概率统计课件【英文】
概率论英文课件：ch5_4 Hypergeometric Distribution
When n is small compared to N, the nature of the N items changes very little in each draw. Thus, the quantity k/N plays the role of the parameter p. As a result, the binomial distribution may be viewed as a large population edition of the hypergeometric distributions.
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Theorem 5.3 Mean and variance of the hypergeometric distribution h(x; N, n, k) are
= nk/N
,
N n k k n (1 ) N 1 N N
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Relationship to the Binomial Distribution
5.4 Hypergeometric Distribution
Hypergeometric Experiment A random sample of size n is selected without replacement from N items. k of the N items may be classified as successes and N – k are classified as failures.
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Example:
A state runs a lottery( 彩 票 ) in which six numbers are randomly selected from 40,without replacement. A player chooses six numbers before the state’s sample is selected. Let X be the numbers selected by the player that match those selected by the state. X has a Hypergeometric Distribution. Question: (a) What is the probability that six numbers chosen by player match all six numbers in state’s sample? N = 40, k = 6, x = 6 (b) What is the probability that five of the six numbers chosen by a player appear in the state’s sample? N = 40, k = 6, x = 5
概率统计英文讲义
Probability Intro
The sample space S of an experiment is the set of all possible outcomes.
• We must understand the sample space in order to determine the probability of each outcome occurring.
Venn Diagrams
Venn Diagrams show various events graphically, and are sometimes helpful in understanding set theory problems.
Standard set theory results hold:
• For three operations? • n1 n2 n3. • How many passwords of length 5 need to be checked by a
password hacking program if only lower case letters are used? • 265 = 11,881,376.
If one operation can be performed in n1 ways, and for each way, a second can be performed in n2 ways, then the two can be performed a total of n1 n2 ways.
• The set of all elements in both A and B. Denoted A B. • B = 2nd or 3rd flip, but not both, are heads. • B = {HHT, HTH, THT, TTH}. A B = ? • A B = {HHT, HTH}
概率论英文课件：ch4_1 Mean of Random Variables
Chapter 4 Mathematical Expectation4.1 Mean of Random Variables4.2 Variance and Covariance4.3 Means and Variances of LinearCombinations of Random variables4.4 Chebyshev’s Theorem13Definition 4.1 Let X be a random variable with probability distribution f(x).The mean or expected value of X isif X is discrete, andif X is continuous.Remark:The mean of a random variable X can be thought of as a measure of the “center of location”in the sense that it indicates where the “center”of the density line. xx xf X E )()( dx x xf X E)()(4Example 4.1, page 89The probability distribution of a random variable X is given byx =0,1,2,3.f(0)=1/35 f(1)=12/35 f(2)=18/35 f(3)=4/3537334)(x x x f 71235435183512351))(3())(2())(1())(0(5ExampleThe probability distribution of a random variable X is given byelsewherex e x f x 00)( 0)(dx e x X E x 00dx e xe xx 1 0 ( )6Example 4.2, page 90In a gambling game a man is paid $5 if he gets all heads or all tails when three coins are tossed, and he will pay out $3 if either one or two heads show. What is his expected gain?Let Y be the amount of gain per bet. The possible values are 5 and –3 dollars.Let X be the number of heads that occur in tossing three coins. The possible values of X are 0, 1, 2, and 3.Solution:P(Y = 5) = P(X = 0 or X = 3) = 1/8 + 1/8 = ¼P(Y = -3) = P(X =1 or X = 2) = 6/8 = ¾= (5)(1/4) + (-3)(3/4) = –1Interpretation : Over the long run, the gambler will, on average, lose $1 per bet. Most likely, the more the gambler plays the games, the more he would lose.7Notice that in the preceding example, there are two random variables, X and Y ; and Y is a function of X , for example if we letE(Y) = E(g(X))= (5)P(Y = 5) + (-3)P(Y = -3)= (5)[P(X = 0) + P(X = 3)] + (-3)[P(X = 1) + P(X = 2)]= (5)P(X = 0) + (5)P(X = 3) + (-3)P(X = 1) + (-3)P(X = 2) = g(0)P(X = 0) + g(3)P(X = 3) + g(1)P(X = 1) + g(2)P(X = 2)= 2,133,05)(X X X g Y x x f x g )()(8Theorem 4.1 Let X be a random variable with probability distribution f(x). The mean or expected value of random variable g(X)isif X