[数学]数学09级1班闻晶晶外文文献翻译
土木工程外文文献翻译

土木工程外文文献翻译数学模型预测水运输混凝土结构中的渗透性eluozo,s.n全文粗骨料细砂的混凝土构件,大孔隙混凝土率确定孔隙率和混凝土结构孔隙比,渗透系数的影响确定率水运混凝土。
数学模型来预测渗透率对水率交通是数学发展,该模型是监测水运输的混凝土率结构。
渗透性建立大孔上构成的影响下一种关系,即由混凝土制成,应用混凝土浇筑的决定渗透性的沉积速率混凝土结构,渗透性建立是大孔的混合物之间的影响下通过水泥净浆,考虑到系统中的变量,数学模型的建立是为了监测水通过通过具体的速度,也确定渗透系数的率对混凝土结构。
关键字:混凝土结构、渗透性和数学模型一、简介混凝土结构的耐久性依赖于通过频密迁移率的熔化的成分。
这种搬迁就是通过磁导率的影响。
在排序该条件混凝土混合物就是通过中存有的基质中的微孔隙的已连续网络混凝土协调比。
其他影响就是通过存有于的界面的孔隙率骨料的级分体式结构。
本研究中,其特征测量的快速和精确度在煮混凝土的渗透性,这包含创建理论模型的叙述渗透性对混凝土结构的影响。
实验中采用的就是瞬时顺利完成渗透性设备监控措施细骨料细沙和水是这种材料例如混凝土称作孔隙率和孔隙率中的组件之间的微孔混凝土结构中,渗透系数的影响确认水的速率运输在混凝土加水物水分搬迁混凝土,设备容许快速和精确测量在混凝土加水水中搬迁。
混凝土就是一种类型的多孔材料做成,并且可以由于在物理上和化学受损其曝露在各种环境中从混凝土浇筑至其使用寿命。
在特别就是,一些外部有毒元素,例如硫酸根,氯离子,和二氧化碳,扩散在混凝土少于长期周期做为溶液或气体状态,并引致物理侵害,由于化学反应。
这些反应可以影响应用领域中钢筋破损具体内容的,这减少了耐热寿命,例如钢筋和力量。
因此,它是非常重要的是插入腐蚀抑制剂为在超过临界恶化元件的情况下钢棒腐蚀的钢筋的位置量[1]。
然而,这是非常困难的保证在使用该应用传统技术钢筋位置的耐腐蚀性腐蚀抑制剂仅在混凝土[2-3]的表面上。
我们班的数学天才英语作文600字初中

我们班的数学天才英语作文600字初中全文共3篇示例,供读者参考篇1The Math Genius in Our ClassYou know that kid in every class who just gets math? Like, really gets it down to the core in a way the rest of us can't even fathom? Well, in my class that genius is Adam Smith. Math seems to flow through his veins more naturally than blood.From the first day of 6th grade when Mr. Patterson worked through some simple algebraic equations on the board, Adam's hand shot up before anyone could even processed what was going on. He explained the steps for solving the equations so clearly and effortlessly that even I, a kid who has always struggled with math, had a Light bulb moment."Oh, that's how you do it!" I thought as Adam walked us through algebraically isolating the variable. It all seemed so obvious when he laid it out systematically.That was just the start of Adam's math wizardry. As we moved into more complex concepts like graphing systems of equations, factoring polynomials, and applying the principles ofgeometry, he became increasingly indispensable. If anyone raised their hand with a question or got stuck on a problem set, we'd all turn our heads eagerly toward Adam waiting for him to share his insight.Despite being shy around school hallways, Adam transforms in math class. He leans forward in his seat, makes deliberate eye contact with the teacher, and articulates his thoughts with impressive coherence and precision. You can almost see the synapses firing in his brain as he works through the logicstep-by-step.Even Mr. Patterson regularly calls on Adam to explain certain problems or theorems to the rest of us when we look puzzled. I've lost count of how many times Adam has dropped a "Big Ah" moment on the class that allowed a breakthrough or clarified something we were all stuck on.What makes Adam even more remarkable is his humility and patience. Despite having a staggering mathematical ability, he never lorded it over anyone or acted arrogant. If someone didn't understand his explanation at first, Adam would re-word it orre-work the problem with a different approach until it clicked for that person.He spends plenty of free periods, study halls, and free time after school walking different classmates through the topics they struggled with. Almost everyone has taken advantage of Adam's tutoring at one point or another. His guidance has allowed so many of us to maintain solid grades in math rather than allowing the subject to become our downfall.While the rest of us dream about getting outdoors as soon as the weather turns nice each year, Adam looks most at peace and in his element when working through a challenge problem set or attempting to decipher a new mathematical proof. Books on advanced mathematics concepts are is his Preference for light reading over young adult fiction.Simply put, Adam is a savant when it comes to numbers, equations, geometry, and the wonderful world of mathematics. We're all just lucky he ended up in our class to shed is brilliant light on the subject and help raise the mathematical excellence of every student. Having a genius in our midst drove the rest of us to live up to our fullest potential as well.篇2My Class's Math GeniusThere's this kid in my class who is, simply put, a total math genius. His name is David and he has an incredible knack for numbers and problem-solving that is really mind-blowing. I'm not exaggerating when I say he could probably end up being one of the great mathematicians of our time if he keeps developing his skills.It's kind of crazy just how advanced David's math abilities are compared to the rest of us. I still have to use a calculator for most complicated calculations, but he can do enormously complex equations entirely in his head in just seconds. Once, when we were learning about algebraic expressions in class, the teacher put a really long and convoluted problem on the board to demonstrate. While the rest of us just stared at it blankly, David's hand shot up and he solved the whole thing, step-by-step, from memory. The teacher was stunned and so were we.David's parents told me he started showing an affinity for math basically from the moment he could talk. Apparently, as a toddler, he would constantly recite numbers and look for patterns and relationships between them. By the age of 4, he could already do basic addition, subtraction, multiplication and division. That's just not normal child development - the kid was clearly some kind of math prodigy from a very young age.In school, he breezes through math assignments and tests like they're nothing. He gets every question right, every time. Our math teacher once told me she plans to start giving David extra advanced coursework because the normal curriculum isn't enough of a challenge for his capabilities.Sometimes he even catches mistakes in the textbooks that the authors and editors all missed.You might think having such a talented student would be great, but it does create some