### two parameter exponential distribution sufficient statistic

., Xn be a random sample from a two-parameter exponential distribution, Xi ~ EXP().a) Assuming it is known that =150, find a pivotal quanitity for the parameter based on the sufficient statistic.. b) Using the data of Exercise 5, find a one-sided lower 95% confidence limit for . What happens if a probability distribution has two parameters, \(\theta_1\) and \(\theta_2\), say, for which we want to find sufficient statistics, \(Y_1\) and \(Y_2\)? Lorem ipsum dolor sit amet, consectetur adipisicing elit. [2]). The two parameter exponential distribution is also a very useful component in reliability engineering. So even if you don't know what the $\theta$ is you can compute those. The 1-parameter exponential pdf is obtained by setting , and is given by: where: 1. Inserting what we know to be the probability density function of a normal random variable with mean \(\theta_1\) and variance \(\theta_2\), the joint p.d.f. 40, 1998, pp. The authors contributed equally to this work. Lett., Vol. 337-349. Note: One should not be surprised that the joint pdf belongs to the exponen-tial family of distribution. Nagaraja, A First Course in Order Statistics, SIAM, Philadelphia, PA, USA, 2008. (20–22), we have, Suppose that counting random variables K−(n,k,a) and K+(n,k,b) be independent. Let's try applying the extended exponential criterion to our previous example. Let \(X_1, X_2, \ldots, X_n\) denote a random sample from a normal distribution \(N(\theta_1, \theta_2\). 197-210. In Chapter 2 we consider the CEM and when the lifetime distributions of the experimental units follow different distributions. 375-395. T ( X 1 n ) = ∑ i = 1 n X i. the function \(h(x_1, ... , x_n)\) does not depend on either of the parameters \(\theta_1\) or \(\theta_2\). 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 \(\theta_2\). That is, \(\theta_1\) denotes the mean \(\mu\) and \(\theta_2\) denotes the variance \(\sigma^2\). = constant rate, in failures per unit of measurement, (e.g., failures per hour, per cycle, etc.) Then, the statistics \(Y_1=\sum_{i=1}^{n}K_1(X_i)\) and \(Y_2=\sum_{i=1}^{n}K_2(X_i)\) are jointly sufficient for \(\theta_1\) and \(\theta_2\). The exponential distribution is often concerned with the amount of time until some specific event occurs. If. 1.1. Order statistics is a kind of statistics distribution commonly used in statistical theory and application of which there are many research [1-6]. The decay parameter is expressed in terms of time (e.g., every 10 mins, every 7 years, etc. Assume that X has exponential distribution. Further, (31) and (32) imply that T2 is a consistent estimator for e−σa. Plan. We conclude that in all examples of a location family of distributions, statistics Yi are ancillary for the location parameter θ. Inf., Vol. 1.1. Further, its performance is compared with the maximum likelihood estimator (MLE) through simulation. [2]. • The partition of a minimal sufficient statistic is the coarsest. Aha! 80, 2010, pp. 1100-1116. Now, we present an asymptotic confidence interval for e−σa based on counting random variable K+(n,k,a) which is stated in the following remark. It is known that only for the exponential distribution, any two non-overlapping spacings will be independent (See, e.g., Arnold et al. We offer world-class services, fast turnaround times and personalised communication. Use the Exponential Criterion to find joint sufficient statistics for \(\theta_1\) and \(\theta_2\). Substituting it in Eqs. Basu’s Theorem. 36, 2007, pp. 3-24. Using (18) and (19), we have. 28, 2001, pp. Exponential distribution. Also, more characterization results of exponential distribution can be seen in Galambos and Kotz [4] and Ahsanullah and Hamedani [5]. 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. An estimator of e−σa is introduced in section 4 and some properties of this estimator are discussed. Plan. Lett., Vol. Two-parameter exponential distribution is often used to model the lifetime of a product. Because the observations are … sufficient statistic whenever and are two data values such that ( ) ( ), then ( ) ( ). Meth., Vol. 37-49. Plan. The exponential distribution is the probability distribution of the time or space between two events in a Poisson process, where the events occur continuously and independently at a constant rate \lambda.. Upcoming Events 2020 Community Moderator Election A.G. Pakes and Y. Li, Stat. That is, \(\theta_1\) denotes the mean \(\mu\) and \(\theta_2\) denotes the variance \(\sigma^2\). 