DOCSLIB.ORG

Bernoulli Distribution

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Bernoulli Distribution Bernoulli Distribution Example: Toss of coin De¯ne X = 1 if head comes up and X = 0 if tail comes up. 1 Both realizations are equally likely: (X = 1) = (X = 0) = 2 Examples: Often: Two outcomes which are not equally likely: Success of medical treatment ± Interviewed person is female ± Student passes exam ± Transmittance of a disease ± Bernoulli distribution (with parameter ) X takes two values, 0 and 1, with probabilities p and 1 p ± ¡ Frequency function of X ± x 1 x (1 ) ¡ for x 0; 1 p(x) = ¡ 2 f g ½ 0 otherwise Often: ± 1 if event A has occured X = ½ 0 otherwise Example: A = blood pressure above 140/90 mm HG. Distributions, Jan 30, 2003 - 1 - Bernoulli Distribution Let X1; : : : ; Xn be independent Bernoulli random variables with same parameter . Frequency function of X1; : : : ; Xn x1+:::+xn n x1 ::: xn p(x ; : : : ; xn) = p(x ) p(xn) = (1 ) ¡ ¡ ¡ 1 1 ¢ ¢ ¢ ¡ for xi 0; 1 and i = 1; : : : ; n 2 f g Example: Paired-Sample Sign Test Study success of new elaborate safety program ± Record average weekly losses in hours of labor due to accidents before ± and after installation of the program in 10 industrial plants Plant 1 2 3 4 5 6 7 8 9 10 Before 45 73 46 124 33 57 83 34 26 17 After 36 60 44 119 35 51 77 29 24 11 De¯ne for the ith plant 1 if ¯rst value is greater than the second Xi = ½ 0 otherwise Result: 1 1 1 1 0 1 1 1 1 1 The Xi's are independently Bernoulli distributed with unknown parameter . Distributions, Jan 30, 2003 - 2 - Binomial Distribution Let X1; : : : ; Xn be independent Bernoulli random variables Often only interested in number of successes ± Y = X1 + : : : + Xn Example: Paired Sample Sign Test (contd) De¯ne for the ith plant 1 if ¯rst value is greater than the second Xi = ½ 0 otherwise n Y = Xi iP=1 Y is the number of plants for which the number of lost hours has decreased after the installation of the safety program We know: Xi is Bernoulli distributed with parameter ± Xi's are independent ± What is the distribution of Y ? Probability of realization x ; : : : ; xn with y successes: ± 1 y n y p(x ; : : : ; xn) = (1 ) ¡ 1 ¡ Number of di®erent realizations with y successes: n ± y ¡ ¢ Distributions, Jan 30, 2003 - 3 - Binomial Distribution Binomial distribution (with parameters n and ) Let X1; : : : ; Xn be independent and Bernoulli distributed with pa- rameter and n Y = Xi: iP=1 Y has frequency function n y n y p(y) = (1 ) ¡ for y 0; : : : ; n µy¶ ¡ 2 f g Y is binomially distributed with parameters n and . We write Y Bin(n; ): » Note that the number of trials is ¯xed, ± the probability of success is the same for each trial, and ± the trials are independent. ± Example: Paired Sample Sign Test (contd) Let Y be the number of plants for which the number of lost hours has decreased after the installation of the safety program. Then Y Bin(n; ) » Distributions, Jan 30, 2003 - 4 - Binomial Distribution Binomial distribution for n = 10 0.4 0.4 θ = 0.1 θ = 0.3 0.3 0.3 0.2 0.2 p(x) p(x) 0.1 0.1 0.0 0.0 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 x x 0.4 0.4 θ = 0.5 θ = 0.8 0.3 0.3 0.2 0.2 p(x) p(x) 0.1 0.1 0.0 0.0 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 x x Distributions, Jan 30, 2003 - 5 - Geometric Distribution Consider a sequence of independent Bernoulli trials. On each trial, a success occurs with probability . ± Let X be the number of trials up to the ¯rst success. ± What is the distribution of X? x 1 Probability of no success in x 1 trials: (1 ) ¡ ± ¡ ¡ Probability of one success in the xth trial: ± The frequency function of X is x 1 p(x) = (1 ) ¡ ; x = 1; 2; 3; : : : ¡ X is geometrically distributed with parameter . Example: 1 Suppose a batter has probability 3 to hit the ball. What is the chance that he misses the ball less than 3 times? The number X of balls up to the ¯rst success is geometrically distributed 1 with parameter 3. Thus 2 1 1 2 1 2 (X 3) = + + = 0:7037: · 3 3 ¢ 3 3³3´ Distributions, Jan 30, 2003 - 6 - Hypergemetric Distribution Example: Quality Control Quality control - sample and examine fraction of produced units N produced units ± M defective units ± n sampled units ± What is the probability that the sample contains x defective units? The frequency function of X is M N M x n¡x p(x) = ¡ ; x = 0; 1; : : : ; n: ¡ ¢¡N ¢ n X is a hypergeometric¡ ¢ random variable with parameters N, M, and n. Example: Suppose that of 100 applicants for a job 50 were women and 50 were men, all equally quali¯ed. If we select 10 applicants at random what is the probability that x of them are female? The number of chosen female applicants is hypergeometrically distributed with parameters 100, 50, and 10. The frequency function is 50 50 x 10 x p(x) = ¡ for x 0; : : : ; n ¡ ¢100¡ ¢ 10 2 f g ¡ ¢ for x = 0; 1; : : : ; 10. Distributions, Jan 30, 2003 - 7 - Poisson Distribution Often we are interested in the number of events which occur in a speci¯c period of time or in a speci¯c area of volume: Number of alpha particles emitted from a radioactive source during a ± given period of time Number of telephone calls coming into an exchange during one unit of ± time Number of diseased trees per acre of a certain woodland ± Number of death claims received per day by an insurance company ± Characteristics Let X be the number of times a certain event occurs during a given unit of time (or in a given area, etc). The probability that the event occurs in a given unit of time is ± the same for all the units. The number of events that occur in one unit of time is inde- ± pendent of the number of events in other units. The mean (or expected) rate is ¸. ± Then X is a Poisson random variable with parameter ¸ and frequency function x ¸ ¸ p(x) = e¡ ; x = 0; 1; 2; : : : x! Distributions, Jan 30, 2003 - 8 - Poisson Approximation The Poisson distribution is often used as an approximation for binomial probabilities when n is large and is small: x n x n x ¸ ¸ p(x) = (1 ) ¡ e¡ µx¶ ¡ ¼ x! with ¸ = n . Example: Fatalities in Prussian cavalry Classical example from von Bortkiewicz (1898). Number of fatalities resulting from being kicked by a horse ± 200 observations (10 corps over a period of 20 years) ± Statistical model: Each soldier is kicked to death by a horse with probability . ± Let Y be the number of such fatalities in one corps. Then ± Y Bin(n; ) » where n is the number of soldiers in one corps. Observation: The data are well approximated by a Poisson distribution with ¸ = 0:61 Deaths per Year Observed Rel. Frequency Poisson Prob. 0 109 0.545 0.543 1 65 0.325 0.331 2 22 0.110 0.101 3 3 0.015 0.021 4 1 0.005 0.003 Distributions, Jan 30, 2003 - 9 - Poisson Approximation Poisson approximation of Bin(40; ) 1.0 1 1.0 1 = 400 ¸ = 10 0.8 0.8 0.6 0.6 p(x) p(x) 0.4 0.4 0.2 0.2 0.0 0.0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 x x 0.5 1 0.5 = 40 ¸ = 1 0.4 0.4 0.3 0.3 p(x) p(x) 0.2 0.2 0.1 0.1 0.0 0.0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 x x 0.2 1 0.2 = 8 ¸ = 5 0.1 0.1 p(x) p(x) 0.0 0.0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 x x 0.2 1 0.2 = 4 ¸ = 10 0.1 0.1 p(x) p(x) 0.0 0.0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 x x Distributions, Jan 30, 2003 - 10 - Continuous Distributions Uniform distribution U(0; ) 40 U(0, θ) Range (0; 1) 30 1 f(x) = 1(0,)(x) 20 Frequency 10 (X) = 2 0 2 var(X) = −2 −1 0 1 2 3 4 12 X 40 Exponential distribution Exp(¸) Exp(λ) 30 Range [0; ) 1 20 f(x) = ¸ exp( ¸x)1[0; )(x) ¡ 1 Frequency 1 10 (X) = ¸ 1 0 var(X) = −2 −1 0 1 2 3 4 ¸2 X 2 40 Normal distribution (¹; ) 2 N N(µ, σ ) 30 Range ¡ 1 1 f x x ¹ 2 20 ( ) = exp 2 ( ) p ¼2 ¡2 ¡ Frequency 2 ³ ´ (X) = ¹ 10 var(X) = 2 0 −2 −1 0 1 2 3 4 X 6 4 2 0 −2 U(0, θ) Exp(λ) N(µ, σ2) Distributions, Jan 30, 2003 - 11 -.
