Statistics (`scipy.stats`)
==========================

.. sectionauthor:: Travis E. Oliphant
.. sectionauthor:: Josef Perktold
.. sectionauthor:: Nicky van Foreest

Introduction
------------

In this tutorial we discuss many, but certainly not all, features of
``scipy.stats``. The intention here is to provide a user with a
working knowledge of this package. We refer to the `reference manual
<https://docs.scipy.org/doc/scipy/reference/stats.html>`_ for further
details.


Note: This documentation is work in progress.


Random Variables
----------------

There are two general distribution classes that have been implemented
for encapsulating :ref:`continuous random variables
<continuous-random-variables>` and :ref:`discrete random variables
<discrete-random-variables>` . Over 80 continuous random variables
(RVs) and 10 discrete random variables have been implemented using
these classes. Besides this, new routines and distributions can easily
added by the end user. (If you create one, please contribute it).

All of the statistics functions are located in the sub-package
:mod:`scipy.stats` and a fairly complete listing of these functions
can be obtained using ``info(stats)``.  The list of the random
variables available can also be obtained from the docstring for the
stats sub-package.

In the discussion below we mostly focus on continuous RVs. Nearly all
applies to discrete variables also, but we point out some differences
here: :ref:`discrete_points_label`.

In the code samples below we assume that the :mod:`scipy.stats` package
is imported as

    >>> from scipy import stats

and in some cases we assume that individual objects are imported as

    >>> from scipy.stats import norm

Getting Help
^^^^^^^^^^^^

First of all, all distributions are accompanied with help
functions. To obtain just some basic information we print the relevant
docstring: ``print(stats.norm.__doc__)``.

To find the support, i.e., upper and lower bound of the distribution,
call:

    >>> print 'bounds of distribution lower: %s, upper: %s' % (norm.a, norm.b)
    bounds of distribution lower: -inf, upper: inf

We can list all methods and properties of the distribution with
``dir(norm)``.  As it turns out, some of the methods are private
methods although they are not named as such (their name does not start
with a leading underscore), for example ``veccdf``, are only available
for internal calculation (those methods will give warnings when one tries to
use them, and will be removed at some point).

To obtain the `real` main methods, we list the methods of the frozen
distribution. (We explain the meaning of a `frozen` distribution
below).

    >>> rv = norm()
    >>> dir(rv)  # 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']

Finally, we can obtain the list of available distribution through
introspection:

    >>> import warnings
    >>> warnings.simplefilter('ignore', DeprecationWarning)
    >>> 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: 95
    >>> print 'number of discrete distributions:  ', len(dist_discrete)
    number of discrete distributions:   13


Common Methods
^^^^^^^^^^^^^^

The main public methods for continuous  RVs 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


Let's take a normal RV as an example.

    >>> norm.cdf(0)
    0.5

To compute the ``cdf`` at a number of points, we can pass a list or a numpy array.

    >>> norm.cdf([-1., 0, 1])
    array([ 0.15865525,  0.5,  0.84134475])
    >>> import numpy as np
    >>> norm.cdf(np.array([-1., 0, 1]))
    array([ 0.15865525,  0.5,  0.84134475])

Thus, the basic methods such as `pdf`, `cdf`, and so on are vectorized
with ``np.vectorize``.

Other generally useful methods are supported too:

    >>> norm.mean(), norm.std(), norm.var()
    (0.0, 1.0, 1.0)
    >>> norm.stats(moments = "mv")
    (array(0.0), array(1.0))

To find the median of a distribution we can use the percent point
function ``ppf``, which is the inverse of the ``cdf``:

    >>> norm.ppf(0.5)
    0.0

To generate a sequence of random variates, use the ``size`` keyword
argument:

    >>> norm.rvs(size=3)
    array([-0.35687759,  1.34347647, -0.11710531])   # random

Note that drawing random numbers relies on generators from
:mod:`numpy.random` package. In the example above, the specific stream of
random numbers is not reproducible across runs. To achieve reproducibility,
you can explicitly seed a global variable

    >>> np.random.seed(1234)

Relying on a global state is not recommended though. A better way is to use
the `random_state` parameter which accepts an instance of
`numpy.random.RandomState` class, or an integer which is then used to seed
an internal `RandomState` object:

    >>> norm.rvs(size=5, random_state=1234)
    array([ 0.47143516, -1.19097569,  1.43270697, -0.3126519 , -0.72058873])

Don't think that ``norm.rvs(5)`` generates 5 variates:

    >>> norm.rvs(5)
    5.471435163732493

Here, ``5`` with no keyword is being interpreted as the first possible
keyword argument, ``loc``, which is the first of a pair of keyword arguments
taken by all continuous distributions.
This brings us to the topic of the next subsection.


