Develop Reference

Chi Squared Analysis on Data sets that don't have matching frequencies

I have 15 data sets each of which I have fitted with a curve. Now I am trying to determine the quality of fit by doing a chi-squared test; however, when I run my code:
chi, p_value = stats.chisquare(n, y)
where n is the actual data and y is the predicted data, I get the error
For each axis slice, the sum of the observed frequencies must agree with the sum of the expected frequencies to a relative tolerance of 1e-08, but the percent differences are:
0.1350785306607008
I can't seem to understand why they have to add up to the same total - are there any ways I can run a chi-squared test without muddling my data?

This chi-squared test for goodness of fit indeed requires the sums of both inputs to be (almost) the same. So, if you want to check whether your model fits the observations n well, you have to adjust the counts y of your model as described e.g. here. This could be done with a small wrapper:
from scipy.stats import chisquare
import numpy as np
def cs(n, y):
return chisquare(n, np.sum(n)/np.sum(y) * y)
Another possibility would be to go for R and use chisq.test.

How to calculate one-sided tolerance interval with scipy

I would like to calculate a one sided tolerance bound based on the normal distribution given a data set with known N (sample size), standard deviation, and mean.
If the interval were two sided I would do the following:
conf_int = stats.norm.interval(alpha, loc=mean, scale=sigma)
In my situation, I am bootstrapping samples, but if I weren't I would refer to this post on stackoverflow: Correct way to obtain confidence interval with scipy and use the following: conf_int = stats.norm.interval(0.68, loc=mean, scale=sigma / np.sqrt(len(a)))
How would you do the same thing, but to calculate this as a one sided bound (95% of values are above or below x<--bound)?

I assume that you are interested in computing one-side tolerance bound using the normal distribution (based on the fact you mention the scipy.stats.norm.interval function as the two-sided equivalent of your need).
Then the good news is that, based on the tolerance interval Wikipedia page:
One-sided normal tolerance intervals have an exact solution in terms of the sample mean and sample variance based on the noncentral t-distribution.
(FYI: Unfortunately, this is not the case for the two-sided setting)
This assertion is based on this paper. Besides paragraph 4.8 (page 23) provides the formulas.
The bad news is that I do not think there is a ready-to-use scipy function that you can safely tweak and use for your purpose.
But you can easily calculate it yourself. You can find on Github repositories that contain such a calculator from which you can find inspiration, for example that one from which I built the following illustrative example:
import numpy as np
from scipy.stats import norm, nct
# sample size
n=1000
# Percentile for the TI to estimate
p=0.9
# confidence level
g = 0.95
# a demo sample
x = np.array([np.random.normal(100) for k in range(n)])
# mean estimate based on the sample
mu_est = x.mean()
# standard deviation estimated based on the sample
sigma_est = x.std(ddof=1)
# (100*p)th percentile of the standard normal distribution
zp = norm.ppf(p)
# gth quantile of a non-central t distribution
# with n-1 degrees of freedom and non-centrality parameter np.sqrt(n)*zp
t = nct.ppf(g, df=n-1., nc=np.sqrt(n)*zp)
# k factor from Young et al paper
k = t / np.sqrt(n)
# One-sided tolerance upper bound
conf_upper_bound = mu_est + (k*sigma_est)

Here is a one-line solution with the openturns library, assuming your data is a numpy array named sample.
import openturns as ot
ot.NormalFactory().build(sample.reshape(-1, 1)).computeQuantile(0.95)
Let us unpack this. NormalFactory is a class designed to fit the parameters of a Normal distribution (mu and sigma) on a given sample: NormalFactory() creates an instance of this class.
The method build does the actual fitting and returns an object of the class Normal which represents the normal distribution with parameters mu and sigma estimated from the sample.
The sample reshape is there to make sure that OpenTURNS understands that the input sample is a collection of one-dimension points, not a single multi-dimensional point.
The class Normal then provides the method computeQuantile to compute any quantile of the distribution (the 95-th percentile in this example).
This solution does not compute the exact tolerance bound because it uses a quantile from a Normal distribution instead of a Student t-distribution. Effectively, that means that it ignores the estimation error on mu and sigma. In practice, this is only an issue for really small sample sizes.
To illustrate this, here is a comparison between the PDF of the standard normal N(0,1) distribution and the PDF of the Student t-distribution with 19 degrees of freedom (this means a sample size of 20). They can barely be distinguished.
deg_freedom = 19
graph = ot.Normal().drawPDF()
student = ot.Student(deg_freedom).drawPDF().getDrawable(0)
student.setColor('blue')
graph.add(student)
graph.setLegends(['Normal(0,1)', 't-dist k={}'.format(deg_freedom)])
graph

How to sample from a custom distribution when parameters are known?

