# Elements of Probability Theory and Statistical Data Analysis



* add math about MVN and define MLE and other quantities

* rewrite about covariance matrix

* add KL theorem

## Domains and probabilities
Consider the following simple example, namely the tossing of two dice, resulting in  the following possible values

$$
\{2,3,4,5,6,7,8,9,10,11,12\}.
$$

These values are called the *domain*. 
To this domain we have the corresponding *probabilities*

$$
\{1/36,2/36/,3/36,4/36,5/36,6/36,5/36,4/36,3/36,2/36,1/36\}.
$$

## Tossing  the dice
The numbers in the domain are the outcomes of the physical process of tossing say two dice.
We cannot tell beforehand whether the outcome is 3 or 5 or any other number in this domain.
This defines the randomness of the outcome, or unexpectedness or any other synonimous word which
encompasses the uncertitude of the final outcome. 

The only thing we can tell beforehand
is that say the outcome 2 has a certain probability.  
If our favorite hobby is to  spend an hour every evening throwing dice and 
registering the sequence of outcomes, we will note that the numbers in the above domain

$$
\{2,3,4,5,6,7,8,9,10,11,12\},
$$

appear in a random order. After 11 throws the results may look like

$$
\{10,8,6,3,6,9,11,8,12,4,5\}.
$$

## Stochastic variables

**Random variables are characterized by a domain which contains all possible values that the random value may take. This domain has a corresponding probability distribution function(PDF)**.




## Stochastic variables and the main concepts, the discrete case
There are two main concepts associated with a stochastic variable. The
*domain* is the set $\mathbb D = \{x\}$ of all accessible values
the variable can assume, so that $X \in \mathbb D$. An example of a
discrete domain is the set of six different numbers that we may get by
throwing of a dice, $x\in\{1,\,2,\,3,\,4,\,5,\,6\}$.

The *probability distribution function (PDF)* is a function
$p(x)$ on the domain which, in the discrete case, gives us the
probability or relative frequency with which these values of $X$
occur

$$
p(x) = \mathrm{Prob}(X=x).
$$

## Stochastic variables and the main concepts, the continuous case
In the continuous case, the PDF does not directly depict the
actual probability. Instead we define the probability for the
stochastic variable to assume any value on an infinitesimal interval
around $x$ to be $p(x)dx$. The continuous function $p(x)$ then gives us
the *density* of the probability rather than the probability
itself. The probability for a stochastic variable to assume any value
on a non-infinitesimal interval $[a,\,b]$ is then just the integral

$$
\mathrm{Prob}(a\leq X\leq b) = \int_a^b p(x)dx.
$$

Qualitatively speaking, a stochastic variable represents the values of
numbers chosen as if by chance from some specified PDF so that the
selection of a large set of these numbers reproduces this PDF.




## The cumulative probability
Of interest to us is the *cumulative probability
distribution function* (**CDF**), $P(x)$, which is just the probability
for a stochastic variable $X$ to assume any value less than $x$

$$
P(x)=\mathrm{Prob(}X\leq x\mathrm{)} =
\int_{-\infty}^x p(x^{\prime})dx^{\prime}.
$$

The relation between a CDF and its corresponding PDF is then

$$
p(x) = \frac{d}{dx}P(x).
$$

## Properties of PDFs

There are two properties that all PDFs must satisfy. The first one is
positivity (assuming that the PDF is normalized)

$$
0 \leq p(x) \leq 1.
$$

Naturally, it would be nonsensical for any of the values of the domain
to occur with a probability greater than $1$ or less than $0$. Also,
the PDF must be normalized. That is, all the probabilities must add up
to unity.  The probability of "anything" to happen is always unity. For
both discrete and continuous PDFs, this condition is

$$
\begin{align*}
\sum_{x_i\in\mathbb D} p(x_i) & =  1,\\
\int_{x\in\mathbb D} p(x)\,dx & =  1.
\end{align*}
$$

## Important distributions, the uniform distribution
The first one
is the most basic PDF; namely the uniform distribution

<!-- Equation labels as ordinary links -->
<div id="eq:unifromPDF"></div>

$$
\begin{equation}
p(x) = \frac{1}{b-a}\theta(x-a)\theta(b-x).
\label{eq:unifromPDF} \tag{1}
\end{equation}
$$

For $a=0$ and $b=1$ we have

$$
\begin{array}{ll}
p(x)dx = dx & \in [0,1].
\end{array}
$$

The latter distribution is used to generate random numbers. For other PDFs, one needs normally a mapping from this distribution to say for example the exponential distribution.