Shifting and Scaling
^^^^^^^^^^^^^^^^^^^^

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 the location is the mean and
the scale is the standard deviation.

    >>> norm.stats(loc = 3, scale = 4, moments = "mv")
    (array(3.0), array(16.0))

In many cases the standardized distribution for a random variable ``X``
is obtained through the transformation ``(X - loc) / scale``.  The
default values are ``loc = 0`` and ``scale = 1``.

Smart use of ``loc`` and ``scale`` can help modify the standard
distributions in many ways. To illustrate the scaling further, the
``cdf`` of an exponentially distributed RV with mean :math:`1/\lambda`
is given by

.. math::

    F(x) = 1 - \exp(-\lambda x)

By applying the scaling rule above, it can be seen that by
taking ``scale  = 1./lambda`` we get the proper scale.

    >>> from scipy.stats import expon
    >>> expon.mean(scale=3.)
    3.0

.. note:: Distributions that take shape parameters may
   require more than simple application of ``loc`` and/or
   ``scale`` to achieve the desired form.  For example, the
   distribution of 2-D vector lengths given a constant vector
   of length :math:`R` perturbed by independent N(0, :math:`\sigma^2`)
   deviations in each component is
   rice(:math:`R/\sigma`, scale= :math:`\sigma`).  The first argument
   is a shape parameter that needs to be scaled along with :math:`x`.

The uniform distribution is also interesting:

    >>> from scipy.stats import uniform
    >>> uniform.cdf([0, 1, 2, 3, 4, 5], loc = 1, scale = 4)
    array([ 0.  ,  0.  ,  0.25,  0.5 ,  0.75,  1.  ])


Finally, recall from the previous paragraph that we are left with the
problem of the meaning of ``norm.rvs(5)``. As it turns out, calling a
distribution like this, the first argument, i.e., the 5, gets passed
to set the ``loc`` parameter. Let's see:

    >>> np.mean(norm.rvs(5, size=500))
    4.983550784784704

Thus, to explain the output of the example of the last section:
``norm.rvs(5)`` generates a single normally distributed random variate with
mean ``loc=5``, because of the default ``size=1``.

We recommend that you set ``loc`` and ``scale`` parameters explicitly, by
passing the values as keywords rather than as arguments. Repetition
can be minimized when calling more than one method of a given RV by
using the technique of `Freezing a Distribution`_, as explained below.


Shape Parameters
^^^^^^^^^^^^^^^^

While a general continuous random variable can be shifted and scaled
with the ``loc`` and ``scale`` parameters, some distributions require
additional shape parameters. For instance, the gamma distribution, with density

.. math::

    \gamma(x, a) = \frac{\lambda (\lambda x)^{a-1}}{\Gamma(a)} e^{-\lambda x}\;,

requires the shape parameter :math:`a`. Observe that setting
:math:`\lambda` can be obtained by setting the ``scale`` keyword to
:math:`1/\lambda`.

Let's check the number and name of the shape parameters of the gamma
distribution. (We know from the above that this should be 1.)

    >>> from scipy.stats import gamma
    >>> gamma.numargs
    1
    >>> gamma.shapes
    'a'

Now we set the value of the shape variable to 1 to obtain the
exponential distribution, so that we compare easily whether we get the
results we expect.

    >>> gamma(1, scale=2.).stats(moments="mv")
    (array(2.0), array(4.0))

Notice that we can also specify shape parameters as keywords:

   >>> gamma(a=1, scale=2.).stats(moments="mv")
   (array(2.0), array(4.0))


Freezing a Distribution
^^^^^^^^^^^^^^^^^^^^^^^

Passing the ``loc`` and ``scale`` keywords time and again can become
quite bothersome. The concept of `freezing` a RV is used to
solve such problems.

    >>> rv = gamma(1, scale=2.)

By using ``rv`` we no longer have to include the scale or the shape
parameters anymore. Thus, distributions can be used in one of two
ways, either by passing all distribution parameters to each method
call (such as we did earlier) or by freezing the parameters for the
instance of the distribution. Let us check this:

    >>> rv.mean(), rv.std()
    (2.0, 2.0)

This is indeed what we should get.


