Difference Between Continuous and Discrete Uniform Distribution

Discrete Lifetime Models

N. Unnikrishnan Nair , ... N. Balakrishnan , in Reliability Modelling and Analysis in Discrete Time, 2018

Discrete Uniform Distribution

The discrete uniform distribution arises from (3.30) when $z=1$ , $s=0$ and $a=1$ , with probability mass function

(3.50) $f(x)=\frac{1}{b},x=1,\dots ,b.$

It has distribution function $F(x)=\frac{x}{b}$ and survival function $S(x)=\frac{b-x+1}{b}$ . Various distributional characteristics are as follows:

$\begin{array}{c}\hfill \mu =\frac{(b+1)(2b+1)}{6},{\mu }_2=\frac{b^2-1}{12},\hfill \\ \hfill P(t)=\frac{t(1-t^b)}{b(1-t)},t\ne 1.\hfill \end{array}$

If $X_1,\dots ,X_n$ are independent random variables with distribution in (3.50), then $X_{(1)}=\min X_i$ and $X_{(n)}=\max X_i$ , $i=1,2,\dots ,n$ have respective probability mass functions

$P(X_{(1)}=x)=\frac{{(b-x+1)}^n-{(b-x)}^n}{b^n}$

and

$P(X_{(n)}=x)=\frac{x^n-{(x-1)}^n}{b^n}.$

Among the reliability functions, the hazard rate is $h(x)=\frac{1}{b-x+1}$ and the mean residual life function is $m(x)=\frac{(b-x+1)}{2}$ , giving

$h(x)m(x)=\frac{1}{2},$

a constant, characterizing the uniform distribution as a special case of the negative hypergeometric distribution (see Chapter 1). The variance residual life is found to be

${\sigma }^2(x)=\frac{(b-x+1)(b-x-1)}{12}=\frac{m(x)(m(x)-1)}{3}.$

Also, the functions in reversed time turns out to be

$\begin{array}{c}\hfill \lambda (x)=\frac{1}{x},\text{ a reciprocal linear function,}\hfill \\ \hfill r(x)=\frac{x}{2},\text{ a linear function of}x\text{,}\hfill \end{array}$

with their product

$\lambda (x)r(x)=\frac{1}{2},$

a constant. Further, the reversed variance residual life becomes

$v(x)=\frac{x(x-2)}{12}=\frac{r(x)(r(x)-1)}{3},$

which is also a characterization of the discrete uniform distribution.

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Discrete Probability Distributions

S. Sinharay , in International Encyclopedia of Education (Third Edition), 2010

Uniform (Discrete) Distribution

In fields such as survey sampling, the discrete uniform distribution often arises because of the assumption that each individual is equally likely to be chosen in the sample on a given draw. The PMF of a discrete uniform distribution is given by $p\left(X=x\right)=\frac{1}{n+1},x=0,1,\dots n$ , which implies that X can take any integer value between 0 and n with equal probability. The mean and variance of the distribution are $\frac{n}{2}$ and $\frac{n\left(n+2\right)}{12}$ .

To generate a random number from the discrete uniform distribution, one can draw a random number R from the U(0, 1) distribution, calculate S  =   (n + 1)R, and take the integer part of S as the draw from the discrete uniform distribution.

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Probability and Sampling Distributions

Rudolf J. Freund , ... Donna L. Mohr , in Statistical Methods (Third Edition), 2010

2.3.3 The Discrete Uniform Distribution

Suppose the possible values of a random variable from an experiment are a set of integer values occurring with the same frequency. That is, the integers 1 through $k$ occur with equal probability. Then the probability of obtaining any particular integer in that range is $1∕k$ and the probability distribution can be written

$\begin{array}{l}\hfill p\left(y\right)=1∕k,y=1,2,\dots ,k.\end{array}$

This is called the discrete uniform (or rectangular) distribution, and may be used for all populations of this type, with $k$ depending on the range of existing values of the variable. Note that we are able to represent many different distributions with one function by using a letter ( $k$ in this case) to represent an arbitrary value of an important characteristic. This characteristic is the only thing that differs between the distributions, and is called aparameterof the distribution. All probability distributions are characterized by one or more parameters, and the descriptive parameters, such as the mean and variance, are known functions of those parameters. For example, for this distribution

$\begin{array}{lll}\hfill \mu =\left(k+1\right)∕2& & \hfill \\ \hfill \end{array}$

and

$\begin{array}{lll}\hfill {\sigma }^2=\left(k^2-1\right)∕12.& & \hfill \\ \hfill \end{array}$

A simple example of an experiment resulting in a random variable having the discrete uniform distribution consists of tossing a fair die. Let $Y$ be the random variable describing the number of spots on the top face of the die. Then

$\begin{array}{lll}\hfill p\left(y\right)=1∕6,y=1,2,\dots ,6,& & \hfill \\ \hfill \end{array}$

which is the discrete uniform distribution with $k=6$ . The mean of $Y$ is

$\begin{array}{lll}\hfill \mu =\left(6+1\right)∕2=3.5,& & \hfill \\ \hfill \end{array}$

and the variance is

$\begin{array}{lll}\hfill {\sigma }^2=\left(36-1\right)∕12=2.917.& & \hfill \\ \hfill \end{array}$

Note that this is an example where the random variable can never take the mean value.

Example 2.5

Simulating a Distribution

The discrete uniform distribution is frequently used in simulation studies. A simulation study is exactly what it sounds like, a study that uses a computer to simulate a real phenomenon or process as closely as possible. The use of simulation studies can often eliminate the need for costly experiments and is also often used to study problems where actual experimentation is impossible.

When the process being simulated requires the use of a probability distribution to describe it, the technique is often referred to as a Monte Carlo method. For example, Monte Carlo methods have been used to simulate collisions between photons and electrons, the decay of radioactive isotopes, and the effect of dropping an atomic bomb on a city.

The basic ingredient of a Monte Carlo simulation is the generation of random numbers (see, for example, Owen, 1962). Random numbers can, for example, be generated to consist of single digits having the discrete uniform distribution with $k=10$ . Using the digits 0 through 9, such random digits can be used to simulate the outcomes of Example 2.2. For each simulated interview we generate a random digit. If the value of the digit is 0 or 1, the outcome is "had childhood measles"; otherwise (digits 2 through 9) the outcome is "did not." The outcome "had" then occurs with a probability of 0.2. The result of the experiment involving a single couple is then simulated by using a pair of such integers, one for each individual.

Solution

Simulation studies usually involve large numbers of simulated events, but for illustration purposes we use only 10 pairs. Assume that we have obtained the following 10 pairs of random numbers (from a table or generated by a computer):

$15386839495419793814$

In the first pair (15), the first digit "1" signifies "had," while the second digit "5" indicates "did not"; hence, for this couple, $y=1$ . For the second pair, $y=0$ , and so forth. The relative frequency distribution for this simulated sample of ten pairs is shown in Table 2.7.