Load more

Recommended publications
  • A Matrix-Valued Bernoulli Distribution Gianfranco Lovison1 Dipartimento Di Scienze Statistiche E Matematiche “S
    Journal of Multivariate Analysis 97 (2006) 1573–1585 www.elsevier.com/locate/jmva A matrix-valued Bernoulli distribution Gianfranco Lovison1 Dipartimento di Scienze Statistiche e Matematiche “S. Vianelli”, Università di Palermo Viale delle Scienze 90128 Palermo, Italy Received 17 September 2004 Available online 10 August 2005 Abstract Matrix-valued distributions are used in continuous multivariate analysis to model sample data matrices of continuous measurements; their use seems to be neglected for binary, or more generally categorical, data. In this paper we propose a matrix-valued Bernoulli distribution, based on the log- linear representation introduced by Cox [The analysis of multivariate binary data, Appl. Statist. 21 (1972) 113–120] for the Multivariate Bernoulli distribution with correlated components. © 2005 Elsevier Inc. All rights reserved. AMS 1991 subject classification: 62E05; 62Hxx Keywords: Correlated multivariate binary responses; Multivariate Bernoulli distribution; Matrix-valued distributions 1. Introduction Matrix-valued distributions are used in continuous multivariate analysis (see, for exam- ple, [10]) to model sample data matrices of continuous measurements, allowing for both variable-dependence and unit-dependence. Their potentials seem to have been neglected for binary, and more generally categorical, data. This is somewhat surprising, since the natural, elementary representation of datasets with categorical variables is precisely in the form of sample binary data matrices, through the 0-1 coding of categories. E-mail address: [email protected] URL: http://dssm.unipa.it/lovison. 1 Work supported by MIUR 60% 2000, 2001 and MIUR P.R.I.N. 2002 grants. 0047-259X/$ - see front matter © 2005 Elsevier Inc. All rights reserved. doi:10.1016/j.jmva.2005.06.008 1574 G.
    [Show full text]
  • Discrete Probability Distributions Uniform Distribution Bernoulli
    Discrete Probability Distributions Uniform Distribution Experiment obeys: all outcomes equally probable Random variable: outcome Probability distribution: if k is the number of possible outcomes, 1 if x is a possible outcome p(x)= k ( 0 otherwise Example: tossing a fair die (k = 6) Bernoulli Distribution Experiment obeys: (1) a single trial with two possible outcomes (success and failure) (2) P trial is successful = p Random variable: number of successful trials (zero or one) Probability distribution: p(x)= px(1 − p)n−x Mean and variance: µ = p, σ2 = p(1 − p) Example: tossing a fair coin once Binomial Distribution Experiment obeys: (1) n repeated trials (2) each trial has two possible outcomes (success and failure) (3) P ith trial is successful = p for all i (4) the trials are independent Random variable: number of successful trials n x n−x Probability distribution: b(x; n,p)= x p (1 − p) Mean and variance: µ = np, σ2 = np(1 − p) Example: tossing a fair coin n times Approximations: (1) b(x; n,p) ≈ p(x; λ = pn) if p ≪ 1, x ≪ n (Poisson approximation) (2) b(x; n,p) ≈ n(x; µ = pn,σ = np(1 − p) ) if np ≫ 1, n(1 − p) ≫ 1 (Normal approximation) p Geometric Distribution Experiment obeys: (1) indeterminate number of repeated trials (2) each trial has two possible outcomes (success and failure) (3) P ith trial is successful = p for all i (4) the trials are independent Random variable: trial number of first successful trial Probability distribution: p(x)= p(1 − p)x−1 1 2 1−p Mean and variance: µ = p , σ = p2 Example: repeated attempts to start
    [Show full text]
  • STATS8: Introduction to Biostatistics 24Pt Random Variables And
    STATS8: Introduction to Biostatistics Random Variables and Probability Distributions Babak Shahbaba Department of Statistics, UCI Random variables • In this lecture, we will discuss random variables and their probability distributions. • Formally, a random variable X assigns a numerical value to each possible outcome (and event) of a random phenomenon. • For instance, we can define X based on possible genotypes of a bi-allelic gene A as follows: 8 < 0 for genotype AA; X = 1 for genotype Aa; : 2 for genotype aa: • Alternatively, we can define a random, Y , variable this way: 0 for genotypes AA and aa; Y = 1 for genotype Aa: Random variables • After we define a random variable, we