## Gaussian distribution
The second one is the Gaussian Distribution

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

with mean value $\mu$ and standard deviation $\sigma$. If $\mu=0$ and $\sigma=1$, it is normally called the **standard normal distribution**

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

The following simple Python code plots the above distribution for different values of $\mu$ and $\sigma$.


```python
%matplotlib inline

import numpy as np
from math import acos, exp, sqrt
from  matplotlib import pyplot as plt
from matplotlib import rc, rcParams
import matplotlib.units as units
import matplotlib.ticker as ticker
rc('text',usetex=True)
rc('font',**{'family':'serif','serif':['Gaussian distribution']})
font = {'family' : 'serif',
        'color'  : 'darkred',
        'weight' : 'normal',
        'size'   : 16,
        }
pi = acos(-1.0)
mu0 = 0.0
sigma0 = 1.0
mu1= 1.0
sigma1 = 2.0
mu2 = 2.0
sigma2 = 4.0

x = np.linspace(-20.0, 20.0)
v0 = np.exp(-(x*x-2*x*mu0+mu0*mu0)/(2*sigma0*sigma0))/sqrt(2*pi*sigma0*sigma0)
v1 = np.exp(-(x*x-2*x*mu1+mu1*mu1)/(2*sigma1*sigma1))/sqrt(2*pi*sigma1*sigma1)
v2 = np.exp(-(x*x-2*x*mu2+mu2*mu2)/(2*sigma2*sigma2))/sqrt(2*pi*sigma2*sigma2)
plt.plot(x, v0, 'b-', x, v1, 'r-', x, v2, 'g-')
plt.title(r'{\bf Gaussian distributions}', fontsize=20)
plt.text(-19, 0.3, r'Parameters: $\mu = 0$, $\sigma = 1$', fontdict=font)
plt.text(-19, 0.18, r'Parameters: $\mu = 1$, $\sigma = 2$', fontdict=font)
plt.text(-19, 0.08, r'Parameters: $\mu = 2$, $\sigma = 4$', fontdict=font)
plt.xlabel(r'$x$',fontsize=20)
plt.ylabel(r'$p(x)$ [MeV]',fontsize=20)

# Tweak spacing to prevent clipping of ylabel                                                                       
plt.subplots_adjust(left=0.15)
plt.savefig('gaussian.pdf', format='pdf')
plt.show()
```

## Exponential distribution
Another important distribution in science is the exponential distribution

$$
p(x) = \alpha\exp{-(\alpha x)}.
$$

## Expectation values
Let $h(x)$ be an arbitrary continuous function on the domain of the stochastic
variable $X$ whose PDF is $p(x)$. We define the *expectation value*
of $h$ with respect to $p$ as follows

<!-- Equation labels as ordinary links -->
<div id="eq:expectation_value_of_h_wrt_p"></div>

$$
\begin{equation}
\langle h \rangle_X \equiv \int\! h(x)p(x)\,dx
\label{eq:expectation_value_of_h_wrt_p} \tag{2}
\end{equation}
$$

Whenever the PDF is known implicitly, like in this case, we will drop
the index $X$ for clarity.  
A particularly useful class of special expectation values are the
*moments*. The $n$-th moment of the PDF $p$ is defined as
follows

$$
\langle x^n \rangle \equiv \int\! x^n p(x)\,dx
$$

## Stochastic variables and the main concepts, mean values
The zero-th moment $\langle 1\rangle$ is just the normalization condition of
$p$. The first moment, $\langle x\rangle$, is called the *mean* of $p$
and often denoted by the letter $\mu$

$$
\langle x\rangle  = \mu \equiv \int x p(x)dx,
$$

for a continuous distribution and

$$
\langle x\rangle  = \mu \equiv \sum_{i=1}^N x_i p(x_i),
$$

for a discrete distribution. 
Qualitatively it represents the centroid or the average value of the
PDF and is therefore simply called the expectation value of $p(x)$.