Table 2.7. Simulation of Measles Probabilities

This result is somewhat different from the theoretical distribution obtained with the use of probability theory because considerable variability is expected in small samples. A sample of 1000 would come much closer but would still not produce the theoretical distribution exactly.

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Introduction

Mark A. Pinsky , Samuel Karlin , in An Introduction to Stochastic Modeling (Fourth Edition), 2011

Problems

1.3.1

Suppose that X has a discrete uniform distribution on the integers 0, 1, …, 9, and Y is independent and has the probability distribution Pr{Y = k} = a_k for k = 0, 1, …. What is the distribution of Z = X + Y (mod 10), their sum modulo 10?

1.3.2

The mode of a probability mass function p (k) is any value k * for which p (k *) ≥ p (k) for all k. Determine the mode(s) for

(a)

The Poisson distribution with parameter λ> 0.

(b)

The binomial distribution with parameters n and p.

1.3.3

Let X be a Poisson random variable with parameterλ. Determine the probability that X is odd.

1.3.4

Let U be a Poisson random variable with meanμ. Determine the expected value of the random variable V = 1/(1 + U).

1.3.5

Let Y = N − X where X has a binomial distribution with parameters N and p. Evaluate the product moment E [XY] and the covariance Cov[X, Y].

1.3.6

Suppose (X ₁, X ₂, X ₃) has a multinomial distribution with parameters M and π_i > 0 for i = 1, 2, 3, with π₁ + π₂ + π₃ = 1.

(a)

Determine the marginal distribution for X _1.

(b)

Find the distribution for N = X ₁ + X ₂.

(c)

What is the conditional probability Pr{X ₁ = k | N = n} for 0 ≤ k ≤ n?

1.3.7

Let X and Y be independent Poisson distributed random variables having means μ and v, respectively. Evaluate the convolution of their mass functions to determine the probability distribution of their sum Z = X + Y.

1.3.8

Let X and Y be independent binomial random variables having parameters (N, p) and (M, p), respectively. Let Z = X + Y.

(a)

Argue that Z has a binomial distribution with parameters (N + M, p) by writing X and Y as appropriate sums of Bernoulli random variables.

(b)

Validate the result in (a) by evaluating the necessary convolution.

1.3.9

Suppose that X and Y are independent random variables with the geometric distribution

$p\left(k\right)=\left(1-\pi \right){\pi }^k\text{for}k=0,1,\mathrm{\dots }.$

Perform the appropriate convolution to identify the distribution of Z = X + Y as a negative binomial.

1.3.10

Determine numerical values to three decimal places for Pr{X = k}, k = 0, 1, 2, when

(a)

X has a binomial distribution with parameters n = 10 and p = 0.1.

(b)

X has a binomial distribution with parameters n = 100 and p = 0.01.

(c)

X has a Poisson distribution with parameter λ = 1.

1.3.11

Let X and Y be independent random variables sharing the geometric distribution whose mass function is

$p\left(k\right)=\left(1-\pi \right){\pi }^k\text{for}k=0,1,\mathrm{\dots },$

where 0 < π < 1. Let U = min{X, Y}, V = max{X, Y}, and W = V − U. Determine the joint probability mass function for U and W and show that U and W are independent.

1.3.12

Suppose that the telephone calls coming into a certain switchboard during a one-minute time interval follow a Poisson distribution with mean λ = 4. If the switchboard can handle at most 6 calls per minute, what is the probability that the switchboard will receive more calls than it can handle during a specified one-minute interval?

1.3.13

Suppose that a sample of 10 is taken from a day's output of a machine that produces parts of which 5% are normally defective. If 100% of a day's production is inspected whenever the sample of 10 gives 2 or more defective parts, then what is the probability that 100% of a day's production will be inspected? What assumptions did you make?

1.3.14

Suppose that a random variable Z has the geometric distribution

$p_Z\left(k\right)=p{\left(1-p\right)}^k\text{for}k=0,1,\mathrm{\dots },$

where p = 0.10.

(a)

Evaluate the mean and variance of Z.

(b)

What is the probability that Z strictly exceeds 10?

1.3.15

Suppose that X is a Poisson distributed random variable with mean λ = 2. Determine Pr{X ≤ λ}.

1.3.16

Consider the generalized geometric distribution defined by

$p_k=b{\left(1-p\right)}^k\text{for}k=1,2,\mathrm{\dots },$

and

$p_0=1-\sum _{k=1}^{\infty }pk,$

where 0 < p < 1 and p ≤ b ≤ p/(1 − p).

(a)

Evaluate p₀ in terms of b and p.

(b)

What does the generalized geometric distribution reduce to when b = p? When b = p/(1 − p)?

(c)

Show that N = X + Z has the generalized geometric distribution when X is a Bernoulli random variable for which Pr{X = 1} =α, 0<α< 1, and Z independently has the usual geometric distribution given in (1.25).

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Flexible regression modelling under shape constraints

Andrew A. Manderson , ... Berwin A. Turlach , in Flexible Bayesian Regression Modelling, 2020

9.4.2 Sampling from the posterior distribution

Given the complete specification of our family of models, we can start to implement an RJMCMC scheme to sample from the posterior distribution of interest. We will specify appropriate proposal distributions for each of the parameters of interest, and discuss the acceptance probability calculation.

9.4.2.1 Dimension proposal

We specify upper and lower bounds for the proposal distribution of the degree of the polynomial, based on the prior values for $q_{\text{min}}$ and $q_{\text{max}}$ . This is to ensure that the Markov chain in polynomial-degree space is neither initialised in an inadmissible region (according to the prior), nor proposes polynomial degrees with zero prior mass. For example, we should never propose polynomials of degree smaller than $q_{\text{min}}$ . The existence of a $q_{\max }$ also allows for straightforward generation of initial values, which will be discussed in Section 9.4.2.2.

When the Markov chain is in either the maximum or the minimum dimension space, the proposal for $q^{⁎}$ is a discrete uniform distribution on that current dimension and the dimension immediately lower or higher than it, respectively. When not in the maximum or minimum dimension states, the proposal is a discrete uniform distribution on its current dimension and the dimensions immediately above and below it. This distribution is best explained by the code used to generate proposals from it, which is shown in Fig. 9.5. This simple proposal distribution admits a straightforward term to the proposal ratio in the acceptance probability expression. It is 1 if the previous degree and proposed degree are not $q_{\max }$ or $q_{\min }$ , and it is $\frac{3}{2}$ or $\frac{2}{3}$ if either the proposed or previous degree are.