difficulties too. Often the concepts we're supposed to be learning go completely over my head because David has already understood it in about 2 seconds. Then he gets bored waiting for the rest of us to catch up and starts distracting the class by joking around. The teacher has had to warn him about it multiple times.Still, David is a pretty humble, down-to-earth guy despite his incredible talents. He never brags or acts arrogant about his superior intellect. I've learned not to feel bad about not being as skilled as him because it's clear his brain is just wired differently for math aptitude. It would be like him feeling bad for not being as good at basketball as one of the players on the school team.For his future plans, David says he's torn between wanting to become a mathematics professor and researcher, or going into amore applied field like engineering or computer science. No matter what path he chooses, I'm confident he'll be hugely successful and make important contributions. David is truly one of the most gifted students in mathematics our school has ever seen. I'm just lucky I get to be in his class and witness his incredible abilities firsthand. Who knows, maybe I'm learning from the next Einstein or Ramanujan.篇3Our Class's Math GeniusYou know that kid in every class who just seems to get math effortlessly? The one who finishes tests in half the time while the rest of us are still struggling? Well, that person in our class is Tony. Tony is, without a doubt, a certified math genius.I've known Tony since we were little kids playing together at the neighborhood park. Even back then, you could tell he just thought differently about numbers and patterns. While the rest of us would count out objects one by one, Tony would immediately see the groupings and patterns. "That's two groups of four!" he would exclaim about a simple pile of rocks. His mind worked that way from the very start.In elementary school, you could already see the math genius emerging. He would race through worksheets of basic operations, making teachers do double-takes. By third grade, he was getting pulled out for an hour each day to work on higher level math meant for older kids. Can you imagine being 8 years old and doing algebra? That was Tony.When we got to middle school, that's when Tony's talents really went into overdrive. I'm in the highest level of math available, and even I can't keep up with that kid half the time. He breezes through complicated equations like a hot knife through butter. Just last week, our math teacher put an extremely tricky problem on the board involving exponential functions and logarithms. The rest of us sat there dumbfounded, but within 30 seconds, Tony's hand shot up with the correct solution fully worked out.Sometimes I think he gets bored in class because the material isn't challenging enough for his incredible mind. He'll start thinking about unsolved math problems or theorems that no one has cracked yet. Once Mrs. Robinson caught him scribbling some sort of crazy number theory conjecture during her lesson. When she asked him about it, he launched into thisintense explanation about prime numbers that went completely over our heads. Even she looked dazed by the end!While Tony's skills are clearly off the charts, he's also a really humble, down-to-earth guy which makes him even more likeable. You'd never know he was a genius unless you saw him do math. In a world where kid geniuses can sometimes be arrogant know-it-alls, Tony is the complete opposite. He never lords his abilities over others or makes us feel dumb in comparison. If someone is struggling, he's the first to try patiently explaining it in a new way. We all wish we had Tony's brain, but are just glad to be his friend.I have no idea what the future holds for our class's math whiz kid. Ivy League universities are definitely in his future, and perhaps he'll go on to win a Fields Medal or formulate the next big theory in mathematics. Wherever life takes him, I'm just happy to be able to witness Tony's incredible talents and be inspired by them. Having a bonafide genius in your midst makes you realize there's so much potential in this world still to be unlocked. Math may seem like a boring subject to some, but seeing it through Tony's eyes shows how exciting and full of possibilities it truly is.。
数学家--中英文对照3篇

数学家--中英文对照数学家数学家是指从事研究数学领域的专业人士。
数学家们通过发表论文、教课、参与研究项目等方式,不断推动数学知识的发展和应用。
数学家们使用符号、公式、模型等工具来描述和解决各种数学问题,如代数、几何、拓扑、概率等。
他们的工作不仅是基础研究,也可以为应用数学提供支持,如在物理、工程、计算机科学等领域中的应用。
数学家们需要具备良好的逻辑思维能力、创造性和耐心,他们需要不断探索数学领域中新的思想和方法。
他们的工作需要深入掌握大量的数学知识,因此需要进行长期的学习和实践。
著名数学家包括欧拉、高斯、牛顿、莫比乌斯、庞加莱等。
他们的工作对数学学科的发展和现代科技的进步起到了重要的作用。
作为未来的数学家,我们需要勤奋学习,探索数学领域中的新思想和方法,并在未来的研究工作中为数学学科的发展和社会的进步做出贡献。
MathematiciansMathematicians are professionals who engage inthe study of mathematics. Mathematicians continue to promote the development and application of mathematical knowledge through publishing papers, teaching, and participating in research projects.Mathematicians use tools such as symbols, formulas, and models to describe and solve a variety of mathematical problems, such as algebra, geometry, topology, and probability theory. Their work is not only basic research, but also provides support for applied mathematics, such as in physics, engineering, and computer science.Mathematicians need to possess good logical thinking ability, creativity, and patience. They need to constantly explore new concepts and methods in the field of mathematics. Their work requires a deep understanding of a large amount of mathematical knowledge, so they need to undergo long-term learning and practice.Famous mathematicians include Euler, Gauss, Newton, Mobius, Poincare, etc. Their work has played an important role in the development of mathematical disciplines and the progress of modern technology.As future mathematicians, we need to study diligently, explore new ideas and methods in the field of mathematics, and contribute to the development of mathematical disciplines and social progress in our future research work.数学家的职业生涯数学家是从事以数学领域为主要研究对象的专业人士,数学家的职业生涯从学生时代开始,到最后可以成为权威,分享了解数学领域中新的思想和技术。
