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# Statistics

## Introduction

SciPy has a tremendous number of basic statistics routines with more easily added by the end user (if you create one please contribute it). All of the statistics functions are located in the sub-package `scipy.stats` and a fairly complete listing of these functions can be had using `info(stats)`.

### Random Variables

There are two general distribution classes that have been implemented for encapsulating continuous random variables and discrete random variables . Over 80 continuous random variables and 10 discrete random variables have been implemented using these classes. The list of the random variables available is in the docstring for the stats sub- package.

Note: The following is work in progress

## Distributions

First some imports

    >>> import numpy as np
    >>> from scipy import stats
    >>> import warnings
    >>> warnings.simplefilter('ignore', DeprecationWarning)

We can obtain the list of available distribution through introspection:

    >>> dist_continu = [d for d in dir(stats) if
    ...                 isinstance(getattr(stats,d), stats.rv_continuous)]
    >>> dist_discrete = [d for d in dir(stats) if
    ...                  isinstance(getattr(stats,d), stats.rv_discrete)]
    >>> print 'number of continuous distributions:', len(dist_continu)
    number of continuous distributions: 84
    >>> print 'number of discrete distributions:  ', len(dist_discrete)
    number of discrete distributions:   12

Distributions can be used in one of two ways, either by passing all distribution parameters to each method call or by freezing the parameters for the instance of the distribution. As an example, we can get the median of the distribution by using the percent point function, ppf, which is the inverse of the cdf:

    >>> print stats.nct.ppf(0.5, 10, 2.5)
    2.56880722561
    >>> my_nct = stats.nct(10, 2.5)
    >>> print my_nct.ppf(0.5)
    2.56880722561

`help(stats.nct)` prints the complete docstring of the distribution. Instead we can print just some basic information:

    >>> print stats.nct.extradoc #contains the distribution specific docs
    Non-central Student T distribution

                                     df**(df/2) * gamma(df+1)
    nct.pdf(x,df,nc) = --------------------------------------------------
                       2**df*exp(nc**2/2)*(df+x**2)**(df/2) * gamma(df/2)
    for df > 0, nc > 0.

    >>> print 'number of arguments: %d, shape parameters: %s'% (stats.nct.numargs,
    ...                                                         stats.nct.shapes)
    number of arguments: 2, shape parameters: df,nc
    >>> print 'bounds of distribution lower: %s, upper: %s' % (stats.nct.a,
    ...                                                        stats.nct.b)
    bounds of distribution lower: -1.#INF, upper: 1.#INF

We can list all methods and properties of the distribution with `dir(stats.nct)`. Some of the methods are private methods, that are not named as such, i.e. no leading underscore, for example veccdf or xa and xb are for internal calculation. The main methods we can see when we list the methods of the frozen distribution:

    >>> print dir(my_nct) #reformatted
        ['__class__', '__delattr__', '__dict__', '__doc__', '__getattribute__',
        '__hash__', '__init__', '__module__', '__new__', '__reduce__', '__reduce_ex__',
        '__repr__', '__setattr__', '__str__', '__weakref__', 'args', 'cdf', 'dist',
        'entropy', 'isf', 'kwds', 'moment', 'pdf', 'pmf', 'ppf', 'rvs', 'sf', 'stats']

The main public methods are:

- rvs: Random Variates
- pdf: Probability Density Function
- cdf: Cumulative Distribution Function
- sf: Survival Function (1-CDF)
- ppf: Percent Point Function (Inverse of CDF)
- isf: Inverse Survival Function (Inverse of SF)
- stats: Return mean, variance, (Fisher’s) skew, or (Fisher’s) kurtosis
- moment: non-central moments of the distribution

The main additional methods of the not frozen distribution are related to the estimation of distrition parameters:

- fit: maximum likelihood estimation of distribution parameters, including location

    and scale

- fit_loc_scale: estimation of location and scale when shape parameters are given

- nnlf: negative log likelihood function

- expect: Calculate the expectation of a function against the pdf or pmf

All continuous distributions take loc and scale as keyword parameters to adjust the location and scale of the distribution, e.g. for the standard normal distribution location is the mean and scale is the standard deviation. The standardized distribution for a random variable x is obtained through `(x - loc) / scale`.

