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Multinomial distribution

From Wikipedia, the free encyclopedia

Multinomial
Parameters

number of trials (integer)
number of mutually exclusive events (integer)

event probabilities, where
Support
PMF
Mean
Variance
Entropy
MGF
CF where
PGF

In probability theory, the multinomial distribution is a generalization of the binomial distribution. For example, it models the probability of counts for each side of a k-sided dice rolled n times. For n independent trials each of which leads to a success for exactly one of k categories, with each category having a given fixed success probability, the multinomial distribution gives the probability of any particular combination of numbers of successes for the various categories.

When k is 2 and n is 1, the multinomial distribution is the Bernoulli distribution. When k is 2 and n is bigger than 1, it is the binomial distribution. When k is bigger than 2 and n is 1, it is the categorical distribution. The term "multinoulli" is sometimes used for the categorical distribution to emphasize this four-way relationship (so n determines the suffix, and k the prefix).

The Bernoulli distribution models the outcome of a single Bernoulli trial. In other words, it models whether flipping a (possibly biased) coin one time will result in either a success (obtaining a head) or failure (obtaining a tail). The binomial distribution generalizes this to the number of heads from performing n independent flips (Bernoulli trials) of the same coin. The multinomial distribution models the outcome of n experiments, where the outcome of each trial has a categorical distribution, such as rolling a k-sided die n times.

Let k be a fixed finite number. Mathematically, we have k possible mutually exclusive outcomes, with corresponding probabilities p1, ..., pk, and n independent trials. Since the k outcomes are mutually exclusive and one must occur we have pi ≥ 0 for i = 1, ..., k and . Then if the random variables Xi indicate the number of times outcome number i is observed over the n trials, the vector X = (X1, ..., Xk) follows a multinomial distribution with parameters n and p, where p = (p1, ..., pk). While the trials are independent, their outcomes Xi are dependent because they must be summed to n.

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  • Multinomial Distributions: Lesson (Basic Probability and Statistics Concepts)