有关数学故事

有关数学故事(中英文实用版)Story: The Detective and the Math Problem故事:侦探与数学问题Once upon a time, there was a famous detective who was known for solving the most difficult mysteries.One day, he received a strange letter from a wealthy collector who claimed that he had been robbed of a priceless artifact.The collector suspected that it was an inside job and that the thief was someone he trusted.从前,有一位著名的侦探,以解决最困难的谜团而闻名。
有一天,他收到了一封来自一位富有收藏家的奇怪信件,信中说他丢失了一件无价的文物。
收藏家怀疑这是一起内部作案,而且窃贼是他所信任的某人。
The collector provided the detective with a list of suspects and a description of the artifact.He also mentioned that the thief must have used some mathematical calculations to determine the best time to strike.Intrigued by the mathematical aspect of the case, the detective decided to use his own math skills to solve it.收藏家向侦探提供了一份嫌疑人名单和文物的描述。
班班幼儿园第二章英文

班班幼儿园第二章英文Chapter 2: English at Banban KindergartenIntroduction:English language learning is an important aspect of education at Banban Kindergarten. In Chapter 2, we will explore the various methods and strategies used to teach English to our young learners. The curriculum is designed to provide a comprehensive and engaging English learning experience, focusing on four key areas: listening, speaking, reading, and writing. Through a combination of interactive activities, games, and songs, children develop their English language skills in a fun and stimulating environment.Listening:At Banban Kindergarten, listening skills are developed through various activities that encourage active listening and comprehension. Teachers use audio materials, such as English stories or songs, to expose children to different accents, intonations, and vocabulary. This helps them to develop their listening skills and comprehension abilities gradually.Additionally, teachers engage students in classroom discussions, group activities, and storytelling sessions, which further enhance their listening skills. By actively participating in these activities, children become more confident in their ability to understand spoken English.Speaking:To foster speaking skills, teachers at Banban Kindergarten use a communicative approach. From the moment children step into the classroom, English is the primary language of instruction. Teachers use simple, clear language and encourage children to express themselves in English. Activities such as role-plays, pair work, and games are incorporated into the curriculum to encourage children to speak confidently and fluently. Through frequent practice and positive reinforcement, children gradually gain confidence in using English to communicate with their teachers and peers.Reading:Reading is an essential part of language development, and at Banban Kindergarten, we aim to cultivate a love for reading in our students. The curriculum includes a wide range of storybooks, picture books, and leveled readers that are suitablefor young learners. Teachers create a welcoming reading corner within the classroom, stocked with books that cater to different reading abilities and interests. Through shared reading sessions, guided reading activities, and independent reading time, children develop reading skills such as phonics, word recognition, and comprehension. The goal is to instill a lifelong love for reading and provide a solid foundation in literacy.Writing:Writing skills are introduced gradually at Banban Kindergarten, beginning with the basics of letter formation and phonics. Teachers provide guided writing activities and encourage children to write simple sentences and short paragraphs. Through interactive writing exercises, children learn to express their ideas and thoughts in English, gradually improving their writing abilities. Teachers provide constructive feedback and assist students in improving their grammar, vocabulary, and sentence structure. The focus is on creating a positive and supportive environment that encourages children to practice and develop their writing skills.Assessment:At Banban Kindergarten, assessment is an integral part of thelearning process. Teachers use a variety of assessment methods to track each child's progress and identify areas for improvement. These methods include observation, informal assessments, and periodic tests. Teachers provide feedback to both students and parents regularly, highlighting areas of strength and areas that need improvement. This feedback helps parents to stay informed about their child's progress and actively engage in their English language learning journey.Conclusion:English language learning at Banban Kindergarten is designedto be an immersive and enjoyable experience for young learners. By focusing on listening, speaking, reading, and writing skills, children develop a solid foundation in English language proficiency. Through a combination of interactive activities, engaging materials, and skilled teachers, students gain confidence in using English to communicate effectively. The goal is to equip children with the necessary language skills to succeed in their future academic endeavors and beyond.。
研究生英语读说写1课文翻译

研究生英语读说写1课文翻译一、A Working Community5、None of us, mind you, was born into these communities. Nor did we move into them, U-Hauling our possessions along with us. None has papers to prove we are card-carrying members of one such group or another. Y et it seems that more and more of us are identified by work these days, rather than by street.值得一提的是,我们没有谁一出生就属于这些社区,也不是后来我们搬了进来。
这些身份是我们随身携带的,没有人可以拿出文件证明我们是这个或那个群体的会员卡持有者。