Discrete distribution have most of the same basic methods, however pdf is replaced the probability mass function pmf, no estimation methods, such as fit, are available, and scale is not a valid keyword parameter. The location parameter, keyword loc can be used to shift the distribution.

The basic methods, pdf, cdf, sf, ppf, and isf are vectorized with `np.vectorize`, and the usual numpy broadcasting is applied. For example, we can calculate the critical values for the upper tail of the t distribution for different probabilites and degrees of freedom.

    >>> stats.t.isf([0.1, 0.05, 0.01], [[10], [11]])
    array([[ 1.37218364,  1.81246112,  2.76376946],
           [ 1.36343032,  1.79588482,  2.71807918]])

Here, the first row are the critical values for 10 degrees of freedom and the second row is for 11 d.o.f., i.e. this is the same as

    >>> stats.t.isf([0.1, 0.05, 0.01], 10)
    array([ 1.37218364,  1.81246112,  2.76376946])
    >>> stats.t.isf([0.1, 0.05, 0.01], 11)
    array([ 1.36343032,  1.79588482,  2.71807918])

If both, probabilities and degrees of freedom have the same array shape, then element wise matching is used. As an example, we can obtain the 10% tail for 10 d.o.f., the 5% tail for 11 d.o.f. and the 1% tail for 12 d.o.f. by

    >>> stats.t.isf([0.1, 0.05, 0.01], [10, 11, 12])
    array([ 1.37218364,  1.79588482,  2.68099799])

### Performance and Remaining Issues

The performance of the individual methods, in terms of speed, varies widely by distribution and method. The results of a method are obtained in one of two ways, either by explicit calculation or by a generic algorithm that is independent of the specific distribution. Explicit calculation, requires that the method is directly specified for the given distribution, either through analytic formulas or through special functions in scipy.special or numpy.random for rvs. These are usually relatively fast calculations. The generic methods are used if the distribution does not specify any explicit calculation. To define a distribution, only one of pdf or cdf is necessary, all other methods can be derived using numeric integration and root finding. These indirect methods can be very slow. As an example, `rgh = stats.gausshyper.rvs(0.5, 2, 2, 2, size=100)` creates random variables in a very indirect way and takes about 19 seconds for 100 random variables on my computer, while one million random variables from the standard normal or from the t distribution take just above one second.

The distributions in scipy.stats have recently been corrected and improved and gained a considerable test suite, however a few issues remain:

- skew and kurtosis, 3rd and 4th moments and entropy are not thoroughly tested and some coarse testing indicates that there are still some incorrect results left.
- the distributions have been tested over some range of parameters, however in some corner ranges, a few incorrect results may remain.
- the maximum likelihood estimation in fit does not work with default starting parameters for all distributions and the user needs to supply good starting parameters. Also, for some distribution using a maximum likelihood estimator might inherently not be the best choice.

The next example shows how to build our own discrete distribution, and more examples for the usage of the distributions are shown below together with the statistical tests.

### Example: discrete distribution rv_discrete

In the following we use stats.rv_discrete to generate a discrete distribution that has the probabilites of the truncated normal for the intervalls centered around the integers.

    >>> npoints = 20 # number of integer support points of the distribution minus 1
    >>> npointsh = npoints / 2
    >>> npointsf = float(npoints)
    >>> nbound = 4 # bounds for the truncated normal
    >>> normbound = (1+1/npointsf) * nbound # actual bounds of truncated normal
    >>> grid = np.arange(-npointsh, npointsh+2, 1) # integer grid
    >>> gridlimitsnorm = (grid-0.5) / npointsh * nbound # bin limits for the truncnorm
    >>> gridlimits = grid - 0.5
    >>> grid = grid[:-1]
    >>> probs = np.diff(stats.truncnorm.cdf(gridlimitsnorm, -normbound, normbound))
    >>> gridint = grid
    >>> normdiscrete = stats.rv_discrete(values = (gridint,
    ...              np.round(probs, decimals=7)), name='normdiscrete')

From the docstring of rv_discrete:
  “You can construct an aribtrary discrete rv where P{X=xk} = pk by passing to the rv_discrete initialization method (through the values= keyword) a tuple of sequences (xk, pk) which describes only those values of X (xk) that occur with nonzero probability (pk).”