Transcription

Let's continue our look at some discrete probability distributions with an introduction to the multinomial distribution. The multinomial distribution is a generalization of the binomial distribution. In the binomial distribution there are only two possible outcomes on any one individual trial, and we labeled those success and failure. In the multinomial distribution, the number of possible outcomes on any one given trial is allowed to be greater than 2. Let's take a look at an example. This is approximately the distribution of blood types in the United States. And suppose we wanted to know the answer to this question: In a random sample of 10 Americans what is the probability 6 have blood type O, 2 have type A, 1 has type B, and 1 has type AB? When any one individual person is sampled, they're going to have one of these four blood types, according to these probabilities. And we're going to be able to answer this question using the multinomial distribution. Suppose we have n independent trials, and each trial results in 1 of k mutually exclusive outcomes, and these k outcomes are exhaustive, so one of them is going to occur. On any single trial these k outcomes comes occur with probabilities p_1 through p_k. And since these outcomes are mutually exclusive and exhaustive, then they must sum to 1. We also need these probabilities to stay constant from trial to trial. We're going to let the random variable X_i represent the number of occurrences of outcome of i. And i is going to take on the values 1 through k, representing those k possible outcomes on any one individual trial. So we're going to have k random variables, representing a count for each of those possible outcomes. Then the probability the random variable X_1 takes on the value little x_1 and all the way up through the random variable X_k taking on the value little x_k is equal to what we have here. Over here on this side we have p_1, the probability of outcome 1 on any one individual trial, raised to the number of times that outcome 1 happens, and all the way up to here, which is the probability of outcome k occurring on any one individual trial, raised to the number of times we need outcome k to occur. And so what we have here is the probability of any one specific ordering of x_1 occurrences of outcome 1 and x_2 occurrences of outcome 2, all the way up through x_k occurrences of outcome k, and what we have over here is the number of possible orderings that give us x_1 occurrences of outcome 1, all the way up through x_k occurrences of outcome k. And so these multiplied together give us the probability of this happening. We really should list out the possible values of X here. The random variable X_1 can take on the possible values 0,1, 2, all the way up through n. And the same is true of X_2 and all the way up through X_k. So this is true for x_i equalling 0,1, all the way up though n. But we know that n things must happen in total, so the sum of all those individual occurrences, or i equalling 1 through k must equal n. And if we think about this a little bit, any one of these random variables, when viewed individually, it will have a binomial distribution. And if you remember our mean and variance for the binomial distribution, we can say that the expectation of X_i is going to be equal to n times p_i. And the variance of the random variable X_i is going to be equal to np_i times (1-p_i). And you might remember that from our discussion of the mean and variance the binomial distribution. Let's return to our example. In a random sample of 10 Americans, what is the probability 6 have blood type O, 2 have type A, 1 has type B, and 1 has type AB? Well, we've got a random sample here, so knowing one person's blood type is going to tell us nothing about the next person's blood type. So that independence assumption is pretty reasonable here. On any one individual trial, we are going to get one of these 4 blood types. and these probabilities are staying constant from trial to trial. So the multinomial distribution is reasonable here. And we want to find out the probability that the random variable X_1, which is representing the number of people in our sample that have blood type O, the probability that takes on the value 6, and our random variable X_2, representing the number with type A, that takes on the value 2, and X_3 takes on the value 1, and X_4 takes on the value 1. And this is going to be equal to n factorial, we've got a sample size 10, so 10! over x_1 factorial, that's just the number with type O and that's 6 factorial, times 2 factorial times, 1 factorial, times 1 factorial, and now it's time for these probabilities. the probability blood type O happens on any one individual person is 0.44. And we need that to happen 6 times, so we're going to raise that to the sixth power. And then we're going to multiply that by the probability of blood type A, 0.42, squared, because we needed that to happen twice, And then we are multiplying that by 0.10, blood type B, raised to the first power, we need that happen once, and then multiplying that by 0.04 raised to the first power. And if we calculate that, we would see that that is 0.01290, when rounded to five decimal places. Let's look at another example here. An urn contains eight red balls, three yellow balls, and nine white balls. Six balls are randomly selected with replacement. What is the probability 2 are red, 1 is yellow, and 3 are white? This "with replacement' is an important notion here. If we are putting the ball back in and shaking it all up and then randomly selecting again, then the individual trials are independent, and the probability of getting a red ball, or a yellow ball, or white ball, those probabilities are staying constant through the different trials. And so the conditions for our multinomial distribution are satisfied here. And we're interested here in the probability that the random variable X_1, which I'm going to let represent the number red balls, the probability the random variable X_1 takes on the value 2, And X_2, the number of yellow balls, takes on the value 1, and X_3, the number of white balls, takes on the value 3. And this is going to be equal to n factorial, so six factorial, over 2 factorial, the number red balls, times 1 factorial, the number of yellow balls, times 3 factorial, the number of white balls. And now to the probabilities. Well we have 8 red balls, and there is 8+3+9, 20 balls in total. So the probability of getting 1 red on any one individual trial is going to be the eight red balls out of the 20 total, and we need that to happen twice, so we're gonna square that. And then we're going to multiply that by the probability of getting a yellow ball on any one individual trial, which is 3 out of 20, raised to the first power, because we need that to happen once. And then times 9/20, getting a white ball on an individual trial, cubed, because that's got to happen three times. and that works out to 0.13122, to 5 decimal places. Had the sampling been done without replacement, then the trials would no longer be independent, and the conditions of the multinomial distribution would no longer be satisfied. We would have to use something called the multivariate hypergeometric distribution to calculate the probability in this case. And, what the heck, let's run through a quick example of that. Here we've got the same problem we just looked at, except I've changed with replacement to without replacement. And there's a fundamental difference there. When I say without replacement, what that means is, if we pull out a red ball, we're putting it aside and it doesn't go back in, and then we randomly select another ball. So if we draw a red ball out on the first trial, say, it's going to be less likely to get a red ball on the second trial, So those trials are no longer independent. We've still got our random variables X_1, X_2 and X_3. And we still want to know the probability that the random variable X_1, the number of red balls, takes on the value 2. And X_2, the number of yellow balls, takes on the value 1, and X_3, the white balls, takes on the value 3. And we're going to do this through the multivariate hypergeometric distribution. On the bottom, we're going to have the number of possible samples. And we are drawing six balls from a total of 20, there are 20 balls altogether. and from those 20 we are choosing 6, so the bottom is going to be the total number of possible samples. In the top, we're going to have the total number of samples that get us what we want. The total number of samples that get 2 red and 1 yellow and 3 white. And for that we're going to say, well we need to pick from those 8 red balls we need to choose 2, so 8 choose 2. And from the 3 yellow balls we need to pick 1, so times 3 choose 1. And from the 9 white balls, we need to pick 3. 8 choose 2, times 3 choose 1, times 9 choose 3, all divided by 20 choose 6. This is the number of ways of getting what we want, over the total number of possible samples of size 6 that can be chosen from 20. And this, if we work this out, to 5 decimal places is 0.18204. Note that's a little bit different from the probability we calculated when it was with replacement