## Stochastic variables and the main concepts, central moments, the variance

A special version of the moments is the set of *central moments*, the n-th central moment defined as

$$
\langle (x-\langle x\rangle )^n\rangle  \equiv \int\! (x-\langle x\rangle)^n p(x)\,dx
$$

The zero-th and first central moments are both trivial, equal $1$ and
$0$, respectively. But the second central moment, known as the
*variance* of $p$, is of particular interest. For the stochastic
variable $X$, the variance is denoted as $\sigma^2_X$ or $\mathrm{Var}(X)$

$$
\begin{align*}
\sigma^2_X &=\mathrm{Var}(X) =  \langle (x-\langle x\rangle)^2\rangle  =
\int (x-\langle x\rangle)^2 p(x)dx\\
& =  \int\left(x^2 - 2 x \langle x\rangle^{2} +\langle x\rangle^2\right)p(x)dx\\
& =  \langle x^2\rangle - 2 \langle x\rangle\langle x\rangle + \langle x\rangle^2\\
& =  \langle x^2 \rangle - \langle x\rangle^2
\end{align*}
$$

The square root of the variance, $\sigma =\sqrt{\langle (x-\langle x\rangle)^2\rangle}$ is called the 
**standard deviation** of $p$. It is the RMS (root-mean-square)
value of the deviation of the PDF from its mean value, interpreted
qualitatively as the "spread" of $p$ around its mean.






## Probability Distribution Functions

The following table collects properties of probability distribution functions.
In our notation we reserve the label $p(x)$ for the probability of a certain event,
while $P(x)$ is the cumulative probability. 


<table border="1">
<thead>
<tr><th align="center">             </th> <th align="center">               Discrete PDF               </th> <th align="center">           Continuous PDF           </th> </tr>
</thead>
<tbody>
<tr><td align="left">   Domain           </td> <td align="center">   $\left\{x_1, x_2, x_3, \dots, x_N\right\}$    </td> <td align="center">   $[a,b]$                                 </td> </tr>
<tr><td align="left">   Probability      </td> <td align="center">   $p(x_i)$                                      </td> <td align="center">   $p(x)dx$                                </td> </tr>
<tr><td align="left">   Cumulative       </td> <td align="center">   $P_i=\sum_{l=1}^ip(x_l)$                      </td> <td align="center">   $P(x)=\int_a^xp(t)dt$                   </td> </tr>
<tr><td align="left">   Positivity       </td> <td align="center">   $0 \le p(x_i) \le 1$                          </td> <td align="center">   $p(x) \ge 0$                            </td> </tr>
<tr><td align="left">   Positivity       </td> <td align="center">   $0 \le P_i \le 1$                             </td> <td align="center">   $0 \le P(x) \le 1$                      </td> </tr>
<tr><td align="left">   Monotonic        </td> <td align="center">   $P_i \ge P_j$ if $x_i \ge x_j$                </td> <td align="center">   $P(x_i) \ge P(x_j)$ if $x_i \ge x_j$    </td> </tr>
<tr><td align="left">   Normalization    </td> <td align="center">   $P_N=1$                                       </td> <td align="center">   $P(b)=1$                                </td> </tr>
</tbody>
</table>





## Probability Distribution Functions
With a PDF we can compute expectation values of selected quantities such as

$$
\langle x^k\rangle=\sum_{i=1}^{N}x_i^kp(x_i),
$$

if we have a discrete PDF or

$$
\langle x^k\rangle=\int_a^b x^kp(x)dx,
$$

in the case of a continuous PDF. We have already defined the mean value $\mu$
and the variance $\sigma^2$.




## The three famous Probability Distribution Functions

There are at least three PDFs which one may encounter. These are the

**Uniform distribution**

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

yielding probabilities different from zero in the interval $[a,b]$.

**The exponential distribution**

$$
p(x)=\alpha \exp{(-\alpha x)},
$$

yielding probabilities different from zero in the interval $[0,\infty)$ and with mean value

$$
\mu = \int_0^{\infty}xp(x)dx=\int_0^{\infty}x\alpha \exp{(-\alpha x)}dx=\frac{1}{\alpha},
$$

with variance

$$
\sigma^2=\int_0^{\infty}x^2p(x)dx-\mu^2 = \frac{1}{\alpha^2}.
$$

## Probability Distribution Functions, the normal distribution
Finally, we have the so-called univariate normal  distribution, or just the **normal distribution**

