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#Randomkit random number generators, wrapped for Torch

NOTE: THIS PACKAGE IS NOT ACTIVELY MAINTAINED

Provides and wraps the Randomkit library, copied from Numpy. Please check-out its website for up-to-date documentation or read below.

##Example

###Single sample

You can call any of the wrapped functions with just the distribution's parameters to generate a single sample and return a number:

require 'randomkit'
randomkit.poisson(5)

###Multiple samples from one distribution

Often, you might want to generate many samples identically distributed. Simply pass as a first argument a tensor of the proper dimension, into which the samples will be stored:

x = torch.Tensor(10000)
randomkit.poisson(x, 5)

The sampler returns the tensor, so you can shorten the above in:

x = randomkit.poisson(torch.Tensor(10000), 5)

###Multiple samples from multiple distributions

Finally, you might want to generate many samples, each from a distribution with different parameters. This is achieved by passing a Tensor as the parameter of the distribution:

many_lambda = torch.Tensor{5, 3, 40, 60}
x = randomkit.poisson(many_lambda)

Of course, this can be combined with passing a result Tensor as an optional first element, to re-use memory and avoid creating a new Tensor at each call:

many_lambda = torch.Tensor{5, 3, 40, 60}
x = torch.Tensor(many_lambda:size())
randomkit.poisson(x, many_lambda)

Note: in the latter case, the size of the result Tensor must correspond to the size of the parameter tensor -- we do not resize the result tensor automatically, yet:

###Getting/setting the seed and the state

Randomkit is transparently integrated with Torch's random stream: just use torch.manualSeed(seed), torch.getRNGState(), and torch.setRNGState(state) as usual. Specifying an (optional) torch.Generator instance as the first argument will only influence the state of that generator, leaving the state of randomkit unchanged.

##Installation

From a terminal:

luarocks install randomkit

##Unit Tests

Last but not least, the unit tests are in the folder luasrc/tests. You can run them from your local clone of the repository with: git clone https://www.github.com/jucor/torch-randomkit

find torch-randomkit/luasrc/tests -name "test*lua" -exec torch {} \;

##Direct access to FFI

randomkit.ffi.*

Functions directly accessible at the top of the randomkit table are Lua wrappers to the actual C functions from Randomkit, with extra error checking. If, for any reason, you want to get rid of this error checking and of a possible overhead, the FFI-wrapper functions can be called directly via randomkit.ffi.myfunction() instead of randomkit.myfunction(). #List of distributions ##beta randomkit.beta([output], a, b)

The Beta distribution over [0, 1].

The Beta distribution is a special case of the Dirichlet distribution, and is related to the Gamma distribution. It has the probability distribution function

$$ f(x; a,b) = \frac{1}{B(\alpha, \beta)} x^{\alpha - 1} (1 - x)^{\beta - 1},$$

where the normalisation, B, is the beta function,

$$ B(\alpha, \beta) = \int_0^1 t^{\alpha - 1} (1 - t)^{\beta - 1} dt.$$

It is often seen in Bayesian inference and order statistics.

####Parameters

####Returns

##binomial randomkit.binomial([output], n, p)

Draw samples from a binomial distribution.

Samples are drawn from a Binomial distribution with specified parameters, n trials and p probability of success where n an integer >= 0 and p is in the interval [0,1]. (n may be input as a float, but it is truncated to an integer in use)

####Parameters

####Returns

####See Also

####Notes The probability density for the Binomial distribution is

$$ P(N) = \binom{n}{N}p^N(1-p)^{n-N},$$

where \(n\) is the number of trials, \(p\) is the probability of success, and \(N\) is the number of successes.

When estimating the standard error of a proportion in a population by using a random sample, the normal distribution works well unless the product pn <=5, where p = population proportion estimate, and n = number of samples, in which case the binomial distribution is used instead. For example, a sample of 15 people shows 4 who are left handed, and 11 who are right handed. Then p = 4/15 = 27%. 0.2715 = 4, so the binomial distribution should be used in this case.

####References

  1. Dalgaard, Peter, "Introductory Statistics with R", Springer-Verlag, 2002.
  2. Glantz, Stanton A. "Primer of Biostatistics.", McGraw-Hill, Fifth Edition, 2002.
  3. Lentner, Marvin, "Elementary Applied Statistics", Bogden and Quigley, 1972.
  4. Weisstein, Eric W. "Binomial Distribution." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/BinomialDistribution.html
  5. Wikipedia, "Binomial-distribution", http://en.wikipedia.org/wiki/Binomial_distribution

####Examples Draw samples from the distribution:

 $ n, p = 10, .5  number of trials, probability of each trial
 $ s = np.random.binomial(n, p, 1000)

result of flipping a coin 10 times, tested 1000 times.

A real world example. A company drills 9 wild-cat oil exploration wells, each with an estimated probability of success of 0.1. All nine wells fail. What is the probability of that happening?

Let's do 20,000 trials of the model, and count the number that generate zero positive results.

 $ sum(np.random.binomial(9,0.1,20000)==0)/20000.

answer = 0.38885, or 38%.

##bytes randomkit.bytes([output], length)

Return random bytes.

####Parameters

####Returns

####Examples

 $ np.random.bytes(10)

' eh\x85\x022SZ\xbf\xa4' random

##chisquare randomkit.chisquare([output], df)

Draw samples from a chi-square distribution.

When df independent random variables, each with standard normal distributions (mean 0, variance 1), are squared and summed, the resulting distribution is chi-square (see Notes). This distribution is often used in hypothesis testing.

####Parameters

####Returns

####Raises ValueError When df <= 0 or when an inappropriate size (e.g. size=-1) is given.

####Notes The variable obtained by summing the squares of df independent, standard normally distributed random variables:

$$ Q = \sum_{i=0}^{\mathtt{df}} X^2_i$$

is chi-square distributed, denoted

$$ Q \sim \chi^2_k.$$

The probability density function of the chi-squared distribution is

$$ p(x) = \frac{(1/2)^{k/2}}{\Gamma(k/2)} x^{k/2 - 1} e^{-x/2},$$

where \(\Gamma\) is the gamma function,

$$ \Gamma(x) = \int_0^{-\infty} t^{x - 1} e^{-t} dt.$$

####References

  1. NIST/SEMATECH e-Handbook of Statistical Methods, http://www.itl.nist.gov/div898/handbook/eda/section3/eda3666.htm

####Examples

 $ np.random.chisquare(2,4)

array([ 1.89920014, 9.00867716, 3.13710533, 5.62318272])

##dirichlet randomkit.dirichlet([output], alpha)

Draw samples from the Dirichlet distribution.

Draw size samples of dimension k from a Dirichlet distribution. A Dirichlet-distributed random variable can be seen as a multivariate generalization of a Beta distribution. Dirichlet pdf is the conjugate prior of a multinomial in Bayesian inference.

####Parameters

####Returns

####Notes $$ X \approx \prod_{i=1}^{k}{x^{\alpha_i-1}_i}$$

Uses the following property for computation: for each dimension, draw a random sample y_i from a standard gamma generator of shape alpha_i, then \(X = \frac{1}{\sum_{i=1}^k{y_i}} (y_1, \ldots, y_n)\) is Dirichlet distributed.

####References

  1. David McKay, "Information Theory, Inference and Learning Algorithms," chapter 23, http://www.inference.phy.cam.ac.uk/mackay/
  2. Wikipedia, "Dirichlet distribution", http://en.wikipedia.org/wiki/Dirichlet_distribution

####Examples Taking an example cited in Wikipedia, this distribution can be used if one wanted to cut strings (each of initial length 1.0) into K pieces with different lengths, where each piece had, on average, a designated average length, but allowing some variation in the relative sizes of the pieces.