然而,不知不觉中人们的身份更倾向于各自所从事的工作,而不是像以往一样由家庭住址来界定。
6、In the past, most Americans live in neighborhoods. W e were members of precincts or parishes or school districts. My dictionary still defines communtiy, first of all in geographic terms, as ―a body of people who live in one place.‖过去大多数彼邻而居的美国人彼此是同一个街区、教区、校区的成员。
今天的词典依然首先从地理的角度来定义社区,称之为“一个由居住在同一地方的人组成的群体”。
7、But today fewer of us do our living in that one place; more of us just use it for sleeping. Now we call our towns ―bedroom suburbs,‖ and many of us, without small children as icebr eakers, would have trouble naming all the people on our street.然而,如今的情况是居住和工作都在同一个地方的人极少,对更多的人来说家成了一个仅仅用来睡觉的地方。
数学与应用数学英文文献及翻译

(外文翻译从原文第一段开始翻译,翻译了约2000字)勾股定理是已知最早的古代文明定理之一。
这个著名的定理被命名为希腊的数学家和哲学家毕达哥拉斯。
毕达哥拉斯在意大利南部的科托纳创立了毕达哥拉斯学派。
他在数学上有许多贡献,虽然其中一些可能实际上一直是他学生的工作。
毕达哥拉斯定理是毕达哥拉斯最著名的数学贡献。
据传说,毕达哥拉斯在得出此定理很高兴,曾宰杀了牛来祭神,以酬谢神灵的启示。
后来又发现2的平方根是不合理的,因为它不能表示为两个整数比,极大地困扰毕达哥拉斯和他的追随者。
他们在自己的认知中,二是一些单位长度整数倍的长度。
因此2的平方根被认为是不合理的,他们就尝试了知识压制。
它甚至说,谁泄露了这个秘密在海上被淹死。
毕达哥拉斯定理是关于包含一个直角三角形的发言。
毕达哥拉斯定理指出,对一个直角三角形斜边为边长的正方形面积,等于剩余两直角为边长正方形面积的总和图1根据勾股定理,在两个红色正方形的面积之和A和B,等于蓝色的正方形面积,正方形三区因此,毕达哥拉斯定理指出的代数式是:对于一个直角三角形的边长a,b和c,其中c是斜边长度。
虽然记入史册的是著名的毕达哥拉斯定理,但是巴比伦人知道某些特定三角形的结果比毕达哥拉斯早一千年。
现在还不知道希腊人最初如何体现了勾股定理的证明。
如果用欧几里德的算法使用,很可能这是一个证明解剖类型类似于以下内容:六^维-论~文.网“一个大广场边a+ b是分成两个较小的正方形的边a和b分别与两个矩形A和B,这两个矩形各可分为两个相等的直角三角形,有相同的矩形对角线c。
四个三角形可安排在另一侧广场a+b中的数字显示。
在广场的地方就可以表现在两个不同的方式:1。
由于两个长方形和正方形面积的总和:2。
作为一个正方形的面积之和四个三角形:现在,建立上面2个方程,求解得因此,对c的平方等于a和b的平方和(伯顿1991)有许多的勾股定理其他证明方法。
一位来自当代中国人在中国现存最古老的含正式数学理论能找到对Gnoman和天坛圆路径算法的经典文本。
取遍数字 数学英文

取遍数字数学英文The Language of Mathematics" with a word count exceeding 1000 words, as requested:Numbers are the fundamental building blocks of our understanding of the world around us. They are the universal language that transcends cultures and barriers, allowing us to communicate, quantify, and make sense of the complexities of our existence. From the ancient civilizations of Mesopotamia and Egypt to the modern technological marvels of our time, the power of numbers has been a constant companion in our journey of discovery and progress.At the heart of this numerical odyssey lies the discipline of mathematics, a field that has evolved alongside human civilization, constantly expanding our horizons and challenging our perceptions. Mathematics is not merely a collection of formulas and equations; it is a way of thinking, a language that allows us to explore the intricate patterns and relationships that govern the universe.One of the most captivating aspects of mathematics is the sheer diversity of the numbers themselves. From the simplicity of the natural numbers to the complexity of irrational and imaginarynumbers, each type of number carries its own unique characteristics and applications. The natural numbers, for instance, form the foundation of our numerical system, allowing us to count, quantify, and compare. These familiar digits, from 1 to 9, and the subsequent powers of 10, have become the backbone of our everyday interactions with numbers, from tallying inventory to calculating the cost of our groceries.As we venture deeper into the realm of mathematics, we encounter numbers that defy the traditional boundaries of our understanding. Fractions, for example, represent the division of a whole into equal parts, enabling us to express and manipulate quantities with greater precision. The introduction of negative numbers, once considered a paradox, has revolutionized our ability to represent and solve complex problems, from tracking financial transactions to mapping the movements of celestial bodies.But the true wonders of mathematics lie in the realm of irrational and imaginary numbers. Irrational numbers, such as the famous pi (π), are numbers that cannot be expressed as a simple fraction, their decimal representations continuing on infinitely without repeating. These enigmatic figures have captivated mathematicians and scientists alike, as they reveal the inherent complexity and beauty of the natural world, from the perfect circles of planets to the intricate patterns of snowflakes.Imaginary numbers, on the other hand, represent a whole new dimension of mathematical exploration. These numbers, denoted by the symbol "i," are defined as the square root of -1, a concept that initially defied logical explanation. Yet, these seemingly abstract constructs have become indispensable in fields ranging from quantum mechanics to electrical engineering, allowing us to model and understand phenomena that defy traditional numerical representations.The power of numbers, however, extends far beyond their mathematical applications. In the realm of language and communication, numbers have become an integral part of our daily lives, serving as a universal medium for expressing ideas and conveying information. From the ubiquitous use of numerical codes in our digital world to the symbolic significance of numbers in various cultural and religious traditions, these seemingly simple entities have become woven into the fabric of human expression.In the realm of art and design, numbers have also played a pivotal role, inspiring and shaping the creative process. The golden ratio, a mathematical proportion found in nature and often employed in art and architecture, has long been celebrated for its aesthetic appeal and its ability to evoke a sense of harmony and balance. Similarly, the use of numerical patterns and symmetries in music and visualarts has been a source of fascination, as artists explore the interplay between the rational and the emotive.As we delve deeper into the world of numbers, we realize that they are not merely abstract constructs, but rather a reflection of the underlying order and structure of the universe. From the microscopic realm of subatomic particles to the vast expanses of the cosmos, numbers serve as the language through which