There are some requirements for this distribution to work. The keyword name is required. The support points of the distribution xk have to be integers. Also, I needed to limit the number of decimals. If the last two requirements are not satisfied an exception may be raised or the resulting numbers may be incorrect.

After defining the distribution, we obtain access to all methods of discrete distributions.

    >>> print 'mean = %6.4f, variance = %6.4f, skew = %6.4f, kurtosis = %6.4f'% \
    ...       normdiscrete.stats(moments =  'mvsk')
    mean = -0.0000, variance = 6.3302, skew = 0.0000, kurtosis = -0.0076

    >>> nd_std = np.sqrt(normdiscrete.stats(moments =  'v'))

Generate a random sample and compare observed frequencies with probabilities

    >>> n_sample = 500
    >>> np.random.seed(87655678) # fix the seed for replicability
    >>> rvs = normdiscrete.rvs(size=n_sample)
    >>> rvsnd = rvs
    >>> f, l = np.histogram(rvs, bins=gridlimits)
    >>> sfreq = np.vstack([gridint, f, probs*n_sample]).T
    >>> print sfreq
    [[ -1.00000000e+01   0.00000000e+00   2.95019349e-02]
     [ -9.00000000e+00   0.00000000e+00   1.32294142e-01]
     [ -8.00000000e+00   0.00000000e+00   5.06497902e-01]
     [ -7.00000000e+00   2.00000000e+00   1.65568919e+00]
     [ -6.00000000e+00   1.00000000e+00   4.62125309e+00]
     [ -5.00000000e+00   9.00000000e+00   1.10137298e+01]
     [ -4.00000000e+00   2.60000000e+01   2.24137683e+01]
     [ -3.00000000e+00   3.70000000e+01   3.89503370e+01]
     [ -2.00000000e+00   5.10000000e+01   5.78004747e+01]
     [ -1.00000000e+00   7.10000000e+01   7.32455414e+01]
     [  0.00000000e+00   7.40000000e+01   7.92618251e+01]
     [  1.00000000e+00   8.90000000e+01   7.32455414e+01]
     [  2.00000000e+00   5.50000000e+01   5.78004747e+01]
     [  3.00000000e+00   5.00000000e+01   3.89503370e+01]
     [  4.00000000e+00   1.70000000e+01   2.24137683e+01]
     [  5.00000000e+00   1.10000000e+01   1.10137298e+01]
     [  6.00000000e+00   4.00000000e+00   4.62125309e+00]
     [  7.00000000e+00   3.00000000e+00   1.65568919e+00]
     [  8.00000000e+00   0.00000000e+00   5.06497902e-01]
     [  9.00000000e+00   0.00000000e+00   1.32294142e-01]
     [  1.00000000e+01   0.00000000e+00   2.95019349e-02]]

(Source code)

(Source code)

Next, we can test, whether our sample was generated by our normdiscrete distribution. This also verifies, whether the random numbers are generated correctly

The chisquare test requires that there are a minimum number of observations in each bin. We combine the tail bins into larger bins so that they contain enough observations.

    >>> f2 = np.hstack([f[:5].sum(), f[5:-5], f[-5:].sum()])
    >>> p2 = np.hstack([probs[:5].sum(), probs[5:-5], probs[-5:].sum()])
    >>> ch2, pval = stats.chisquare(f2, p2*n_sample)

    >>> print 'chisquare for normdiscrete: chi2 = %6.3f pvalue = %6.4f' % (ch2, pval)
    chisquare for normdiscrete: chi2 = 12.466 pvalue = 0.4090

The pvalue in this case is high, so we can be quite confident that our random sample was actually generated by the distribution.