 $ s = np.random.dirichlet((10, 5, 3), 20).transpose()


 $ plt.barh(range(20), s[0])
 $ plt.barh(range(20), s[1], left=s[0], color='g')
 $ plt.barh(range(20), s[2], left=s[0]+s[1], color='r')
 $ plt.title("Lengths of Strings")

##exponential randomkit.exponential([output], scale)

Exponential distribution.

Its probability density function is

$$ f(x; \frac{1}{\beta}) = \frac{1}{\beta} \exp(-\frac{x}{\beta}),$$

for x > 0 and 0 elsewhere. \(\beta\) is the scale parameter, which is the inverse of the rate parameter \(\lambda = 1/\beta\). The rate parameter is an alternative, widely used parameterization of the exponential distribution [3].

The exponential distribution is a continuous analogue of the geometric distribution. It describes many common situations, such as the size of raindrops measured over many rainstorms [1], or the time between page requests to Wikipedia [2].

####Parameters

####References

  1. Peyton Z. Peebles Jr., "Probability, Random Variables and Random Signal Principles", 4th ed, 2001, p. 57.
  2. "Poisson Process", Wikipedia, http://en.wikipedia.org/wiki/Poisson_process
  3. "Exponential Distribution, Wikipedia, http://en.wikipedia.org/wiki/Exponential_distribution

##f randomkit.f([output], dfnum, dfden)

Draw samples from a F distribution.

Samples are drawn from an F distribution with specified parameters, dfnum (degrees of freedom in numerator) and dfden (degrees of freedom in denominator), where both parameters should be greater than zero.

The random variate of the F distribution (also known as the Fisher distribution) is a continuous probability distribution that arises in ANOVA tests, and is the ratio of two chi-square variates.

####Parameters

####Returns

####See Also

####Notes The F statistic is used to compare in-group variances to between-group variances. Calculating the distribution depends on the sampling, and so it is a function of the respective degrees of freedom in the problem. The variable dfnum is the number of samples minus one, the between-groups degrees of freedom, while dfden is the within-groups degrees of freedom, the sum of the number of samples in each group minus the number of groups.

####References

  1. Glantz, Stanton A. "Primer of Biostatistics.", McGraw-Hill, Fifth Edition, 2002.
  2. Wikipedia, "F-distribution", http://en.wikipedia.org/wiki/F-distribution

####Examples An example from Glantz[1], pp 47-40. Two groups, children of diabetics (25 people) and children from people without diabetes (25 controls). Fasting blood glucose was measured, case group had a mean value of 86.1, controls had a mean value of 82.2. Standard deviations were 2.09 and 2.49 respectively. Are these data consistent with the null hypothesis that the parents diabetic status does not affect their children's blood glucose levels? Calculating the F statistic from the data gives a value of 36.01.

Draw samples from the distribution:

 $ dfnum = 1.  between group degrees of freedom
 $ dfden = 48.  within groups degrees of freedom
 $ s = np.random.f(dfnum, dfden, 1000)

The lower bound for the top 1% of the samples is :

 $ sort(s)[-10]

7.61988120985

So there is about a 1% chance that the F statistic will exceed 7.62, the measured value is 36, so the null hypothesis is rejected at the 1% level.

##gamma randomkit.gamma([output], shape, scale)

Draw samples from a Gamma distribution.

Samples are drawn from a Gamma distribution with specified parameters, shape (sometimes designated "k") and scale (sometimes designated "theta"), where both parameters are > 0.

####Parameters

####Returns

####See Also

####Notes The probability density for the Gamma distribution is

$$ p(x) = x^{k-1}\frac{e^{-x/\theta}}{\theta^k\Gamma(k)},$$

where \(k\) is the shape and \(\theta\) the scale, and \(\Gamma\) is the Gamma function.

The Gamma distribution is often used to model the times to failure of electronic components, and arises naturally in processes for which the waiting times between Poisson distributed events are relevant.

####References

  1. Weisstein, Eric W. "Gamma Distribution." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/GammaDistribution.html
  2. Wikipedia, "Gamma-distribution", http://en.wikipedia.org/wiki/Gamma-distribution

####Examples Draw samples from the distribution:

 $ shape, scale = 2., 2.  mean and dispersion
 $ s = np.random.gamma(shape, scale, 1000)

Display the histogram of the samples, along with the probability density function:

 $ import matplotlib.pyplot as plt
 $ import scipy.special as sps
 $ count, bins, ignored = plt.hist(s, 50, normed=True)
 $ y = bins**(shape-1)*(np.exp(-bins/scale) /
                      (sps.gamma(shape)*scale**shape))
 $ plt.plot(bins, y, linewidth=2, color='r')
 $ plt.show()

##geometric randomkit.geometric([output], p)

Draw samples from the geometric distribution.

Bernoulli trials are experiments with one of two outcomes: success or failure (an example of such an experiment is flipping a coin). The geometric distribution models the number of trials that must be run in order to achieve success. It is therefore supported on the positive integers, k = 1, 2, ....

The probability mass function of the geometric distribution is

$$ f(k) = (1 - p)^{k - 1} p$$

where p is the probability of success of an individual trial.

####Parameters

####Returns

####Examples Draw ten thousand values from the geometric distribution, with the probability of an individual success equal to 0.35:

 $ z = np.random.geometric(p=0.35, size=10000)

How many trials succeeded after a single run?

 $ (z == 1).sum() / 10000.

0.34889999999999999 random

##gumbel randomkit.gumbel([output], loc, scale)

Gumbel distribution.

Draw samples from a Gumbel distribution with specified location and scale. For more information on the Gumbel distribution, see Notes and References below.

####Parameters

####Returns

####See Also scipy.stats.gumbel_l scipy.stats.gumbel_r scipy.stats.genextreme probability density function, distribution, or cumulative density function, etc. for each of the above weibull

####Notes The Gumbel (or Smallest Extreme Value (SEV) or the Smallest Extreme Value Type I) distribution is one of a class of Generalized Extreme Value (GEV) distributions used in modeling extreme value problems. The Gumbel is a special case of the Extreme Value Type I distribution for maximums from distributions with "exponential-like" tails.

The probability density for the Gumbel distribution is

$$ p(x) = \frac{e^{-(x - \mu)/ \beta}}{\beta} e^{ -e^{-(x - \mu)/ \beta}},$$

where \(\mu\) is the mode, a location parameter, and \(\beta\) is the scale parameter.

The Gumbel (named for German mathematician Emil Julius Gumbel) was used very early in the hydrology literature, for modeling the occurrence of flood events. It is also used for modeling maximum wind speed and rainfall rates. It is a "fat-tailed" distribution - the probability of an event in the tail of the distribution is larger than if one used a Gaussian, hence the surprisingly frequent occurrence of 100-year floods. Floods were initially modeled as a Gaussian process, which underestimated the frequency of extreme events.

It is one of a class of extreme value distributions, the Generalized Extreme Value (GEV) distributions, which also includes the Weibull and Frechet.

The function has a mean of \(\mu + 0.57721\beta\) and a variance of \(\frac{\pi^2}{6}\beta^2\).

####References Gumbel, E. J., Statistics of Extremes, New York: Columbia University Press, 1958.

Reiss, R.-D. and Thomas, M., Statistical Analysis of Extreme Values from Insurance, Finance, Hydrology and Other Fields, Basel: Birkhauser Verlag, 2001.