can find the probabilities for its possible values based on the probabilities for its underlying random phenomenon. • This way, instead of talking about the probabilities for different outcomes and events, we can talk about the probability of different values for a random variable. • For example, suppose P(AA) = 0:49, P(Aa) = 0:42, and P(aa) = 0:09. • Then, we can say that P(X = 0) = 0:49, i.e., X is equal to 0 with probability of 0.49. • Note that the total probability for the random variable is still 1. Random variables • The probability distribution of a random variable specifies its possible values (i.e., its range) and their corresponding probabilities. • For the random variable X defined based on genotypes, the probability distribution can be simply specified as follows: 8 < 0:49 for x = 0; P(X = x) = 0:42 for x = 1; : 0:09 for x = 2: Here, x denotes a specific value (i.e., 0, 1, or 2) of the random variable.
    [Show full text]
  • Chapter 5 Sections
    Chapter 5 Chapter 5 sections Discrete univariate distributions: 5.2 Bernoulli and Binomial distributions Just skim 5.3 Hypergeometric distributions 5.4 Poisson distributions Just skim 5.5 Negative Binomial distributions Continuous univariate distributions: 5.6 Normal distributions 5.7 Gamma distributions Just skim 5.8 Beta distributions Multivariate distributions Just skim 5.9 Multinomial distributions 5.10 Bivariate normal distributions 1 / 43 Chapter 5 5.1 Introduction Families of distributions How: Parameter and Parameter space pf /pdf and cdf - new notation: f (xj parameters ) Mean, variance and the m.g.f. (t) Features, connections to other distributions, approximation Reasoning behind a distribution Why: Natural justification for certain experiments A model for the uncertainty in an experiment All models are wrong, but some are useful – George Box 2 / 43 Chapter 5 5.2 Bernoulli and Binomial distributions Bernoulli distributions Def: Bernoulli distributions – Bernoulli(p) A r.v. X has the Bernoulli distribution with parameter p if P(X = 1) = p and P(X = 0) = 1 − p. The pf of X is px (1 − p)1−x for x = 0; 1 f (xjp) = 0 otherwise Parameter space: p 2 [0; 1] In an experiment with only two possible outcomes, “success” and “failure”, let X = number successes. Then X ∼ Bernoulli(p) where p is the probability of success. E(X) = p, Var(X) = p(1 − p) and (t) = E(etX ) = pet + (1 − p) 8 < 0 for x < 0 The cdf is F(xjp) = 1 − p for 0 ≤ x < 1 : 1 for x ≥ 1 3 / 43 Chapter 5 5.2 Bernoulli and Binomial distributions Binomial distributions Def: Binomial distributions – Binomial(n; p) A r.v.
    [Show full text]
  • Lecture 6: Special Probability Distributions
    The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di Lecture 6: Special Probability Distributions Assist. Prof. Dr. Emel YAVUZ DUMAN MCB1007 Introduction to Probability and Statistics Istanbul˙ K¨ult¨ur University The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di Outline 1 The Discrete Uniform Distribution 2 The Bernoulli Distribution 3 The Binomial Distribution 4 The Negative Binomial and Geometric Distribution The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di Outline 1 The Discrete Uniform Distribution 2 The Bernoulli Distribution 3 The Binomial Distribution 4 The Negative Binomial and Geometric Distribution The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di The Discrete Uniform Distribution If a random variable can take on k different values with equal probability, we say that it has a discrete uniform distribution; symbolically, Definition 1 A random variable X has a discrete uniform distribution and it is referred to as a discrete uniform random variable if and only if its probability distribution is given by 1 f (x)= for x = x , x , ··· , xk k 1 2 where xi = xj when i = j. The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di The mean and the variance of this distribution are Mean: k k 1 x1 + x2 + ···+ xk μ = E[X ]= xi f (xi ) = xi · = , k k i=1 i=1 1/k Variance: k 2 2 2 1 σ = E[(X − μ) ]= (xi − μ) · k i=1 2 2 2 (x − μ) +(x − μ) + ···+(xk − μ) = 1 2 k The Discrete Uniform Distribution The Bernoulli Distribution The Binomial Distribution The Negative Binomial and Geometric Di In the special case where xi = i, the discrete uniform distribution 1 becomes f (x)= k for x =1, 2, ··· , k, and in this from it applies, for example, to the number of points we roll with a balanced die.
