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import numpy as np

def correlation_from_covariance(covariance):
    v = np.sqrt(np.diag(covariance))
    outer_v = np.outer(v, v)
    correlation = covariance / outer_v
    correlation[covariance == 0] = 0
    return correlation

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def loop_iterator(iterator):
  object = iterator.Next()
  while object:
    yield object
    object = iterator.Next()

def iter_collection(rooAbsCollection):
  iterator = rooAbsCollection.createIterator()
  return loop_iterator(iterator)

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import numpy as np
import tensorflow as tf

# Newton's optimization method for multivariate function in tensorflow

def cons(x):
    return tf.constant(x, dtype=tf.float32)

def compute_hessian(fn, vars):
    mat = []

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def safe_factory(func):
    def wrapper(self, *args):
        result = func(self, *args)
        if not result:
            raise ValueError('invalid factory input "%s"' % args)
        return result
    return wrapper

ROOT.RooWorkspace.factory = safe_factory(ROOT.RooWorkspace.factory)

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# from http://kvfrans.com/simple-algoritms-for-solving-cartpole/

import gym
from gym import wrappers
import numpy as np

env = gym.make('CartPole-v0')

def run_episode(env, parameters):  
    observation = env.reset()

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The data have been binned in such a way that every bin contains more than 10 events. For every bin the integral of the S+B postfit pdf has been computed ($E_i$).

In the table the value of the Pearson-$\chi^2 = \sum_i (E_i - O_i)^2 / E_i$ is reported with the number of bins of  $m_{\gamma\gamma}$.

Note that the $\chi^2$ is taking into account only the physical pdf, and not the product of constraints. In fact it is difficult to compute the number of degrees of freedom. We have 5 NPs for the background (4 "$\alpha$" + normalization) plus all the NPs for the statistical fluctuations (100+). Since these parameters are constrained they don't count -1 in the sum of the degree of freedom (something between 0 and -1). To try to evaluate their contribution we can imagine to add the constraint pdf to the computation of the $\chi^2$. This means to add 1 "bin", to subtract 1 dof, and to add a contribution to the $\chi^2$. This can (?) be evaluated as

$$-2\log (pdf(x | x_{true})) + 2\log(pdf(x_{true}|x_{true}))$$

for e