## Analysing One Sample

First, we create some random variables. We set a seed so that in each run we get identical results to look at. As an example we take a sample from the Student t distribution:

    >>> np.random.seed(282629734)
    >>> x = stats.t.rvs(10, size=1000)

Here, we set the required shape parameter of the t distribution, which in statistics corresponds to the degrees of freedom, to 10. Using size=100 means that our sample consists of 1000 independently drawn (pseudo) random numbers. Since we did not specify the keyword arguments loc and scale, those are set to their default values zero and one.

### Descriptive Statistics

x is a numpy array, and we have direct access to all array methods, e.g.

    >>> print x.max(), x.min()  # equivalent to np.max(x), np.min(x)
    5.26327732981 -3.78975572422
    >>> print x.mean(), x.var() # equivalent to np.mean(x), np.var(x)
    0.0140610663985 1.28899386208

How do the some sample properties compare to their theoretical counterparts?

    >>> m, v, s, k = stats.t.stats(10, moments='mvsk')
    >>> n, (smin, smax), sm, sv, ss, sk = stats.describe(x)

    >>> print 'distribution:',
    distribution:
    >>> sstr = 'mean = %6.4f, variance = %6.4f, skew = %6.4f, kurtosis = %6.4f'
    >>> print sstr %(m, v, s ,k)
    mean = 0.0000, variance = 1.2500, skew = 0.0000, kurtosis = 1.0000
    >>> print 'sample:      ',
    sample:
    >>> print sstr %(sm, sv, ss, sk)
    mean = 0.0141, variance = 1.2903, skew = 0.2165, kurtosis = 1.0556

Note: stats.describe uses the unbiased estimator for the variance, while np.var is the biased estimator.

For our sample the sample statistics differ a by a small amount from their theoretical counterparts.

### T-test and KS-test

We can use the t-test to test whether the mean of our sample differs in a statistcally significant way from the theoretical expectation.

    >>> print 't-statistic = %6.3f pvalue = %6.4f' %  stats.ttest_1samp(x, m)
    t-statistic =  0.391 pvalue = 0.6955

The pvalue is 0.7, this means that with an alpha error of, for example, 10%, we cannot reject the hypothesis that the sample mean is equal to zero, the expectation of the standard t-distribution.

As an exercise, we can calculate our ttest also directly without using the provided function, which should give us the same answer, and so it does:

    >>> tt = (sm-m)/np.sqrt(sv/float(n))  # t-statistic for mean
    >>> pval = stats.t.sf(np.abs(tt), n-1)*2  # two-sided pvalue = Prob(abs(t)>tt)
    >>> print 't-statistic = %6.3f pvalue = %6.4f' % (tt, pval)
    t-statistic =  0.391 pvalue = 0.6955

The Kolmogorov-Smirnov test can be used to test the hypothesis that the sample comes from the standard t-distribution

    >>> print 'KS-statistic D = %6.3f pvalue = %6.4f' % stats.kstest(x, 't', (10,))
    KS-statistic D =  0.016 pvalue = 0.9606

Again the p-value is high enough that we cannot reject the hypothesis that the random sample really is distributed according to the t-distribution. In real applications, we don’t know what the underlying distribution is. If we perform the Kolmogorov-Smirnov test of our sample against the standard normal distribution, then we also cannot reject the hypothesis that our sample was generated by the normal distribution given that in this example the p-value is almost 40%.

    >>> print 'KS-statistic D = %6.3f pvalue = %6.4f' % stats.kstest(x,'norm')
    KS-statistic D =  0.028 pvalue = 0.3949

However, the standard normal distribution has a variance of 1, while our sample has a variance of 1.29. If we standardize our sample and test it against the normal distribution, then the p-value is again large enough that we cannot reject the hypothesis that the sample came form the normal distribution.

    >>> d, pval = stats.kstest((x-x.mean())/x.std(), 'norm')
    >>> print 'KS-statistic D = %6.3f pvalue = %6.4f' % (d, pval)
    KS-statistic D =  0.032 pvalue = 0.2402

Note: The Kolmogorov-Smirnov test assumes that we test against a distribution with given parameters, since in the last case we estimated mean and variance, this assumption is violated, and the distribution of the test statistic on which the p-value is based, is not correct.