Figure 9.5. The R code that proposes a new polynomial degree.

9.4.2.2 Regression coefficient proposal

As discussed in Section 9.2, we define our models in terms of the orthogonal regression coefficients γ . This is advantageous as all the coefficients are orthogonal to each other, and as such a random walk with diagonal covariance matrix can be used to produce acceptable proposals. In the monomial space this would not be true, as the covariance structure for all β coefficients is hard to determine, and the probable values of β would also change considerably each time the degree of the polynomial changed.

When the proposed dimension is the same as that of the previous dimension, generating the proposal for the regression coefficients is straightforward:

(9.18) ${\mathit{\gamma }}^{⁎}\sim \mathrm{N}({\mathit{\gamma }}^{[t]},{\sigma }_{\gamma ,\mathrm{innov}}^2{\mathcal{I}}_{q^{[t]}+1}),$

where ${\sigma }_{\gamma ,\mathrm{innov}}^2$ is the innovation variance associated with the proposal distribution of γ , and ${\mathcal{I}}_{q+1}$ is the $(q+1)\times (q+1)$ identity matrix.

When the proposed dimension does not match the current dimension, things are more difficult. Consider a proposed dimension one greater than the previous dimension, $q^{{⁎}^{\prime }}=q^{[t]}+1$ . We could use a random walk proposal for the first $q^{[t]}+1$ components of ${\mathit{\gamma }}^{{⁎}^{\prime }}$ ; however, the ( ${(q)}^{{⁎}^{\prime }}+1$ )th component does not exist in ${\mathit{\gamma }}^{[t]}$ . As such we need a different proposal distribution for this specific coefficient. It would be preferable to use a random walk with mean equal to the value of the last time we were in the $q^{{⁎}^{\prime }}$ space, i.e.

${\gamma }_{q^{{⁎}^{\prime }}}^{{⁎}^{\prime }}\sim \mathrm{N}({\gamma }_{q^{{⁎}^{\prime }}}^{[s]},{\sigma }_{\gamma ,\mathrm{innov}}^2),$

where s is the time point where the chain was last in state $q^{{⁎}^{\prime }}$ . However, as this depends upon a time point prior to $t-1$ it constructs a sampling scheme that is no longer Markovian, and as such breaks the detailed balance required for the Metropolis–Hastings algorithm. As a result, we use an independent proposal for ${\gamma }_{q^{{⁎}^{\prime }}+1}^{{⁎}^{\prime }}$ , with a very particular mean to ensure it proposes in the right region of the parameter space. To obtain this mean we first fit a polynomial of degree $q_{\max }$ using either [10] or the R package MonoPoly by [17]. The estimates for β are then converted to their corresponding orthonormal coefficients γ for use as means of independent proposal distributions. Mathematically,

(9.19) ${\mathit{\gamma }}^{{⁎}^{\prime }}\sim \mathrm{N}({[{\mathit{\gamma }}^{[t]},{\mu }_{\gamma ,q^{{⁎}^{\prime }}+1}]}^{\top },{\sigma }_{\gamma ,\mathrm{innov}}^2{\mathcal{I}}_{q^{{⁎}^{\prime }}+1}),$

where ${\mu }_{\gamma ,q^{{⁎}^{\prime }}+1}$ includes the ( $q^{{⁎}^{\prime }}+1$ )th component of the aforementioned initial $q_{\max }$ fit. There is an acceptance probability implication of this choice, as the independently proposed term no longer cancels in the ratio of proposal distributions. This will be discussed in more detail in Section 9.4.3.

9.4.2.3 Variance proposal

We use a log-normal random walk to propose values for ${({\sigma }^2)}^{⁎}$ , i.e.

(9.20) ${({\sigma }_y^2)}^{⁎}\sim \mathrm{Log}-\mathrm{normal}(\mathrm{Log}({({\sigma }_y^2)}^{[t]}),{\sigma }_{{\sigma }^2,\mathrm{innov}}^2),$

as we need the value of ${\sigma }_y^2$ to remain positive.

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Cognitive Computing: Theory and Applications

S. Chakraborty , in Handbook of Statistics, 2016

2.2 Prior Distribution

In this section we specify priors distribution on the model parameters as described in Table 1.

Prior distributions on $\left(T_k^j,M_k^j\right)$ : We will follow the prior distribution for $\left(T_k^j,M_k^j\right)$ as suggested by Chipman et al. (2010). The prior distribution of the tree parameters for the j-th component conditional on the number of trees m _j are assigned as

(6) $\begin{array}{l}\hfill \pi \left(\left(T_1^j,M_1^j\right),\dots ,\left(T_{m_j}^j,M_{m_j}^j\right)|m_j\right)=\prod _{k=1}^{m_j}\pi \left(M_k^j|T_k^j\right)\pi \left(T_k^j\right)\\ \hfill \mathrm{and}\pi \left(M_k^j|T_k^j\right)=\prod _l\pi \left({\mu }_{\mathit{kl}}^j|T_k^j\right)\text{.}\end{array}$

This prior structure implies independent prior on each individual trees. The priors for $\pi \left(T_k^j\right)$ in (6) are assigned following the suggestion in Chipman et al. (2010) such that (i) the distribution on the assignment of the splitting variable at each internal node is a uniform distribution over all available variables corresponding to the particular response component. That means we pick one variable out of p _j variables corresponding to the j -th component for splitting with equal probability; (ii) the splitting rule in each interior node (conditional on the splitting variable) follows a discrete uniform distribution over the set of available splitting values corresponding to the particular response component; and (iii) On all the nodes across all j-components we assign iid probability that a node at depth h is nonterminal is given by a(1+h)^−c , where 0 < a < 1 and $0\le c<\infty$ are the parameters that controls the size of the tree. Conditional on $T_k^j$ , on the ${\mu }_{\mathit{kl}}^j$ parameter we assign conjugate normal distribution $N\left(0,\frac{0.25}{k^2{m}_j}\right)$ . This particular prior induces the fact that the leaf nodes will be shrunk to zero and as we increase the number of trees m _j the contribution from one individual tree decreases. This mimics the concept of weak leathering strategy (Schapire, 1990). In our chapter we fixed k  =   2, a  =   0.5, c  =   2. This ensures each individual tree in our BART-SUR model is small and have moderate shrinkage.

Prior distributions on Σ: On the error variance–covariance matrix Σ we put an inverse Wishart distribution as Σ ∼ IW(r, R). Where r > d and R is a d × d positive definite matrix. A good default choice for the hyperparameters is to set r  = d  +   1 and R  = r I , where I is the identity matrix. This choice of the prior parameter produces the least informative prior structure on Σ.