    [Show full text]
  • 11. Parameter Estimation
    11. Parameter Estimation Chris Piech and Mehran Sahami May 2017 We have learned many different distributions for random variables and all of those distributions had parame- ters: the numbers that you provide as input when you define a random variable. So far when we were working with random variables, we either were explicitly told the values of the parameters, or, we could divine the values by understanding the process that was generating the random variables. What if we don’t know the values of the parameters and we can’t estimate them from our own expert knowl- edge? What if instead of knowing the random variables, we have a lot of examples of data generated with the same underlying distribution? In this chapter we are going to learn formal ways of estimating parameters from data. These ideas are critical for artificial intelligence. Almost all modern machine learning algorithms work like this: (1) specify a probabilistic model that has parameters. (2) Learn the value of those parameters from data. Parameters Before we dive into parameter estimation, first let’s revisit the concept of parameters. Given a model, the parameters are the numbers that yield the actual distribution. In the case of a Bernoulli random variable, the single parameter was the value p. In the case of a Uniform random variable, the parameters are the a and b values that define the min and max value. Here is a list of random variables and the corresponding parameters. From now on, we are going to use the notation q to be a vector of all the parameters: Distribution Parameters Bernoulli(p) q = p Poisson(l) q = l Uniform(a,b) q = (a;b) Normal(m;s 2) q = (m;s 2) Y = mX + b q = (m;b) In the real world often you don’t know the “true” parameters, but you get to observe data.
    [Show full text]
  • Chapter 3. Discrete Random Variables and Probability Distributions
    Chapter 3. Discrete random variables and probability distributions. ::::Defn: A :::::::random::::::::variable (r.v.) is a function that takes a sample point in S and maps it to it a real number. That is, a r.v. Y maps the sample space (its domain) to the real number line (its range), Y : S ! R. ::::::::Example: In an experiment, we roll two dice. Let the r.v. Y = the sum of two dice. The r.v. Y as a function: (1; 1) ! 2 (1; 2) ! 3 . (6; 6) ! 12 Discrete random variables and their distributions ::::Defn: A :::::::discrete::::::::random :::::::variable is a r.v. that takes on a ::::finite or :::::::::countably ::::::infinite number of different values. ::::Defn: The probability mass function (pmf) of a discrete r.v Y is a function that maps each possible value of Y to its probability. (It is also called the probability distribution of a discrete r.v. Y ). pY : Range(Y ) ! [0; 1]: pY (v) = P(Y = v), where ‘Y=v’ denotes the event f! : Y (!) = vg. ::::::::Notation: We use capital letters (like Y ) to denote a random variable, and a lowercase letters (like v) to denote a particular value the r.v. may take. Cumulative distribution function ::::Defn: The cumulative distribution function (cdf) is defined as F(v) = P(Y ≤ v): pY : Range(Y ) ! [0; 1]: Example: We toss two fair coins. Let Y be the number of heads. What is the pmf of Y ? What is the cdf? Quiz X a random variable. values of X: 1 3 5 7 cdf F(a): .5 .75 .9 1 What is P(X ≤ 3)? (A) .15 (B) .25 (C) .5 (D) .75 Quiz X a random variable.