####Examples Draw samples from the distribution:

 $ mu, beta = 0, 0.1  location and scale
 $ s = np.random.gumbel(mu, beta, 1000)

Display the histogram of the samples, along with the probability density function:

 $ import matplotlib.pyplot as plt
 $ count, bins, ignored = plt.hist(s, 30, normed=True)
 $ plt.plot(bins, (1/beta)*np.exp(-(bins - mu)/beta)
          * np.exp( -np.exp( -(bins - mu) /beta) ),
          linewidth=2, color='r')
 $ plt.show()

Show how an extreme value distribution can arise from a Gaussian process and compare to a Gaussian:

 $ means = []
 $ maxima = []
 $ for i in range(0,1000) :
    a = np.random.normal(mu, beta, 1000)
    means.append(a.mean())
    maxima.append(a.max())
 $ count, bins, ignored = plt.hist(maxima, 30, normed=True)
 $ beta = np.std(maxima)*np.pi/np.sqrt(6)
 $ mu = np.mean(maxima) - 0.57721*beta
 $ plt.plot(bins, (1/beta)*np.exp(-(bins - mu)/beta)
          * np.exp(-np.exp(-(bins - mu)/beta)),
          linewidth=2, color='r')
 $ plt.plot(bins, 1/(beta * np.sqrt(2 * np.pi))
          * np.exp(-(bins - mu)**2 / (2 * beta**2)),
          linewidth=2, color='g')
 $ plt.show()

##hypergeometric randomkit.hypergeometric([output], ngood, nbad, nsample)

Draw samples from a Hypergeometric distribution.

Samples are drawn from a Hypergeometric distribution with specified parameters, ngood (ways to make a good selection), nbad (ways to make a bad selection), and nsample = number of items sampled, which is less than or equal to the sum ngood + nbad.

####Parameters

####Returns

####See Also

####Notes The probability density for the Hypergeometric distribution is

$$ P(x) = \frac{\binom{m}{n}\binom{N-m}{n-x}}{\binom{N}{n}},$$

where \(0 \le x \le m\) and \(n+m-N \le x \le n\)

for P(x) the probability of x successes, n = ngood, m = nbad, and N = number of samples.

Consider an urn with black and white marbles in it, ngood of them black and nbad are white. If you draw nsample balls without replacement, then the Hypergeometric distribution describes the distribution of black balls in the drawn sample.

Note that this distribution is very similar to the Binomial distribution, except that in this case, samples are drawn without replacement, whereas in the Binomial case samples are drawn with replacement (or the sample space is infinite). As the sample space becomes large, this distribution approaches the Binomial.

####References

  1. Lentner, Marvin, "Elementary Applied Statistics", Bogden and Quigley, 1972.
  2. Weisstein, Eric W. "Hypergeometric Distribution." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/HypergeometricDistribution.html
  3. Wikipedia, "Hypergeometric-distribution", http://en.wikipedia.org/wiki/Hypergeometric-distribution

####Examples Draw samples from the distribution:

 $ ngood, nbad, nsamp = 100, 2, 10

number of good, number of bad, and number of samples $ s = np.random.hypergeometric(ngood, nbad, nsamp, 1000) $ hist(s) note that it is very unlikely to grab both bad items

Suppose you have an urn with 15 white and 15 black marbles. If you pull 15 marbles at random, how likely is it that 12 or more of them are one color?

 $ s = np.random.hypergeometric(15, 15, 15, 100000)
 $ sum(s>=12)/100000. + sum(s<=3)/100000.

answer = 0.003 pretty unlikely!

##laplace randomkit.laplace([output], loc, scale)

Draw samples from the Laplace or double exponential distribution with specified location (or mean) and scale (decay).

The Laplace distribution is similar to the Gaussian/normal distribution, but is sharper at the peak and has fatter tails. It represents the difference between two independent, identically distributed exponential random variables.

####Parameters

####Notes It has the probability density function

$$ f(x; \mu, \lambda) = \frac{1}{2\lambda} \exp\left(-\frac{|x - \mu|}{\lambda}\right).$$

The first law of Laplace, from 1774, states that the frequency of an error can be expressed as an exponential function of the absolute magnitude of the error, which leads to the Laplace distribution. For many problems in Economics and Health sciences, this distribution seems to model the data better than the standard Gaussian distribution

####References

  1. Abramowitz, M. and Stegun, I. A. (Eds.). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, 1972.

  2. The Laplace distribution and generalizations By Samuel Kotz, Tomasz J. Kozubowski, Krzysztof Podgorski, Birkhauser, 2001.

  3. Weisstein, Eric W. "Laplace Distribution." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/LaplaceDistribution.html

  4. Wikipedia, "Laplace distribution", http://en.wikipedia.org/wiki/Laplace_distribution

####Examples Draw samples from the distribution

 $ loc, scale = 0., 1.
 $ s = np.random.laplace(loc, scale, 1000)

Display the histogram of the samples, along with the probability density function:

 $ import matplotlib.pyplot as plt
 $ count, bins, ignored = plt.hist(s, 30, normed=True)
 $ x = np.arange(-8., 8., .01)
 $ pdf = np.exp(-abs(x-loc/scale))/(2.*scale)
 $ plt.plot(x, pdf)

Plot Gaussian for comparison:

 $ g = (1/(scale * np.sqrt(2 * np.pi)) *
      np.exp( - (x - loc)**2 / (2 * scale**2) ))
 $ plt.plot(x,g)

##logistic randomkit.logistic([output], loc, scale)

Draw samples from a Logistic distribution.

Samples are drawn from a Logistic distribution with specified parameters, loc (location or mean, also median), and scale (>0).

####Parameters

####Returns

####See Also

####Notes The probability density for the Logistic distribution is

$$ P(x) = P(x) = \frac{e^{-(x-\mu)/s}}{s(1+e^{-(x-\mu)/s})^2},$$

where \(\mu\) = location and \(s\) = scale.

The Logistic distribution is used in Extreme Value problems where it can act as a mixture of Gumbel distributions, in Epidemiology, and by the World Chess Federation (FIDE) where it is used in the Elo ranking system, assuming the performance of each player is a logistically distributed random variable.

####References

  1. Reiss, R.-D. and Thomas M. (2001), Statistical Analysis of Extreme Values, from Insurance, Finance, Hydrology and Other Fields, Birkhauser Verlag, Basel, pp 132-133.
  2. Weisstein, Eric W. "Logistic Distribution." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/LogisticDistribution.html
  3. Wikipedia, "Logistic-distribution", http://en.wikipedia.org/wiki/Logistic-distribution

####Examples Draw samples from the distribution:

 $ loc, scale = 10, 1
 $ s = np.random.logistic(loc, scale, 10000)
 $ count, bins, ignored = plt.hist(s, bins=50)

plot against distribution

 $ def logist(x, loc, scale):
     return exp((loc-x)/scale)/(scale*(1+exp((loc-x)/scale))**2)
 $ plt.plot(bins, logist(bins, loc, scale)*count.max()/\
 logist(bins, loc, scale).max())
 $ plt.show()

##lognormal randomkit.lognormal([output], mean, sigma)

Return samples drawn from a log-normal distribution.

Draw samples from a log-normal distribution with specified mean, standard deviation, and array shape. Note that the mean and standard deviation are not the values for the distribution itself, but of the underlying normal distribution it is derived from.

####Parameters

####Returns

####See Also

####Notes A variable x has a log-normal distribution if log(x) is normally distributed. The probability density function for the log-normal distribution is:

$$ p(x) = \frac{1}{\sigma x \sqrt{2\pi}} e^{(-\frac{(ln(x)-\mu)^2}{2\sigma^2})}$$

where \(\mu\) is the mean and \(\sigma\) is the standard deviation of the normally distributed logarithm of the variable. A log-normal distribution results if a random variable is the product of a large number of independent, identically-distributed variables in the same way that a normal distribution results if the variable is the sum of a large number of independent, identically-distributed variables.