Prior distribution on m _j : In the original BART model (Chipman et al., 2010) the authors kept the number of trees fixed and argued that the BART model fits well as long as m is large enough. Increasing m further caused only a gradual increase in the out-of-sample mean squared error (MSE). Since the amount of shrinkage on the leaf-node values depends on m _j , increasing the number of trees m _j lessens the effect of over-fitting. While it may be relatively safe to set m _j to a very large value as far as model fitting and prediction is concerned, unfortunately, a value too large will unnecessarily cause the algorithm to run much slower. To overcome that problem we put prior distributions on the number of trees m _j for each of the j-components of our BART-SUR model as follow

(7) $\begin{array}{ll}\hfill P\left[m_j=k\right]& =p_{jk}\text{,}\hfill \end{array}$

for $k=m_{j,min},\dots ,m_{j,max}\text{.}$ Sufficiently large support for m _j ensures that the model can grow freely as required from the data. In this chapter, we recommend to use the default prior for the number of trees as $m_{j,min}=max\left\{2,⌈\frac{10}{d-1}⌉\right\}$ and $m_{j,max}=max\left\{2,⌈\frac{300}{d-1}⌉\right\}$ with the prior probabilities p _jk being proportional to $2^{-m_j}$ to aggressively shrink the number of trees down. This encourages our BART-SUR model to select small ensemble of trees and thus produces much faster result than the original BART model. Nevertheless, if the data have a strong need of large number of trees, the prior shrinkage on the number of trees will be overwhelmed and the model will end up selecting sufficiently large number of trees.

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Selected rank tests

Jaroslav Hájek , ... Pranab K. Sen , in Theory of Rank Tests (Second Edition), 1999

4.10.2 Multivariate signed-rank tests

Consider now signed-rank statistics for testing the null hypothesis of diagonal symmetry in a genuine multivariate setup. Recall that a random p-variate vector X has a distribution function F(x; θ) diagonally symmetric about a p-variate constant θ if both X - θ and θ - X both have the same distribution function. If we write

(1) $F\left(\text{x}:\theta \right)=F\left(\text{x}-\theta \right),$

then whenever F has a density f(x), we have an equivalent characterization of diagonal symmetry:

(2) $\begin{array}{cc}f\left(\text{x}\right)=f\left(\left(-1\right)\text{x}\right),& \text{for all}p-\text{pvariate}\text{x}.\end{array}$

Suppose now that. X ₁,…,X _N is a random sample from the distribution function F(x; θ) that is diagonally symmetric about θ. We want to test the null hypothesis

(3) $H_0:\theta =0,\text{against}{H}_1:\theta \ne 0.$

For the univariate case (i.e., for p = 1). signed-rank tests were shown to be EDF for this hypothesis testing problem. However, if the coordinates of X are not stochastically independent, the vector of coordinatewi.se signed-ranks may not be EDF even under the null hypothesis. This may easily be verified by considering the particular case of p = 2, N = 2. To eliminate this impasse, we consider the following sign-invariance principle, due to Sen and Puri (1967), that renders CDF rank tests.

Let us denote the sample point by E _N = (X ₁,…, X _N), so that the corresponding sample space is ε_N = pN-dimensional Euclidean space. Let then

(4) $g_N{\text{E}}_N=\left({\left(-1\right)}^{i_1}{\text{X}}_1,\dots {\left(-1\right)}^{i_N}{X}_N\right),g_N\in {\mathcal{G}}_N,$

where i_j = 0,1, for j = 1,…, N, so that G_N is the group of 2 ^N possible sign-inversions. Now, for any observed E _N , the group G_N generates an orbit O _N of 2 ^N points in ε _N , given by the set of realizations in (4). The distribution of E _N may generally depend on the unknown F through the unknown association pattern (even under H ₀). On the other hand, under H ₀, each element of O _N has the same distribution, and hence if we denote the conditional probability law of E _N on O _N by P_N , it follows that P_N has the discrete uniform distribution with the common probability mass 2 ^−N at each of the points of O _N. This conditional probability law (under H ₀) does not depend on the unknown F, and hence test statistics based on P_N will be CDF under H _0.

As in the univariate case, we consider here the jth row of E _N , and denote these elements by X_1j,…, X_Nj , for j = 1,…, p. Let then

(5) $\begin{array}{ccc}{R}_{ij}^{+}=\sum _{r=1}^N\mathrm{I}\left(\left|X_{rj}\right|\le \left|X_{ij}\right|\right),& i=1,\dots ,N;& j=1,\dots ,P.\end{array}$

This way, we generate a rank-collection matrix

(6) ${\mathcal{R}}_N={\left(\left(R_{ij}^{+}\right)\right)}_{j=1,\dots ,p:i=1,\dots ,N.}$

Also, we let S_ij = sign X_ij , for all i, j, and denote the sign-collection matrix by

(7) ${\mathcal{S}}_N={\left(\left(S_{ij}\right)\right)}_{j=1,\dots ,p;i=1,\dots N.}$

Further, for each j(= 1,…,p), we consider a set of scores a_Nj(k), k = 1,…, N, defined as in the univariate case, and construct a signed-rank statistic

(8) $\begin{array}{cc}{T}_{N_j}=N^{-1}\sum _{i=1}^N{S}_{ij}{a}_{Nj}\left(R_{ij}^{+}\right),& j=1,\dots ,p;\end{array}$

we let T _N = (T_N1,…, T_Np)′. It follows from the composition of P_N that

(9) $\text{E}\left({\text{T}}_N|{\mathcal{P}}_N\right)=0\left(a.\text{e},{\mathcal{P}}_N\right);$

(10) $N\text{E}\left({\text{T}}_N{{\text{T}}^{\prime }}_N|{\mathcal{P}}_N\right)=V_N={\left(\left(v_{Njs}\right)\right)}_{j,s=1,\dots ,p,}$

where

(11) $v_{Njs}=N^{-1}\sum _{i=1}^N{a}_{Nj}\left(R_{ij}^{+}\right)a_{Ns}\left(R_{is}^{+}\right)S_{ij}{S}_{is},$

for j, s = 1,…, p: note that v_Njj = N⁻¹ Σ_i=1 ^N a_Nj ²(i) is non-random, for each j = 1,…, p, but the off-diagonal elements are generally random. By analogy with the classical Hotelling's T ²-test (based on the vector of sample means and the sample covariance matrix), we consider here the following quadratic form as a suitable signed-rank test statistic

(12) ${\mathcal{L}}_N=N{\text{T}}_N^{\text{'}}{\text{V}}_N^{-}{\text{T}}_N,$

where V ⁻ _N is a suitable generalized inverse of V _N - The conditional distribution of L_N , under P_N , over the set O _N generates a CDF test for H ₀ against H ₁ in (3). This test statistic is a direct multivariate extension of the univariate linear signed-rank statistic, and for small (to moderate) values of N, the exact conditional distribution of L_N can be obtained by enumerating all possible 2 ^N realizations of T _N (under P_N ) note that V _N is P_N -invariant, and hence it does not vary over O _N.