we can understand and quantify the fundamental principles that govern our existence.In the field of science, numbers have been instrumental in unlocking the mysteries of the natural world. The precise measurement and analysis of physical phenomena, from the speed of light to the half-life of radioactive isotopes, have been made possible through the rigorous application of mathematical principles. These numerical insights have not only expanded our knowledge but have also enabled us to harness the power of nature for the betterment of humanity, from the development of life-saving medical treatments to the harnessing of renewable energy sources.As we continue to push the boundaries of our understanding, the role of numbers in shaping our future becomes ever more apparent. In the realm of artificial intelligence and machine learning, numbers are the foundation upon which complex algorithms and models are built, allowing us to make sense of vast troves of data and uncoverhidden patterns that were once beyond our reach.In the end, the true power of numbers lies not just in their ability to quantify and analyze, but in their capacity to inspire wonder, foster creativity, and unlock the secrets of our universe. As we take a tour of the numerical landscape, we are reminded of the enduring legacy of mathematics, a discipline that has been a constant companion in our quest to understand and shape the world around us. It is a journey of discovery that continues to captivate and challenge us, inviting us to explore the infinite possibilities that lie within the realm of numbers.。
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河南理工大学本科毕业设计(论文)外文文献资料翻译院(系部)数学信息科学学院专业名称数学与应用数学年级班级 2009级01班学生姓名闻晶晶学生学号 3109110101082013年6月3 日一类负相伴随机阵列部分和的精致大偏差汪世界 王伟 王文胜(安徽大学数学科学院,合肥,230039) (华东师范大学金融统计学院,上海,200241)摘要本文在一些适当的条件下得到了多风险模型中负相伴随机阵列的精致大偏差,推广了一些已知的结果,同时表明在多风险模型中负相伴结构对精致大偏差同样不具有敏感性.关键词:负相伴随机阵列,大偏差,一致变化尾 学科分类号:O212.3.§1. 引言近年来,很多学者都总结出重尾分布和的精致大偏差,因为用大偏差概率的损失过程来描述破产概率的估计,是一个非常重要的目标风险管理.为此,我们参阅了一些最新文献,如Ng et al.(2004),Tang (2006),Wang et al.(2006),Liu,(2007),Chen and Zhang (2007),,Yang et al.(2009),Liu(2009)等.然而,他们只研究单一类型的风险,即他们总是假定保险公司只提供一种保险合同.在实际生活中,这种假设是不存在的,所以,研究多风险模型的大偏差问题是很有价值的.为此,Wang and Wang (2007)首次把精致大偏差的相关结论扩展到独立索赔多风险模型中.显然,Wang and Wang (2007)的独立性假设是极其不符合现实的.Alam and Saxena (1981)及Joag-Dev and Proschan (1983)中介绍到这种较弱的结构是负相关的.定义1.1 d 是正整数,{}dn n X N∈;是有限的实值随机变量.我们称一维随机变量是负相伴的,如果对dN 任意两个不相交的非空子集T S ,都成立()()()0j ;,;≤∈∈T X g S i X f Cov j i其中()S i X f i ∈;和()T j X g j ∈;是任意两个使得协方差存在且对任意变量都增加的函数.在本文中,我们称{}ki j X 1ij 1,=≥是NA 序列,其中{}1,≥j X ij ,k i ,,1 =,表示关于i 的同分布损失函数()x F i ,满足∞<=i ij EX μ,()()01i >-=x F x F i ,()∞∞-∈,x .我们同样可以假定,对任意k i ,,1 =,C F ∈i ,如果满足()()1limliminf1y x F xy F x →∞= 或 ()()1lim limsup1y x F xy F x →∞=,我们说分布函数F 属于重尾子集C ,其中分布函数F 具有一致变化尾.Cline et al.(1994)也曾研究过重尾子集C ,他称其为‘中间正规变量’.另一个著名的重尾子集被称为控制变量集(D 族). 一个分布函数F 支撑在(),-∞∞上且属于D ,当且仅当对任意0y<1<(或某些()0,1y ∈),()()l i m s u p→∞<∞x F xy F x 成立.对于像R ,S ,L 等其他重尾子集的更多细节,参考文献Ng et al.(2004)或者Wang and Wang (2007).集合()log =inf ,1log **⎧⎫->⎨⎬⎩⎭:FF y J y y ,其中,()()()liminf x F y F xy F x *→∞=.在Tang (2006)的专业用语中,F J *被称为F 的上 Matuszewska指数.{},1,,=i n i k 是k 正整数序列.为方便起见,令∑==ii n j ij n X S 1,k i ,,1 =,()111;,,===∑∑in kk ij i i S k n n X .(){},1,,=i N t i k 是一列关于索赔次数的独立非负整数计数过程,我们假定{}1,1kij i X j =≥和(){},1,,=i N t i k 是相互独立的,且当()1,,→∞=t i k 时,()()i i EN t t λ=→∞.令()()11;,0i N t kij i i S k t X t ===≥∑∑,Tang(2006)研究了带有一致变化尾的负相伴随机变量和的精致大偏差,Chen et al.(2007)和Liu (2007)把Tang (2006)的研究结果扩展到负相伴随机变量的随机和,它们各自具有一致变化尾.在本文中,我们研究多风险模型中的负相伴随机阵列部分和的精致大偏差.我们对一些已知的结论进行推广,发现在多风险模型中精致大偏差的渐近同样呈现负相伴结构.后面的章节安排如下:在第二节中,我们介绍一些预备知识,主要的结果和证明将在第三章节给出,第四章将会给出一个应用程序的主要结果.§2预备知识在这一章节,我们按照惯例用符号n 1nii S X==∑,()()1t N t N ii S X==∑以及G F ≈表示0liminflimsup <≤<∞F FG G. 显然,如果F G ∈,那么,对任意0>c ,()()≈F cx F x .这在T ang and Yan (2002)中同样也可以看到.下面我们给出一些证明定理的引理,引理2.1是对Joag-Dev 和Proschan(1983)的轻微调整.引理2.1 设{},1k X k n ≤≤为一NA 随机变量序列,1,,m A A 为{}1,,n 的任意一列两两不交子集.如果{},1,,=i f i m 为对每个分量不降(或不增)函数,()()11,,,,∈∈j m j m f X j A f X j A 仍为NA 序列,且对任意 ,2,1=n 以及12,,,n x x x ,有()11==⎛⎫>≤> ⎪⎝⎭∏n ni i k k k i P X x P X x 以及()11n ni i k k k i P X x P X x ==⎛⎫≤≤≤ ⎪⎝⎭∏引理2.2 设{},1,2,=k X k 是一列同分布的NA 随机变量,共同发布()F x D ∈,期望为 μ,且()()(),.F x F x x ο-=→∞如果存在某1>r ,使得()1-<∞rE X ,{}11max 0,X X -=-.则对任意给定的常数0γ>,当n →∞时,对x n γ≥一致地有()()(),n P S n x nF x μο-≤=对x n γ≥一致成立,即()()limsup 0n n x nP S n x nF x γμ→∞≥-≤-=.证明:由于{},1,2,=k X k 为NA 序列,根据定义,{},1,2,-=k X k 同样是NA 序列.由Tang(2006)的引理2.3得,对任意0γ>,*>F p J ,必存在某正常数0υ与C ,使得对任意γ>x n ,1,2,=n 有()()()10μμμυ--≤-=-+≥≤-+>+p n n P S n x P S n x nP X Cx()0pnF Cxυμ-≤-++. (2.1)显而易见,对任意给定的*>F p J ,则当x →∞时,有()()p x F x ο-=;对于较大的x ,()0F v x()F x .在(2.1)中,利用条件()()()F x F x ο-=,我们得到()()()()0p n P S n x n x Cx nF x nF x μυμ--≤--++≤()()()0pn F Cx nF x ου-+=()()()1ο⎛⎫≈+ ⎪⎝⎭F x C n F x . 从而引理2.2证毕.注1(1)在引理2.2的证明中,对任意0ε>,用x ε替换x ,当n →∞时,()()()n P S n x nF x μεο-≤-= (2.2) 对x n γ≥一致成立.(2)设{}1,1kij i X j =≥是负相伴序列,且()()1,,=i F x i k 满足定理2.2 的条件.我们可以用数学归纳法证明,对任意0ε>,当i n →∞时,()111i k kk n i i i i i i i P S n x n F x μεο===⎛⎫⎛⎫-≤-= ⎪ ⎪⎝⎭⎝⎭∑∑∑ (2.3)对所有{}max ,1,,γ≥=i x n i k 一致成立.事实上,对2k =和任意()0,12δ∈,由引理2.1,引理2.2和负相伴性质,有2211i n i i i i P S n x με==⎛⎫-≤- ⎪⎝⎭∑∑()()()()12112211n n P S n x P S n x μδεμδε≤-≤--+-≤--()()121122n n P S n x P S n x μδεμδε+-≤--≤- ()()()()()()()()11221122n F x n F x n F x n F x οοοο=++()()()1122n F x n F x ο=+ (2.4) 因此,(2.3)可以直接由(2.4)用归纳假设证出.§3 主要结论及其证明定理 3.1 设{}1,1kij i X j =≥为NA 随机阵列,对任意{}1,,,,1=≥ij i k X j 具有相同的分布()i F x C ∈,有限期望为i μ,且满足()(),i i xF x F x x -=→∞.{},1,2,,=i n i k 为任意给定的k 个正整数,如果对任意的1,,=i k ,存在某1>r 使得<∞rij E X .则对任意给定的0>r ,对所有的1,,=i k ,当n →∞时,有()()111;,,μ==⎛⎫-> ⎪⎝⎭∑∑kkk i i i ii i P S k n n n x n F x , (3.1)对所有{}()max ,1,,k γ≥==∆:i x n i k 一致成立.注2 假定所有()()1,,=i F x i k 是同分布函数,那么(3.1)可以推出Tang(2006)的定理1.1.特别的,如果我们已知{}1,1kij i X j =≥是非负随机变量序列,很容易可以验证定理3.1的条件一定成立.因此,(3.1)验证Liu(2007)的定理2.1.