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). This is an exponential family distribution so T = X2 1 + + X2 n is a complete su cient statistic; moreover, since it’s a scale parameter problem, U= X2 1 =(X 2 1 + + X n) is an ancillary statistic. Stat. 179-193. Theor. 138, 2008, pp. 69, 2004, pp. Lett., Vol. There exists a unique relationship between the exponential distribution and the Poisson distribution. Therefore the sum ∑ni = 1(yi − k) is sufficient if k is known. A.G. Pakes, Adv. Browse other questions tagged self-study mathematical-statistics sufficient-statistics or ask your own question. Similar to the proof of Theorem 2.2, F¯(x)=ce−σx is the most general solution of (17) and this completes the proof. 54, 2009, pp. Stat., Vol. The parameters . Math. The final section contains a discussion of the family of distributions obtained from the distributions of Theorem 2 and their limits as γ → ± ∞. . We refer the reader to Higgins [25] for Hilbert space and complete sequence function. We have just extended the Factorization Theorem. J. Aczél, Lectures on Functional Equations and Their Applications, Academic Press, London, England, New York, NY, USA, 1966. Excepturi aliquam in iure, repellat, fugiat illum The exponential distribution. Let \(X_1, X_2, \ldots, X_n\) denote random variables with a joint p.d.f. We have factored the joint p.d.f. S. Müller, Methodol. J. This study considers the … Nagaraja, Order Statistics, John Wiley-Sons, New York, NY, USA, 2003. J. Then the cumulative distribution function (CDF) of X is, According to (1), the probability mass function (pmf) of K+(n,k,a) for any j=0,1,⋯,n−k, have been obtained as (See Dembińska et al. ), which is a reciprocal (1/λ) of the rate (λ) in Poisson. into two functions, one (ϕ) being only a function of the statistic Y = X ¯ and the other (h) not depending on the parameter μ: Therefore, the Factorization Theorem tells us that Y = X ¯ is a sufficient statistic for μ. Stat., Vol. So, the obtained results show that with choosing appropriate k, the estimator T2 can be considered as a good estimator for parameter e−σa. Y. Li and A. Pakes, Insur. Therefore, K+(n,k,a) is a sufficient and complete statistic for e−σa. 2. a dignissimos. Comput. 1. The other factor, the exponential function, depends on y1, …, yn only through the given sum. 21, 2017, pp. Thus, a sufficient and complete statistics function for θ, is : ∑ i = 1 n ln (x i − 1) ⟶ T (x) = ∑ i = 1 n ln 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. are also joint sufficient statistics for \(\theta_1\) and \(\theta_2\). = operating time, life, or age, in hours, cycles, miles, actuations… In the next theorem, we show an another characterization for exponential distribution based on independent near-order statistics. According to Müntz theorem that is stated in Theorem 2.1, the all results of this section are true for any increasing subsequence {nj,j≥1} which satisfies in (9) instead of for all n≥1. The trick is to look at -Statistic examples are sample mean, min, max, median, order statistics... etc. The sequence {xn, n≥1} is the most important complete sequence function. 85-97. A sequence {Φn}n≥1 of elements of a Hilbert space H is called complete if the only element which is orthogonal to every {Φn} is the null element, that is. {\displaystyle T (X_ {1}^ {n})=\sum _ {i=1}^ {n}X_ {i}} is a sufficient statistic for θ. It is shown that its probability mass function and its first moment can characterize the exponential distribution. sufﬁcient statistic is characterized in the following result. The densi ties of the two exponential distributions are written as . 