Definitions

Probability mass function

Suppose one does an experiment of extracting n balls of k different colors from a bag, replacing the extracted balls after each draw. Balls of the same color are equivalent. Denote the variable which is the number of extracted balls of color i (i = 1, ..., k) as Xi, and denote as pi the probability that a given extraction will be in color i. The probability mass function of this multinomial distribution is:

for non-negative integers x1, ..., xk.

The probability mass function can be expressed using the gamma function as:

This form shows its resemblance to the Dirichlet distribution, which is its conjugate prior.

Example

Suppose that in a three-way election for a large country, candidate A received 20% of the votes, candidate B received 30% of the votes, and candidate C received 50% of the votes. If six voters are selected randomly, what is the probability that there will be exactly one supporter for candidate A, two supporters for candidate B and three supporters for candidate C in the sample?

Note: Since we’re assuming that the voting population is large, it is reasonable and permissible to think of the probabilities as unchanging once a voter is selected for the sample. Technically speaking this is sampling without replacement, so the correct distribution is the multivariate hypergeometric distribution, but the distributions converge as the population grows large in comparison to a fixed sample size[1].

Properties

Normalization

The multinomial distribution is normalized according to:

where the sum is over all permutations of such that .

Expected value and variance

The expected number of times the outcome i was observed over n trials is

The covariance matrix is as follows. Each diagonal entry is the variance of a binomially distributed random variable, and is therefore

The off-diagonal entries are the covariances:

for i, j distinct.

All covariances are negative because for fixed n, an increase in one component of a multinomial vector requires a decrease in another component.

When these expressions are combined into a matrix with i, j element the result is a k × k positive-semidefinite covariance matrix of rank k − 1. In the special case where k = n and where the pi are all equal, the covariance matrix is the centering matrix.

The entries of the corresponding correlation matrix are

Note that the number of trials n drops out of this expression.

Each of the k components separately has a binomial distribution with parameters n and pi, for the appropriate value of the subscript i.

The support of the multinomial distribution is the set

Its number of elements is

Matrix notation

In matrix notation,

and

with pT = the row vector transpose of the column vector p.

Visualization

As slices of generalized Pascal's triangle

Just like one can interpret the binomial distribution as (normalized) one-dimensional (1D) slices of Pascal's triangle, so too can one interpret the multinomial distribution as 2D (triangular) slices of Pascal's pyramid, or 3D/4D/+ (pyramid-shaped) slices of higher-dimensional analogs of Pascal's triangle. This reveals an interpretation of the range of the distribution: discretized equilateral "pyramids" in arbitrary dimension—i.e. a simplex with a grid.[citation needed]

As polynomial coefficients

Similarly, just like one can interpret the binomial distribution as the polynomial coefficients of when expanded, one can interpret the multinomial distribution as the coefficients of when expanded, noting that just the coefficients must sum up to 1.

Large deviation theory

Asymptotics

By Stirling's formula, at the limit of , we have

where relative frequencies in the data can be interpreted as probabilities from the empirical distribution , and is the Kullback–Leibler divergence.

This formula can be interpreted as follows.

Consider , the space of all possible distributions over the categories . It is a simplex. After independent samples from the categorical distribution (which is how we construct the multinomial distribution), we obtain an empirical distribution .

By the asymptotic formula, the probability that empirical distribution deviates from the actual distribution decays exponentially, at a rate . The more experiments and the more different is from , the less likely it is to see such an empirical distribution.

If is a closed subset of , then by dividing up into pieces, and reasoning about the growth rate of on each piece , we obtain Sanov's theorem, which states that

Concentration at large n

Due to the exponential decay, at large , almost all the probability mass is concentrated in a small neighborhood of . In this small neighborhood, we can take the first nonzero term in the Taylor expansion of , to obtain

This resembles the gaussian distribution, which suggests the following theorem:

Theorem. At the limit, converges in distribution to the chi-squared distribution .