    [Show full text]
  • Some Discrete Distributions.Pdf
    CHAPTER 6 Some discrete distributions 6.1. Examples: Bernoulli, binomial, Poisson, geometric distributions Bernoulli distribution A random variable X such that P(X = 1) = p and P(X = 0) = 1 − p is said to be a Bernoulli random variable with parameter p. Note EX = p and EX2 = p, so Var X = p − p2 = p(1 − p). We denote such a random variable by X ∼ Bern (p). Binomial distribution A random variable X has a binomial distribution with parameters n and p if P(X = n k n−k. k) = k p (1 − p) We denote such a random variable by X ∼ Binom (n; p). The number of successes in n Bernoulli trials is a binomial random variable. After some cumbersome calculations one can derive EX = np. An easier way is to realize that if X is binomial, then X = Y1 + ··· + Yn, where the Yi are independent Bernoulli variables, so . EX = EY1 + ··· + EYn = np We have not dened yet what it means for random variables to be independent, but here we mean that the events such as (Yi = 1) are independent. Proposition 6.1 Suppose , where n are independent Bernoulli random variables X := Y1 +···+Yn fYigi=1 with parameter p, then EX = np; Var X = np(1 − p): Proof. First we use the denition of expectation to see that n n X n X n X = i pi(1 − p)n−i = i pi(1 − p)n−i: E i i i=0 i=1 Then 81 82 6. SOME DISCRETE DISTRIBUTIONS n X n! X = i pi(1 − p)n−i E i!(n − i)! i=1 n X (n − 1)! = np pi−1(1 − p)(n−1)−(i−1) (i − 1)!((n − 1) − (i − 1))! i=1 n−1 X (n − 1)! = np pi(1 − p)(n−1)−i i!((n − 1) − i)! i=0 n−1 X n − 1 = np pi(1 − p)(n−1)−i = np; i i=0 where we used the Binomial Theorem (Theorem 1.1).
    [Show full text]
  • Chapter 2: Random Variables Example 1
    ECE511: Analysis of Random Signals Fall 2016 Lecture 1 - 10, 24, 2016 Dr. Salim El Rouayheb Scribe: Peiwen Tian, Lu Liu Chapter 2: Random Variables Example 1. Tossing a fair coin twice: Ω = fHH;HT;TH;TT g: Define for any ! 2 Ω , X(!)=number of heads in !. X(!) is a random variable. Definition 1. A random variable (RV) is a function X: Ω ! R. Definition 2 (Cumulative distribution function(CDF)). F (χ) = P (X ≤ χ): (1) F(x) 1 3 4 0:5 1 4 x −2 −1 1 2 3 4 . Figure 1: Cumulative distribution function of x Example 2. The cumulative distribution function of x is as (Figure 1) 8 > >0 x < 0; <> 1 0 ≤ x < 1; F (x) = 4 X > 3 > 1 ≤ x < 2; :> 4 1 x ≥ 2: Lemma 1. Properties of CDF (1) lim FX (x) = 0 (2) x→−∞ lim FX (x) = 1; (3) x!+1 1 (2) FX (x) is non-decreasing: x1 ≤ x2 =) FX (x1) ≤ FX (x2) (4) (3) FX (x)is continuous from the right lim FX (x + ϵ) = FX (x); ϵ > 0 (5) ϵ!0 (4) P (a ≤ X ≤ b) = P (X ≤ b) − P (X ≤ a) + P (X = a) (6) = FX (b) − FX (a) + P (X = a) (7) (5) P (X = a) = lim FX (a) − FX (a − "); " > 0 (8) "!0 Definition 3. If random variable X has finite or countable number of values, X is called discrete. Example 3. Non-countable example:<: A set S is countable if you can find a bijection of f. f : S ! N: Definition 4. X is continuous if FX (x) is continuous.
    [Show full text]
  • Discrete Distributions: Empirical, Bernoulli, Binomial, Poisson
    Empirical Distributions An empirical distribution is one for which each possible event is assigned a probability derived from experimental observation. It is assumed that the events are independent and the sum of the probabilities is 1. An empirical distribution may represent either a continuous or a discrete distribution. If it represents a discrete distribution, then sampling is done “on step”. If it represents a continuous distribution, then sampling is done via “interpolation”. The way the table is described usually determines if an empirical distribution is to be handled discretely or continuously; e.g., discrete description continuous description value probability value probability 10 .1 0 – 10- .1 20 .15 10 – 20- .15 35 .4 20 – 35- .4 40 .3 35 – 40- .3 60 .05 40 – 60- .05 To use linear interpolation for continuous sampling, the discrete points on the end of each step need to be connected by line segments. This is represented in the graph below by the green line segments. The steps are represented in blue: rsample 60 50 40 30 20 10 0 x 0 .5 1 In the discrete case, sampling on step is accomplished by accumulating probabilities from the original table; e.g., for x = 0.4, accumulate probabilities until the cumulative probability exceeds 0.4; rsample is the event value at the point this happens (i.e., the cumulative probability 0.1+0.15+0.4 is the first to exceed 0.4, so the rsample value is 35). In the continuous case, the end points of the probability accumulation are needed, in this case x=0.25 and x=0.65 which represent the points (.25,20) and (.65,35) on the graph.
