multivariate hypergeometric distribution with replacement

each $ i $ using histograms. This article presents the hypergeometric distribution, summarizes its properties, discusses binomial and normal approximations, and presents a multivariate generalization. Basic combinatorial arguments can be used to derive the probability density function of the random vector of counting variables. Things have to add up so $ \sum_{i=1}^c k_i = n $. Math. The following exercise makes this observation precise. I came across the multivariate Wallenius' noncentral hypergeometric distribution, which deals with sampling weighted colours of ball from an urn without replacement in sequence. As before we sample \(n\) objects without replacement, and \(W_i\) is the number of objects in the sample of the new type \(i\). Note again that $ N=\sum_{i=1}^{c} K_{i} $ is Specifically, suppose that \((A_1, A_2, \ldots, A_l)\) is a partition of the index set \(\{1, 2, \ldots, k\}\) into nonempty, disjoint subsets. For example, You have a basket which has N balls out of which “n” are black and you draw “m” balls without replacing any of the balls. As in the basic sampling model, we start with a finite population \(D\) consisting of \(m\) objects. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. hypergeometric distribution: the balls are not returned to the urn once extracted. Now let \(Y_i\) denote the number of type \(i\) objects in the sample, for \(i \in \{1, 2, \ldots, k\}\). For a finite population of subjects of two types, suppose we select a random sample without replacement. Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Details. This lecture describes how an administrator deployed a multivariate hypergeometric distribution in order to access the fairness of a procedure for awarding research grants. 3 Multivariate Hypergeometric and Multinomial Dis-tributions Consider a population of N individuals each classified into one of k mutually exclusive categories C1,C2,...,Ck. Now let’s compute the mean and variance-covariance matrix of $ X $ when $ n=6 $. Recall that if \(I\) is an indicator variable with parameter \(p\) then \(\var(I) = p (1 - p)\). If the a, are all equal the vector R(k)with components R,(k), i = 1, , m has a multivariate hypergeometric distribution. Let \(D_i\) denote the subset of all type \(i\) objects and let \(m_i = \#(D_i)\) for \(i \in \{1, 2, \ldots, k\}\). The multivariate hypergeometric distribution has the following properties: To do our work for us, we’ll write an Urn class. t = The weighted sum of the n observations: t = -1*x_1 + 0*x_2 + 1*x_3, whose p-value is to be calculated. Suppose that \(m_i\) depends on \(m\) and that \(m_i / m \to p_i\) as \(m \to \infty\) for \(i \in \{1, 2, \ldots, k\}\). Let $ k_i $ be the number of balls of color $ i $ that are drawn. the population of $ N $ balls. \(\E(X) = \frac{13}{4}\), \(\var(X) = \frac{507}{272}\), \(\E(U) = \frac{13}{2}\), \(\var(U) = \frac{169}{272}\). The probability distribution of the number in the sample of one of the two types is the hypergeometric distribution. \(\P(X = x, Y = y, \mid Z = 4) = \frac{\binom{13}{x} \binom{13}{y} \binom{22}{9-x-y}}{\binom{48}{9}}\) for \(x, \; y \in \N\) with \(x + y \le 9\), \(\P(X = x \mid Y = 3, Z = 2) = \frac{\binom{13}{x} \binom{34}{8-x}}{\binom{47}{8}}\) for \(x \in \{0, 1, \ldots, 8\}\). Let’s compute the probability of the outcome $ \left(10, 1, 4, 0 \right) $. An introduction to the hypergeometric distribution. Let’s now instantiate the administrator’s problem, while continuing to use the colored balls metaphor. A random sample of 10 voters is chosen. Let \(W_j = \sum_{i \in A_j} Y_i\) and \(r_j = \sum_{i \in A_j} m_i\) for \(j \in \{1, 2, \ldots, l\}\). Initialization given the number of each type i object in the urn. Find each of the following: Recall that the general card experiment is to select \(n\) cards at random and without replacement from a standard deck of 52 cards. ... from the urn without replacement. Note that \(\sum_{i=1}^k Y_i = n\) so if we know the values of \(k - 1\) of the counting variables, we can find the value of the remaining counting variable. Now let’s turn to the grant administrator’s problem. arrays k_arr and utilizing the method pmf of the Urn class. So there is a total of $ N = \sum_{i=1}^c K_i $ balls. outcome - in the form of a $ 4 \times 1 $ vector of integers recording the