10, 2007, pp. The sufficient statistic of a set of independent identically distributed data observations is simply the sum of individual sufficient statistics, and encapsulates all the information needed to describe the posterior distribution of the parameters, given the data (and hence to derive any desired estimate of the parameters). Let \(X_1, X_2, \ldots, X_n\) be a random sample from a distribution with a p.d.f. CHARACTERIZATION BASED ON DEPENDENCY ASSUMPTIONS, 4. Arcu felis bibendum ut tristique et egestas quis: In each of the examples we considered so far in this lesson, there is one and only one parameter. 42, 1971, pp. P(X = x | T(X) = t) does (or joint p.m.f. 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: \(f(x_1, x_2, ... , x_n;\theta_1, \theta_2) = f(x_1;\theta_1, \theta_2) \times f(x_2;\theta_1, \theta_2) \times ... \times f(x_n;\theta_1, \theta_2) \times \). 39, 1997, pp. It is shown that the joint distribution of m-generalized order statistics has a representation as a regular exponential family in the model parameters, as it is the case for the comprising model. Let's try the extended theorem out for size on an example. Econ., Vol. Econ., Vol. But it is difficult to calculate MSE of T1 theoretically. J. The results are proved through properties of completeness sequence function. E. Hashorva, Stat. Math. H.A. Stat. Probab. 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. 199-210. The quantity (26) shows that the spacings W1 and W2 are independent. Key Definitions: Sufficient, Complete, and Ancillary Statistics. Desu, Ann. Let X be a random variable having two-parameter exponential distribution with parameters μ and σ, denoted by Exp(μ,σ). In this section, we will show that Eqs. MSE of two estimators T1 and T2 with respect to e−σa under n=50, a=1, μ=3 and different k. Our simulation results demonstrate that the performance of T1 and T2 has little differences with increasing a. J.R. Higgins, Completeness and Basis Properties of Sets of Special Functions, Cambridge University Press, New York, NY, USA, 1977. (4), we conclude easily that K+(n,k,a) has binomial distribution with parameters (n−k) and (1−e−σa), that is. Odit molestiae mollitia is: \(f(x_1, x_2, ... , x_n;\theta_1, \theta_2) = \dfrac{1}{\sqrt{2\pi\theta_2}} \text{exp} \left[-\dfrac{1}{2}\dfrac{(x_1-\theta_1)^2}{\theta_2} \right] \times ... \times = \dfrac{1}{\sqrt{2\pi\theta_2}} \text{exp} \left[-\dfrac{1}{2}\dfrac{(x_n-\theta_1)^2}{\theta_2} \right] \). 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. Y. Nikitin, ACUTM, Vol. A. Dembińska, Statistics, Vol. 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. 1.6 Organization of the monograph. Except where otherwise noted, content on this site is licensed under a CC BY-NC 4.0 license. Atlantis Press is a professional publisher of scientific, technical and medical (STM) proceedings, journals and books. The probability density function of a normal random variable with mean \(\theta_1\) and variance \(\theta_2\) can be written in exponential form as: Therefore, the statistics \(Y_1=\sum_{i=1}^{n}X^{2}_{i}\) and \(Y_2=\sum_{i=1}^{n}X_i\) are joint sufficient statistics for \(\theta_1\) and \(\theta_2\). N. Balakrishnan and A. Stepanov, J. Stat. At first, Pakes and Stutel [6] defined the number of observations within a of the sample maximum Xn:n as, Then, this definition was developed for the number of observations falling in the open left and right a–vicinity of the kth order statistics by Pakes and Li [7] and Balakrishnan and Stepanov [8], respectively. the Fisher–Neyman factorization theorem implies is a sufficient statistic for . See, Nikitin [27] for more details on application of characterization in goodness-of-fit test. In this paper, we have shown some applications of counting random variable K+(n,k,a) for two-parameter exponential distribution. 18.1 One Parameter Exponential Family Exponential families can have any ﬂnite number of parameters. Desu [3] proved that distribution of population is exponential if and only if nX1:n=dX1, for all n≥1, where the notation =d states the equality in distribution. Exponential distribution [edit | edit source] If are independent and exponentially distributed with expected value θ (an unknown real-valued positive parameter), then is a sufficient statistic for θ. In summary, we have factored the joint p.d.f. To see this, consider the joint probability density function of . Conversely, let (10) holds, then, The above inequality shows that η(u)∈L2(0,1). 117-128. Look at that! 