Broadcasting
^^^^^^^^^^^^

The basic methods ``pdf`` and so on satisfy the usual numpy broadcasting rules. For
example, we can calculate the critical values for the upper tail of
the t distribution for different probabilities 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 for 11 degrees of freedom (d.o.f.). Thus, the
broadcasting rules give the same result of calling ``isf`` twice:

    >>> 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 the array with probabilities, i.e., ``[0.1, 0.05, 0.01]`` and the
array of degrees of freedom i.e., ``[10, 11, 12]``, 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 calling

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


.. _discrete_points_label:

Specific Points for Discrete Distributions
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

Discrete distribution have mostly the same basic methods as the
continuous distributions.  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 still be used to shift the
distribution.

The computation of the cdf requires some extra attention. In the case
of continuous distribution the cumulative distribution function is in
most standard cases strictly monotonic increasing in the bounds (a,b)
and has therefore a unique inverse. The cdf of a discrete
distribution, however, is a step function, hence the inverse cdf,
i.e., the percent point function, requires a different definition:

::

    ppf(q) = min{x : cdf(x) >= q, x integer}

For further info, see the docs `here
<https://docs.scipy.org/doc/scipy/reference/tutorial/stats/discrete.html#percent-point-function-inverse-cdf>`__.


We can look at the hypergeometric distribution as an example

    >>> from scipy.stats import hypergeom
    >>> [M, n, N] = [20, 7, 12]

If we use the cdf at some integer points and then evaluate the ppf at those
cdf values, we get the initial integers back, for example

    >>> x = np.arange(4)*2
    >>> x
    array([0, 2, 4, 6])
    >>> prb = hypergeom.cdf(x, M, n, N)
    >>> prb
    array([ 0.0001031991744066,  0.0521155830753351,  0.6083591331269301,
            0.9897832817337386])
    >>> hypergeom.ppf(prb, M, n, N)
    array([ 0.,  2.,  4.,  6.])

If we use values that are not at the kinks of the cdf step function, we get
the next higher integer back:

    >>> hypergeom.ppf(prb + 1e-8, M, n, N)
    array([ 1.,  3.,  5.,  7.])
    >>> hypergeom.ppf(prb - 1e-8, M, n, N)
    array([ 0.,  2.,  4.,  6.])


Fitting Distributions
^^^^^^^^^^^^^^^^^^^^^

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


.. _performance_issues_label:

Performance Issues and Cautionary Remarks
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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, on the one hand, 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, on the other hand, 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. However, 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.


Remaining Issues
^^^^^^^^^^^^^^^^

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

* 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.


Building Specific Distributions
-------------------------------

The next examples shows how to build your own distributions.  Further
examples show the usage of the distributions and some statistical
tests.


Making a Continuous Distribution, i.e., Subclassing ``rv_continuous``
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

Making continuous distributions is fairly simple.

    >>> from scipy import stats
    >>> class deterministic_gen(stats.rv_continuous):
    ...     def _cdf(self, x):
    ...         return np.where(x < 0, 0., 1.)
    ...     def _stats(self):
    ...         return 0., 0., 0., 0.

    >>> deterministic = deterministic_gen(name="deterministic")
    >>> deterministic.cdf(np.arange(-3, 3, 0.5))
    array([ 0.,  0.,  0.,  0.,  0.,  0.,  1.,  1.,  1.,  1.,  1.,  1.])

Interestingly,  the ``pdf`` is now computed automatically:

    >>> deterministic.pdf(np.arange(-3, 3, 0.5))
    array([  0.00000000e+00,   0.00000000e+00,   0.00000000e+00,
             0.00000000e+00,   0.00000000e+00,   0.00000000e+00,
             5.83333333e+04,   4.16333634e-12,   4.16333634e-12,
             4.16333634e-12,   4.16333634e-12,   4.16333634e-12])


Be aware of the performance issues mentions in
:ref:`performance_issues_label`. The computation of unspecified
common methods can become very slow, since only general methods are
called which, by their very nature, cannot use any specific
information about the distribution. Thus, as a cautionary example:

    >>> from scipy.integrate import quad
    >>> quad(deterministic.pdf, -1e-1, 1e-1)
    (4.163336342344337e-13, 0.0)

But this is not correct: the integral over this pdf should be 1. Let's make the
integration interval smaller:

    >>> quad(deterministic.pdf, -1e-3, 1e-3)  # warning removed
    (1.000076872229173, 0.0010625571718182458)

This looks better. However, the problem originated from the fact that
the pdf is not specified in the class definition of the deterministic
distribution.


Subclassing ``rv_discrete``
^^^^^^^^^^^^^^^^^^^^^^^^^^^

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

**General Info**

From the docstring of rv_discrete, ``help(stats.rv_discrete)``,

  "You can construct an arbitrary 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)."

Next to this, there are some further requirements for this approach to
work:

* The keyword `name` is required.
* The support points of the distribution xk have to be integers.
* The number of significant digits (decimals) needs to be specified.