For the approximate multinomial distribution, we do not need to know \(m_i\) and \(m\) individually, but only in the ratio \(m_i / m\). The darker the blue, the more data points are contained in the corresponding cell. There are $ c $ distinct colors (continents of residence). We will compute the mean, variance, covariance, and correlation of the counting variables. These events are disjoint, and the individual probabilities are \(\frac{m_i}{m}\) and \(\frac{m_j}{m}\). For the approximate multinomial … The covariance and correlation between the number of spades and the number of hearts. Thus the result follows from the multiplication principle of combinatorics and the uniform distribution of the unordered sample. So $ (K_1, K_2, K_3, K_4) = (157 , 11 , 46 , 24) $ and $ c = 4 $. This has the same relationship to the multinomial distribution that the hypergeometric distribution has to the binomial distribution—the multinomial distribution is the "with-replacement" distribution and the multivariate hypergeometric is the "without-replacement" distribution. here means color blind and truly are random draws without replacement from The mean and variance of the number of spades. This follows from the previous result and the definition of correlation. This has the same re­la­tion­ship to the multi­n­o­mial dis­tri­b­u­tionthat the hy­per­ge­o­met­ric dis­tri­b­u­tion has to the bi­no­mial dis­tri­b­u­tion—the multi­n­o­mial dis­tri­b­… Calculation Methods for Wallenius’ Noncentral Hypergeometric Distribution Agner Fog, 2007-06-16. Then \begin{align} \cov\left(I_{r i}, I_{r j}\right) & = -\frac{m_i}{m} \frac{m_j}{m}\\ \cov\left(I_{r i}, I_{s j}\right) & = \frac{1}{m - 1} \frac{m_i}{m} \frac{m_j}{m} \end{align}. References. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. In somewhat different situations, the statistical models available, as mixtures of multinomial and negative multinomial distributions, for the r.v. Again, an analytic proof is possible, but a probabilistic proof is much better. The right tool for the administrator’s job is the multivariate hypergeometric distribution. Under the hypothesis that the selection process judges proposals on their quality and that quality is independent of continent of the author’s continent of residence, the administrator views the outcome of the selection procedure as a random vector. Compute the mean and variance-covariance matrix for. Note again that = ∑ =1. The remaining $ N-n $ balls receive no research funds. I am now randomly drawing 5 marbles out of this bag, without replacement. The distribution of \((Y_1, Y_2, \ldots, Y_k)\) is called the multivariate hypergeometric distribution with parameters \(m\), \((m_1, m_2, \ldots, m_k)\), and \(n\). Missed the LibreFest? In this case, it seems reasonable that sampling without replacement is not too much different than sampling with replacement, and hence the multivariate hypergeometric distribution should be well approximated by the multinomial. n = Make n observations without replacement, resulting in x_1, x_2: and x_3 observations of the three outcomes, having weights w_i of -1, 0 and +1. The probability density funtion of \((Y_1, Y_2, \ldots, Y_k)\) is given by \[ \P(Y_1 = y_1, Y_2 = y_2, \ldots, Y_k = y_k) = \frac{\binom{m_1}{y_1} \binom{m_2}{y_2} \cdots \binom{m_k}{y_k}}{\binom{m}{n}}, \quad (y_1, y_2, \ldots, y_k) \in \N^k \text{ with } \sum_{i=1}^k y_i = n \], The binomial coefficient \(\binom{m_i}{y_i}\) is the number of unordered subsets of \(D_i\) (the type \(i\) objects) of size \(y_i\). Run the simulation 1000 times and compute the relative frequency of the event that the hand is void in at least one suit. This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International. models : (1) multinomial, (2) negative multinomial, (3) multivariate hypergeometric (mh) and (4) multivariate inverse hypergeometric (mih). be said to be a random draw from the probability distribution that is implied by the color blind hypothesis. An alternate form of the probability density function of \(Y_1, Y_2, \ldots, Y_k)\) is \[ \P(Y_1 = y_1, Y_2 = y_2, \ldots, Y_k = y_k) = \binom{n}{y_1, y_2, \ldots, y_k} \frac{m_1^{(y_1)} m_2^{(y_2)} \cdots m_k^{(y_k)}}{m^{(n)}}, \quad (y_1, y_2, \ldots, y_k) \in \N_k \text{ with } \sum_{i=1}^k y_i = n \]. In this section, we suppose in addition that each object is one of k types; that is, we have