In passing, we may note that the LMPR test characterization of univariate signed-rank statistics established in Section 3.4 may not be generally-tenable in the multivariate case. The main reason for this feature is the fact that θ in (3) is a p-variate vector, so that the power function, even locally, for θ → 0, depends on p arguments. In that vein, if we consider the scores

(13) $\begin{array}{ll}\left(\partial /\partial \theta \right)logf\left({\text{X}}_i,\theta \right){|}_{\theta =0},\hfill & i=1,\dots ,N,\hfill \end{array}$

they may not correspond to a vector of marginal scores (that is, the (∂/∂θ_j) log f_j(X_ij,θ_j)|_θj=0) or a linear combination of them; here the f_j stand for the marginal densities.) In fact, unlike the Hotelling T ²-statistic, L_N is not affine invariant (i.e., invariant under arbitrary non-singular linear transformations on the X _i), and hence the best-invariance property may not be tenable for rank statistics. We shall discuss this further later on.

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Random Algebraic Geometry, Attractors and Flux Vacua

M.R. Douglas , in Encyclopedia of Mathematical Physics, 2006

Asymptotic Counting Formulas

We have just defined two classes of physically preferred points in the complex structure moduli space of Calabi–Yau 3-folds, the attractor points and the flux vacua. Both have simple definitions in terms of Hodge structure, eqn [8] and eqn [16], and both are also critical points of integral periods of the holomorphic 3-form.

This second phrasing of the problem suggests the following language. We define a random period of the holomorphic 3-form to be the period for a randomly chosen cycle in $H_3\left(M,\mathbb{Z}\right)$ of the types we just discussed (real or complex, and with the appropriate control parameters). We are then interested in the expected distribution of critical points for a random period. This brings our problem into the framework of random algebraic geometry. Before proceeding to use this framework, let us first point out some differences with the toy problems we discussed. First, while eqn [12] and eqn [17] are sums of the form eqn [6], we take not an orthonormal basis but instead a basis s _i of integral periods of Ω. Second, the coefficients c _i are not normally distributed but instead drawn from a discrete uniform distribution, that is, correspond to a choice of γ in $H^3\left(M,\mathbb{Z}\right)$ or F as in eqn [15], satisfying the bounds on |Z| or L. Finally, we do not normalize the distribution (which is thus not a probability measure) but instead take each choice with unit weight.

These choices can of course be modified, but are made in order to answer the question, "how many attractor points (or flux vacua) sit within a specified region of moduli space?" The answer we will get is a density $\mu \left(Z_{\text{max}}\right)$ or $\mu \left(L_{\text{max}}\right)$ on moduli space, such that as the control parameter becomes large, the number of critical points within a region R asymptotes to

$\mathcal{N}\left(R;Z_{\text{max}}\right)\sim {\int }_R\mu \left(Z_{\text{max}}\right)$

The key observation is that to get such asymptotics, we can start with a Gaussian random element s of $H^3\left(M,\mathbb{R}\right)$ or $H^3\left(M,\mathbb{C}\right)$ . In other words, we neglect the integral quantization of the charge or flux. Intuitively, this might be expected to make little difference in the limit that the charge or flux is large, and in fact one can prove that this simplification reproduces the leading large L or |Z| asymptotics for the density of critical points, using standard ideas in lattice point counting.

This justifies starting with a two-point function like eqn [7]. While the integral periods s_i of Ω can be computed in principle (and have been in many examples) by solving a system of linear PDEs, the Picard–Fuchs equations, it turns out that one does not need such detailed results. Rather, one can use the following ansatz for the two-point function,

$\begin{array}{c}G\left(z_1,{{\displaystyle \overline{z}}}_2\right)={\displaystyle \sum _{I=1}^{b_3}}{\eta }^{IJ}{s}_I\left(z_1\right)s_J^{*}\left({{\displaystyle \overline{z}}}_2\right)\\ ={\int }_M\Omega \left(z_1\right)\text{Λ}{\displaystyle \overline{\Omega }}\left({{\displaystyle \overline{z}}}_2\right)\\ =\text{exp}-K\left(z_1,{{\displaystyle \overline{z}}}_2\right)\end{array}$

In words, the two-point function is the formal continuation of the Kähler potential on ${\mathcal{M}}_c\left(M\right)$ to independent holomorphic and antiholomorphic variables. This incorporates the quadratic form appearing in eqn [18] and can be used to count sections with such a bound.

We can now follow the same strategy as before, by introducing an expected density of critical points,

[19] $\text{d}\mu \left(z\right)=\mathbb{E}\left[{\delta }^{\left(n\right)}\left(D_{i}s\left(z\right)\right){\delta }^{\left(n\right)}\left({{\displaystyle \overline{D}}}_i{\displaystyle \overline{s}}\left({\displaystyle \overline{z}}\right)\right)\left|\underset{1\le i,j\le 2n}{\text{det}}{H}_{ij}\right|\right]$

where the "complex Hessian" H is the $2n\times 2n$ matrix of second derivatives

[20] $H\equiv \left(\begin{array}{cc}{\partial }_i{{\displaystyle \overline{D}}}_{{\displaystyle \overline{j}}}{\displaystyle \overline{s}}\left(z\right)& {\partial }_i{D}_{j}s\left(z\right)\\ {{\displaystyle \overline{\partial }}}_{{\displaystyle \overline{i}}}{{\displaystyle \overline{D}}}_{{\displaystyle \overline{j}}}{\displaystyle \overline{s}}\left(z\right)& {{\displaystyle \overline{\partial }}}_{{\displaystyle \overline{i}}}{D}_{j}s\left(z\right)\end{array}\right)$

(note that $\partial Ds=DDs$ at a critical point). One can then compute this density along the same lines. The holomorphy of s implies that ${\partial }_i{{\displaystyle \overline{D}}}_{{\displaystyle \overline{j}}}s={\omega }_{i{\displaystyle \overline{j}}}s$ , which is one simplification. Other geometric simplifications follow from the fact that eqn [19] depends only on s and a finite number of its derivatives at the point z.

