N. Balakrishnan and A. Stepanov, J. Stat. = constant rate, in failures per unit of measurement, (e.g., failures per hour, per cycle, etc.) Nagaraja, A First Course in Order Statistics, SIAM, Philadelphia, PA, USA, 2008. 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. Atlantis Press is a professional publisher of scientific, technical and medical (STM) proceedings, journals and books. It is shown that its probability mass function and its first moment can characterize the exponential distribution. The final section contains a discussion of the family of distributions obtained from the distributions of Theorem 2 and their limits as γ → ± ∞. 117-128. Probab. The densi ties of the two exponential distributions are written as . S(X) is a statistic if it does NOT depend on any unknown quantities including $\theta$, which means you can actually compute S(X). Stat., Vol. J. J. A. Dembińska, Statistics, Vol. Hamedani, Exponential Distribution—Theory and Methods, Nova Science Publications Inc., New York, NY, USA, 2009. But it is difficult to calculate MSE of T1 theoretically. Proof. N. Balakrishnan and A. Stepanov, J. Stat. the Fisher–Neyman factorization theorem implies is a sufficient statistic for . Basu’s Theorem. The exponential distribution is often concerned with the amount of time until some specific event occurs. One-parameter exponential distribution has been considered by different authors since the work of … See, Nikitin [27] for more details on application of characterization in goodness-of-fit test. Debasis Kundu, Ayon Ganguly, in Analysis of Step-Stress Models, 2017. 1010-1020. Probab. 134, 2005, pp. For example, the amount of time (beginning now) until an earthquake occurs has an exponential distribution. Recently, the problem of number of observations near the order statistics is considered. In this study, we explore the MSE of T1 and T2 under different μ, a and k which are stated in Figure 1. Conversely, let (10) holds, then, The above inequality shows that η(u)∈L2(0,1). The other factor, the exponential function, depends on y1, …, yn only through the given sum. S. Müller, Methodol. A. Dembińska, Aust. We refer the reader to Higgins [25] for Hilbert space and complete sequence function. What's Sufficient Statistic? So, the obtained results show that with choosing appropriate k, the estimator T2 can be considered as a good estimator for parameter e−σa. 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 \(\theta_2\). 138, 2008, pp. Let's try the extended theorem out for size on an example. 37-49. Relationship between the Poisson and the Exponential Distribution. Substituting in Eq. the function \(h(x_1, ... , x_n)\) does not depend on either of the parameters \(\theta_1\) or \(\theta_2\). 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. 142, 2012, pp. So, the proof is completed. [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). Inf., Vol. 42, 1971, pp. So, we compare them numerically. 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: \(f(x_1, x_2, ... , x_n;\theta_1, \theta_2) = f(x_1;\theta_1, \theta_2) \times f(x_2;\theta_1, \theta_2) \times ... \times f(x_n;\theta_1, \theta_2) \times \). 40, 1998, pp. From (6), the pmf of K+(n,k,a) can be written as, Eq. f t t i i i i ( ) = exp − , , = 1 12 θ θ. Econ., Vol. 32, 2000, pp. 375-395. or p.m.f. H.A. of the exponential form: \( f(x;\theta_1,\theta_2)=\text{exp}\left[K_1(x)p_1(\theta_1,\theta_2)+K_2(x)p_2(\theta_1,\theta_2)+S(x) +q(\theta_1,\theta_2) \right] \). 508-523. That is, the data contain no more information than the estimators \(\bar{X}\) and \(S^2\) do about the parameters \(\mu\) and \(\sigma^2\)! So, the 100(1−α)% interval confidence for e−σa is given by. In this paper, we have shown some applications of counting random variable K+(n,k,a) for two-parameter exponential distribution. 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. Also, this results are obtained based on 2000 bootstrap samples. 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. Suppose that the time that elapses between two successive events follows the exponential distribution with a mean of \(\mu\) units of time. Further, its performance is compared with the maximum likelihood estimator (MLE) through simulation. (or joint p.m.f. are also joint sufficient statistics for \(\theta_1\) and \(\theta_2\). In Chapter 2 we consider the CEM and when the lifetime distributions of the experimental units follow different distributions. Recall that the Poisson distribution with parameter \(\theta \in (0, \infty)\) is a discrete distribution on \( \N \) with probability density function \( g \) defined by \[ g(x) = e^{-\theta} \frac{\theta^x}{x! The proceedings and journals on our platform are Open Access and generate millions of downloads every month. 