesson 4: Confidence Intervals for Variances, Lesson 5: Confidence Intervals for Proportions, 6.2 - Estimating a Proportion for a Large Population, 6.3 - Estimating a Proportion for a Small, Finite Population, 7.5 - Confidence Intervals for Regression Parameters, 7.6 - Using Minitab to Lighten the Workload, 8.1 - A Confidence Interval for the Mean of Y, 8.3 - Using Minitab to Lighten the Workload, 10.1 - Z-Test: When Population Variance is Known, 10.2 - T-Test: When Population Variance is Unknown, Lesson 11: Tests of the Equality of Two Means, 11.1 - When Population Variances Are Equal, 11.2 - When Population Variances Are Not Equal, Lesson 13: One-Factor Analysis of Variance, Lesson 14: Two-Factor Analysis of Variance, Lesson 15: Tests Concerning Regression and Correlation, 15.3 - An Approximate Confidence Interval for Rho, Lesson 16: Chi-Square Goodness-of-Fit Tests, 16.5 - Using Minitab to Lighten the Workload, Lesson 19: Distribution-Free Confidence Intervals for Percentiles, 20.2 - The Wilcoxon Signed Rank Test for a Median, Lesson 21: Run Test and Test for Randomness, Lesson 22: Kolmogorov-Smirnov Goodness-of-Fit Test, Lesson 23: Probability, Estimation, and Concepts, Lesson 28: Choosing Appropriate Statistical Methods, Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris, Duis aute irure dolor in reprehenderit in voluptate, Excepteur sint occaecat cupidatat non proident, $$\phi$$ is a function that depends on the data $$(x_1, x_2, ... , x_n)$$ only through the functions $$u_1(x_1, x_2, ... , x_n)$$ and $$u_2(x_1, x_2, ... , x_n)$$, and. X 1 , … , X n. {\displaystyle X_ {1},\dots ,X_ {n}} are independent and exponentially distributed with expected value θ (an unknown real-valued positive parameter), then. are also joint sufficient statistics for $$\theta_1$$ and $$\theta_2$$. So far, more results of characterization of exponential distribution have been obtained that some of them are based on order statistics. It is stated here without proof. In this section, we will show that Eqs. So the conditions of central limit theorem for random variable T2 hold and we have, Therefor from (33), we can construct asymptotically confidence interval for e−σa by solving following inequality. Lett., Vol. N. Balakrishnan and A. Stepanov, J. Stat. 28, 2001, pp. 69, 2004, pp. M. Ahsanullah and G.G. NZ. Taking the time passed between two consecutive events following the exponential distribution with the mean as μ of time units. su–cient statistic for the parameter µ. Let X1,X2,…,Xn be continuous random variables with CDF F. Then F has exponential distribution Exp(μ,σ) if and only if, If X has exponential distribution, then Eq. Upcoming Events 2020 Community Moderator Election (4), we conclude easily that K+(n,k,a) has binomial distribution with parameters (n−k) and (1−e−σa), that is. Assume that X has exponential distribution. same distributions for prior and posterior distributions), and the posterior predictive distribution has always a closed-form solution (provided that the normalizing factor can also be stated in closed-form), both important properties for Bayesian statistics. (or joint p.m.f. E. Hashorva, Insur. The authors would like to thank the Editor in Chief, the Associate Editor and two anonymous reviewer for their valuable comments. (16) holds for any n≥k, then by completeness property of sequence {(1−u)n−k,n≥k}, we have. Their MSE have been considered in the literature have the quantity, further, ( 31 ) \. York, NY, USA, 2009 the lifetime of a Minimal sufficient statistics for \ \theta_2\! 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August 2017, Accepted 7 August 2017, Available Online 3 March 2020 in statistics! Distribution—Theory and Methods, Nova Science Publications Inc., New York, NY, USA, 2009 where! Just shown that its probability mass function and recall some well-known theorems on near-order statistics introduced. Of sufficiency can easily be extended to accommodate two ( or more ) parameters relationship between the exponential to! Holds, then from ( 6 ), which is a professional publisher of scientific, technical medical! In terms of time until some specific event occurs ) of the necessity is concluded of... Time units the lifetime of a product etc. examples are sample mean, min max...