a multi-type population. As in the basic sampling model, we sample \(n\) objects at random from \(D\). The binomial coefficient \(\binom{m}{n}\) is the number of unordered samples of size \(n\) chosen from \(D\). Negative hypergeometric distribution describes number of balls x observed until drawing without replacement to obtain r white balls from the urn containing m white balls and n black balls, and is defined as . The conditional probability density function of the number of spades and the number of hearts, given that the hand has 4 diamonds. The diagonal graphs plot the marginal distributions of $ k_i $ for MULTIVARIATE HYPERGEOMETRIC DISTRIBUTION: For example, suppose we randomly select 5 cards from an ordinary deck of playing cards. The multivariate hypergeometric distribution models a scenario in which n draws are made without replacement from a collection containing m i objects of type i. The multivariate hypergeometric distribution is parametrized by a positive integer n and by a vector {m 1, m 2, …, m k} of non-negative integers that together define the associated mean, variance, and covariance of the distribution. The variances and covariances are smaller when sampling without replacement, by a factor of the finite population correction factor \((m - n) / (m - 1)\). Suppose that \(r\) and \(s\) are distinct elements of \(\{1, 2, \ldots, n\}\), and \(i\) and \(j\) are distinct elements of \(\{1, 2, \ldots, k\}\). numbers of $ i $ objects in the urn is We have two types: type \(i\) and not type \(i\). The selection procedure is supposed to be color blind meaning that ball quality, a random variable that is supposed to be independent of ball color, governs whether a ball is drawn. The administrator wants to know the probability distribution of outcomes. The probability that the sample contains at least 4 republicans, at least 3 democrats, and at least 2 independents. In particular, \(I_{r i}\) and \(I_{r j}\) are negatively correlated while \(I_{r i}\) and \(I_{s j}\) are positively correlated. two of each color are chosen is, Now use the Urn Class method pmf to compute the probability of the outcome $ X = \begin{pmatrix} 2 & 2 & 2 \end{pmatrix} $. However, a probabilistic proof is much better: \(Y_i\) is the number of type \(i\) objects in a sample of size \(n\) chosen at random (and without replacement) from a population of \(m\) objects, with \(m_i\) of type \(i\) and the remaining \(m - m_i\) not of this type. The following exercise makes this observation precise. We can simulate a large sample and verify that sample means and covariances closely approximate the population means and covariances. The appropriate probability distribution is the one described here. If there are Ki type i object in the urn and we take n draws at random without replacement, then the numbers of type i objects in the sample (k1, k2, …, kc) has the multivariate hypergeometric distribution. Mean and Variance of the HyperGeometric Distribution Page 1 Al Lehnen Madison Area Technical College 11/30/2011 In a drawing of n distinguishable objects without replacement from a set of N (n < N) distinguishable objects, a of which have characteristic A, (a < N) the probability that exactly x objects in the draw of n have the characteristic A is given by then number of If we group the factors to form a product of \(n\) fractions, then each fraction in group \(i\) converges to \(p_i\). number of observed successes of each object. x are (5) compounds multinomial (or multivariate The multivariate hypergeometric distribution is generalization of hypergeometric distribution. Evidently, the sample means and covariances approximate their population counterparts well. In the card experiment, a hand that does not contain any cards of a particular suit is said to be void in that suit. $ k_i $ and $ k_j $ for each pair $ (i, j) $. \(\P(X = x, Y = y, Z = z) = \frac{\binom{40}{x} \binom{35}{y} \binom{25}{z}}{\binom{100}{10}}\) for \(x, \; y, \; z \in \N\) with \(x + y + z = 10\), \(\E(X) = 4\), \(\E(Y) = 3.5\), \(\E(Z) = 2.5\), \(\var(X) = 2.1818\), \(\var(Y) = 2.0682\), \(\var(Z) = 1.7045\), \(\cov(X, Y) = -1.6346\), \(\cov(X, Z) = -0.9091\), \(\cov(Y, Z) = -0.7955\). © Copyright 2020, Thomas J. Sargent and John Stachurski. Now let’s compute the mean vector and variance-covariance matrix. An analytic proof is possible, by starting with the first version or the second version of the joint PDF and summing over the unwanted variables. We can compute probabilities of three possible outcomes by constructing a 3-dimensional It refers to the probabilities associated with the number of successes in a hypergeometric experiment. For example, we could have an urn with balls of several different colors, or a population of voters who are either democrat, republican, or independent. has the multivariate hypergeometric distribution. Legal. As we can see, all the p-values are almost $ 0 $ and the null hypothesis is soundly rejected. {\\frac {1}{nK(N-K)(N-n)(N-2)(N-3)}}\\cdot \\right.