    [Show full text]
  • Hand-Book on STATISTICAL DISTRIBUTIONS for Experimentalists
    Internal Report SUF–PFY/96–01 Stockholm, 11 December 1996 1st revision, 31 October 1998 last modification 10 September 2007 Hand-book on STATISTICAL DISTRIBUTIONS for experimentalists by Christian Walck Particle Physics Group Fysikum University of Stockholm (e-mail: [email protected]) Contents 1 Introduction 1 1.1 Random Number Generation .............................. 1 2 Probability Density Functions 3 2.1 Introduction ........................................ 3 2.2 Moments ......................................... 3 2.2.1 Errors of Moments ................................ 4 2.3 Characteristic Function ................................. 4 2.4 Probability Generating Function ............................ 5 2.5 Cumulants ......................................... 6 2.6 Random Number Generation .............................. 7 2.6.1 Cumulative Technique .............................. 7 2.6.2 Accept-Reject technique ............................. 7 2.6.3 Composition Techniques ............................. 8 2.7 Multivariate Distributions ................................ 9 2.7.1 Multivariate Moments .............................. 9 2.7.2 Errors of Bivariate Moments .......................... 9 2.7.3 Joint Characteristic Function .......................... 10 2.7.4 Random Number Generation .......................... 11 3 Bernoulli Distribution 12 3.1 Introduction ........................................ 12 3.2 Relation to Other Distributions ............................. 12 4 Beta distribution 13 4.1 Introduction .......................................
    [Show full text]
  • Discrete Probability Distributions
    Machine Learning Srihari Discrete Probability Distributions Sargur N. Srihari 1 Machine Learning Srihari Binary Variables Bernoulli, Binomial and Beta 2 Machine Learning Srihari Bernoulli Distribution • Expresses distribution of Single binary-valued random variable x ε {0,1} • Probability of x=1 is denoted by parameter µ, i.e., p(x=1|µ)=µ – Therefore p(x=0|µ)=1-µ • Probability distribution has the form Bern(x|µ)=µ x (1-µ) 1-x • Mean is shown to be E[x]=µ Jacob Bernoulli • Variance is var[x]=µ (1-µ) 1654-1705 • Likelihood of n observations independently drawn from p(x|µ) is N N x 1−x p(D | ) = p(x | ) = n (1− ) n µ ∏ n µ ∏µ µ n=1 n=1 – Log-likelihood is N N ln p(D | ) = ln p(x | ) = {x ln +(1−x )ln(1− )} µ ∑ n µ ∑ n µ n µ n=1 n=1 • Maximum likelihood estimator 1 N – obtained by setting derivative of ln p(D|µ) wrt m equal to zero is = x µML ∑ n N n=1 • If no of observations of x=1 is m then µML=m/N 3 Machine Learning Srihari Binomial Distribution • Related to Bernoulli distribution • Expresses Distribution of m – No of observations for which x=1 • It is proportional to Bern(x|µ) Histogram of Binomial for • Add up all ways of obtaining heads N=10 and ⎛ ⎞ m=0.25 ⎜ N ⎟ m N−m Bin(m | N,µ) = ⎜ ⎟µ (1− µ) ⎝⎜ m ⎠⎟ Binomial Coefficients: ⎛ ⎞ N ! ⎜ N ⎟ • Mean and Variance are ⎜ ⎟ = ⎝⎜ m ⎠⎟ m!(N −m)! N E[m] = ∑mBin(m | N,µ) = Nµ m=0 Var[m] = N (1− ) 4 µ µ Machine Learning Srihari Beta Distribution • Beta distribution a=0.1, b=0.1 a=1, b=1 Γ(a + b) Beta(µ |a,b) = µa−1(1 − µ)b−1 Γ(a)Γ(b) • Where the Gamma function is defined as ∞ Γ(x) = ∫u x−1e−u du
    [Show full text]