is discrete, andif X is continuous.xX g x f x g X g E )()()]([)(dxx f x g X g E X g )()()]([)(9ExampleLet X denote the length in minutes of a long-distance telephone conversation. Assume that the density for X is given byFind E(X) and E(2X+3)Solution:E(X )= = = 10 E(2X+3)= = 2(10) + 3 = 23 .0)(10/101x e x f x dx e x x 10/0101)32( dxx f x )(dx e x x 10/010110Definition 4.2 Let X and Y be random variables with joint probability distribution f(x, y).The mean or expected value of the random variable g(X, Y)isif X and Y are discrete, andif X and Y are continuous. ),(),(),(),()],([y x Y X g y x f y x g Y X g E dydx y x f y x g Y X g E Y X g),(),()],([),( Extension11ExampleExample:Suppose two dice are rolled, one red and one white. Let X be the number on the top face of the red die, and Y be the number on the top face of the white one. Find E(X+Y).E[X + Y]= == ==3.5 + 3.5 = 7 ),(),()(y x y Y x X P y x 6161),()(x y y Y x X P y x 61616161),(),(x y x y y Y x X yP y Y x X xP 61616161)36/1()36/1(x y x y y x12Example 4.7, page 93. Find E[Y/X]for the densitySolution:= = elsewhere ,0 2020 ,16),(33 y , x y x y x f dydx y x f y x g Y X g E Y X g),(),()],([),( dy dx y x x y 20332016dy dx y x 2042201613In generalIf X and Y are two random variables, f(x, y)is thejoint density function, then:E(X)= = (discrete case) E(X)= (continuous case)E(Y)= = (discrete case)E(Y)= (continuous case)g(x)and h(y)are marginal probability distributions of X and Y , respectively. x y y x xf ),( xx xg )( dxx xg dxdy y x xf )(),( y x y x yf ),( y y yh )( dyy yh dxdy y x yf )(),(。
概率统计(英文)chapter3
Introduction
The concept of a random variable allows us to pass from the experimental outcomes themselves to a numerical function of the outcomes. There are two fundamentally different types of random variables----discrete random variables and continuous random variables. In this chapter, we examine the basic properties and discuss the most important examples of discrete variables, and we will study continuous variables in Chapter 4.
Example 3.3 Example 2.3 described an experiment in which the
number of pumps in use at each of two gas stations was determined. Define rv’s X,Y, and U by X=the total number of pumps in use at the two stations Y=the difference between the number of pumps in use at station 1 and the number in use at station 2 U=the maximum of the numbers of pumps in use at the two stations
概率与统计(英文)chapter 1(2011) Probability and Statistic
The probability & statistics is a science of studying statistic law of random phenomena . This science generated from 17th century, it comes of gambling and is applied in gambling.But now it is the foundation of many sciences, for example, econometrics, control science, information science, decision theory, game theory. Especially, in finance and accounting.
The set of measurements collected for a particular element is called an observation.
• Branched of Statistics
Descriptive statistics
An investigator who has collected data may wish simply to summarize and describe important features of the data. This entails using methods from descriptive statistics. Some of these methods are graphical in nature; the construction of histograms, boxplots, and scatter plots are primary examples. Other descriptive methods involve calculation of numerical summary measures, such as means, standard deviations, and correlations coefficients
概率论英文课件：ch2_5,6,7 Additive Rules
B-A A
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Example: Rich and famous In a certain population, 10% of the people are rich,5% are famous, 3% are rich and famous: P(r)= 10% P(f)= 5% P(rf)= 3% For a person picked at random from the population, (a) What is the probability that the person is not rich? P(r’)=1- P(r)= 90% (b) What is the probability that the person is rich but not famous? P(r-f)=P(r-rf)=P(r)-P(rf)=10% - 3%= 7% (c) What is the probability that the person is either rich or famous? P(r f)=P(r)+P(f)-P(rf) = 10% +5% -3%= 12%
A AB B
S
By Axiom 4 of probability, P(AB ) =P (AB’) + P(A B) + P(A’ B) P(A) = P(AB’) + P(A B) and P(B) = P(BA’) + P(B A). Combine the above two equations to obtain Additive Rule.
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Definition 2.10: Two events A and B are independent if and only if P(B | A) = P(B) or P(A|B) = P(A) . Otherwise, A and B are dependent.