The target is to get samples from a distribution whose parameters is known.
For example, the self-defined distribution is p(X|theta), where theta the parameter vector of K dimensions and X is the random vector of N dimensions.
Now we know (1) the theta is known; (2) p(X|theta) is NOT known, but I know p(X|theta) ∝ f(X,theta), and f is a known function.
Can pymc3 do such sampling from p(X|theta), and how?
The purpose is not sampling from posterior distribution of parameters, but want to samples from a self-defined distribution.
Starting from a simple example of sampling from a Bernoulli distribution. I did the following:
import pymc3 as pm
import numpy as np
import scipy.stats as stats
import pandas as pd
import theano.tensor as tt
with pm.Model() as model1:
p=0.3
density = pm.DensityDist('density',
lambda x1: tt.switch( x1, tt.log(p), tt.log(1 - p) ),
) #tt.switch( x1, tt.log(p), tt.log(1 - p) ) is the log likelihood from pymc3 source code
with model1:
step = pm.Metropolis()
samples = pm.sample(1000, step=step)
I expect the result is 1000 binary digits, with the proportion of 1 is about 0.3. However, I got strange results where very large numbers occur in the output.
I know something is wrong. Please help on how to correctly write pymc3 codes for such non-posterior MCMC sampling questions.

Prior predictive sampling (for which you should be using pm.sample_prior_predictive()) involves only using the RNGs provided by the RandomVariable objects in your compute graph. By default, DensityDist does not implement a RNG, but does provide the random parameter for this purpose, so you'll need to use that. The log-likelihood is only evaluated with respect to observables, so it plays no role here.
A simple way to generate a valid RNG for an arbitrary distribution is to use inverse transform sampling. In this case, one samples a uniform distribution on the unit interval and then transforms it through the inverse CDF of the desired function. For the Bernoulli case, the inverse CDF partitions the unit line based on the probability of success, assigning 0 to one part and 1 to the other.
Here is a factory-like implementation that creates a Bernoulli RNG compatible with pm.DensityDist's random parameter (i.e., accepts point and size kwargs).
def get_bernoulli_rng(p=0.5):
def _rng(point=None, size=1):
# Bernoulli inverse CDF, given p (prob of success)
_icdf = lambda q: np.uint8(q < p)
return _icdf(pm.Uniform.dist().random(point=point, size=size))
return _rng
So, to fill out the example, it would go something like
with pm.Model() as m:
p = 0.3
y = pm.DensityDist('y', lambda x: tt.switch(x, tt.log(p), tt.log(1-p)),
random=get_bernoulli_rng(p))
prior = pm.sample_prior_predictive(random_seed=2019)
prior['y'].mean() # 0.306
Obviously, this could equally be done with random=pm.Bernoulli.dist(p).random, but the above illustrates generically how one could do this with arbitrary distributions, given their inverse CDF, i.e., you only need to modify _icdf and the parameters.

How to choose between a two tailed and one tailed T-test with scipy?

Scipy website seems to be down again :(
I am conducting a paired t test. But I am not sure how to choose between two-tailed and one-tailed (I need to use two-tailed). And also the ttest_reul only return t and p values. Anyway to find out degrees of freedom? Many thanks
from scipy import stats
t_val_Input, p_val_Input = stats.ttest_rel(c1,c2)

From the docstring of ttest_rel:
ttest_rel(a, b, axis=0, nan_policy='propagate')
Calculates the T-test on TWO RELATED samples of scores, a and b.
This is a two-sided test for the null hypothesis that 2 related or
repeated samples have identical average (expected) values.
"Two-sided" means two-tailed.
The degrees of freedom is n - 1, where n is the number of pairs (i.e. len(a) or len(b)).