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: \(f(x_1, x_2, ... , x_n;\theta_1, \theta_2) =\phi\left[u_1(x_1, ... , x_n), u_2(x_1, ... , x_n);\theta_1, \theta_2 \right] h(x_1, ... , x_n)\). Plan. 837-838. 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. 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. 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. David and H.N. B.C. J.R. Higgins, Completeness and Basis Properties of Sets of Special Functions, Cambridge University Press, New York, NY, USA, 1977. laudantium assumenda nam eaque, excepturi, soluta, perspiciatis cupiditate sapiente, adipisci quaerat odio 48, 2014, pp. -Statistic examples are sample mean, min, max, median, order statistics... etc. Appl. We offer world-class services, fast turnaround times and personalised communication. Rewriting the first factor, and squaring the quantity in parentheses, and distributing the summation, in the second factor, we get: \(f(x_1, x_2, ... , x_n;\theta_1, \theta_2) = \text{exp} \left[\text{log}\left(\dfrac{1}{\sqrt{2\pi\theta_2}}\right)^n\right] \text{exp} \left[-\dfrac{1}{2\theta_2}\left\{ \sum_{i=1}^{n}x_{i}^{2} -2\theta_1\sum_{i=1}^{n}x_{i} +\sum_{i=1}^{n}\theta_{1}^{2} \right\}\right] \), \(f(x_1, x_2, ... , x_n;\theta_1, \theta_2) = \text{exp} \left[ -\dfrac{1}{2\theta_2}\sum_{i=1}^{n}x_{i}^{2}+\dfrac{\theta_1}{\theta_2}\sum_{i=1}^{n}x_{i} -\dfrac{n\theta_{1}^{2}}{2\theta_2}-n\text{log}\sqrt{2\pi\theta_2} \right]\). Math. Let's try applying the extended exponential criterion to our previous example. Theor. The confusion starts when you see the term “decay parameter”, or even worse, the term “decay rate”, which is frequently used in exponential distribution. 10, 2007, pp. Finally, the exponential families have conjugate priors (i.e. Using (18) and (19), we have. A.G. Pakes, Adv. Also, more characterization results of exponential distribution can be seen in Galambos and Kotz [4] and Ahsanullah and Hamedani [5]. 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. ), which is a reciprocal (1/λ) of the rate (λ) in Poisson. 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. Arnold, N. Balakrishnan, and H.N. So far, more results of characterization of exponential distribution have been obtained that some of them are based on order statistics. 199-210. Math. The exponential distribution is a probability distribution which represents the time between events in a Poisson process. 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 \(Y_2\)? So far, more results of characterization of exponential distribution have been obtained that some of them are based on order statistics. In the next theorem, we show an another characterization for exponential distribution based on independent near-order statistics. If. Also, an estimator based on near-order statistics is introduced for tail thickness of exponential distribution. (20–22), we have, Suppose that counting random variables K−(n,k,a) and K+(n,k,b) be independent. Substituting its CDF into (27), imply that, It is well-known that the MLE of unknown scale parameter σ is n∑i=1n(Xi−X1:n)−1, when the underlying distribution is Exp(μ,σ). Except where otherwise noted, content on this site is licensed under a CC BY-NC 4.0 license. 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 results are proved through properties of completeness sequence function. Meth., Vol. Simplifying by collecting like terms, we get: \(f(x_1, x_2, ... , x_n;\theta_1, \theta_2) = \left(\dfrac{1}{\sqrt{2\pi\theta_2}}\right)^n \text{exp} \left[-\dfrac{1}{2}\dfrac{\sum_{i=1}^{n}(x_i-\theta_1)^2}{\theta_2} \right] \). Therefore, K+(n,k,a) is a sufficient and complete statistic for e−σa. 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. Use the Factorization Theorem to find joint sufficient statistics for \(\theta_1\) and \(\theta_2\). Conversely, Let us assume that, On the other hand, from (4) the first moment of K+(n,k,a) is given by, If Eq. 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. 151-160. The corresponding order statistics are the Xi's arranged in non-decreasing order, denoted by X1:n 0 is the lifetime. 6 ) and \ ( \theta_2\ ) of statistics distribution commonly used in statistical theory and application of characterization goodness-of-fit! Of time until some specific event occurs a probability distribution which represents the time length 't ' independent! 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