####References Limpert, E., Stahel, W. A., and Abbt, M., "Log-normal Distributions across the Sciences: Keys and Clues," BioScience, Vol. 51, No. 5, May, 2001. http://stat.ethz.ch/~stahel/lognormal/bioscience.pdf

Reiss, R.D. and Thomas, M., Statistical Analysis of Extreme Values, Basel: Birkhauser Verlag, 2001, pp. 31-32.

####Examples Draw samples from the distribution:

 $ mu, sigma = 3., 1.  mean and standard deviation
 $ s = np.random.lognormal(mu, sigma, 1000)

Display the histogram of the samples, along with the probability density function:

 $ import matplotlib.pyplot as plt
 $ count, bins, ignored = plt.hist(s, 100, normed=True, align='mid')


 $ x = np.linspace(min(bins), max(bins), 10000)
 $ pdf = (np.exp(-(np.log(x) - mu)**2 / (2 * sigma**2))
        / (x * sigma * np.sqrt(2 * np.pi)))

 $ plt.plot(x, pdf, linewidth=2, color='r')
 $ plt.axis('tight')
 $ plt.show()

Demonstrate that taking the products of random samples from a uniform distribution can be fit well by a log-normal probability density function.

 $  Generate a thousand samples: each is the product of 100 random
 $  values, drawn from a normal distribution.
 $ b = []
 $ for i in range(1000):
    a = 10. + np.random.random(100)
    b.append(np.product(a))

 $ b = np.array(b) / np.min(b)  scale values to be positive
 $ count, bins, ignored = plt.hist(b, 100, normed=True, align='center')
 $ sigma = np.std(np.log(b))
 $ mu = np.mean(np.log(b))


 $ x = np.linspace(min(bins), max(bins), 10000)
 $ pdf = (np.exp(-(np.log(x) - mu)**2 / (2 * sigma**2))
        / (x * sigma * np.sqrt(2 * np.pi)))

 $ plt.plot(x, pdf, color='r', linewidth=2)
 $ plt.show()

##logseries randomkit.logseries([output], p)

Draw samples from a Logarithmic Series distribution.

Samples are drawn from a Log Series distribution with specified parameter, p (probability, 0 < p < 1).

####Parameters

####Returns

####See Also

####Notes The probability density for the Log Series distribution is

$$ P(k) = \frac{-p^k}{k \ln(1-p)},$$

where p = probability.

The Log Series distribution is frequently used to represent species richness and occurrence, first proposed by Fisher, Corbet, and Williams in 1943 [2]. It may also be used to model the numbers of occupants seen in cars [3].

####References

  1. Buzas, Martin A.; Culver, Stephen J., Understanding regional species diversity through the log series distribution of occurrences: BIODIVERSITY RESEARCH Diversity & Distributions, Volume 5, Number 5, September 1999 , pp. 187-195(9).
  2. Fisher, R.A,, A.S. Corbet, and C.B. Williams. 1943. The relation between the number of species and the number of individuals in a random sample of an animal population. Journal of Animal Ecology, 12:42-58.
  3. D. J. Hand, F. Daly, D. Lunn, E. Ostrowski, A Handbook of Small Data Sets, CRC Press, 1994.
  4. Wikipedia, "Logarithmic-distribution", http://en.wikipedia.org/wiki/Logarithmic-distribution

####Examples Draw samples from the distribution:

 $ a = .6
 $ s = np.random.logseries(a, 10000)
 $ count, bins, ignored = plt.hist(s)

plot against distribution

 $ def logseries(k, p):
     return -p**k/(k*log(1-p))
 $ plt.plot(bins, logseries(bins, a)*count.max()/

logseries(bins, a).max(), 'r')

 $ plt.show()

##multivariate_normal randomkit.multivariate_normal([output], mean, cov[])

Draw random samples from a multivariate normal distribution.

The multivariate normal, multinormal or Gaussian distribution is a generalization of the one-dimensional normal distribution to higher dimensions. Such a distribution is specified by its mean and covariance matrix. These parameters are analogous to the mean (average or "center") and variance (standard deviation, or "width," squared) of the one-dimensional normal distribution.

####Parameters

####Returns

In other words, each entry out[i,j,...,:] is an N-dimensional value drawn from the distribution.

####Notes The mean is a coordinate in N-dimensional space, which represents the location where samples are most likely to be generated. This is analogous to the peak of the bell curve for the one-dimensional or univariate normal distribution.

Covariance indicates the level to which two variables vary together. From the multivariate normal distribution, we draw N-dimensional samples, \(X = [x_1, x_2, x_N]\). The covariance matrix element \(C_{ij}\) is the covariance of \(x_i\) and \(x_j\). The element \(C_{ii}\) is the variance of \(x_i\) (i.e. its "spread").

Instead of specifying the full covariance matrix, popular approximations include:

####- Spherical covariance (cov is a multiple of the identity matrix) Diagonal covariance (cov has non-negative elements, and only on the diagonal)

This geometrical property can be seen in two dimensions by plotting generated data-points:

 $ mean = [0,0]
 $ cov = [[1,0],[0,100]]  diagonal covariance, points lie on x or y-axis


 $ import matplotlib.pyplot as plt
 $ x,y = np.random.multivariate_normal(mean,cov,5000).T
 $ plt.plot(x,y,'x'); plt.axis('equal'); plt.show()

Note that the covariance matrix must be non-negative definite.

####References Papoulis, A., Probability, Random Variables, and Stochastic Processes, 3rd ed., New York: McGraw-Hill, 1991.

Duda, R. O., Hart, P. E., and Stork, D. G., Pattern Classification, 2nd ed., New York: Wiley, 2001.

####Examples

 $ mean = (1,2)
 $ cov = [[1,0],[1,0]]
 $ x = np.random.multivariate_normal(mean,cov,(3,3))
 $ x.shape

(3, 3, 2)

The following is probably true, given that 0.6 is roughly twice the standard deviation:

 $ print list( (x[0,0,:] - mean) < 0.6 )

[True, True]

##negative_binomial randomkit.negative_binomial([output], n, p)

Draw samples from a negative_binomial distribution.

Samples are drawn from a negative_Binomial distribution with specified parameters, n trials and p probability of success where n is an integer > 0 and p is in the interval [0, 1].

####Parameters

####Returns

####Notes The probability density for the Negative Binomial distribution is

$$ P(N;n,p) = \binom{N+n-1}{n-1}p^{n}(1-p)^{N},$$

where \(n-1\) is the number of successes, \(p\) is the probability of success, and \(N+n-1\) is the number of trials.

The negative binomial distribution gives the probability of n-1 successes and N failures in N+n-1 trials, and success on the (N+n)th trial.

If one throws a die repeatedly until the third time a "1" appears, then the probability distribution of the number of non-"1"s that appear before the third "1" is a negative binomial distribution.

####References

  1. Weisstein, Eric W. "Negative Binomial Distribution." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/NegativeBinomialDistribution.html
  2. Wikipedia, "Negative binomial distribution", http://en.wikipedia.org/wiki/Negative_binomial_distribution

####Examples Draw samples from the distribution:

A real world example. A company drills wild-cat oil exploration wells, each with an estimated probability of success of 0.1. What is the probability of having one success for each successive well, that is what is the probability of a single success after drilling 5 wells, after 6 wells, etc.?

 $ s = np.random.negative_binomial(1, 0.1, 100000)
 $ for i in range(1, 11):
    probability = sum(s<i) / 100000.
    print i, "wells drilled, probability of one success =", probability

##noncentral_chisquare randomkit.noncentral_chisquare([output], df, nonc)

Draw samples from a noncentral chi-square distribution.