P-value from Chi sq test statistic in Python

I have computed a test statistic that is distributed as a chi square with 1 degree of freedom, and want to find out what P-value this corresponds to using python.
I'm a python and maths/stats newbie so I think what I want here is the probability denisty function for the chi2 distribution from SciPy. However, when I use this like so:
from scipy import stats
stats.chi2.pdf(3.84 , 1)
0.029846
However some googling and talking to some colleagues who know maths but not python have said it should be 0.05.
Any ideas?
Cheers,
Davy

Quick refresher here:
Probability Density Function: think of it as a point value; how dense is the probability at a given point?
Cumulative Distribution Function: this is the mass of probability of the function up to a given point; what percentage of the distribution lies on one side of this point?
In your case, you took the PDF, for which you got the correct answer. If you try 1 - CDF:
>>> 1 - stats.chi2.cdf(3.84, 1)
0.050043521248705147
PDF
CDF

To calculate probability of null hypothesis given chisquared sum, and degrees of freedom you can also call chisqprob:
>>> from scipy.stats import chisqprob
>>> chisqprob(3.84, 1)
0.050043521248705189
Notice:
chisqprob is deprecated! stats.chisqprob is deprecated in scipy 0.17.0; use stats.distributions.chi2.sf instead

Update: as noted, chisqprob() is deprecated for scipy version 0.17.0 onwards. High accuracy chi-square values can now be obtained via scipy.stats.distributions.chi2.sf(), for example:
>>>from scipy.stats.distributions import chi2
>>>chi2.sf(3.84,1)
0.050043521248705189
>>>chi2.sf(1424,1)
1.2799986253099803e-311
While stats.chisqprob() and 1-stats.chi2.cdf() appear comparable for small chi-square values, for large chi-square values the former is preferable. The latter cannot provide a p-value smaller than machine epsilon,and will give very inaccurate answers close to machine epsilon. As shown by others, comparable values result for small chi-squared values with the two methods:
>>>from scipy.stats import chisqprob, chi2
>>>chisqprob(3.84,1)
0.050043521248705189
>>>1 - chi2.cdf(3.84,1)
0.050043521248705147
Using 1-chi2.cdf() breaks down here:
>>>1 - chi2.cdf(67,1)
2.2204460492503131e-16
>>>1 - chi2.cdf(68,1)
1.1102230246251565e-16
>>>1 - chi2.cdf(69,1)
1.1102230246251565e-16
>>>1 - chi2.cdf(70,1)
0.0
Whereas chisqprob() gives you accurate results for a much larger range of chi-square values, producing p-values nearly as small as the smallest float greater than zero, until it too underflows:
>>>chisqprob(67,1)
2.7150713219425247e-16
>>>chisqprob(68,1)
1.6349553217245471e-16
>>>chisqprob(69,1)
9.8463440314253303e-17
>>>chisqprob(70,1)
5.9304458500824782e-17
>>>chisqprob(500,1)
9.505397766554137e-111
>>>chisqprob(1000,1)
1.7958327848007363e-219
>>>chisqprob(1424,1)
1.2799986253099803e-311
>>>chisqprob(1425,1)
0.0

You meant to do:
>>> 1 - stats.chi2.cdf(3.84, 1)
0.050043521248705147

Some of the other solutions are deprecated. Use scipy.stats.chi2 Survival Function. Which is the same as 1 - cdf(chi_statistic, df)
Example:
from scipy.stats import chi2
p_value = chi2.sf(chi_statistic, df)

If you want to understand the math, the p-value of a sample, x (fixed), is
P[P(X) <= P(x)] = P[m(X) >= m(x)] = 1 - G(m(x)^2)
where,
P is the probability of a (say k-variate) normal distribution w/ known covariance (cov) and mean,
X is a random variable from that normal distribution,
m(x) is the mahalanobis distance = sqrt( < cov^{-1} (x-mean), x-mean >. Note that in 1-d this is just the absolute value of the z-score.
G is the CDF of the chi^2 distribution w/ k degrees of freedom.
So if you're computing the p-value of a fixed observation, x, then you compute m(x) (generalized z-score), and 1-G(m(x)^2).
for example, it's well known that if x is sampled from a univariate (k = 1) normal distribution and has z-score = 2 (it's 2 standard deviations from the mean), then the p-value is about .046 (see a z-score table)
In [7]: from scipy.stats import chi2
In [8]: k = 1
In [9]: z = 2
In [10]: 1-chi2.cdf(z**2, k)
Out[10]: 0.045500263896358528

For ultra-high precision, when scipy's chi2.sf() isn't enough, bring out the big guns:
>>> import numpy as np
>>> from rpy2.robjects import r
>>> np.exp(np.longdouble(r.pchisq(19000, 2, lower_tail=False, log_p=True)[0]))
1.5937563168532229629e-4126
Update by another user (WestCoastProjects) When using the values from the OP we get:
np.exp(np.longdouble(r.pchisq(3.84,1, lower_tail=False, log_p=True)[0]))
Out[5]: 0.050043521248705198928
So there's that 0.05 you were looking for.