如果{},1,1,,≥=ij X j i k 是独立随机阵列,由(3.1)推出Wang and Wang(2007)的引理3.1.证明 我们用数学归纳法证明(3.1).当2k =时,首先,显然有()()()()122121,212212;,lim inf 1μ=→∞≥∆⎛⎫-> ⎪⎝⎭≥+∑i i i n n x P S n n n x n F x n F x . (3.2)注意,对任意01ε<<,任意0>x ,()()1211222;,μμ-->P S n n n n x()()1211221,μεμε≥->+->-n n P S n x S n x()()2122111,μεμε+->+->-n n P S n x S n x()()()1211221,1μεμε-->+->+n n P Sn x S n x123:I I I =+-. (3.3)先估计1I ,注意到,()()12112221,μεμε=->+->-n n I P S n x S n x()()()1211221μεμε≥->+--≤-n n P S n x P S n x . (3.4) 由Tang (2006)定理2.1得,对任意01δ<<,当1n →∞时, ()()()1111111sup11γμεδε≥->+-<+n x n P S n xn F x. (3.5)又()()()22xF x F x ο-=,则更有()()()22F x F x ο-=成立,由引理2.2,对2γ>x n 一致的有,()()()22222n P S n x n F x μεο-≤-=.综合以上各式,对充分大的1n ,2n ,()()()()()1112211I n F x n F x δεο≥-++ (3.6) 对()2x ≥∆一致成立.同理亦有对充分大的12,n n ,()()()()222111I n F x n F x εο≥++.对()2x ≥∆一致成立.最后我们估计3I ,由于{}12111212,,,,,n n X X X X 为NA ,则由Wang andWang(2007)得,()()()01lim lim sup10,1,2i ii n x ni F x i F x εγε↓→∞≥+-== (3.7)注意到{}12111212,,,,,n n X X X X 是NA ,1n S ,2n S 也是NA.因此,由T ang(2006)的引理2.1和(3.11)得,()()()()123112211μεμε≥->+->+n n I P S n x P S n x()()()()()21122111n F x n F x δεε≤+++()()()211221n F x n F x δ≤+ (3.8) 联合(3.3)-(3.8)得,当12,n n →∞时,对()2x ≥∆一致地有, ()()1211222;,μμ-->P S n n n n x()()()()()()()2112211221n F x n F x n F x n F x δο≥-+++此外,令0δ↓,我们得到(3.2).下面,我们再证()()()()()12121122,211222;,limsup sup1μμ→∞≥∆-->≤+n n x P S n n n n x n F x n F x . (3.9)任意给定()210,∈ε以及0>x ,由NA 性质、引理2.1和T ang (2006)的定理2.1,有, ()()1211222;,μμ-->P S n n n n x()()()()12112211μεμε≤->-+->-n n P S n x P S n x()()121122μεμε+->->n n P S n x P S n x()()()()()()11221122ο≈++++n F x n F x n F x n F x (3.10) 从而(3.9)成立.这样(3.1)对2k =时成立.假定(3.1)对1k -时成立,下面往证结果对k 时也成立.我们采用类似(3.3)的分解法,可得到()11;,...,μ=⎛⎫-> ⎪⎝⎭∑kk i P S k n n n x()1111,μεμε--==⎛⎫≥->+->- ⎪⎝⎭∑∑i k k i k n i i n k k i i P S n x S n x()1111,μεμε--==⎛⎫+->+->- ⎪⎝⎭∑∑k i k i k n k k n i i i i P S n x S n x()()⎪⎭⎫ ⎝⎛+>-+>--∑∑-=-=x n S x n S P i i n k i i i i k i n i i εμεμ1,1111由NA 性质,注1和归纳假设得,()()()111,...,1;,,liminf inf1μ=→∞≥∆=⎛⎫-> ⎪⎝⎭≥∑∑k kk i i i kn n x k i ii P S k n n n x n F x .(3.12)另一方面,利用归纳假设表明,()()()111,...,1;,,limsup sup1μ=→∞≥∆=⎛⎫-> ⎪⎝⎭≤∑∑k kk i i i kn n x k i ii P S k n n n x n F x (3.13)结合(3.12)(3.13),定理证明成立.定理3.2 设{}1,1kij i X j =≥为一负相伴随机阵列,对1,,=i k ,具有相同的分布()i F x C ∈,期望为0μ>i ,且满足()(),i i xF x F x x -=→∞,如果对任意的1,,=i k ,存在某1>r 使得<∞rij E X .再令(){}1ki i N t =为一列相互独立的非负正整数值计数过程()()λ=→∞i EN t t ,()n i ,,2,1 =,且{}1,1kij i X j =≥与(){}1ki i N t =相互独立.如果(){}1ki i N t =满足:对任意0δ>,均存在+>Fp J ,当t →∞时,使得 ()()()()()()()1+t δλολ>=iipii N t EN t I t . (3.14)则对任意固定的{}max ,1,2,,γμ>=i i k ,当t →∞时,有()()()()11;μλλ==⎛⎫-> ⎪⎝⎭∑∑kki i iii i P S k t t x t F t (3.15)对(){}()max,1,,γλ≥==Γ:i x t i k k 一致成立.注3 如果假定所有的()()1,,=i F x i k 为同一分布,则由(3.15)可推出Chen 和Zhang 定理1.2.特别地,如果我们假定{},1,1,,≥=ij X j i k 是非负随机变量序列,可以很轻易的看出满足定理3.2的条件.所以,(3.15)验证了Liu (2007)定理2.2.如果假定{},1,1,,≥=ij X j i k 是一列相互独立的序列,可由(3.15)证出Wang 和Wang (2007)的定理4.1.证明 我们仍然采用数学归纳法证明本定理的结论,其证明思路与定理3.1完全相同,为简洁起见,这里我们只证明2k =情形.为此,我们首先证 ()()()()()()()()()1122211222;liminf inf1λμλμλλ→∞≥Γ-->≥+t x P S t t t x t F x t F x . (3.16)同理,对任意01ε<<以及0>x ,()()()()11222;λμλμ-->P S t t t x()()()()()()1211221,λμελμε≥->+->-t t N N P S t x S t x()()()()()()2122111,λμελμε+->+->-t t N N P S t x S t x()()()()()()()1211221,1λμελμε-->+->+t t N N P S t x S t x123J J J =+-: (3.17)先估计1J ,由于()()()()()()()12111221λμελμε≥->+--≤-t tN N J P S t x P S t x . (3.18)由Chen 和Zhang (2007)的定理1.2易知()()()()()()()1111111lim sup101γλλμελε→∞≥->+-=+t N t x t P S t xt F x. (3.19)现在对任意()0,βε∈,令0μ<,()()()222λμε-≤-tN P S t x()()()()()22221n n P S n x n t P N t n μελμ∞==-≤-+-=∑()()()()2222ελμβελμβ-+-≤--+->-=+∑∑x n txx n tx12:K K =+ (3.20) 首先,运用引理2.2,我们得到()()()()()()22222n x n t xK P S t x P N t n ελμβλμβ-+-≤-≤-≤-=∑()()()()()()2222x n t xnF x P N t n ελμβο-+-≤-==∑()()()()()()()22221n F x n PN t n t F x οολ∞=≤==∑ (3.21) 现在我们估计2K ,为简单起见,我们声明在下文中()2εβμ-属于C .事实上,对任意2*>Fp J ,20μ<,用Tchebychef 不等式,我们可得出()()()()()()222222ελμβλμε-+->-=-≤-=∑n x n t xK P S t x P N t n()()()()(){}()()222+2222λεβλμλ>⎛⎫-≤>+≤ ⎪ ⎪+⎝⎭pN t Cx t p EN t I P N t x t Cx t ()()()(){}()()()22212γλλ>+--≤=O p N t C t p p pEN t I C x t Cx()()22t F x ολ=. (3.22)由Tang (2006)引理2.1中的()()2p x F x ο-=,最后一个等式成立.联合(3.18)-(3,22)得,对任意0δ>都有()()()()()()()1112211J t F x t F x δλεολ≥-++. (3.23)同理可得()()()()()()()2221111J t F x t F x δλεολ≥-++.最后我们估计3J,类比(3.7)我们易得()()()()1limlim sup 10,1,2.ii t x t i F x i F x εγλε↓∞→∞≥±-== (3.24)注意到(){}21i i N t =相互独立以及{}21iji X =为NA 序列,由引理2.1,Chen 和Zhang (2007)以及(3.24)可得,()()()()()()()()123112211λμελμε≤->+->+N t N t J P S t x P S t x()()()()()()112211t F x t F x λελε++()()()()()1122t F x t F x ολλ=+. (3.25) 所以,用(3.23)-(3.25),令0δ↓,对任意充分大的t ,()2x ≥Γ,由()()()()()()()()()1122211222;liminf inf1λμλμλλ→∞≥Γ-->≥+t x P S t t t x t F x t F x证出(3.16)另一方面,我们再证明 ()()()()()()()()()1122211222;limsup sup1λμλμλλ→∞≥Γ-->≤+t x P S t t t x t F x t F x . (3.26)注意到对任意()0,12ε∈以及0>x 并利用NA 性质和(3.10)相同的方法,当(),2t x →∞≥Γ时,有()()()()11222;λμλμ-->P S t t t x()()()()()()()()12112211λμελμε≥->-+->-t t N N P S t x P S t x()()()()()()121122λμελμε+->->t t N N P St x P S t x()()()()()()()()()()1122112211t F x t F x t F x t F x λελελελε-+-+()()()()()()()()()11221122t F x t F x t F x t F x λλολλ++ (3.27)这样我们得到(3.26).联合(3.16)(3.26)(3.15)定理对2k =时成立.定理3.2证明完毕.§4应用本节我们给出一个例子对本章主要结果加以应用.假定某保险公司经营着两个不同险种,而与第一个险种对应的索赔额记为{},1j X X j =≥,为一列独立同分布的非负随机变量,共同分布为F C ∈,有限期望为μ.若该索赔到来的时刻{},1j j σ≥为一更新过程(){}sup 1,σ=≥≤i N t n t ,0>t 其中对任意0t ≥,满足()()λ=<∞i i t EN t .再令{},1j I j ≥为一列Bernoulli 随机变量序列(即,(1)j I j ≥服从两点分布),且j I 的期望为q ,其中01<≤q ,q 表示第j 个索赔到来的概率.假定与公司第二个险种相应的索赔额为{},1j Y j ≥,为另外一列独立同分布的非负随机变量,共同分布为()G F C ≠∈,有限期望为ν.再令()()()2N t N t =Λ为一Cox 过程,其中()N t 为由一列独立同分布的非负随机变量序列{},1j Z j ≥生成的更新过程,且满足1j EZ =,令(),0t t Λ≥为另一个右连续的非降的随机过程,且满足()00Λ=.若()t Λ与()2N t 相互独立,对任意0t ≥()()1Λ<∞=P t .假定上述随机变量序列{}{}{},1,,1,,1j j j X j I j Y j ≥≥≥以及(){}(){}12,0,,0N t t N t t ≥≥相互独立,{},1j I j ≥为NA 序列,则可以看出到t 时刻时公司的累计索赔额为 ()()()1211,0.N t N t jj j j j S t XI Y t ===+≥∑∑ (4.1)这里我们假定公司同时经营着两种不同的险种,因此该模型是Denuit 等(2002)与Ng (2004)所研究的一维风险模型的推广.