### Tails of the distribution

Finally, we can check the upper tail of the distribution. We can use the percent point function ppf, which is the inverse of the cdf function, to obtain the critical values, or, more directly, we can use the inverse of the survival function

    >>> crit01, crit05, crit10 = stats.t.ppf([1-0.01, 1-0.05, 1-0.10], 10)
    >>> print 'critical values from ppf at 1%%, 5%% and 10%% %8.4f %8.4f %8.4f'% (crit01, crit05, crit10)
    critical values from ppf at 1%, 5% and 10%   2.7638   1.8125   1.3722
    >>> print 'critical values from isf at 1%%, 5%% and 10%% %8.4f %8.4f %8.4f'% tuple(stats.t.isf([0.01,0.05,0.10],10))
    critical values from isf at 1%, 5% and 10%   2.7638   1.8125   1.3722

    >>> freq01 = np.sum(x>crit01) / float(n) * 100
    >>> freq05 = np.sum(x>crit05) / float(n) * 100
    >>> freq10 = np.sum(x>crit10) / float(n) * 100
    >>> print 'sample %%-frequency at 1%%, 5%% and 10%% tail %8.4f %8.4f %8.4f'% (freq01, freq05, freq10)
    sample %-frequency at 1%, 5% and 10% tail   1.4000   5.8000  10.5000

In all three cases, our sample has more weight in the top tail than the underlying distribution. We can briefly check a larger sample to see if we get a closer match. In this case the empirical frequency is quite close to the theoretical probability, but if we repeat this several times the fluctuations are still pretty large.

    >>> freq05l = np.sum(stats.t.rvs(10, size=10000) > crit05) / 10000.0 * 100
    >>> print 'larger sample %%-frequency at 5%% tail %8.4f'% freq05l
    larger sample %-frequency at 5% tail   4.8000

We can also compare it with the tail of the normal distribution, which has less weight in the tails:

    >>> print 'tail prob. of normal at 1%%, 5%% and 10%% %8.4f %8.4f %8.4f'% \
    ...       tuple(stats.norm.sf([crit01, crit05, crit10])*100)
    tail prob. of normal at 1%, 5% and 10%   0.2857   3.4957   8.5003

The chisquare test can be used to test, whether for a finite number of bins, the observed frequencies differ significantly from the probabilites of the hypothesized distribution.

    >>> quantiles = [0.0, 0.01, 0.05, 0.1, 1-0.10, 1-0.05, 1-0.01, 1.0]
    >>> crit = stats.t.ppf(quantiles, 10)
    >>> print crit
    [       -Inf -2.76376946 -1.81246112 -1.37218364  1.37218364  1.81246112
      2.76376946         Inf]
    >>> n_sample = x.size
    >>> freqcount = np.histogram(x, bins=crit)[0]
    >>> tprob = np.diff(quantiles)
    >>> nprob = np.diff(stats.norm.cdf(crit))
    >>> tch, tpval = stats.chisquare(freqcount, tprob*n_sample)
    >>> nch, npval = stats.chisquare(freqcount, nprob*n_sample)
    >>> print 'chisquare for t:      chi2 = %6.3f pvalue = %6.4f' % (tch, tpval)
    chisquare for t:      chi2 =  2.300 pvalue = 0.8901
    >>> print 'chisquare for normal: chi2 = %6.3f pvalue = %6.4f' % (nch, npval)
    chisquare for normal: chi2 = 64.605 pvalue = 0.0000

We see that the standard normal distribution is clearly rejected while the standard t-distribution cannot be rejected. Since the variance of our sample differs from both standard distribution, we can again redo the test taking the estimate for scale and location into account.

The fit method of the distributions can be used to estimate the parameters of the distribution, and the test is repeated using probabilites of the estimated distribution.