For the attractor problem, using the identity

$D_i{D}_{j}s={\mathcal{F}}_{ijk}{\omega }^{k{\displaystyle \overline{k}}}{{\displaystyle \overline{D}}}_{{\displaystyle \overline{k}}}s=0$

from special geometry of Calabi–Yau 3-folds, the Hessian becomes trivial, and $\text{det}H=|s{|}^{2n}$ . One thus finds (Denef and Douglas 2004) that the asymptotic density of attractor points with large $\left|Z\right|\le Z_{\text{max}}$ in a region R is

$\mathcal{N}\left(R,\left|Z\right|\le Z_{\text{max}}\right)\sim \frac{2^{n+1}}{\left(n+1\right){\pi }^n}{Z}_{\text{max}}^{n+1}\cdot \text{vol}\left(R\right)$

where $\text{vol}\left(R\right)={\int }_R{\omega }^n/n!$ is the volume of R in the Weil–Peterson metric. The total volume is known to be finite for Calabi–Yau 3-fold moduli spaces, and thus so is the number of attractor points under this bound.

The flux vacuum problem is complicated by the fact that DDs is nonzero and thus the determinant of the Hessian does not take a definite sign, and implementing the absolute value in eqn [19] is nontrivial. The result (Douglas, et al. 2004) is

$\begin{array}{c}\mu \left(z\right)\sim \frac{1}{b_3!\sqrt{\text{det}\text{Λ}\left(z\right)}}{\int }_{\mathcal{H}\left(z\right)\times \mathbb{C}}\left|\text{det}\left(HH*-|x{|}^2\cdot 1\right)\right|\\ \times {\text{e}}^{H^t\text{Λ}{\left(z\right)}^{-1}H-|x{|}^2}\text{d}H\text{d}x\end{array}$

where $\mathcal{H}\left(z\right)$ is the subspace of Hessian matrices eqn [20] obtainable from periods at the point z, and $\text{Λ}\left(z\right)$ is a covariance matrix computable from the period data.

A simpler lower bound for the number of solutions can be obtained by instead computing the index density

[21] ${\mu }_I\left(z\right)=\mathbb{E}\left[{\delta }^{\left(n\right)}\left(D_{i}s\right){\delta }^{\left(n\right)}\left({{\displaystyle \overline{D}}}_i{\displaystyle \overline{s}}\right)\underset{1\le i,j\le 2n}{\text{det}}{H}_{ij}\right]$

so-called because it weighs the vacua with a Morse–Witten sign factor. This admits a simple explicit formula (Ashok and Douglas 2004),

[22] $\begin{array}{c}{I}_{\text{vac}}\left(R,L\le L_{\text{max}}\right)\\ \sim \frac{{\left(2\pi L_{\text{max}}\right)}^{b_3}}{{\pi }^{n+1}{b}_3!}{\int }_R\text{det}\left(\mathcal{R}+\omega \cdot 1\right)\end{array}$

where $\mathcal{R}$ is the $\left(n+1\right)\times \left(n+1\right)$ -dimensional matrix of curvature 2-forms for the Weil–Peterson metric.

One might have guessed this density by the following reasoning. If s had been a single-valued section on a compact ${\mathcal{M}}_c$ (it is not), topological arguments determine the total index to be $\left[c_{n+1}\left(L\otimes ,T*\mathcal{M}\right)\right]$ , and this is the simplest density constructed solely from the metric and curvatures in the same cohomology class.

It is not in general known whether this integral over Calabi–Yau moduli space is finite, though this is true in examples studied so far. One can also control $|W{|}^2$ as well as other observables, and one finds that the distribution of $|W{|}^2$ among flux vacua is to a good approximation uniform. Considering explicit examples, the prefactor in eqn [22] is of order ${10}^{100}–{10}^{300}$ , so assuming that this factor dominates the integral, we have justified the Bousso–Polchinski solution to the cosmological constant problem in these models.

The finite L corrections to these formulas can be estimated using van der Corput techniques, and are suppressed by better than the naive $L^{-1/2}$ or $|Z{|}^{-1}$ one might have expected. However the asymptotic formulas for the numbers of flux vacuum break down in certain limits of moduli space, such as the large complex structure limit. This is because eqn [18] is an indefinite quadratic form, and the fact that it bounds the number of solutions at all is somewhat subtle. These points are discussed at length in (Douglas et al. 2005).

Similar results have been obtained for a wide variety of flux vacuum counting problems, with constraints on the value of the effective potential at the minimum, on the masses of scalar fields, on scales of supersymmetry breaking, and so on. And in principle, this is just the tip of an iceberg, as the study of more or less any class of superstring vacua leads to similar questions of counting and distribution, less well understood at present. Some of these are discussed in Douglas (2003), Acharya et al. (2005), Denef and Douglas (2005), Blumenhagen et al. (2005).

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Four decades of research on the open-shop scheduling problem to minimize the makespan

Mohammad Mahdi Ahmadian , ... T.C.E. Cheng , in European Journal of Operational Research, 2021

Benchmark instances

Three well-known sets of benchmark instances are available for the open-shop scheduling problem. These three sets (and sometimes extended variants of them) have been utilized by many researchers for assessing the performance of the algorithms and the solution techniques. The benchmark instances are due to Taillard (1993), Brucker, Hurink, Jurisch, and Wöstmann (1997), and Guéret and Prins (1999).

Taillard (1993) introduced the first set of benchmark instances for the open-shop scheduling problem, in which an instance is characterized by the pair ( $n,m$ ) and consists of six different sizes as follows: (4,4), (5,5), (7,7), (10,10), (15,15), and (20,20), where the processing times are randomly generated from the discrete uniform distribution in the range $U[1,99]$ . For each size, they generated ten instances that resulted in a total of 60 instances. This set of instances is considered easy because the trivial lower bound $L{B}_0$ (see below) is equal to the optimal makespan for 40 of the larger instances (Malapert et al., 2012). The trivial lower bound on the makespan can be calculated as the maximum of jobs' durations and machines' loads, denoted as $L{B}_0,$ as follows:

(1) $L{B}_0=\max \{\underset{j}{\max }\left\{\sum _i{p}_{ij}\right\},\underset{i}{\max }\left\{\sum _j{p}_{ij}\right\}\}.$

The first term in the lower bound represents the longest job duration and the second term defines the maximum machine load $l_{\max },$ i.e., $l_{\max }=\max \left\{l_i\right\},$ where $l_i={\sum }_j{p}_{ij}$ represents the load of machine $i$ . It should be noted that the first non-trivial lower bound for the open-shop scheduling problem is due to Guéret and Prins (1999). The lower bound is calculated by determining the optimal makespan for a relaxed version of the problem. We refer the interested reader to Guéret and Prins (1999) for the details.