If we sample from the multinomial distribution , and plot the heatmap of the samples within the 2-dimensional simplex (here shown as a black triangle), we notice that as , the distribution converges to a gaussian around the point , with the contours converging in shape to ellipses, with radii converging as . Meanwhile, the separation between the discrete points converge as , and so the discrete multinomial distribution converges to a continuous gaussian distribution.
[Proof]

The space of all distributions over categories is a simplex: , and the set of all possible empirical distributions after experiments is a subset of the simplex: . That is, it is the intersection between and the lattice .

As increases, most of the probability mass is concentrated in a subset of near , and the probability distribution near becomes well-approximated by

From this, we see that the subset upon which the mass is concentrated has radius on the order of , but the points in the subset are separated by distance on the order of , so at large , the points merge into a continuum. To convert this from a discrete probability distribution to a continuous probability density, we need to multiply by the volume occupied by each point of in . However, by symmetry, every point occupies exactly the same volume (except a negligible set on the boundary), so we obtain a probability density , where is a constant.

Finally, since the simplex is not all of , but only within a -dimensional plane, we obtain the desired result.

Conditional concentration at large n

The above concentration phenomenon can be easily generalized to the case where we condition upon linear constraints. This is the theoretical justification for Pearson's chi-squared test.

Theorem. Given frequencies observed in a dataset with points, we impose independent linear constraints

(notice that the first constraint is simply the requirement that the empirical distributions sum to one), such that empirical satisfy all these constraints simultaneously. Let denote the -projection of prior distribution on the sub-region of the simplex allowed by the linear constraints. At the limit, sampled counts from the multinomial distribution conditional on the linear constraints are governed by which converges in distribution to the chi-squared distribution .
[Proof]

An analogous proof applies in this Diophantine problem of coupled linear equations in count variables ,[2] but this time is the intersection of with and hyperplanes, all linearly independent, so the probability density is restricted to a -dimensional plane. In particular, expanding the KL divergence around its minimum (the -projection of on ) in the constrained problem ensures by the Pythagorean theorem for -divergence that any constant and linear term in the counts vanishes from the conditional probability to multinationally sample those counts.

Notice that by definition, every one of must be a rational number, whereas may be chosen from any real number in and need not satisfy the Diophantine system of equations. Only asymptotically as , the 's can be regarded as probabilities over .

Away from empirically observed constraints (such as moments or prevalences) the theorem can be generalized:

Theorem.

  • Given functions , such that they are continuously differentiable in a neighborhood of , and the vectors are linearly independent;
  • given sequences , such that asymptotically for each ;
  • then for the multinomial distribution conditional on constraints , we have the quantity converging in distribution to at the limit.

In the case that all are equal, the Theorem reduces to the concentration of entropies around the Maximum Entropy.[3][4]

Related distributions

In some fields such as natural language processing, categorical and multinomial distributions are synonymous and it is common to speak of a multinomial distribution when a categorical distribution is actually meant. This stems from the fact that it is sometimes convenient to express the outcome of a categorical distribution as a "1-of-k" vector (a vector with one element containing a 1 and all other elements containing a 0) rather than as an integer in the range ; in this form, a categorical distribution is equivalent to a multinomial distribution over a single trial.

Statistical inference

Equivalence tests for multinomial distributions

The goal of equivalence testing is to establish the agreement between a theoretical multinomial distribution and observed counting frequencies. The theoretical distribution may be a fully specified multinomial distribution or a parametric family of multinomial distributions.

Let denote a theoretical multinomial distribution and let be a true underlying distribution. The distributions and are considered equivalent if for a distance and a tolerance parameter . The equivalence test problem is versus . The true underlying distribution is unknown. Instead, the counting frequencies are observed, where is a sample size. An equivalence test uses to reject . If can be rejected then the equivalence between and is shown at a given significance level. The equivalence test for Euclidean distance can be found in text book of Wellek (2010).[5] The equivalence test for the total variation distance is developed in Ostrovski (2017).[6] The exact equivalence test for the specific cumulative distance is proposed in Frey (2009).[7]

The distance between the true underlying distribution and a family of the multinomial distributions is defined by . Then the equivalence test problem is given by and . The distance is usually computed using numerical optimization. The tests for this case are developed recently in Ostrovski (2018).[8]

Confidence intervals for the difference of two proportions

In the setting of a multinomial distribution, constructing confidence intervals for the difference between the proportions of observations from two events, , requires the incorporation of the negative covariance between the sample estimators and .