In fact, if the last two requirements are not satisfied an exception
may be raised or the resulting numbers may be incorrect.

**An Example**

Let's do the work. First

    >>> 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   # used later in the analysis
    >>> grid = grid[:-1]
    >>> probs = np.diff(stats.truncnorm.cdf(gridlimitsnorm, -normbound, normbound))
    >>> gridint = grid

And finally we can subclass ``rv_discrete``:

    >>> normdiscrete = stats.rv_discrete(values=(gridint,
    ...              np.round(probs, decimals=7)), name='normdiscrete')

Now that we have defined the distribution, we have access to all
common 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'))

**Testing the Implementation**

Let's generate a random sample and compare observed frequencies with
the 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]]


.. plot:: tutorial/examples/normdiscr_plot1.py
   :align: center
   :include-source: 0


.. plot:: tutorial/examples/normdiscr_plot2.py
   :align: center
   :include-source: 0


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=1000 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 statistically 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 probabilities 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)
    >>> crit
    array([-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 = %6.2f pvalue = %6.4f' % (tch, tpval)
    chisquare for t:      chi2 =  2.30 pvalue = 0.8901
    >>> print 'chisquare for normal: chi2 = %6.2f pvalue = %6.4f' % (nch, npval)
    chisquare for normal: chi2 = 64.60 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 probabilities 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 = %6.2f pvalue = %6.4f' % (tch, tpval)
    chisquare for t:      chi2 =  1.58 pvalue = 0.9542
    >>> print 'chisquare for normal: chi2 = %6.2f pvalue = %6.4f' % (nch, npval)
    chisquare for normal: chi2 = 11.08 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)


Kernel Density Estimation
-------------------------

A common task in statistics is to estimate the probability density function
(PDF) of a random variable from a set of data samples.  This task is called
density estimation.  The most well-known tool to do this is the histogram.
A histogram is a useful tool for visualization (mainly because everyone
understands it), but doesn't use the available data very efficiently.  Kernel
density estimation (KDE) is a more efficient tool for the same task.  The
:func:`gaussian_kde` estimator can be used to estimate the PDF of univariate as
well as multivariate data.  It works best if the data is unimodal.


Univariate estimation
^^^^^^^^^^^^^^^^^^^^^

We start with a minimal amount of data in order to see how :func:`gaussian_kde`
works, and what the different options for bandwidth selection do.  The data
sampled from the PDF is show as blue dashes at the bottom of the figure (this
is called a rug plot):

.. plot::

    >>> from scipy import stats
    >>> import matplotlib.pyplot as plt

    >>> x1 = np.array([-7, -5, 1, 4, 5], dtype=np.float)
    >>> kde1 = stats.gaussian_kde(x1)
    >>> kde2 = stats.gaussian_kde(x1, bw_method='silverman')

    >>> fig = plt.figure()
    >>> ax = fig.add_subplot(111)

    >>> ax.plot(x1, np.zeros(x1.shape), 'b+', ms=20)  # rug plot
    >>> x_eval = np.linspace(-10, 10, num=200)
    >>> ax.plot(x_eval, kde1(x_eval), 'k-', label="Scott's Rule")
    >>> ax.plot(x_eval, kde2(x_eval), 'r-', label="Silverman's Rule")

    >>> plt.show()

We see that there is very little difference between Scott's Rule and
Silverman's Rule, and that the bandwidth selection with a limited amount of
data is probably a bit too wide.  We can define our own bandwidth function to
get a less smoothed out result.

    >>> def my_kde_bandwidth(obj, fac=1./5):
    ...     """We use Scott's Rule, multiplied by a constant factor."""
    ...     return np.power(obj.n, -1./(obj.d+4)) * fac

    >>> fig = plt.figure()
    >>> ax = fig.add_subplot(111)

    >>> ax.plot(x1, np.zeros(x1.shape), 'b+', ms=20)  # rug plot
    >>> kde3 = stats.gaussian_kde(x1, bw_method=my_kde_bandwidth)
    >>> ax.plot(x_eval, kde3(x_eval), 'g-', label="With smaller BW")

    >>> plt.show()

.. plot:: tutorial/stats/plots/kde_plot2.py
   :align: center
   :include-source: 0

We see that if we set bandwidth to be very narrow, the obtained estimate for
the probability density function (PDF) is simply the sum of Gaussians around
each data point.

We now take a more realistic example, and look at the difference between the
two available bandwidth selection rules.  Those rules are known to work well
for (close to) normal distributions, but even for unimodal distributions that
are quite strongly non-normal they work reasonably well.  As a non-normal
distribution we take a Student's T distribution with 5 degrees of freedom.