The noncentral \(\chi^2\) distribution is a generalisation of the \(\chi^2\) distribution.

####Parameters

####Notes The probability density function for the noncentral Chi-square distribution is

$$ P(x;df,nonc) = \sum^{\infty}{i=0} \frac{e^{-nonc/2}(nonc/2)^{i}}{i!}P{Y_{df+2i}}(x),$$

where \(Y_{q}\) is the Chi-square with q degrees of freedom.

In Delhi (2007), it is noted that the noncentral chi-square is useful in bombing and coverage problems, the probability of killing the point target given by the noncentral chi-squared distribution.

####References

  1. Delhi, M.S. Holla, "On a noncentral chi-square distribution in the analysis of weapon systems effectiveness", Metrika, Volume 15, Number 1 / December, 1970.
  2. Wikipedia, "Noncentral chi-square distribution" http://en.wikipedia.org/wiki/Noncentral_chi-square_distribution

####Examples Draw values from the distribution and plot the histogram

 $ import matplotlib.pyplot as plt
 $ values = plt.hist(np.random.noncentral_chisquare(3, 20, 100000),
                   bins=200, normed=True)
 $ plt.show()

Draw values from a noncentral chisquare with very small noncentrality, and compare to a chisquare.

 $ plt.figure()
 $ values = plt.hist(np.random.noncentral_chisquare(3, .0000001, 100000),
                   bins=np.arange(0., 25, .1), normed=True)
 $ values2 = plt.hist(np.random.chisquare(3, 100000),
                    bins=np.arange(0., 25, .1), normed=True)
 $ plt.plot(values[1][0:-1], values[0]-values2[0], 'ob')
 $ plt.show()

Demonstrate how large values of non-centrality lead to a more symmetric distribution.

 $ plt.figure()
 $ values = plt.hist(np.random.noncentral_chisquare(3, 20, 100000),
                   bins=200, normed=True)
 $ plt.show()

##noncentral_f randomkit.noncentral_f([output], dfnum, dfden, nonc)

Draw samples from the noncentral F distribution.

Samples are drawn from an F distribution with specified parameters, dfnum (degrees of freedom in numerator) and dfden (degrees of freedom in denominator), where both parameters > 1. nonc is the non-centrality parameter.

####Parameters

####Returns

####Notes When calculating the power of an experiment (power = probability of rejecting the null hypothesis when a specific alternative is true) the non-central F statistic becomes important. When the null hypothesis is true, the F statistic follows a central F distribution. When the null hypothesis is not true, then it follows a non-central F statistic.

####References Weisstein, Eric W. "Noncentral F-Distribution." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/NoncentralF-Distribution.html

Wikipedia, "Noncentral F distribution", http://en.wikipedia.org/wiki/Noncentral_F-distribution

####Examples In a study, testing for a specific alternative to the null hypothesis requires use of the Noncentral F distribution. We need to calculate the area in the tail of the distribution that exceeds the value of the F distribution for the null hypothesis. We'll plot the two probability distributions for comparison.

 $ dfnum = 3  between group deg of freedom
 $ dfden = 20  within groups degrees of freedom
 $ nonc = 3.0
 $ nc_vals = np.random.noncentral_f(dfnum, dfden, nonc, 1000000)
 $ NF = np.histogram(nc_vals, bins=50, normed=True)
 $ c_vals = np.random.f(dfnum, dfden, 1000000)
 $ F = np.histogram(c_vals, bins=50, normed=True)
 $ plt.plot(F[1][1:], F[0])
 $ plt.plot(NF[1][1:], NF[0])
 $ plt.show()

##normal randomkit.normal([output], loc, scale)

Draw random samples from a normal (Gaussian) distribution.

The probability density function of the normal distribution, first derived by De Moivre and 200 years later by both Gauss and Laplace independently [2], is often called the bell curve because of its characteristic shape (see the example below).

The normal distributions occurs often in nature. For example, it describes the commonly occurring distribution of samples influenced by a large number of tiny, random disturbances, each with its own unique distribution [2].

####Parameters

####See Also

####Notes The probability density for the Gaussian distribution is

$$ p(x) = \frac{1}{\sqrt{ 2 \pi \sigma^2 }} e^{ - \frac{ (x - \mu)^2 } {2 \sigma^2} },$$

where \(\mu\) is the mean and \(\sigma\) the standard deviation. The square of the standard deviation, \(\sigma^2\), is called the variance.

The function has its peak at the mean, and its "spread" increases with the standard deviation (the function reaches 0.607 times its maximum at \(x + \sigma\) and \(x - \sigma\) [2]). This implies that numpy.random.normal is more likely to return samples lying close to the mean, rather than those far away.

####References

  1. Wikipedia, "Normal distribution", http://en.wikipedia.org/wiki/Normal_distribution
  2. P. R. Peebles Jr., "Central Limit Theorem" in "Probability, Random Variables and Random Signal Principles", 4th ed., 2001, pp. 51, 51, 125.

####Examples Draw samples from the distribution:

 $ mu, sigma = 0, 0.1  mean and standard deviation
 $ s = np.random.normal(mu, sigma, 1000)

Verify the mean and the variance:

 $ abs(mu - np.mean(s)) < 0.01

True

 $ abs(sigma - np.std(s, ddof=1)) < 0.01

True

Display the histogram of the samples, along with the probability density function:

 $ import matplotlib.pyplot as plt
 $ count, bins, ignored = plt.hist(s, 30, normed=True)
 $ plt.plot(bins, 1/(sigma * np.sqrt(2 * np.pi)) *
                np.exp( - (bins - mu)**2 / (2 * sigma**2) ),
          linewidth=2, color='r')
 $ plt.show()

##pareto randomkit.pareto([output], a)

Draw samples from a Pareto II or Lomax distribution with specified shape.

The Lomax or Pareto II distribution is a shifted Pareto distribution. The classical Pareto distribution can be obtained from the Lomax distribution by adding the location parameter m, see below. The smallest value of the Lomax distribution is zero while for the classical Pareto distribution it is m, where the standard Pareto distribution has location m=1. Lomax can also be considered as a simplified version of the Generalized Pareto distribution (available in SciPy), with the scale set to one and the location set to zero.

The Pareto distribution must be greater than zero, and is unbounded above. It is also known as the "80-20 rule". In this distribution, 80 percent of the weights are in the lowest 20 percent of the range, while the other 20 percent fill the remaining 80 percent of the range.

####Parameters

####See Also

####Notes The probability density for the Pareto distribution is

$$ p(x) = \frac{am^a}{x^{a+1}}$$

where \(a\) is the shape and \(m\) the location

The Pareto distribution, named after the Italian economist Vilfredo Pareto, is a power law probability distribution useful in many real world problems. Outside the field of economics it is generally referred to as the Bradford distribution. Pareto developed the distribution to describe the distribution of wealth in an economy. It has also found use in insurance, web page access statistics, oil field sizes, and many other problems, including the download frequency for projects in Sourceforge [1]. It is one of the so-called "fat-tailed" distributions.