Some of the literature on the subject focused on the use-case of matched-pairs binary data, which requires careful attention when translating the formulas to the general case of for any multinomial distribution. Formulas in the current section will be generalized, while formulas in the next section will focus on the matched-pairs binary data use-case.

Wald's standard error (SE) of the difference of proportion can be estimated using:[9]: 378 [10]

For a approximate confidence interval, the margin of error may incorporate the appropriate quantile from the standard normal distribution, as follows:

[Proof]

As the sample size () increases, the sample proportions will approximately follow a multivariate normal distribution, thanks to the multidimensional central limit theorem (and it could also be shown using the Cramér–Wold theorem). Therefore, their difference will also be approximately normal. Also, these estimators are weakly consistent and plugging them into the SE estimator makes it also weakly consistent. Hence, thanks to Slutsky's theorem, the pivotal quantity approximately follows the standard normal distribution. And from that, the above approximate confidence interval is directly derived.

The SE can be constructed using the calculus of the variance of the difference of two random variables:

A modification which includes a continuity correction adds to the margin of error as follows:[11]: 102–3 

Another alternative is to rely on a Bayesian estimator using Jeffreys prior which leads to using a dirichlet distribution, with all parameters being equal to 0.5, as a prior. The posterior will be the calculations from above, but after adding 1/2 to each of the k elements, leading to an overall increase of the sample size by . This was originally developed for a multinomial distribution with four events, and is known as wald+2, for analyzing matched pairs data (see the next section for more details).[12]

This leads to the following SE:

[Proof]

Which can just be plugged into the original Wald formula as follows:

Occurrence and applications

Confidence intervals for the difference in matched-pairs binary data (using multinomial with k=4)

For the case of matched-pairs binary data, a common task is to build the confidence interval of the difference of the proportion of the matched events. For example, we might have a test for some disease, and we may want to check the results of it for some population at two points in time (1 and 2), to check if there was a change in the proportion of the positives for the disease during that time.

Such scenarios can be represented using a two-by-two contingency table with the number of elements that had each of the combination of events. We can use small f for sampling frequencies: , and capital F for population frequencies: . These four combinations could be modeled as coming from a multinomial distribution (with four potential outcomes). The sizes of the sample and population can be n and N respectively. And in such a case, there is an interest in building a confidence interval for the difference of proportions from the marginals of the following (sampled) contingency table:

Test 2 positive Test 2 negative Row total
Test 1 positive
Test 1 negative
Column total

In this case, checking the difference in marginal proportions means we are interested in using the following definitions: , . And the difference we want to build confidence intervals for is:

Hence, a confidence intervals for the marginal positive proportions () is the same as building a confidence interval for the difference of the proportions from the secondary diagonal of the two-by-two contingency table ().

Calculating a p-value for such a difference is known as McNemar's test. Building confidence interval around it can be constructed using methods described above for Confidence intervals for the difference of two proportions.

The Wald confidence intervals from the previous section can be applied to this setting, and appears in the literature using alternative notations. Specifically, the SE often presented is based on the contingency table frequencies instead of the sample proportions. For example, the Wald confidence intervals, provided above, can be written as:[11]: 102–3 

Further research in the literature has identified several shortcomings in both the Wald and the Wald with continuity correction methods, and other methods have been proposed for practical application.[11]

One such modification includes Agresti and Min’s Wald+2 (similar to some of their other works[13]) in which each cell frequency had an extra added to it.[12] This leads to the Wald+2 confidence intervals. In a Bayesian interpretation, this is like building the estimators taking as prior a dirichlet distribution with all parameters being equal to 0.5 (which is, in fact, the Jeffreys prior). The +2 in the name wald+2 can now be taken to mean that in the context of a two-by-two contingency table, which is a multinomial distribution with four possible events, then since we add 1/2 an observation to each of them, then this translates to an overall addition of 2 observations (due to the prior).

This leads to the following modified SE for the case of matched pairs data:

Which can just be plugged into the original Wald formula as follows:

Other modifications include Bonett and Price’s Adjusted Wald, and Newcombe’s Score.

Computational methods

Random variate generation

First, reorder the parameters such that they are sorted in descending order (this is only to speed up computation and not strictly necessary). Now, for each trial, draw an auxiliary variable X from a uniform (0, 1) distribution. The resulting outcome is the component

{Xj = 1, Xk = 0 for k ≠ j } is one observation from the multinomial distribution with and n = 1. A sum of independent repetitions of this experiment is an observation from a multinomial distribution with n equal to the number of such repetitions.