$$
p(x)=\frac{1}{b\sqrt{2\pi}}\exp{\left(-\frac{(x-a)^2}{2b^2}\right)}
$$

with probabilities different from zero in the interval $(-\infty,\infty)$.
The integral $\int_{-\infty}^{\infty}\exp{\left(-(x^2\right)}dx$ appears in many calculations, its value
is $\sqrt{\pi}$,  a result we will need when we compute the mean value and the variance.
The mean value is

$$
\mu = \int_0^{\infty}xp(x)dx=\frac{1}{b\sqrt{2\pi}}\int_{-\infty}^{\infty}x \exp{\left(-\frac{(x-a)^2}{2b^2}\right)}dx,
$$

which becomes with a suitable change of variables

$$
\mu =\frac{1}{b\sqrt{2\pi}}\int_{-\infty}^{\infty}b\sqrt{2}(a+b\sqrt{2}y)\exp{-y^2}dy=a.
$$

## Probability Distribution Functions, the normal distribution
Similarly, the variance becomes

$$
\sigma^2 = \frac{1}{b\sqrt{2\pi}}\int_{-\infty}^{\infty}(x-\mu)^2 \exp{\left(-\frac{(x-a)^2}{2b^2}\right)}dx,
$$

and inserting the mean value and performing a variable change we obtain

$$
\sigma^2 = \frac{1}{b\sqrt{2\pi}}\int_{-\infty}^{\infty}b\sqrt{2}(b\sqrt{2}y)^2\exp{\left(-y^2\right)}dy=
\frac{2b^2}{\sqrt{\pi}}\int_{-\infty}^{\infty}y^2\exp{\left(-y^2\right)}dy,
$$

and performing a final integration by parts we obtain the well-known result $\sigma^2=b^2$.
It is useful to introduce the standard normal distribution as well, defined by $\mu=a=0$, viz. a distribution
centered around zero and with a variance $\sigma^2=1$, leading to

<!-- Equation labels as ordinary links -->
<div id="_auto1"></div>

$$
\begin{equation}
   p(x)=\frac{1}{\sqrt{2\pi}}\exp{\left(-\frac{x^2}{2}\right)}.
\label{_auto1} \tag{3}
\end{equation}
$$

## Probability Distribution Functions, the cumulative distribution

The exponential and uniform distributions have simple cumulative functions,
whereas the normal distribution does not, being proportional to the so-called
error function $erf(x)$, given by

$$
P(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^x\exp{\left(-\frac{t^2}{2}\right)}dt,
$$

which is difficult to evaluate in a quick way.





## Probability Distribution Functions, other important distribution

Some other PDFs which one encounters often in the natural sciences are the binomial distribution

$$
p(x) = \left(\begin{array}{c} n \\ x\end{array}\right)y^x(1-y)^{n-x} \hspace{0.5cm}x=0,1,\dots,n,
$$

where $y$ is the probability for a specific event, such as the tossing of a coin or moving left or right
in case of a random walker. Note that $x$ is a discrete stochastic variable. 

The sequence of binomial trials is characterized by the following definitions

  * Every experiment is thought to consist of $N$ independent trials.

  * In every independent trial one registers if a specific situation happens or not, such as the  jump to the left or right of a random walker.

  * The probability for every outcome in a single trial has the same value, for example the outcome of tossing (either heads or tails) a coin is always $1/2$.




## Probability Distribution Functions, the binomial distribution

In order to compute the mean and variance we need to recall Newton's binomial
formula

$$
(a+b)^m=\sum_{n=0}^m \left(\begin{array}{c} m \\ n\end{array}\right)a^nb^{m-n},
$$

which can be used to show that

$$
\sum_{x=0}^n\left(\begin{array}{c} n \\ x\end{array}\right)y^x(1-y)^{n-x} = (y+1-y)^n = 1,
$$

the PDF is normalized to one. 
The mean value is

$$
\mu = \sum_{x=0}^n x\left(\begin{array}{c} n \\ x\end{array}\right)y^x(1-y)^{n-x} =
\sum_{x=0}^n x\frac{n!}{x!(n-x)!}y^x(1-y)^{n-x},
$$

resulting in

$$
\mu = 
\sum_{x=0}^n x\frac{(n-1)!}{(x-1)!(n-1-(x-1))!}y^{x-1}(1-y)^{n-1-(x-1)},
$$

which we rewrite as

$$
\mu=ny\sum_{\nu=0}^n\left(\begin{array}{c} n-1 \\ \nu\end{array}\right)y^{\nu}(1-y)^{n-1-\nu} =ny(y+1-y)^{n-1}=ny.
$$

The variance is slightly trickier to get. It reads $\sigma^2=ny(1-y)$. 