    >>> tdof, tloc, tscale = stats.t.fit(x)
    >>> nloc, nscale = stats.norm.fit(x)
    >>> tprob = np.diff(stats.t.cdf(crit, tdof, loc=tloc, scale=tscale))
    >>> nprob = np.diff(stats.norm.cdf(crit, loc=nloc, scale=nscale))
    >>> tch, tpval = stats.chisquare(freqcount, tprob*n_sample)
    >>> nch, npval = stats.chisquare(freqcount, nprob*n_sample)
    >>> print 'chisquare for t:      chi2 = %6.3f pvalue = %6.4f' % (tch, tpval)
    chisquare for t:      chi2 =  1.577 pvalue = 0.9542
    >>> print 'chisquare for normal: chi2 = %6.3f pvalue = %6.4f' % (nch, npval)
    chisquare for normal: chi2 = 11.084 pvalue = 0.0858

Taking account of the estimated parameters, we can still reject the hypothesis that our sample came from a normal distribution (at the 5% level), but again, with a p-value of 0.95, we cannot reject the t distribution.

### Special tests for normal distributions

Since the normal distribution is the most common distribution in statistics, there are several additional functions available to test whether a sample could have been drawn from a normal distribution

First we can test if skew and kurtosis of our sample differ significantly from those of a normal distribution:

    >>> print 'normal skewtest teststat = %6.3f pvalue = %6.4f' % stats.skewtest(x)
    normal skewtest teststat =  2.785 pvalue = 0.0054
    >>> print 'normal kurtosistest teststat = %6.3f pvalue = %6.4f' % stats.kurtosistest(x)
    normal kurtosistest teststat =  4.757 pvalue = 0.0000

These two tests are combined in the normality test

    >>> print 'normaltest teststat = %6.3f pvalue = %6.4f' % stats.normaltest(x)
    normaltest teststat = 30.379 pvalue = 0.0000

In all three tests the p-values are very low and we can reject the hypothesis that the our sample has skew and kurtosis of the normal distribution.

Since skew and kurtosis of our sample are based on central moments, we get exactly the same results if we test the standardized sample:

    >>> print 'normaltest teststat = %6.3f pvalue = %6.4f' % \
    ...                      stats.normaltest((x-x.mean())/x.std())
    normaltest teststat = 30.379 pvalue = 0.0000

Because normality is rejected so strongly, we can check whether the normaltest gives reasonable results for other cases:

    >>> print 'normaltest teststat = %6.3f pvalue = %6.4f' % stats.normaltest(stats.t.rvs(10, size=100))
    normaltest teststat =  4.698 pvalue = 0.0955
    >>> print 'normaltest teststat = %6.3f pvalue = %6.4f' % stats.normaltest(stats.norm.rvs(size=1000))
    normaltest teststat =  0.613 pvalue = 0.7361

When testing for normality of a small sample of t-distributed observations and a large sample of normal distributed observation, then in neither case can we reject the null hypothesis that the sample comes from a normal distribution. In the first case this is because the test is not powerful enough to distinguish a t and a normally distributed random variable in a small sample.

## Comparing two samples

In the following, we are given two samples, which can come either from the same or from different distribution, and we want to test whether these samples have the same statistical properties.

### Comparing means

Test with sample with identical means:

    >>> rvs1 = stats.norm.rvs(loc=5, scale=10, size=500)
    >>> rvs2 = stats.norm.rvs(loc=5, scale=10, size=500)
    >>> stats.ttest_ind(rvs1, rvs2)
    (-0.54890361750888583, 0.5831943748663857)

Test with sample with different means:

    >>> rvs3 = stats.norm.rvs(loc=8, scale=10, size=500)
    >>> stats.ttest_ind(rvs1, rvs3)
    (-4.5334142901750321, 6.507128186505895e-006)

### Kolmogorov-Smirnov test for two samples ks_2samp

For the example where both samples are drawn from the same distribution, we cannot reject the null hypothesis since the pvalue is high

    >>> stats.ks_2samp(rvs1, rvs2)
    (0.025999999999999995, 0.99541195173064878)

In the second example, with different location, i.e. means, we can reject the null hypothesis since the pvalue is below 1%

    >>> stats.ks_2samp(rvs1, rvs3)
    (0.11399999999999999, 0.0027132103661283141)

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