Sampling using repeated conditional binomial samples

Given the parameters and a total for the sample such that , it is possible to sample sequentially for the number in an arbitrary state , by partitioning the state space into and not-, conditioned on any prior samples already taken, repeatedly.

Algorithm: Sequential conditional binomial sampling

S = n
rho = 1
for i in [1,k-1]:
    if rho != 0:
        X[i] ~ Binom(S,p[i]/rho)
    else X[i] = 0
    S = S - X[i]
    rho = rho - p[i]
X[k] = S

Heuristically, each application of the binomial sample reduces the available number to sample from and the conditional probabilities are likewise updated to ensure logical consistency.[14]

Software implementations

  • The MultinomialCI R package allows the computation of simultaneous confidence intervals for the probabilities of a multinomial distribution given a set of observations.[15]

See also

Further reading

  • Evans, Morton; Hastings, Nicholas; Peacock, Brian (2000). Statistical Distributions (3rd ed.). New York: Wiley. pp. 134–136. ISBN 0-471-37124-6.
  • Weisstein, Eric W. "Multinomial Distribution". MathWorld. Wolfram Research.

References

  1. ^ "probability - multinomial distribution sampling". Cross Validated. Retrieved 2022-07-28.
  2. ^ Loukas, Orestis; Chung, Ho Ryun (2023). "Total Empiricism: Learning from Data". arXiv:2311.08315 [math.ST].
  3. ^ Loukas, Orestis; Chung, Ho Ryun (April 2022). "Categorical Distributions of Maximum Entropy under Marginal Constraints". arXiv:2204.03406.
  4. ^ Loukas, Orestis; Chung, Ho Ryun (June 2022). "Entropy-based Characterization of Modeling Constraints". arXiv:2206.14105.
  5. ^ Wellek, Stefan (2010). Testing statistical hypotheses of equivalence and noninferiority. Chapman and Hall/CRC. ISBN 978-1439808184.
  6. ^ Ostrovski, Vladimir (May 2017). "Testing equivalence of multinomial distributions". Statistics & Probability Letters. 124: 77–82. doi:10.1016/j.spl.2017.01.004. S2CID 126293429.Official web link (subscription required). Alternate, free web link.
  7. ^ Frey, Jesse (March 2009). "An exact multinomial test for equivalence". The Canadian Journal of Statistics. 37: 47–59. doi:10.1002/cjs.10000. S2CID 122486567.Official web link (subscription required).
  8. ^ Ostrovski, Vladimir (March 2018). "Testing equivalence to families of multinomial distributions with application to the independence model". Statistics & Probability Letters. 139: 61–66. doi:10.1016/j.spl.2018.03.014. S2CID 126261081.Official web link (subscription required). Alternate, free web link.
  9. ^ Fleiss, Joseph L.; Levin, Bruce; Paik, Myunghee Cho (2003). Statistical Methods for Rates and Proportions (3rd ed.). Hoboken, N.J: J. Wiley. p. 760. ISBN 9780471526292.
  10. ^ Newcombe, R. G. (1998). "Interval Estimation for the Difference Between Independent Proportions: Comparison of Eleven Methods". Statistics in Medicine. 17 (8): 873–890. doi:10.1002/(SICI)1097-0258(19980430)17:8<873::AID-SIM779>3.0.CO;2-I. PMID 9595617.
  11. ^ a b c "Confidence Intervals for the Difference Between Two Correlated Proportions" (PDF). NCSS. Retrieved 2022-03-22.
  12. ^ a b Agresti, Alan; Min, Yongyi (2005). "Simple improved confidence intervals for comparing matched proportions" (PDF). Statistics in Medicine. 24 (5): 729–740. doi:10.1002/sim.1781. PMID 15696504.
  13. ^ Agresti, A.; Caffo, B. (2000). "Simple and effective confidence intervals for proportions and difference of proportions result from adding two successes and two failures". The American Statistician. 54 (4): 280–288. doi:10.1080/00031305.2000.10474560.
  14. ^ "11.5: The Multinomial Distribution". Statistics LibreTexts. 2020-05-05. Retrieved 2023-09-13.
  15. ^ "MultinomialCI - Confidence Intervals for Multinomial Proportions". CRAN. 11 May 2021. Retrieved 2024-03-23.
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