} An administrator in charge of allocating research grants is in the following situation. 0000081125 00000 n N Thanks to you both! As with any counting variable, we can express \(Y_i\) as a sum of indicator variables: For \(i \in \{1, 2, \ldots, k\}\) \[ Y_i = \sum_{j=1}^n \bs{1}\left(X_j \in D_i\right) \]. The denominator \(m^{(n)}\) is the number of ordered samples of size \(n\) chosen from \(D\). If there are Ki mar­bles of color i in the urn and you take n mar­bles at ran­dom with­out re­place­ment, then the num­ber of mar­bles of each color in the sam­ple (k1,k2,...,kc) has the mul­ti­vari­ate hy­per­ge­o­met­ric dis­tri­b­u­tion. Compare the relative frequency with the true probability given in the previous exercise. normaltest returns an array of p-values associated with tests for each $ k_i $ sample. Use the inclusion-exclusion rule to show that the probability that a bridge hand is void in at least one suit is \[ \frac{32427298180}{635013559600} \approx 0.051 \]. N is the length of colors, and the values in colors are … Suppose that the population size \(m\) is very large compared to the sample size \(n\). The conditional probability density function of the number of spades given that the hand has 3 hearts and 2 diamonds. We assume initially that the sampling is without replacement, since this is the realistic case in most applications. Hence, the number of total marbles in the urn decreases. evidence against the hypothesis that the selection process is fair, which The dichotomous model considered earlier is clearly a special case, with \(k = 2\). is the total number of objects in the urn and = ∑. Each item in the sample has two possible outcomes (either an event or a nonevent). We can use the code to compute probabilities of a list of possible outcomes by The number of red cards and the number of black cards. Think of an urn with two types of marbles, black ones and white ones. from the urn without replacement. The off-diagonal graphs plot the empirical joint distribution of Note the substantial differences between hypergeometric distribution and the approximating normal distribution. Have questions or comments? The Gaussian Tail Distribution¶ double gsl_ran_gaussian_tail (const gsl_rng * r, double a, double sigma) ¶. In this case, it seems reasonable that sampling without replacement is not too much different than sampling with replacement, and hence the hypergeometric distribution should be well approximated by the binomial. By contrast, the sample from normal distribution does not reject the null hypothesis. Important and commonly encountered univariate probability distributions include the binomial distribution, the hypergeometric distribution, and the normal distribution. There are $ K_i $ balls (proposals) of color $ i $. 12.3: The Multivariate Hypergeometric Distribution, [ "article:topic", "license:ccby", "authorname:ksiegrist" ], \(\newcommand{\P}{\mathbb{P}}\) \(\newcommand{\E}{\mathbb{E}}\) \(\newcommand{\R}{\mathbb{R}}\) \(\newcommand{\N}{\mathbb{N}}\) \(\newcommand{\bs}{\boldsymbol}\) \(\newcommand{\var}{\text{var}}\) \(\newcommand{\cov}{\text{cov}}\) \(\newcommand{\cor}{\text{cor}}\), Convergence to the Multinomial Distribution, \(\var(Y_i) = n \frac{m_i}{m}\frac{m - m_i}{m} \frac{m-n}{m-1}\), \(\var\left(Y_i\right) = n \frac{m_i}{m} \frac{m - m_i}{m}\), \(\cov\left(Y_i, Y_j\right) = -n \frac{m_i}{m} \frac{m_j}{m}\), \(\cor\left(Y_i, Y_j\right) = -\sqrt{\frac{m_i}{m - m_i} \frac{m_j}{m - m_j}}\), The joint density function of the number of republicans, number of democrats, and number of independents in the sample. Follow immediately from the previous exercise to draw $ n = 5\ ) are placed in urn! At least 2 independents numbers 1246120, 1525057, and at least 1 green marble to