## Probability Distribution Functions, Poisson's  distribution

Another important distribution with discrete stochastic variables $x$ is  
the Poisson model, which resembles the exponential distribution and reads

$$
p(x) = \frac{\lambda^x}{x!} e^{-\lambda} \hspace{0.5cm}x=0,1,\dots,;\lambda > 0.
$$

In this case both the mean value and the variance are easier to calculate,

$$
\mu = \sum_{x=0}^{\infty} x \frac{\lambda^x}{x!} e^{-\lambda} = \lambda e^{-\lambda}\sum_{x=1}^{\infty}
\frac{\lambda^{x-1}}{(x-1)!}=\lambda,
$$

and the variance is $\sigma^2=\lambda$.






## Probability Distribution Functions, Poisson's  distribution
An example of applications of the Poisson distribution could be the counting
of the number of $\alpha$-particles emitted from a radioactive source in a given time interval.
In the limit of $n\rightarrow \infty$ and for small probabilities $y$, the binomial distribution
approaches the Poisson distribution. Setting $\lambda = ny$, with $y$ the probability for an event in
the binomial distribution we can show that

$$
\lim_{n\rightarrow \infty}\left(\begin{array}{c} n \\ x\end{array}\right)y^x(1-y)^{n-x} e^{-\lambda}=\sum_{x=1}^{\infty}\frac{\lambda^x}{x!} e^{-\lambda}.
$$

## Meet the  covariance!
An important quantity in a statistical analysis is the so-called covariance. 

Consider the set $\{X_i\}$ of $n$
stochastic variables (not necessarily uncorrelated) with the
multivariate PDF $P(x_1,\dots,x_n)$. The *covariance* of two
of the stochastic variables, $X_i$ and $X_j$, is defined as follows

<!-- Equation labels as ordinary links -->
<div id="_auto2"></div>

$$
\begin{equation}
\mathrm{Cov}(X_i,\,X_j)  = \langle (x_i-\langle x_i\rangle)(x_j-\langle x_j\rangle)\rangle 
\label{_auto2} \tag{4}
\end{equation}
$$

<!-- Equation labels as ordinary links -->
<div id="eq:def_covariance"></div>

$$
\begin{equation} 
=\int\cdots\int (x_i-\langle x_i\rangle)(x_j-\langle x_j\rangle)P(x_1,\dots,x_n)\,dx_1\dots dx_n,
\label{eq:def_covariance} \tag{5}
\end{equation}
$$

with

$$
\langle x_i\rangle =
\int\cdots\int x_i P(x_1,\dots,x_n)\,dx_1\dots dx_n.
$$

## Meet the  covariance in matrix disguise
If we consider the above covariance as a matrix

$$
C_{ij} =\mathrm{Cov}(X_i,\,X_j),
$$

then the diagonal elements are just the familiar
variances, $C_{ii} = \mathrm{Cov}(X_i,\,X_i) = \mathrm{Var}(X_i)$. It turns out that
all the off-diagonal elements are zero if the stochastic variables are
uncorrelated.




## Covariance


```python
# Importing various packages
from math import exp, sqrt
from random import random, seed
import numpy as np
import matplotlib.pyplot as plt

def covariance(x, y, n):
    sum = 0.0
    mean_x = np.mean(x)
    mean_y = np.mean(y)
    for i in range(0, n):
        sum += (x[(i)]-mean_x)*(y[i]-mean_y)
    return  sum/n

n = 10

x=np.random.normal(size=n)
y = 4+3*x+np.random.normal(size=n)
covxy = covariance(x,y,n)
print(covxy)
z = np.vstack((x, y))
c = np.cov(z.T)

print(c)
```