####References

  1. Francis Hunt and Paul Johnson, On the Pareto Distribution of Sourceforge projects.
  2. Pareto, V. (1896). Course of Political Economy. Lausanne.
  3. Reiss, R.D., Thomas, M.(2001), Statistical Analysis of Extreme Values, Birkhauser Verlag, Basel, pp 23-30.
  4. Wikipedia, "Pareto distribution", http://en.wikipedia.org/wiki/Pareto_distribution

####Examples Draw samples from the distribution:

 $ a, m = 3., 1.  shape and mode
 $ s = np.random.pareto(a, 1000) + m

Display the histogram of the samples, along with the probability density function:

 $ import matplotlib.pyplot as plt
 $ count, bins, ignored = plt.hist(s, 100, normed=True, align='center')
 $ fit = a*m**a/bins**(a+1)
 $ plt.plot(bins, max(count)*fit/max(fit),linewidth=2, color='r')
 $ plt.show()

##poisson randomkit.poisson([output], lam)

Draw samples from a Poisson distribution.

The Poisson distribution is the limit of the Binomial distribution for large N.

####Parameters

####Notes The Poisson distribution

$$ f(k; \lambda)=\frac{\lambda^k e^{-\lambda}}{k!}$$

For events with an expected separation \(\lambda\) the Poisson distribution \(f(k; \lambda)\) describes the probability of \(k\) events occurring within the observed interval \(\lambda\).

Because the output is limited to the range of the C long type, a ValueError is raised when lam is within 10 sigma of the maximum representable value.

####References

  1. Weisstein, Eric W. "Poisson Distribution." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/PoissonDistribution.html
  2. Wikipedia, "Poisson distribution", http://en.wikipedia.org/wiki/Poisson_distribution

####Examples Draw samples from the distribution:

 $ import numpy as np
 $ s = np.random.poisson(5, 10000)

Display histogram of the sample:

 $ import matplotlib.pyplot as plt
 $ count, bins, ignored = plt.hist(s, 14, normed=True)
 $ plt.show()

##power randomkit.power([output], a)

Draws samples in [0, 1] from a power distribution with positive exponent a - 1.

Also known as the power function distribution.

####Parameters

####Returns

####Raises ValueError If a<1.

####Notes The probability density function is

$$ P(x; a) = ax^{a-1}, 0 \le x \le 1, a>0.$$

The power function distribution is just the inverse of the Pareto distribution. It may also be seen as a special case of the Beta distribution.

It is used, for example, in modeling the over-reporting of insurance claims.

####References

  1. Christian Kleiber, Samuel Kotz, "Statistical size distributions in economics and actuarial sciences", Wiley, 2003.
  2. Heckert, N. A. and Filliben, James J. (2003). NIST Handbook 148: Dataplot Reference Manual, Volume 2: Let Subcommands and Library Functions", National Institute of Standards and Technology Handbook Series, June 2003. http://www.itl.nist.gov/div898/software/dataplot/refman2/auxillar/powpdf.pdf

####Examples Draw samples from the distribution:

 $ a = 5.  shape
 $ samples = 1000
 $ s = np.random.power(a, samples)

Display the histogram of the samples, along with the probability density function:

 $ import matplotlib.pyplot as plt
 $ count, bins, ignored = plt.hist(s, bins=30)
 $ x = np.linspace(0, 1, 100)
 $ y = a*x**(a-1.)
 $ normed_y = samples*np.diff(bins)[0]*y
 $ plt.plot(x, normed_y)
 $ plt.show()

Compare the power function distribution to the inverse of the Pareto.

 $ from scipy import stats
 $ rvs = np.random.power(5, 1000000)
 $ rvsp = np.random.pareto(5, 1000000)
 $ xx = np.linspace(0,1,100)
 $ powpdf = stats.powerlaw.pdf(xx,5)


 $ plt.figure()
 $ plt.hist(rvs, bins=50, normed=True)
 $ plt.plot(xx,powpdf,'r-')
 $ plt.title('np.random.power(5)')


 $ plt.figure()
 $ plt.hist(1./(1.+rvsp), bins=50, normed=True)
 $ plt.plot(xx,powpdf,'r-')
 $ plt.title('inverse of 1 + np.random.pareto(5)')


 $ plt.figure()
 $ plt.hist(1./(1.+rvsp), bins=50, normed=True)
 $ plt.plot(xx,powpdf,'r-')
 $ plt.title('inverse of stats.pareto(5)')

##randint randomkit.randint(low, high)

Return random integers from low (inclusive) to high (inclusive).

Return random integers from the "discrete uniform" distribution in the closed interval [low, high].

Note: This function behaves differently from the numpy version shown in examples.

####Parameters

####Returns

####See Also

####Examples

 $ np.random.randint(2, size=10)

array([1, 0, 0, 0, 1, 1, 0, 0, 1, 0])

 $ np.random.randint(1, size=10)

array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0])

Generate a 2 x 4 array of ints between 0 and 4, inclusive:

 $ np.random.randint(5, size=(2, 4))

array([[4, 0, 2, 1], [3, 2, 2, 0]])

##random randomkit.random_sample([output], )

Return random floats in the half-open interval [0.0, 1.0).

Results are from the "continuous uniform" distribution over the stated interval. To sample \(Unif[a, b), b > a\) multiply the output of random_sample by (b-a) and add a::

(b - a) * random_sample() + a

####Parameters

####Returns

####Examples

 $ np.random.random_sample()

0.47108547995356098

 $ type(np.random.random_sample())

<type 'float'>

 $ np.random.random_sample((5,))

array([ 0.30220482, 0.86820401, 0.1654503 , 0.11659149, 0.54323428])

Three-by-two array of random numbers from [-5, 0):

 $ 5 * np.random.random_sample((3, 2)) - 5

array([[-3.99149989, -0.52338984], [-2.99091858, -0.79479508], [-1.23204345, -1.75224494]])

##random_sample randomkit.random_sample([output], )

Return random floats in the half-open interval [0.0, 1.0).

Results are from the "continuous uniform" distribution over the stated interval. To sample \(Unif[a, b), b > a\) multiply the output of random_sample by (b-a) and add a::

(b - a) * random_sample() + a

####Parameters

####Returns

####Examples

 $ np.random.random_sample()

0.47108547995356098

 $ type(np.random.random_sample())

<type 'float'>

 $ np.random.random_sample((5,))

array([ 0.30220482, 0.86820401, 0.1654503 , 0.11659149, 0.54323428])

Three-by-two array of random numbers from [-5, 0):

 $ 5 * np.random.random_sample((3, 2)) - 5

array([[-3.99149989, -0.52338984], [-2.99091858, -0.79479508], [-1.23204345, -1.75224494]])

##rayleigh randomkit.rayleigh([output], scale)

Draw samples from a Rayleigh distribution.

The \(\chi\) and Weibull distributions are generalizations of the Rayleigh.

####Parameters

####Notes The probability density function for the Rayleigh distribution is

$$ P(x;scale) = \frac{x}{scale^2}e^{\frac{-x^2}{2 \cdotp scale^2}}$$

The Rayleigh distribution arises if the wind speed and wind direction are both gaussian variables, then the vector wind velocity forms a Rayleigh distribution. The Rayleigh distribution is used to model the expected output from wind turbines.

####References

  1. Brighton Webs Ltd., Rayleigh Distribution, http://www.brighton-webs.co.uk/distributions/rayleigh.asp
  2. Wikipedia, "Rayleigh distribution" http://en.wikipedia.org/wiki/Rayleigh_distribution

####Examples Draw values from the distribution and plot the histogram

 $ values = hist(np.random.rayleigh(3, 100000), bins=200, normed=True)

Wave heights tend to follow a Rayleigh distribution. If the mean wave height is 1 meter, what fraction of waves are likely to be larger than 3 meters?

 $ meanvalue = 1
 $ modevalue = np.sqrt(2 / np.pi) * meanvalue
 $ s = np.random.rayleigh(modevalue, 1000000)

The percentage of waves larger than 3 meters is:

 $ 100.*sum(s>3)/1000000.