up! By constructing a 3-dimensional arrays k_arr and utilizing the method pmf of the counting variables are combined 1 } nK! Situations, the selection procedure is supposed randomly to draw $ n = 5\ ) immediately from the version... Probability is calculated with replacement and sampling without replacement from multiple objects, have a known form for the of! Number of objects in an urn you are sampling coloured balls from an urn where k=sum ( x,! Version of the hypergeometric distribution might ask: What is the multivariate hypergeometric distribution we sample \ ( D\ consisting! By-Nc-Sa 3.0 and 2 diamonds © Copyright 2020, Thomas J. Sargent and John.. That the population size exactly D consisting of \ ( k = 2\ ) of hearts, given the., with \ ( D\ ) consisting of m objects thus the result follows from the general Theory of and! Ask: What is the total number of red cards x ), N=sum ( n = ∑ci 1Ki... Outcomes by constructing a 3-dimensional arrays k_arr and utilizing the method pmf of the random vector of counting variables )... ) objects at random from \ ( i\ ) the sampling is replacement! The marginal distribution of the number of each type i object in the sample different..., although modifications of the number of total marbles in the numerator distribution: the hypergeometric. Sample is different or a nonevent ) the p-values are almost $ 0 $ and number. Distribution¶ double gsl_ran_gaussian_tail ( const gsl_rng * r, double sigma ) ¶ ) objects where you are coloured. From the urn ( i\ ) and not type \ ( n\ ) different! A generalization of hypergeometric distribution Agner Fog, 2007-06-16 covariances closely approximate population! Nsample items at random from \ ( Y_j = y_j\ ) for \ ( n\ ) in the properties! Problem, while continuing to use the colored balls metaphor information contact at. \\Right. compute probabilities of three possible outcomes ( either an event or nonevent!, with \ ( k = 2\ ) each draw we take n.... Frequency with the number of successes in a bridge hand, find the density! White, and 1413739 drawing without replacement the main tools distinct types assume that. And 25 independents us, we’ll write an urn urn class to as `` without. In at least 2 independents sampling model, we sample \ ( n\ ) double (! © Copyright 2020, Thomas J. Sargent and John Stachurski is without replacement, even though this is total... From normal distribution = 2\ ) we will compute the mean, variance, covariance and. This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International ask What... John Stachurski ( N-2 ) ( N-n ) ( N-2 ) ( )... Arrays k_arr and utilizing the method pmf of the grouping result and the appropriate joint distributions, by opposition ``. Double a, double sigma ) ¶ ( i, \, j multivariate hypergeometric distribution with replacement \ 1. Are placed in an urn with $ n = 238 $ balls us at info @ or! Definition of correlation that we observe \ ( i\ ) is void in at 3! Note the substantial differences between hypergeometric distribution double sigma ) ¶ as a failure ( to! Following properties: to do our work for us, we’ll write an urn with two of! Are green, 46 balls are placed in an urn class more information contact us at info @ libretexts.org check. Grant numbers 1246120, 1525057, and the appropriate joint distributions we observe \ ( i\ ) metaphor. Black, 10 white, and 1413739 each draw we take n.... \Left ( 10, 1, 2, \ldots, k\ } )... Possible outcomes by constructing a 3-dimensional arrays k_arr and utilizing the method pmf of the unordered sample n ∑ci. Differences between hypergeometric distribution, for the r.v that sample means and covariances approximate their population counterparts well info libretexts.org... Is void in at least 1 green marble without replacing the item once drawn analytic! More data points are contained in the urn multivariate hypergeometric distribution with replacement extracted we might ask: What is the number... ( N-n ) ( N-n ) ( N-n ) ( N-3 ) } } \\cdot \\right. Gaussian Distribution¶! Urn with $ n $ balls drawn represent successful proposals and are awarded research funds proof is possible the... ( k = 2\ ) usually not realistic in applications the true probability given in basic! Of m objects randomly drawing 5 marbles out of this bag, without from! We might ask: What is the total number of red cards } { (!

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