    2.599907486910053
    [[ 0.2319431   1.78702425  2.33555576  1.01612686  0.80805005  1.10075065
       1.70915039  0.60318841  0.62019593  1.49305147]
     [ 1.78702425 13.76827179 17.99447691  7.82883097  6.22568649  8.48082175
      13.16828638  4.64731357  4.77834938 11.5033349 ]
     [ 2.33555576 17.99447691 23.51792615 10.23191002  8.13667637 11.08403099
      17.21032448  6.07381797  6.24507555 15.03431202]
     [ 1.01612686  7.82883097 10.23191002  4.45158225  3.54001199  4.82231328
       7.48767091  2.64252717  2.71703596  6.54095632]
     [ 0.80805005  6.22568649  8.13667637  3.54001199  2.81510801  3.83482678
       5.95438729  2.10140515  2.16065644  5.20153566]
     [ 1.10075065  8.48082175 11.08403099  4.82231328  3.83482678  5.22391907
       8.11124963  2.86259878  2.94331271  7.08569196]
     [ 1.70915039 13.16828638 17.21032448  7.48767091  5.95438729  8.11124963
      12.59444676  4.44479575  4.57012136 11.00204953]
     [ 0.60318841  4.64731357  6.07381797  2.64252717  2.10140515  2.86259878
       4.44479575  1.56864447  1.61287402  3.88281152]
     [ 0.62019593  4.77834938  6.24507555  2.71703596  2.16065644  2.94331271
       4.57012136  1.61287402  1.65835067  3.9922914 ]
     [ 1.49305147 11.5033349  15.03431202  6.54095632  5.20153566  7.08569196
      11.00204953  3.88281152  3.9922914   9.61098936]]


## Meet the  covariance, uncorrelated events

Consider the stochastic variables $X_i$ and $X_j$, ($i\neq j$). We have

$$
\begin{align*}
Cov(X_i,\,X_j) &= \langle (x_i-\langle x_i\rangle)(x_j-\langle x_j\rangle)\rangle\\
&=\langle x_i x_j - x_i\langle x_j\rangle - \langle x_i\rangle x_j + \langle x_i\rangle\langle x_j\rangle\rangle\\
&=\langle x_i x_j\rangle - \langle x_i\langle x_j\rangle\rangle - \langle \langle x_i\rangle x_j \rangle +
\langle \langle x_i\rangle\langle x_j\rangle\rangle \\
&=\langle x_i x_j\rangle - \langle x_i\rangle\langle x_j\rangle - \langle x_i\rangle\langle x_j\rangle +
\langle x_i\rangle\langle x_j\rangle \\
&=\langle x_i x_j\rangle - \langle x_i\rangle\langle x_j\rangle
\end{align*}
$$

If $X_i$ and $X_j$ are independent (assuming $i \neq j$), we have that

$$
\langle x_i x_j\rangle = \langle x_i\rangle\langle x_j\rangle,
$$

leading to

$$
Cov(X_i, X_j) = 0 \hspace{0.1cm} (i\neq j).
$$

## Numerical experiments and the covariance

Now that we have constructed an idealized mathematical framework, let
us try to apply it to empirical observations. Examples of relevant
physical phenomena may be spontaneous decays of nuclei, or a purely
mathematical set of numbers produced by some deterministic
mechanism. It is the latter we will deal with, using so-called pseudo-random
number generators.  In general our observations will contain only a limited set of
observables. We remind the reader that
a *stochastic process* is a process that produces sequentially a
chain of values

$$
\{x_1, x_2,\dots\,x_k,\dots\}.
$$

## Numerical experiments and the covariance
We will call these
values our *measurements* and the entire set as our measured
*sample*.  The action of measuring all the elements of a sample
we will call a stochastic *experiment* (since, operationally,
they are often associated with results of empirical observation of
some physical or mathematical phenomena; precisely an experiment). We
assume that these values are distributed according to some 
PDF $p_X^{\phantom X}(x)$, where $X$ is just the formal symbol for the
stochastic variable whose PDF is $p_X^{\phantom X}(x)$. Instead of
trying to determine the full distribution $p$ we are often only
interested in finding the few lowest moments, like the mean
$\mu_X^{\phantom X}$ and the variance $\sigma_X^{\phantom X}$.





## Numerical experiments and the covariance, actual situations
In practical situations however, a sample is always of finite size. Let that
size be $n$. The expectation value of a sample $\alpha$, the **sample mean**, is then defined as follows

$$
\langle x_{\alpha} \rangle \equiv \frac{1}{n}\sum_{k=1}^n x_{\alpha,k}.
$$

The *sample variance* is:

$$
\mathrm{Var}(x) \equiv \frac{1}{n}\sum_{k=1}^n (x_{\alpha,k} - \langle x_{\alpha} \rangle)^2,
$$

with its square root being the *standard deviation of the sample*.