0.087300000000000003

##standard_cauchy randomkit.standard_cauchy([output], )

Standard Cauchy distribution with mode = 0.

Also known as the Lorentz distribution.

####Parameters

####Returns

####Notes The probability density function for the full Cauchy distribution is

$$ P(x; x_0, \gamma) = \frac{1}{\pi \gamma \bigl[ 1+ (\frac{x-x_0}{\gamma})^2 \bigr] }$$

and the Standard Cauchy distribution just sets \(x_0=0\) and \(\gamma=1\)

The Cauchy distribution arises in the solution to the driven harmonic oscillator problem, and also describes spectral line broadening. It also describes the distribution of values at which a line tilted at a random angle will cut the x axis.

When studying hypothesis tests that assume normality, seeing how the tests perform on data from a Cauchy distribution is a good indicator of their sensitivity to a heavy-tailed distribution, since the Cauchy looks very much like a Gaussian distribution, but with heavier tails.

####References

  1. NIST/SEMATECH e-Handbook of Statistical Methods, "Cauchy Distribution", http://www.itl.nist.gov/div898/handbook/eda/section3/eda3663.htm
  2. Weisstein, Eric W. "Cauchy Distribution." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/CauchyDistribution.html
  3. Wikipedia, "Cauchy distribution" http://en.wikipedia.org/wiki/Cauchy_distribution

####Examples Draw samples and plot the distribution:

 $ s = np.random.standard_cauchy(1000000)
 $ s = s[(s>-25) & (s<25)]   truncate distribution so it plots well
 $ plt.hist(s, bins=100)
 $ plt.show()

##standard_exponential randomkit.standard_exponential([output], )

Draw samples from the standard exponential distribution.

standard_exponential is identical to the exponential distribution with a scale parameter of 1.

####Parameters

####Returns

####Examples Output a 3x8000 array:

 $ n = np.random.standard_exponential((3, 8000))

##standard_gamma randomkit.standard_gamma([output], shape)

Draw samples from a Standard Gamma distribution.

Samples are drawn from a Gamma distribution with specified parameters, shape (sometimes designated "k") and scale=1.

####Parameters

####Returns

####See Also

####Notes The probability density for the Gamma distribution is

$$ p(x) = x^{k-1}\frac{e^{-x/\theta}}{\theta^k\Gamma(k)},$$

where \(k\) is the shape and \(\theta\) the scale, and \(\Gamma\) is the Gamma function.

The Gamma distribution is often used to model the times to failure of electronic components, and arises naturally in processes for which the waiting times between Poisson distributed events are relevant.

####References

  1. Weisstein, Eric W. "Gamma Distribution." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/GammaDistribution.html
  2. Wikipedia, "Gamma-distribution", http://en.wikipedia.org/wiki/Gamma-distribution

####Examples Draw samples from the distribution:

 $ shape, scale = 2., 1.  mean and width
 $ s = np.random.standard_gamma(shape, 1000000)

Display the histogram of the samples, along with the probability density function:

 $ import matplotlib.pyplot as plt
 $ import scipy.special as sps
 $ count, bins, ignored = plt.hist(s, 50, normed=True)
 $ y = bins**(shape-1) * ((np.exp(-bins/scale))/ \
                       (sps.gamma(shape) * scale**shape))
 $ plt.plot(bins, y, linewidth=2, color='r')
 $ plt.show()

##standard_normal randomkit.standard_normal([output], )

Returns samples from a Standard Normal distribution (mean=0, stdev=1).

####Parameters

####Returns

####Examples

 $ s = np.random.standard_normal(8000)
 $ s

####array([ 0.6888893 , 0.78096262, -0.89086505, ..., 0.49876311, random0.38672696, -0.4685006 ]) random

 $ s.shape

(8000,)

 $ s = np.random.standard_normal(size=(3, 4, 2))
 $ s.shape

(3, 4, 2)

##standard_t randomkit.standard_t([output], df)

Standard Student's t distribution with df degrees of freedom.

A special case of the hyperbolic distribution. As df gets large, the result resembles that of the standard normal distribution (standard_normal).

####Parameters

####Returns

####Notes The probability density function for the t distribution is

$$ P(x, df) = \frac{\Gamma(\frac{df+1}{2})}{\sqrt{\pi df} \Gamma(\frac{df}{2})}\Bigl( 1+\frac{x^2}{df} \Bigr)^{-(df+1)/2}$$

The t test is based on an assumption that the data come from a Normal distribution. The t test provides a way to test whether the sample mean (that is the mean calculated from the data) is a good estimate of the true mean.

The derivation of the t-distribution was forst published in 1908 by William Gisset while working for the Guinness Brewery in Dublin. Due to proprietary issues, he had to publish under a pseudonym, and so he used the name Student.

####References

  1. Dalgaard, Peter, "Introductory Statistics With R", Springer, 2002.
  2. Wikipedia, "Student's t-distribution" http://en.wikipedia.org/wiki/Student's_t-distribution

####Examples From Dalgaard page 83 [1], suppose the daily energy intake for 11 women in Kj is:

 $ intake = np.array([5260., 5470, 5640, 6180, 6390, 6515, 6805, 7515, \
                    7515, 8230, 8770])

Does their energy intake deviate systematically from the recommended value of 7725 kJ?

We have 10 degrees of freedom, so is the sample mean within 95% of the recommended value?

 $ s = np.random.standard_t(10, size=100000)
 $ np.mean(intake)

6753.636363636364

 $ intake.std(ddof=1)

1142.1232221373727

Calculate the t statistic, setting the ddof parameter to the unbiased value so the divisor in the standard deviation will be degrees of freedom, N-1.

 $ t = (np.mean(intake)-7725)/(intake.std(ddof=1)/np.sqrt(len(intake)))
 $ import matplotlib.pyplot as plt
 $ h = plt.hist(s, bins=100, normed=True)

For a one-sided t-test, how far out in the distribution does the t statistic appear?

 $      $ np.sum(s<t) / float(len(s))

0.0090699999999999999 random

So the p-value is about 0.009, which says the null hypothesis has a probability of about 99% of being true.

##triangular randomkit.triangular([output], left, mode, right)

Draw samples from the triangular distribution.

The triangular distribution is a continuous probability distribution with lower limit left, peak at mode, and upper limit right. Unlike the other distributions, these parameters directly define the shape of the pdf.

####Parameters

####Returns

####Notes The probability density function for the Triangular distribution is

$$ P(x;l, m, r) = \begin{cases} \frac{2(x-l)}{(r-l)(m-l)}& \text{for $l \leq x \leq m$},\ \frac{2(m-x)}{(r-l)(r-m)}& \text{for $m \leq x \leq r$},\ 0& \text{otherwise}. \end{cases}$$

The triangular distribution is often used in ill-defined problems where the underlying distribution is not known, but some knowledge of the limits and mode exists. Often it is used in simulations.

####References

  1. Wikipedia, "Triangular distribution" http://en.wikipedia.org/wiki/Triangular_distribution

####Examples Draw values from the distribution and plot the histogram:

 $ import matplotlib.pyplot as plt
 $ h = plt.hist(np.random.triangular(-3, 0, 8, 100000), bins=200,
              normed=True)
 $ plt.show()

##uniform randomkit.uniform([output], low, high)

Draw samples from a uniform distribution.

Samples are uniformly distributed over the half-open interval [low, high) (includes low, but excludes high). In other words, any value within the given interval is equally likely to be drawn by uniform.

####Parameters

####Returns

####See Also

####Notes The probability density function of the uniform distribution is

$$ p(x) = \frac{1}{b - a}$$

anywhere within the interval [a, b), and zero elsewhere.