这里我们假设随机过程()t Λ满足,对任意0t ≥,()():λ*=Λ<∞t E t ,且当t →∞时,()t λ*→∞,对任意0θ>,存在+>Gp J 使得, ()()()()()()()11.θλλ**Λ>+Λ=O p t t E t t记{}1sup ,1,0n n N t I t σ*=≥=≥,则(){}1,0N t t *≥表示在[]0,t 时刻内发生索赔的真实次数.易见()()111N t j j N t I *==∑,以及()()1,0EN t q t t λ*=≥.因此(4.1)式可以被重新改写为 ()()()1211N t N t jjj j S t XY*===+∑∑.用和Wang 等(2007)中第五节相同的方法和定理3.2,我们得到,当t →∞时,有()()()()()()()()11λμνλλλ**-->+P S t q t t x q t F x t G x对任意0γ>,及()(){}1max ,x t t γλγλ*≥一致成立.参考文献[1] Alem, K. and Sexena, K.M.L., Positive dependence in multivariate distribution, Comm. 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(Chinese), 52(2009), 289-300.ψ-混合相依变量线性形式的强稳定性杨延召 刘妍岩(青岛科技大学数学系,青岛,266061)(武汉大学数学与统计学院,武汉,430072)线性形式的强稳定性在科学技术上存在着广泛应用.本文讨论了ψ-混合随机变量列线性形式的强稳定性.通过对ψ-混合随机变量列运用截尾术,借助于ψ-混合随机变量的性质以及 Borel- Cantelli 引理,得到了ψ-混合随机变量线性形式具有强稳健性的充分条件.同时也给出了一些其它形式的结果.关键词:强稳定性,ψ-混合,线性形式. 学科分类号:O211.4§1 序言概率密度估计,非参数非线性回归可能是研究最为广泛的非参数估计问题.许多研究方法已经在独立的观察下独立发展起来.近年来,一些论文因为广泛存在的独立随机变量产生的如强稳定性的线性形式等大量概率问题,就把这些方法扩展到不独立的情况.强稳定的线性形式在生态学、分子生物学、生物化学等领域都有应用.研究线性的强稳定性被大量的定律推动,在线性模型的兼容的最小平方估计中很有用.因此,研究线性强稳定性的重要性是毋庸置疑的.2004年,Gan(2004)研究了几乎收敛的-ρ混合随机变量.对于严平稳序列,-ψ很合序列首次在Blum 等(1963)中首次被提出.-ψ混合序列包括一些被广泛应用的例子,比如可数状态空间马尔可夫过程,在Blum 等(1963)中可以发现更多的-ψ混合序,列的例子.众所周知,极少的关于-ψ混合序列的研究可以被找出.在本文中,我们首先通过使用终止来研究变量,然后通过Broel-Cantelli 引理和ψ-混合序列的性质找到通常情况下ψ-混合序列的强稳定线性形式的充分条件,基于以上结果,我们给出在ψ-混合序列中其他线性形式的一些结果.接下来,我们证明ψ-混合序列强稳定线性形式的一些结果.本文的其他部分组织如下:在第二节中,我们陈述和证明主要的结论,然后在第三节中,我们证明ψ-混合序列中强稳定性的其他线性形式.§2.n X 的强稳定性线性形式在我们叙述主要结论之前,我们先复习几定义个下文即将用到的定义.定义2.1 设{}1,≥n X n 是定义在概率空间(Ω,F ,P)的一列稳定变量.分别用nm F F ,表示σ代数生成的{}m x X i ≤≤1,和{}n i X i ≥,.令()()()()()()1supsup,1-=≠∈∈≥+B P A P AB P r B P A P F B F A p pr p ψ, 如果当0→r 时,()0→r ψ,我们就说{}1,≥n X n 是ψ-混合随机序列.()r ψ是ψ-混合相关系数.定义2.2 一随机变量序列{},1≥n X n ,是强稳定的,如果存在两列常数{}{}∞↑<n n n b d b 0,,则01→--n n d X b ..s a (2.1) 定义2.3一随机变量序列{}1≥n X n ,被非负变量X 所控制,如果存在整数0>c ,则有 ()()1,0,≥∀>∀>≤>n t t X cP t X P n , (2.2) 记为{}X X n <.除特别说明外,全文假定{}1≥n X n ,是ψ-混合随机变量序列,相应的混合系数满足()∞<∑∞=1r r ψ. (2.3)下面的定理是对ψ-混合序列{}n X 线性强稳定性的总结.定理2.1 设{}1,≥n X n 是一列零均值ψ-混合随机变量,{}n b 是一列正数∞↑n b ,若存在某个∞<≤≤∑∞=121n pn pn b X E p ,,则..,011s a Xbi i→∑∞=-为证明定理2.1,需先介绍以下引理.引理2.1([3],引理1.2.11)设{}1,≥n X n 是ψ-混合随机变量,r k kF Y F X +∈∈,,∞<∞<Y E X E ,,则()Y E X E EXEY EXY XY E τψ≤-∞<,.引理2.2 设{}1,≥n X n 是一列零均值ψ-混合随机变量,∑==ni in XS 1,∞<2n EX ,1≥∀n .则0>∀ε,有()2121141max εψε∑∑=∞=≤≤⎥⎦⎤⎢⎣⎡+≤⎪⎭⎫ ⎝⎛>n i il j n j EX l S P . 证明:对0>∀ε,令(){}εωω>=Λ≤≤j nj S 1max :.对任意Λ∈ω,有()(){}εωω>≤≤=j S n j j v ,1:min ,(){}k v k ==Λωω:,当1=k 时,()0max 1=≤≤ωj kj S ,这样j i j i ≠Φ=Λ⋂Λ,且 nk k1=Λ=Λ.由()[]∑⎰∑⎰⎰=Λ=ΛΛ-+==nk k n k k n k nnkkdP S S S S dP S dP S 1221222,进行如[3]和引理2.1同样的讨论,得到()()∑∑∑=∞==≤-ni i l nk k nkEX l S SES 1211ψ,可得()()()Λ≥≥+≥+∑⎰∑∑⎰∑∑=Λ=∞=Λ=∞=P dP S EX l dP S EX l ES nk n ni il nni il k2121212121222εψψ (2.4) 由引理2.1,可得()()∑∑∑∑=∞==⎥⎦⎤⎢⎣⎡+≤-+≤ni i l j i j i ni inEX l X E X E i j EX ES 121122212ψψ ,由以上公式得()()212141εψ∑∑=∞=⎥⎦⎤⎢⎣⎡+≤Λn i i l EX l P . 引理2.3 设{}1,≥n X n 是一列ψ-混合随机变量,满足(2.3)的条件.若 (i )∞<∑∞=1n nEX;(ii )()∞<∑∞=1n nX Var ,则序列∑=nk kX1收敛.证明:对序列{}j j EX X -由引理2.2得知,对任意正整数m,21n n ≤有()()()∑∑∑=∞==≤≤⎥⎦⎤⎢⎣⎡+≤⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧≥-2112112411max n n i i l kn j j j n k n X Var l m m EX X P ψ, 由(ii )知,对任意m 有,()01max lim 12121,=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧≥-∑=≤≤∞→∞→m EX X P kn j jj nk n n n . 所以{}∑-nn nEX X尾部收敛到0 a.s.,即{}∑-nn n EX X 收敛,由(i)知∑nn X 收敛.定理2.1的证明 设()x F n 是n X 的分布函数,令()n n n n b X I X Y ≤=,()⋅I 是示性函数,则()()∑∑⎰∑⎰∑≤≤=⎪⎪⎭⎫ ⎝⎛≤≤n p n pn n b x n n nb x n n n n b X E x dF b x x dF b x b Y E n n 222222,于是∞<≤⎪⎪⎭⎫⎝⎛≤⎪⎪⎭⎫ ⎝⎛∑∑∑n pnpn n n n n n n b X E b Y E b Y Var 22.因此,由定理2.3得,∑-nn nn b EY Y 收敛a.s. (2.5) 因为0=n EX ,故()()∑⎰∑⎰∑==≤nn b x n nn b x n nnn b x xdF b x xdF b EY nn()()x dF bxx dF b xn nb x pnpn nb x nnn∑⎰∑⎰≤≤∞<≤∑np npn bX E . (2.6)由(2.5)和(2.6)知∑nn nb Y一致收敛.且()()()∞<≤≤=≠∑∑⎰∑⎰∑np npn n nb x pnpnb x n nn nbX E x dF bxx dF Y XP nn由Borel-Cantelli 定理,∑nn nb X收敛a.s..应用Kronecker 定理,在每一个概率为1的集合上的任意一个样本点ω有()011→∑=-ωni inX b故有011→∑=-ni inXba.s.§3 其他线性形式的稳定性在这一节中,我们将给出ψ-混合随机变量的其他线性形式的稳定性.所有的证明建立在定理2.1的结果中.定理3.1 设{}{}n n b a 是两列正数,∞↑=n n n n b a b c ,,{}1≥n X n ,是一列ψ-混合随机变量,{}X X n <,令(){}R x x c n Card x N n ∈≤=,:.若下列条件满足(a)()∞<X EN (b)()()()21101≤≤∞<>⎰⎰∞+∞-p dydt y y N t X P ttp p ,则存在R d n ∈ ,,21=N 有011→-∑=-ni n ii nd Xa b a.s.证明 设()∑∑====≤=ni ii nni iin n n n n Y a T X a S c X I X Y 11,,,则()()()()∞<≤>≤>=≠∑∑∑∞=∞=∞=X cEN c X P c c X P Y XP n n n n n n n n111.由Broel-Cantelli 引理知:对任意实数列{}n d ,{}n n n d T b --1和{}n n n d S b --1在相同的集合上收敛到相同的极限.只须证明()011→-∑=-iini inEY Y a b,a.s.就有了定理的∑=-=ni iinnEY a bd11由于(){}1,≥-n EY Y a n n n 是一列零均值ψ-混合随机变量且()∞<∑∞=1l l ψ,得()∑∑∞=--∞=≤-11n pnpn n pn pn n n Y E c c b EY Y a E()∑⎰∞=-->≤`101n c n p p n ndt t X P t pc c(){}⎰∑∞>-->≤0:1tc n pnp n dt c t X P t cp()()⎰⎰∞+∞->≤tp p dydt y y N t X P t cp 1012最后不等式成立是由于以下事实{}{}()⎰∑∑-∞←<<-∞←>-==utp u u c t n p n u t c n p n y dN y c c n n lim lim::()()()⎥⎦⎤⎢⎣⎡+-=⎰≤≤+--∞→u y t p p pu dy y y N p t N t t N u1lim且()()∞→→≤⎰∞+-u dy y t N p u N u up p ,01由条件(2)和定理2.1即可证得.定理3.2 如果我们用如下条件替换定理3.1的条件(1)(2): (3)()∞<X EN ; (4)()∞<⎰∞1ds s X EN ;(5)()n ccnj p jpjnj ο=∑∞=-≤≤1max ;此外假定0=n EX ,得011→∑=-ni ii nXa ba.s.证明:由定理3.1的n n n T S Y ,,.同理有()()∞<≤≠∑∞=X cEN Y XP n n n1.为了证明所要的结果只需须证明()..,011s a EY Y a bni iii→-∑=-.由条件(3)(4)易证011→∑=-ni ii Y a b .这样,我们只需证()011→-∑=-ni iiinEY Y a b.因为(){}n n n EY Y a -是一列ψ-混合随机变量且满足(2.3)得()()∑∑∞=-∞=≤≤-11n n n pn pn n ppn n n c X I X E c c b EY Y a E ()()()∑∞=-≤+>≤1n npnp npnc X I EX c X P c c c()()∑∑∞=-∞=≤+>=11n n p p nn nc X I EX cc c X P c.令0,max 01==≤≤d c d j nj n ,则()()()j j p n nj pn n n ppnn n pp nd X d I EX c d X I EX c c X I EX c≤≤=≤≤≤-∞==-∞=-∞=-∑∑∑∑11111()()()()()∞<⎪⎪⎭⎫⎝⎛>+≤>=<<≤<<≤<<=∑∑∑∑∑∑∑∞=∞=-∞=-∞=∞=--∞=-∞=-1111111111j j j j j j j j jn p np jj j jn p nj j j pc X P cd X P c d X djP ccdd X d P cd X d I EX由定理2.1得知,()∑=-→-ni iiinEY Y a b110证毕.