北邮概率统计课件3
在统计学、决策理论、机器学习等领域中经常用到条件期望，特别是在回归分析和预测中。
应用场景
离散型条件方差
应用场景
在统计学、决策理论、机器学习等领域中经常用到条件方差，特别是在回归分析和预测中。
定义
在离散型随机变量中，给定某些信息后，随机变量的方差称为条件方差。
公式
$Var(X|Y=y) = sum (x-mu)^2 times P(X=x|Y=y)$，其中 $mu$ 是条件期望 $E(X|Y=y)$。
当事件A和事件B相互独立时，条件概率等于联合概率，即P(B|A)=P(B)。
条件分布与联合分布的关系
联合分布描述了多个事件同时发生的概率，而条件分布描述了在某个特定事件发生时，其他事件发生的概率。 联合分布和条件分布在描述事件之间的概率关系时是互补的，它们一起构成了完整的概率描述。 在实际应用中，条件分布在贝叶斯推断、统计决策等领域有广泛的应用。
在统计推断、贝叶斯推断、马尔科夫链蒙特卡洛方法等领域中，条件概率密度函数有着广泛的应用。
连续型条件期望
条件期望是在给定某个随机变量或随机向量取值的条件下，另一个随机变量的期望值。对于连续型随机变量，条件期望描述了在给定另一随机变量值的条件下，该随机变量的期望值。
定义
在统计推断、回归分析、时间序列分析等领域中，条件期望有着广泛的应用。
PART TWO
离散型条件分布
3.1关键技术 3.2技术难点 3.3案例分析
离散型条件概率
定义
在离散型随机变量中，给定某些信息后，随机事件发生的概率称为条件概率。
公式
$P(A|B) = frac{P(A cap B)}{P(B)}$，其中 $P(A cap B)$ 是事件 $A$ 和 $B$ 同时发生的概率，$P(B)$ 是事件 $B$ 发生的概率。
概率论与数理统计教程第一章精品PPT课件
4.互不相容（互斥）事件 AB
5.事件的和（并） AB
A1,A2, ,An 的并，记作
n
A i.
i 1
6.对立事件(互逆事件)
若AB ,且AB ,
则B为A的对立事件，记A为 。
7.差事件 AB A B AAB
事件的运算(Operation of Events)
样本点简记为： wi ={直到第i次才击中目标}, i = 1,2,…。
则样本空间可记为 Ω={w1，w2，…} 。
随机事件(Random Events)
在随机试验中可能的结果称为随机事件， 简称事件. 如在掷色子试验中，观察掷出的点数 .
“掷出1点”
"掷出奇数点"
事件就是由样本点组成的某个集合.
（1）事件“A与B发生，C不发生”可表示成
ABC
（2）事件“A，B，C中至少有一个发生”可表示成
ABC
（3）事件“A，B，C中恰好有一个发生”可表示成
A B C A B C A B C
A={w2，w4，w6，w8 , w10}
85 1946 7 2 3 10
B~"取出的球号大于8" B={w9，w10} C~"取出的球号大于10" D~"取出的球号不大于10"
事件间的关系 (Relation of Events)
1.事件的包含 AB
2.事件的相等 AB
3.事件的积（交） AB
n
机事件吗？
两个特殊的事件：
然
即在试验中必定发生的事件，记为Ω ;
可
即在一次试验中不可能发生的事件，记为φ 。
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Event – set of outcomes
Venn Diagrams
Outcomes are mutually exclusive – disjoint
S
1
4 Event A
2 5
3
6
Outcomes
An Example from Card Games
What is the probability of drawing two of the same card in a row in a shuffled deck of cards?
Communications Speech and Image Processing Machine Learning Decision Making Network Systems Artificial Intelligence
Used in many undergraduate courses (every grad course)
By their average behavior By the likelihood of particular outcomes
Allows us to build models for many physical behaviors
Speech, images, traffic …
Applications
Introduction to Probability: Counting Methods
Rutgers University Discrete Mathematics for ECE 14:332:202
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Why Probability?
We can describe processes for which the outcome is uncertain
Experiment
Roll a dice Roll a six 1,2,…6 Dice rolled is odd
Outcome – any possible observation of an exp.
Sample Space – the set of all possible outcomes
# outcom es _ in _ event _ space P(Event) = # outcom es _ in _ sam ple_ space
Expressed as the ratio of favorable outcomes to total outcomes
-- Only when all outcomes are EQUALLY LIKELY
Probabilities from Combinations
Rule of Product:
Total number of two card combinations? We need to find all the combinations of suit and value that describe our event set: use rule of product to find the number of combinations First, we find number of values – 13 choices, and choices of suits: 4 4!
Combinatorics
Number of ways to arrange n distinct objects n! Number of ways to obtain an ordered sequence of k objects from a set of n: n!/(n-k)! -- k permutation Number of ways to choose k objects out of n distinguishable objects:
n n! k k!(n k )!
This one comes up a lot!
Set Theory and Probability
We use the same ideas from set theory in our study of probability
Coin flipping Dice rolling Card Games
Combinatorics
Mathematical tools to help us count:
How many ways can 12 distinct objects be arranged? How many different sets of 4 objects be chosen from a group of 20 objects? -- Extend this to find probabilities …
Methods of Counting
One way of interpreting probability is by the ratio of favorable to total outcomes Means we need to be able to count both the desired and the total outcomes For illustration, we explore only the most important applications:
Event Space
Sample Space
Sample Space/Event Space
Venn Diagram
Event Space (set of favorable outcomes)
S all possible outcomes
{A,A} {K,2}
Calculating the Probability
Experiment
Pulling two cards from the deck All outcomes that describe our event: Two cards are the same All Possible Outcomes All combinations of 2 cards from a deck of 52