####Examples Draw samples from the distribution:

 $ s = np.random.uniform(-1,0,1000)

All values are within the given interval:

 $ np.all(s >= -1)

True

 $ np.all(s < 0)

True

Display the histogram of the samples, along with the probability density function:

 $ import matplotlib.pyplot as plt
 $ count, bins, ignored = plt.hist(s, 15, normed=True)
 $ plt.plot(bins, np.ones_like(bins), linewidth=2, color='r')
 $ plt.show()

##vonmises randomkit.vonmises([output], mu, kappa)

Draw samples from a von Mises distribution.

Samples are drawn from a von Mises distribution with specified mode (mu) and dispersion (kappa), on the interval [-pi, pi].

The von Mises distribution (also known as the circular normal distribution) is a continuous probability distribution on the unit circle. It may be thought of as the circular analogue of the normal distribution.

####Parameters

####Returns

####See Also

####Notes The probability density for the von Mises distribution is

$$ p(x) = \frac{e^{\kappa cos(x-\mu)}}{2\pi I_0(\kappa)},$$

where \(\mu\) is the mode and \(\kappa\) the dispersion, and \(I_0(\kappa)\) is the modified Bessel function of order 0.

The von Mises is named for Richard Edler von Mises, who was born in Austria-Hungary, in what is now the Ukraine. He fled to the United States in 1939 and became a professor at Harvard. He worked in probability theory, aerodynamics, fluid mechanics, and philosophy of science.

####References Abramowitz, M. and Stegun, I. A. (ed.), Handbook of Mathematical Functions, New York: Dover, 1965.

von Mises, R., Mathematical Theory of Probability and Statistics, New York: Academic Press, 1964.

####Examples Draw samples from the distribution:

 $ mu, kappa = 0.0, 4.0  mean and dispersion
 $ s = np.random.vonmises(mu, kappa, 1000)

Display the histogram of the samples, along with the probability density function:

 $ import matplotlib.pyplot as plt
 $ import scipy.special as sps
 $ count, bins, ignored = plt.hist(s, 50, normed=True)
 $ x = np.arange(-np.pi, np.pi, 2*np.pi/50.)
 $ y = -np.exp(kappa*np.cos(x-mu))/(2*np.pi*sps.jn(0,kappa))
 $ plt.plot(x, y/max(y), linewidth=2, color='r')
 $ plt.show()

##wald randomkit.wald([output], mean, scale)

Draw samples from a Wald, or Inverse Gaussian, distribution.

As the scale approaches infinity, the distribution becomes more like a Gaussian.

Some references claim that the Wald is an Inverse Gaussian with mean=1, but this is by no means universal.

The Inverse Gaussian distribution was first studied in relationship to Brownian motion. In 1956 M.C.K. Tweedie used the name Inverse Gaussian because there is an inverse relationship between the time to cover a unit distance and distance covered in unit time.

####Parameters

####Returns

####Notes The probability density function for the Wald distribution is

$$ P(x;mean,scale) = \sqrt{\frac{scale}{2\pi x^3}}e^ \frac{-scale(x-mean)^2}{2\cdotp mean^2x}$$

As noted above the Inverse Gaussian distribution first arise from attempts to model Brownian Motion. It is also a competitor to the Weibull for use in reliability modeling and modeling stock returns and interest rate processes.

####References

  1. Brighton Webs Ltd., Wald Distribution, http://www.brighton-webs.co.uk/distributions/wald.asp
  2. Chhikara, Raj S., and Folks, J. Leroy, "The Inverse Gaussian
  2. Wikipedia, "Wald distribution" http://en.wikipedia.org/wiki/Wald_distribution

####Examples Draw values from the distribution and plot the histogram:

 $ import matplotlib.pyplot as plt
 $ h = plt.hist(np.random.wald(3, 2, 100000), bins=200, normed=True)
 $ plt.show()

##weibull randomkit.weibull([output], a)

Weibull distribution.

Draw samples from a 1-parameter Weibull distribution with the given shape parameter a.

$$ X = (-ln(U))^{1/a}$$

Here, U is drawn from the uniform distribution over (0,1].

The more common 2-parameter Weibull, including a scale parameter \(\lambda\) is just \(X = \lambda(-ln(U))^{1/a}\).

####Parameters

####See Also scipy.stats.distributions.weibull_max scipy.stats.distributions.weibull_min scipy.stats.distributions.genextreme gumbel

####Notes The Weibull (or Type III asymptotic extreme value distribution for smallest values, SEV Type III, or Rosin-Rammler distribution) is one of a class of Generalized Extreme Value (GEV) distributions used in modeling extreme value problems. This class includes the Gumbel and Frechet distributions.

The probability density for the Weibull distribution is

$$ p(x) = \frac{a} {\lambda}(\frac{x}{\lambda})^{a-1}e^{-(x/\lambda)^a},$$

where \(a\) is the shape and \(\lambda\) the scale.

The function has its peak (the mode) at \(\lambda(\frac{a-1}{a})^{1/a}\).

When a = 1, the Weibull distribution reduces to the exponential distribution.

####References

  1. Waloddi Weibull, Professor, Royal Technical University, Stockholm, 1939 "A Statistical Theory Of The Strength Of Materials", Ingeniorsvetenskapsakademiens Handlingar Nr 151, 1939, Generalstabens Litografiska Anstalts Forlag, Stockholm.
  2. Waloddi Weibull, 1951 "A Statistical Distribution Function of Wide Applicability", Journal Of Applied Mechanics ASME Paper.
  3. Wikipedia, "Weibull distribution", http://en.wikipedia.org/wiki/Weibull_distribution

####Examples Draw samples from the distribution:

 $ a = 5.  shape
 $ s = np.random.weibull(a, 1000)

Display the histogram of the samples, along with the probability density function:

 $ import matplotlib.pyplot as plt
 $ x = np.arange(1,100.)/50.
 $ def weib(x,n,a):
     return (a / n) * (x / n)**(a - 1) * np.exp(-(x / n)**a)

 $ count, bins, ignored = plt.hist(np.random.weibull(5.,1000))
 $ x = np.arange(1,100.)/50.
 $ scale = count.max()/weib(x, 1., 5.).max()
 $ plt.plot(x, weib(x, 1., 5.)*scale)
 $ plt.show()

##zipf randomkit.zipf([output], a)

Draw samples from a Zipf distribution.

Samples are drawn from a Zipf distribution with specified parameter a > 1.

The Zipf distribution (also known as the zeta distribution) is a continuous probability distribution that satisfies Zipf's law: the frequency of an item is inversely proportional to its rank in a frequency table.

####Parameters

####Returns

####See Also

####Notes The probability density for the Zipf distribution is

$$ p(x) = \frac{x^{-a}}{\zeta(a)},$$

where \(\zeta\) is the Riemann Zeta function.

It is named for the American linguist George Kingsley Zipf, who noted that the frequency of any word in a sample of a language is inversely proportional to its rank in the frequency table.

####References Zipf, G. K., Selected Studies of the Principle of Relative Frequency in Language, Cambridge, MA: Harvard Univ. Press, 1932.

####Examples Draw samples from the distribution:

 $ a = 2.  parameter
 $ s = np.random.zipf(a, 1000)

Display the histogram of the samples, along with the probability density function:

 $ import matplotlib.pyplot as plt
 $ import scipy.special as sps

Truncate s values at 50 so plot is interesting

 $ count, bins, ignored = plt.hist(s[s<50], 50, normed=True)
 $ x = np.arange(1., 50.)
 $ y = x**(-a)/sps.zetac(a)
 $ plt.plot(x, y/max(y), linewidth=2, color='r')
 $ plt.show()