在下文中,令()++→R R x :α为正的不增函数.令()n n nni i n n a b ca b n a ===∑=,,1α,假定(Ⅰ)()()∞<≤≤-∞→-∞→n n n n n n c c n c c n log sup lim log inf lim 011αα;(Ⅱ)对0>x ,()x x +log α是不增的. 在条件(Ⅰ)和(Ⅱ)下,我们有如下定理:定理3.3设{}n X 与()∞<+11log X X E α同分布,则存在n d ,使011→-∑=-n ni ii d Xa ba.s.证明 由n n n c b a ,,的定义及假定(Ⅱ)知,存在0,0,0>>∈βαN m ,对任意0m n ≥,有()n n n c c n βαα≤≤log .故()()1log -≥n n c n c αα,它能确保对任意0m m ≥都有()m c cm mj j222log αα≤∑∞=-.仿照定理3.1,令()n n n n c X I X Y ≤=,则当0m m ≥时,()()()()()()()()()()()()()()()()∞<≤≤+≤≤≤+≤≤≤+≤≤≤+≤⎥⎦⎤⎢⎣⎡≤≤+≤=≤=≤≤--∞=+--∞=--∞=--∞==--=--∞=-∞=-∞=-∞=∑∑∑∑∑∑∑∑∑∑i i m i i i m i ii i mi i m j jmi i i jjm i i i m m j jj mj j j j mj j jmj jj jjc X c I X X E c c X cI EX c c c X c I EX c i c c X c I EX cc c X c I EX c X I EX c c c X I X E c c c X I X E c c b EY Ya E 111121121211212121121211211121212122222log1log 1log 11αβαοαβαοααοο根据定理2.1,得到()011→-∑=-ni iiinEY Y a ba.s.另一方面,()()()()()()()()()()()∞<≥+-≤≥+-≤>+>=>=≠∑∑∑∑∑∑∞=+∞=+∞=-=∞=∞=0000log 1log log 1001111m i i i m i i i i i m i i i m i i ii i i i i ii X X P m c c X X P m c X P c XP c X P Y XP αααα由Borel-Cantelli 定理知,定理3.3的结论正确 ,ini inn EY a b d ∑=-=11.致谢 作者在此呜谢相关文献作者,他们做出的极有价值的研究.参考文献[1] Gan, S.X., Almost sure convergence for ½-mixing random variable sequences, Statist. Prob. L 67(4)(2004), 289-298.[2] Blum, J.R., Hanson, D.L. and Koopmans, L., On the strong law of large number for a cla stochastic processes, Z. Wahrsch., Verw. Gebiete, 2(1963), 1-11.[3] Lu, C.Y . and Lin, Z.Y ., The Limiting Theory of Mixing-Dependent Random Variables, China demic Press, 1997.[4] Chung, K.L., A Course in Probability Theory (2nd Ed.), Academic Press, New York, 1974.三参数威布尔、对数正态及伽马分布下的估计Russell F. KAPPENMAN西北和阿拉斯加渔业中心,国家海洋渔业局,国家海洋气象局,美国华盛顿州西雅图981121984年4月收到 1984年11月修订摘要:威布尔分布、对数正态分布、伽马分布的的位置、尺度、形状参数的新估计是发达的.估计是在封闭的形式,他们并不需要同步的非线性方程组求解.模拟研究结果和其他人已经提出的新的估计的性能进行比较.这些研究表明,新的估计更好,至少考虑到可能的偏差和均方误差.关键词:位置、尺度和形状参数,最大似然估计法、矩估计法,参数估计,仿真§1引言概率密度函数下三参数威布尔、对数正态和伽马分布的形式分别为:()()()[]()[]{}()()[]()[]{}()()()[]()[]()[]b a x b a xc b c b a x f c b a x a x c c b a x f b a x a x c c b a x f c c ---Γ=---=--=--exp 1,,;2ln exp 21,,;ln exp ,,;12221π 在每种情况下,a 是位置参数,b 是尺度参数,c 是形状参数.在这里所考虑的问题是给定一个随机从这些分布中观测的样本,来评估a,b,c.我们提出各个分布的参数估计,进行模拟仿真,然后和其他人提出的估计量的性能作对比.在一系列的论文中,Cohen and Whitten [2-4]报道他们运用最大似然法、改进的最大似然法和矩估计法估计三参数威布尔分布、对数正态分布、伽马分布的研究成果.在考虑多种不同的可能性,他们提出对每个分布的估计建议.对于对数正态分布的位置、尺度、形状参数估计,已经由Munro and Wixley [11]提出及LaRiccia 和Kindermann[10]进一步研究.检查估计在这些论文中设计使用迭代或搜索过程,来同时解决三个非线性方程.迭代或搜索过程有时无法找到方程的解.因此,它很有可能,在任何给定情况下,该程序将无法得到参数估计,即使样品来自假定分布.尤其是来自一个真实的中小程度大小的样本.本文提出的估计有三个重要优势,相比那些仅仅是引用的.首先,它们是相对比较简单的,也就是说,它们是在封闭形式而涉及非线性方程的数值解.第二,估计总是能被发现.最后,最重要的是,这些估计量似乎完成一些东西比那些到这个时候提出的好,至少考虑到偏差和均方误差.这里考虑的三种分布的参数估计都是来自一种相似的方式中.然而,不同的情况可以根据基本的开发方案稍作修改或调整.估计量的发展本质上源于 Wyckoff, Bain, and Engelhardt [12],因他们的工作单单只是威布尔案例.他们的发展是稍做改进的威布尔分布产生较好的估计值,并产生了类似的对数正态分布和伽马分布的参数估计量.对于三个分布中任何一个,我们产生参数估计通过(1)以最初的、非参数估计量的位置为初始位置;(2)假设等于它的初始估计,并寻找形状参数的初始估计;(3)设置第一个次序统计量等于其预期值;(4)在形状参数被其初始估计替换后方程的解为位置参数,位置参数的一个函数替换尺度参数;(5)估计的尺度和形状参数通过假定位置参数等于(4)中得到的参数值.在以下三部分我们将详细介绍,我们将轮流检验威布尔分布、对数正态分布和伽马分布.在下文中,x1,x2,…,X ,N ,表示一个随机样本的N 个观测的分布的参数估计量的次序统计量,Y 代表样本均值,2s 表示样本的方差,F 表示经验分布函数.通常,100p 百分位数是样品q x ,其中np q =,若np 是整数,则[]np q +=1,否则,[]np 表示不超过np 的最大整数.100p 分布百分位上是Up ,()p Up F =.§2威布尔分布Wyckoff, Bain, and Engelhardt [12]提出了下列程序估计这三参数威布尔分布.第一阶段的统计作为位置参数的初始估计.形状参数的初始估计通过假设1x a =,利用估计的c 由 Dubey[6]得到两参数威布尔分布.这个估计为其中,k x 和h x 分别是第94和第17样本百分位数.则威布尔分布为的新的估计解用aˆ来表示,假定a a ˆ=,然后估计参数b 和c,根据Engelhardt and Bain [7],利用b 和c 的估计推出两参数威布尔分布.注意,对于这个程序,b 本质上是由其重新估计的矩估计量替代的,似乎三个参数的估计都因此略有改善;反之,b 由a x m -替代,其中m x 是第63样本百分位数.a 的函数通过等同于第63百分位数样本和分布的百分比得到,后者是b a +.由此得到的a ,b ,c 分别为()[]()[]111ˆ1ˆˆc r c r x x am --= (2.2a ) ()[]()⎭⎬⎫⎩⎨⎧----=∑∑=+=s i in s i i n a x a x s n s nk c 11)ˆln(ˆln ˆ (2.2b ) ()()⎭⎬⎫⎩⎨⎧⎥⎦⎤⎢⎣⎡-+=∑=n a x c b n i i 1ˆln ˆ5772.0exp ˆ (2.2c )数.我们可以根据Engelhardt and Bain [7]中的一个表确定k 值.Cohen and Whitten [4] 研究了威布尔分布参数的最大似然估计法和矩估计法并进行了改进.他们建议在求解非线性方程组的同时找到估计.方程组的前两个方程通过使似然函数关于b 和c 的偏导数为零得到.第三个方程是第一次序统计量与其期望值相等,或()1x F 等于期望值()11+n . Cohen and Whitten [4]提出以比较估计的性能(2.2)和改进的最大似然估计(MMLE )进行模拟统计研究.完成一个相当全面的统计量相对性能的研究所需的工作量,可以利用分布上的()c b b a aˆ,ˆ,ˆ-与a ,b 的取值无关而显著减少,其中c b a ˆ,ˆ,ˆ表示的估计量在(2.1)或最大似然估计中已经给出.这样,为了比较统计量的相对性,需要假定a=0和b=1.形状参数值c=0.5、1、1.5、2、2.5、3.5常被用于仿真研究中.对c 的每一个取值,分别由生a=0,b=1,形状参数相同的威布尔分布生成500个随机样本,每个样本容量20.对于每一个样本,可得到最大似然估计和(2.2)中的估计.在这一次研究中,我们第一次尝试通过1x 与其期望值相等求得最大似然估计.如果这些估计无法找到,我们将使用其他求解最大似然估计的方法.如果没有方法成功找到所求估计,样品将被舍弃.我们使用和Cohen and Whitten [4] 同样的标准,决定是否停止某一个方案.经过500次的实验,对于每个形状参数值,有如下结果:当c=3.5时,有77次未能找到估计值;当c=2.5时,27次未找到;当c=2,3,1.5时,有8次未找到估计值.表1给出了基于未丢弃的样本的最大似然估计和(2.2)估计的偏差和均方误差.a 和c 估计量(2.2)的均方误差总是远远小于那些用最大似然估计得到的a 和c 的估计值的均方误差.这同样适用与b 估计,除了c=0.5的情况.这些结论也基本上适用于偏差比较,除了c=3.5的情况.最大似然估计的偏差在这种情况下较小.表一偏差和均方误差的威布尔分布参数估计使用从a= 0,b= 1的情况下威布尔分布的500个随机样本,样本容量为20.然而,我们再次指出,在c=3.5的情况下,77个样本被丢弃,因此,这并不有助于表1.此外,当a ,b ,c 利用废弃的样本(2.2)的估计结果结合不采用废弃样本的估计结果,(2.2)估计的偏差大大降低了.同样适用于均方误差,除了c =3.5的情况,c 的估计的均方误差略为从1.6599增加到1.8398,这仍然是远远小于其MMLE 的情况.§3对数正态分布对一个对数正态分布的位置参数的初始估计,我们令∑==ni ii x a a 1ˆ, (3.1) 其中()[]n n a 1111-+=,()[]()[]nn i n i n i a 111----= n i ,...,2=.这种随机变量下界的非参数估计是由Cooke [5]推导出来的,它似乎能实现我们猜想最好的几个可能性.对于形状参数的初始估计c ,有()()[]{}()k k n k z E a x a x c 2ˆˆln ˆ1111--=+-, (3.2) 其中1ˆa由(3.1)给出,k x 和1+-k n x 分别是第25和第75个样本百分位数,k z 是一个正态分布下一个样本容量为n 的随机抽样的第二十五样本百分位数.标准正态分布的次序统计量已经被Harter [9]提出.统计量c 的估计是通过假设a 已知,则())ln(ln 1a x a x r k k n ---=+-是来自一个以b ln 为均值,c 为标准偏差的正态分布下四分位变化的随机样本.进一步()()k z cE r E 2-=.如果我们假设1ˆaa =,r 等同于它的期望值,求解c ,所得解即为(3.2).重新估计a ,设()a x -1ln 等同于其自身期望值)(ln 1z cE b +.1z 是标准正态分布下一个样本量为n 的随机样本的第一次序统计量.在生成的方程中,我们用1ˆc替换c ,a 的函数替换b.对我们而言,最佳的函数是y-a ,其中,y 是样本中位数;即()1+=n x y ,如果n 是奇数;()[]2122+-=n n x x y ,如果n 是偶数.这个函数是通过等同()a y -ln 和它的期望值b ln 得到的,求解b.a 修正后的估计可由求解方程的解a 得到.如果这个解是2ˆa ,则b 和c 的估计可以有假定2ˆaa =和寻找这些参数的最大似然估计求得,分别表示参数估计11ˆ,ˆcb .事实证明,对于对数正态分布,如果我们更进一步的话,可以改善估计量a ,b ,c.另一种修订的估计量a 可以通过用1ˆb 和2ˆc 分别替换b 和c ,及()a x -1ln 等于其期望值的等式求得.如果使用最大似然估计和假定aa ˆ=可以重新估计a ,b ,c 表示. 由上可知,最终得到的a ,b ,c 分别为。