Brucker et al. (1997) proposed a few measures to capture the hardness of an instance and applied them to Taillard (1993)'s instances. Since they observed that Taillard (1993)'s instances are easy to solve, they generated their own instances. They proposed a set of 52 challenging instances. Similar to the instances of Taillard (1993), they considered an equal number of jobs and machines to generate the instances. They proposed eight instances for $n=m\in \{3,8\}$ and nine instances for $n=m\in \{4,\cdots ,7\}$ . The third benchmark is that of Guéret and Prins (1999), in which there are ten instances for each size of $n=m\in \{3,\cdots ,10\},$ leading to 80 instances. It should be noted that the benchmarks proposed by Brucker et al. (1997), and Guéret and Prins (1999) are more challenging than those of Taillard (1993) because the advanced GA by Prins (2000), ACO by Blum (2005), and PSO by Sha and Hsu (2008) struggle to optimally solve all the instances of Brucker et al. (1997) and Guéret and Prins (1999). We refer the interested reader to Malapert et al. (2012) for the details on the performance of exact and heuristic methods in solving the instances of Taillard (1993), Brucker et al. (1997), and Guéret and Prins (1999).

Next, we review the available exact methods, followed by the heuristic and meta-heuristic algorithms that were proposed to solve the general open-shop scheduling problem.

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A new way of quantifying the symmetry of a random variable: Estimation and hypothesis testing

Gil González-Rodríguez , ... M. Asunción Lubiano , in Journal of Statistical Planning and Inference, 2012

5 Empirical comparisons

The performance of the tests described in Section 4 is analyzed and compared to other methods by simulations.

Each simulation is the result of 10,000 iterations of the test at a nominal significance level 0.05. The number of bootstrap replications has been 1000, and we have considered m=10,000 in the asymptotic case. In addition, for the distance we have chosen $\phi$ to be the Lebesgue measure on [0, 1] and $\theta =1/3$ . This particular election reduces to the $(W,\phi )\mathrm{-}\mathrm{distance}$ introduced by Bertoluzza et al. (1995) being W and $\phi$ the Lebesgue measure on [0, 1].

In this first example we have considered both the asymptotic and the bootstrap versions of the test just for comparative purposes under H ₀.

Example 5.1

We have developed tests for symmetry about the mean value of different symmetric distributions: some typical continuous distributions as Normal, Cauchy or t -Student distributions, and some discrete distributions as binomial and discrete uniform distributions. In all cases the symmetry point is $\tau =0$ , excepting for the binomial one ( $\tau =5)$ .

Table 2 shows the test sizes at the nominal significance level 0.05 and sample sizes of n=30, 100. The behaviour is very similar for all distributions (discrete and continuous). Both the bootstrap approach and the asymptotic one provide us with an empirical percentage of rejections close to the nominal one. The results are more similar as the sample size increases. For sample sizes smaller that n=30 the asymptotic procedure should not be used in practice (for instance, for a Gaussian distribution and n=20 the percentage of rejections for the asymptotic procedure is around 7.33 whereas for the bootstrap one is close to 4.69). For this reason, from now on in order to establish comparisons the bootstrap procedures will be considered.

Table 2. Test sizes at significance level $\alpha =0.05$ .

Distribution	n=30		n=100
	Asymptotic/Inv. Asymp.	Bootstrap/Inv. Boot.	Asymptotic/Inv. Asymp.	Bootstrap/Inv. Boot.
N(0,1)	5.06/5.03	4.76/4.82	5.1/5.09	4.77/4.75
Cauchy	5.13/5.54	5.04/1.17	5.08/5.49	4.94/0.81
t 3	5.14/5.28	4.72/3.88	5.03/4.99	5.03/4.65
$\mathcal{B}(10,0.5)$	4.94/4.92	4.42/4.76	5.04/5.00	4.86/4.8
$\mathcal{U}\{-2,-\mathrm{1,1},2\}$	5.12/5.07	4.5/4.49	5.04/5.02	4.75/4.68

The problem of testing the symmetry of a distribution about a given value against two-sided alternatives has been widely studied in the literature. Many of the proposed procedures have been compared with the Wilcoxon signed rank test and the Cramér–von Mises type test which have been shown to be more efficient than others. In the next examples we will compare our approach to these tests by considering, specifically, the Wilcoxon signed rank test, and the Rothman and Woodroofe CVM-type test.

In the following example we will compare the different approaches for continuous distributions.

Example 5.2

First of all the test sizes at the nominal significance level 0.05 for sample sizes of n=30 and n=100 are collected in Table 3 for different continuous symmetric distributions. Namely, the standard normal, the Laplace and the Cauchy distributions, a t-distribution with 3 degrees of freedom and some k-mixtures of normal distributions denoted by $M(k,\mu )$ whose densities are

$f_m(x)={\displaystyle \frac{1}{k}}\sum _{i=0}^{k-1}\varphi (x-i\mu +(k-1)\mu /2-m),$

where $\varphi$ denotes the standard normal density, are considered. All these distributions are symmetric about $\tau =0$ .

Table 3. Test sizes at significance level $\alpha =0.05$ for symmetric τ=0 continuous distributions.

Distribution	n	Bootstrap	Inv. Boot.	Wilcoxon	CVM
N(0,1)	30	4.76	4.82	4.92	4.80
	100	4.67	4.75	4.32	4.53
Laplace	30	4.72	4.57	5.45	4.76
	100	4.86	4.82	4.38	4.93
Cauchy	30	5.04	1.17	4.63	4.77
	100	4.94	0.81	4.68	4.68
t 3	30	4.72	3.88	4.71	4.52
	100	5.03	4.65	5.51	5.01
M(3,5)	30	4.91	5.02	5.12	4.93
	100	4.97	4.89	5.04	4.91
M(3,3)	30	4.64	4.81	5.00	4.90
	100	4.85	4.71	4.79	4.84

In general the sizes are close to the nominal ones for all the considered approaches, and they are closer for larger sample size. There is one exception: the proposed invariant Bootstrap method does not perform well for the Cauchy distribution, which was expectable since the Cauchy distribution has no finite order moments.

To complete the empirical analysis, the power of the different approaches under several alternative distributions is analyzed. Basically, two families of asymmetric distributions are considered: Chi-squared and Skew Normal distributions (for which the "amount of asymmetry" varies as a function of its parameters). In addition a mixture of uniform distributions in the following forms (see Fig. 4) is considered:

$0.7\times \mathcal{U}(-0.85;-0.15)+0.15\times \mathcal{U}(0.0;0.15)+0.15\times \mathcal{U}(0.85;1.0).$

In all cases the symmetry point has been chosen as the median of the corresponding distribution.

Fig. 4. Mixture of uniform distributions.

The percentage of rejections for the different situations is gathered in Table 4. The invariant bootstrap outperforms the non-invariant bootstrap in almost all the considered situations. In addition, both the non-invariant as well as the invariant bootstrap outperforms the usual benchmark tests (Wilcoxon and CVM) in most of the situations.

Table 4. Empirical percentage of rejections at significance level $\alpha =0.05$ for non-symmetric continuous distributions.