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. Theor. The authors would like to thank the Editor in Chief, the Associate Editor and two anonymous reviewer for their valuable comments. Stat., Vol. NZ. Theorem 6.2.24 (Basu’s theorem) Let V and T be two statistics of X from a population indexed by q 2 . 46, 2012, pp. We have just shown that the intuitive estimators of \(\mu\) and \(\sigma^2\) are also sufficient estimators. Substituting in Eq. M.M. So, we compare them numerically. By the way, can you propose several other ancillary statistics? In this paper, we will prove some characterization results of two-parameter exponential distribution based on these counting random variables which are stated in sections 2 and 3. Probab. A. Dembińska, Stat. 309-323. What's Sufficient Statistic? of the exponential form: \( f(x;\theta_1,\theta_2)=\text{exp}\left[K_1(x)p_1(\theta_1,\theta_2)+K_2(x)p_2(\theta_1,\theta_2)+S(x) +q(\theta_1,\theta_2) \right] \). minimal statistic for θ is given by T(X,Y) m j=1 X2 j, n i=1 Y2 i, m j=1 X , n i=1 Y i. A. Dembińska, J. Stat. So, the proof is completed. Partition Interpretation for Minimal Sufficient Statistics: • Any sufficient statistic introduces a partition on the sample space. Also, this results are obtained based on 2000 bootstrap samples. M. Ahsanullah and G.G. It is important to know the probability density function, the distribution function and the quantile function of the exponential distribution. So far, more results of characterization of exponential distribution have been obtained that some of them are based on order statistics. For example, Lawless [ 1 151-160. Since the time length 't' is independent, it cannot affect the times between the current events. Steutel, Aust. 34, 2005, pp. Relationship between the Poisson and the Exponential Distribution. A continuous random variable x (with scale parameter λ > 0) is said to have an exponential distribution only if its probability density function can be expressed by multiplying the scale parameter to the exponential function of minus scale parameter and x for all x greater than or equal to zero, otherwise the probability density function is equal to zero. Let X1,X2,⋯Xn be independent and continuous random variables. f t t i i i i ( ) = exp − , , = 1 12 θ θ. It is enough to show that joint pgf of K−(n,k,a) and K+(n,k,b) is equal with multiplication of their pgfs. Now, the Exponential Criterion can also be extended to accommodate two (or more) parameters. Let's start by extending the Factorization Theorem. Appl., Vol. 54, 2012, pp. Also, an estimator based on near-order statistics is introduced for tail thickness of exponential distribution. 1. That seems like a good thing! S(X) is a statistic if it does NOT depend on any unknown quantities including $\theta$, which means you can actually compute S(X). voluptates consectetur nulla eveniet iure vitae quibusdam? A. Dembińska, Aust. 837-838. Then, the statistics \(Y_1=u_1(X_1, X_2, ... , X_n)\) and \(Y_2=u_2(X_1, X_2, ... , X_n)\) are joint sufficient statistics for \(\theta_1\) and \(\theta_2\) if and only if: \(f(x_1, x_2, ... , x_n;\theta_1, \theta_2) =\phi\left[u_1(x_1, ... , x_n), u_2(x_1, ... , x_n);\theta_1, \theta_2 \right] h(x_1, ... , x_n)\). Finally, the exponential families have conjugate priors (i.e. According to distribution of K+(n,k,a), it can be considered as sum of independent and identically distributed random variables from binomial 1,1−e−σa. So, the 100(1−α)% interval confidence for e−σa is given by. 9.Write X i = Z i where Z i ˘N(0;1). voluptate repellendus blanditiis veritatis ducimus ad ipsa quisquam, commodi vel necessitatibus, harum quos θ. i. are interpreted as the average failure times, the mean time to failure (MTTF), or the mean time between failures (MTBF) of the two groups. Simplifying by collecting like terms, we get: \(f(x_1, x_2, ... , x_n;\theta_1, \theta_2) = \left(\dfrac{1}{\sqrt{2\pi\theta_2}}\right)^n \text{exp} \left[-\dfrac{1}{2}\dfrac{\sum_{i=1}^{n}(x_i-\theta_1)^2}{\theta_2} \right] \). The authors declare that there is no potential conflict of interest related to this study. 