.. plot:: tutorial/stats/plots/kde_plot3.py
   :align: center
   :include-source: 1

We now take a look at a bimodal distribution with one wider and one narrower
Gaussian feature.  We expect that this will be a more difficult density to
approximate, due to the different bandwidths required to accurately resolve
each feature.

    >>> from functools import partial

    >>> loc1, scale1, size1 = (-2, 1, 175)
    >>> loc2, scale2, size2 = (2, 0.2, 50)
    >>> x2 = np.concatenate([np.random.normal(loc=loc1, scale=scale1, size=size1),
    ...                      np.random.normal(loc=loc2, scale=scale2, size=size2)])

    >>> x_eval = np.linspace(x2.min() - 1, x2.max() + 1, 500)

    >>> kde = stats.gaussian_kde(x2)
    >>> kde2 = stats.gaussian_kde(x2, bw_method='silverman')
    >>> kde3 = stats.gaussian_kde(x2, bw_method=partial(my_kde_bandwidth, fac=0.2))
    >>> kde4 = stats.gaussian_kde(x2, bw_method=partial(my_kde_bandwidth, fac=0.5))

    >>> pdf = stats.norm.pdf
    >>> bimodal_pdf = pdf(x_eval, loc=loc1, scale=scale1) * float(size1) / x2.size + \
    ...               pdf(x_eval, loc=loc2, scale=scale2) * float(size2) / x2.size

    >>> fig = plt.figure(figsize=(8, 6))
    >>> ax = fig.add_subplot(111)

    >>> ax.plot(x2, np.zeros(x2.shape), 'b+', ms=12)
    >>> ax.plot(x_eval, kde(x_eval), 'k-', label="Scott's Rule")
    >>> ax.plot(x_eval, kde2(x_eval), 'b-', label="Silverman's Rule")
    >>> ax.plot(x_eval, kde3(x_eval), 'g-', label="Scott * 0.2")
    >>> ax.plot(x_eval, kde4(x_eval), 'c-', label="Scott * 0.5")
    >>> ax.plot(x_eval, bimodal_pdf, 'r--', label="Actual PDF")

    >>> ax.set_xlim([x_eval.min(), x_eval.max()])
    >>> ax.legend(loc=2)
    >>> ax.set_xlabel('x')
    >>> ax.set_ylabel('Density')
    >>> plt.show()

.. plot:: tutorial/stats/plots/kde_plot4.py
   :align: center
   :include-source: 0

As expected, the KDE is not as close to the true PDF as we would like due to
the different characteristic size of the two features of the bimodal
distribution.  By halving the default bandwidth (``Scott * 0.5``) we can do
somewhat better, while using a factor 5 smaller bandwidth than the default
doesn't smooth enough.  What we really need though in this case is a
non-uniform (adaptive) bandwidth.


Multivariate estimation
^^^^^^^^^^^^^^^^^^^^^^^

With :func:`gaussian_kde` we can perform multivariate as well as univariate
estimation.  We demonstrate the bivariate case.  First we generate some random
data with a model in which the two variates are correlated.

    >>> def measure(n):
    ...     """Measurement model, return two coupled measurements."""
    ...     m1 = np.random.normal(size=n)
    ...     m2 = np.random.normal(scale=0.5, size=n)
    ...     return m1+m2, m1-m2

    >>> m1, m2 = measure(2000)
    >>> xmin = m1.min()
    >>> xmax = m1.max()
    >>> ymin = m2.min()
    >>> ymax = m2.max()

Then we apply the KDE to the data:

    >>> X, Y = np.mgrid[xmin:xmax:100j, ymin:ymax:100j]
    >>> positions = np.vstack([X.ravel(), Y.ravel()])
    >>> values = np.vstack([m1, m2])
    >>> kernel = stats.gaussian_kde(values)
    >>> Z = np.reshape(kernel.evaluate(positions).T, X.shape)

Finally we plot the estimated bivariate distribution as a colormap, and plot
the individual data points on top.

    >>> fig = plt.figure(figsize=(8, 6))
    >>> ax = fig.add_subplot(111)

    >>> ax.imshow(np.rot90(Z), cmap=plt.cm.gist_earth_r,
    ...           extent=[xmin, xmax, ymin, ymax])
    >>> ax.plot(m1, m2, 'k.', markersize=2)

    >>> ax.set_xlim([xmin, xmax])
    >>> ax.set_ylim([ymin, ymax])

    >>> plt.show()

.. plot:: tutorial/stats/plots/kde_plot5.py
   :align: center
   :include-source: 0