Distribution	n	Bootstrap	Inv. Boot.	Wilcoxon	CVM
${\chi }_1^2$	30	77.91	70.11	53.78	53.77
(Me=1−2/3)	100	99.98	100	96.63	99.83
${\chi }_2^2$	30	17.96	30.83	17.58	15.63
(Me=2−2/3)	100	56.88	93.24	50.88	63.64
${\chi }_3^2$	30	10.13	20.28	12.77	10.46
(Me=3−2/3)	100	27.15	75.58	31.43	35.11
SN(2,0,1)	30	8.75	9.94	7.71	6.45
(Me=0.555)	100	20.97	29.73	14.22	11.9
SN(3, 0, 1)	30	10.43	12.17	8.73	6.95
(Me=0.611)	100	28.38	42.1	17.12	16.25
MixtureUnif	30	16.36	20.46	9.23	8.13
(Me=−0.35)	100	57.16	75.55	21.92	25.82

We have carried out some additional simulations for continuous distributions in order to show that, as it is expectable, the proposed techniques are not "uniformly" the best ones. For that, we have compared them to some tests developed for some specific situations.

Example 5.3

In this example we will examine the behaviour of our approaches to test the symmetry of continuous distributions and we will compare to other specific methods given in Mizushima and Nagao (1998): The sign test, the signed rank test, and the test I _n based on density estimates. The study focuses on some of the continuous symmetric (w.r.t. 0) distributions considered in the above-mentioned paper. As in Mizushima and Nagao (1998), in order to establish the comparisons regarding the power of the tests, we have also tested the symmetry about $\tau =0.1$ and $\tau =0.5$ .

Table 5 shows the obtained percentage of rejections at the nominal significance level 0.05 and sample size n=100. The tests considered in this paper have better power than the other ones in the case of normal distributions. When the density is an $M(\mathrm{3,5})$ , the I _n test is more powerful than the other methods, but with other densities as $M(\mathrm{3,3})$ or $M(\mathrm{2,3})$ the I _n test has lower power than others.

Table 5. Empirical percentage of rejections at significance level $\alpha =0.05$ for symmetric continuous distributions and different symmetry points.

Dist.	$\tau$	Bootstrap	Inv. Boot.	Sign	Signed rank	I n
$\mathcal{N}(\mathrm{0,1})$	0	4.67	4.75	5.61	4.9	5.02
	0.1	15.71	16.21	13.72	15.94	11.22
	0.5	99.81	99.85	97.90	99.75	98.38
$M(\mathrm{3,5})$	0	4.94	4.94	5.48	4.95	6.44
	0.1	5.55	5.49	6.43	6.03	8.32
	0.5	16.96	20.41	26.92	32.76	79.18
$M(\mathrm{3,3})$	0	4.99	4.72	5.47	4.96	5.97
	0.1	5.94	6.64	6.43	6.38	6.57
	0.5	31.97	44.35	28.37	43.09	32.90

In the following example the behaviour of the different techniques for discrete distributions is analyzed.

Example 5.4

The percentage of rejections at a nominal significance level of 0.05 for sample sizes of n=30 and n=100 and different discrete symmetric distributions is collected in Table 6. We have considered three finite discrete distributions taking the values $(-2,-\mathrm{1,1},2)$ and with corresponding probabilities $(0.25,0.25,0.25,0.25)$ for the Uniform case, $(0.375,0.125,0.125,0.375)$ for the Convex case and $(0.125,0.375,0.375,0.125)$ for the Concave case. In addition as non-finite discrete distributions we have considered: a mixture of $\mathcal{P}(1)$ distributions

$0.5\times \mathcal{P}(1)+0.5\times (-\mathcal{P}(1)),$

(analogous with $\mathcal{P}(2)$ and $\mathcal{P}(3)$ ).

Table 6. Test sizes at significance level $\alpha =0.05$ for symmetric discrete distributions.

Distribution	n	Bootstrap	Inv. Boot.	Wilcoxon	CVM
Discrete	30	4.50	4.49	4.10	99.95
Uniform	100	4.75	4.68	4.26	100
Discrete	30	4.82	4.89	4.01	100
Concave	100	5.00	5.10	3.72	100
Discrete	30	4.61	4.87	4.02	99.84
Convex	100	5.37	5.25	4.20	100
Mixture	30	4.89	4.96	11.79	98.87
P(1)	100	5.06	5.11	23.50	100
Mixture	30	4.69	4.87	4.81	92.09
P(2)	100	4.63	4.86	5.49	99.66
Mixture	30	4.85	4.88	4.70	87.95
P(3)	100	4.94	5.01	4.86	98.77

The main conclusion is that neither Wilcoxon signed Rank nor CVM tests are reliable when working with discrete distributions: Cramér–von Mises is clearly not consistent since the percentage of rejections under H ₀ is much higher than the real significance level $\alpha =0.05$ . For the Wilcoxon signed rank test, the percentage of rejections is considerably smaller than 5% in several cases, whereas in other cases, the percentage of rejections is clearly higher than the nominal significance level. With respect to the proposed bootstrap techniques, the percentage of rejections is close to the nominal one in all cases.

There are not many tests for symmetry for discrete distributions. The unique specific study seems to be the one of Vorlickova (1972), in which ties in rank tests are handled by randomization or by averaging scores. Unfortunately, that work is focused on the theoretic aspects of the statistics. The asymptotic distributions under the null hypothesis depend on some unknown population constants and it cannot be applied in practice. Consequently, the power comparison can be established only between the two proposed bootstrap procedures (see Table 7). The results are gathered in Table 7 for several non-symmetric discrete distributions: some Poisson distributions as well as some finite distributions taking on values from 1 to 10 and with corresponding probabilities $(0.1,0.1,0.05,0.05,0.05,0.075,0.075,0.1,0.15,0.25)$ —Discrete A distribution, $(0.25,0.15,0.1,0.075,0.075,0.05,0.05,0.05,0.1,0.1)$ —Discrete B distribution and $(0.25,0.15,0.15,0.10,0.075,0.05,0.05,0.075,0.05,0.05)$ —Discrete C distribution. The results show again a generally better power for the invariant bootstrap procedure than for the non-invariant one.

Table 7. Empirical percentage of rejections at significance level $\alpha =0.05$ for non-symmetric discrete distributions.

Distribution	n	Bootstrap	Inv. Boot.
$P(0.05)$	30	38.31	38.51
(Me=172)	100	82.08	81.67
$P(0.1)$	30	57.28	57.61
(Me=172)	100	99.4	99.4
Finite A	30	25.95	61.66
(Me=8)	100	74.1	99.64
Finite B	30	4.65	8.47
(Me=4)	100	5.15	25.36
Finite C	30	13.84	35.61
(Me=3)	100	41.15	91.96

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