1-14. Similarly we have the quantity, Further, using Minkowski's inequality for the quantity. Inf., Vol. Let X1,X2,…,Xn be continuous random variables with CDF F. Then F has Exp(μ,σ) if and only if K−(n,k,a) and K+(n,k,b) be independent for a fixed k≥1 and for any a>0 and b>0. Math. So far, more results of characterization of exponential distribution have been obtained that some of them are based on order statistics. Let \(X_1, X_2, \ldots, X_n\) denote a random sample from a normal distribution \(N(\theta_1, \theta_2)\). One-parameter exponential distribution has been considered by different authors since the work of … After that, following two random variables have been considered in the literature. J. Galambos and S. Kotz, Characterizations of Probability Distributions, Springer-Verlag, New York, NY, USA, 1978. https://books.google.com/books?id=BkcRRgAACAAJ. 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. If V is ancillary and T is boundedly complete and sufﬁcient for q, then V and T are independent with respect to Pq for any q 2 . In this study, we explore the MSE of T1 and T2 under different μ, a and k which are stated in Figure 1. Fortunately, the definitions of sufficiency can easily be extended to accommodate two (or more) parameters. A.G. Pakes, Extremes, Vol. Let us define two spacings W1 and W2 as follows, From (1) and (2), one can obtain easily the probability generating functions (pgf) of K−(n,k,a) and K+(n,k,b) as follows (see Balakrishnan and Stepanov [8]), It also follows that the joint pgf K−(n,k,a) and K+(n,k,b) is. Two-parameter exponential distribution is the simplest lifetime distributions that is useable in survival analysis and reliability theory. (6) and (7) can characterize exponential distribution. Use the Factorization Theorem to find joint sufficient statistics for \(\theta_1\) and \(\theta_2\). Inf., Vol. Received 24 January 2017, Accepted 7 August 2017, Available Online 3 March 2020. So, we firstly define complete sequence function and recall some well-known theorems. 5, 2003, pp. 32, 2003, pp. An exact confidence interval for e−σa when a is known can be obtained by this fact that a confidence interval is available for σ in two-parameter exponential distribution. The two-parameter exponential distribution with density: 1 (; , ) = − e x p − , (1. If the parameters of a two-parameter exponential family of distributions may be taken to be location and scale parameters, then the distributions must be normal. Recently, the problem of number of observations near the order statistics is considered. So, one estimator for e−σa based on MLE can be considered as, Following, we introduce an estimator for e−σa based on near-order statistic. 142, 2012, pp. or p.m.f. Arnold, N. Balakrishnan, and H.N. Stat., Vol. Let X 1, X 2, ⋯ X n be independent and continuous random variables. That is, the data contain no more information than the estimators \(\bar{X}\) and \(S^2\) do about the parameters \(\mu\) and \(\sigma^2\)! For more information, please contact us at: Department of Statistics, University of Mazandaran, Babolsar, Mazandaran 47416-95447, Iran, Department of Statistics, University of Birjand, Birjand, 97175-615, Iran, This is an open access article distributed under the CC BY-NC 4.0 license (. Hamedani, Exponential Distribution—Theory and Methods, Nova Science Publications Inc., New York, NY, USA, 2009. CHARACTERIZATION BASED ON DISTRIBUTIONAL RESULTS, 3. 1) where < is the threshold parameter, and > 0 is the scale parameter, is widely used in applied statistics. A.G. Pakes, Aust. Math., Vol. It is stated here without proof. The probability density function of a normal random variable with mean θ 1 and variance θ 2 can be written in exponential form as: Therefore, the statistics Y 1 = ∑ i = 1 n X i 2 and Y 2 = ∑ i = 1 n X i are joint sufficient statistics for θ 1 and θ 2. AN ESTIMATOR BASED ON NEAR-ORDER STATISTIC, https://doi.org/10.2991/jsta.d.200224.001, http://creativecommons.org/licenses/by-nc/4.0/. [10]), 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). G. Iliopoulos, A. Dembińska, and N. Balakrishnan, Statistics, Vol. 851-867. 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. which depends on the parameters \(\theta_1\) and \(\theta_2\). 134, 2005, pp. Higgins ([25], p. 95) The set {xn1,xn2,⋯;1≤n1

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