## Numerical experiments and the covariance, our observables
You can think of the above observables as a set of quantities which define
a given experiment. This experiment is then repeated several times, say $m$ times.
The total average is then

<!-- Equation labels as ordinary links -->
<div id="eq:exptmean"></div>

$$
\begin{equation}
\langle X_m \rangle= \frac{1}{m}\sum_{\alpha=1}^mx_{\alpha}=\frac{1}{mn}\sum_{\alpha, k} x_{\alpha,k},
\label{eq:exptmean} \tag{6}
\end{equation}
$$

where the last sums end at $m$ and $n$.
The total variance is

$$
\sigma^2_m= \frac{1}{mn^2}\sum_{\alpha=1}^m(\langle x_{\alpha} \rangle-\langle X_m \rangle)^2,
$$

which we rewrite as

<!-- Equation labels as ordinary links -->
<div id="eq:exptvariance"></div>

$$
\begin{equation}
\sigma^2_m=\frac{1}{m}\sum_{\alpha=1}^m\sum_{kl=1}^n (x_{\alpha,k}-\langle X_m \rangle)(x_{\alpha,l}-\langle X_m \rangle).
\label{eq:exptvariance} \tag{7}
\end{equation}
$$

## Numerical experiments and the covariance, the sample variance

We define also the sample variance $\sigma^2$ of all $mn$ individual experiments as

<!-- Equation labels as ordinary links -->
<div id="eq:sampleexptvariance"></div>

$$
\begin{equation}
\sigma^2=\frac{1}{mn}\sum_{\alpha=1}^m\sum_{k=1}^n (x_{\alpha,k}-\langle X_m \rangle)^2.
\label{eq:sampleexptvariance} \tag{8}
\end{equation}
$$

These quantities, being known experimental values or the results from our calculations, 
may differ, in some cases
significantly,  from the similarly named
exact values for the mean value $\mu_X$, the variance $\mathrm{Var}(X)$
and the covariance $\mathrm{Cov}(X,Y)$.




## Numerical experiments and the covariance, central limit theorem

The central limit theorem states that the PDF $\tilde{p}(z)$ of
the average of $m$ random values corresponding to a PDF $p(x)$ 
is a normal distribution whose mean is the 
mean value of the PDF $p(x)$ and whose variance is the variance
of the PDF $p(x)$ divided by $m$, the number of values used to compute $z$.

The central limit theorem leads then to the well-known expression for the
standard deviation, given by

$$
\sigma_m=
\frac{\sigma}{\sqrt{m}}.
$$

In many cases the above estimate for the standard deviation, in particular if correlations are strong, may be too simplistic.  We need therefore a more precise defintion of the error and the variance in our results.




## Definition of Correlation Functions and Standard Deviation
Our estimate of the true average $\mu_{X}$ is the sample mean $\langle X_m \rangle$

$$
\mu_{X}^{\phantom X} \approx X_m=\frac{1}{mn}\sum_{\alpha=1}^m\sum_{k=1}^n x_{\alpha,k}.
$$

We can then use Eq. ([7](#eq:exptvariance))

$$
\sigma^2_m=\frac{1}{mn^2}\sum_{\alpha=1}^m\sum_{kl=1}^n (x_{\alpha,k}-\langle X_m \rangle)(x_{\alpha,l}-\langle X_m \rangle),
$$

and rewrite it as

$$
\sigma^2_m=\frac{\sigma^2}{n}+\frac{2}{mn^2}\sum_{\alpha=1}^m\sum_{k<l}^n (x_{\alpha,k}-\langle X_m \rangle)(x_{\alpha,l}-\langle X_m \rangle),
$$

where the first term is the sample variance of all $mn$ experiments divided by $n$
and the last term is nothing but the covariance which arises when $k\ne l$.








## Definition of Correlation Functions and Standard Deviation
Our estimate of the true average $\mu_{X}$ is the sample mean $\langle X_m \rangle$

If the 
observables are uncorrelated, then the covariance is zero and we obtain a total variance
which agrees with the central limit theorem. Correlations may often be present in our data set, resulting in a non-zero covariance.  The first term is normally called the uncorrelated 
contribution.
Computationally the uncorrelated first term is much easier to treat
efficiently than the second.
We just accumulate separately the values $x^2$ and $x$ for every
measurement $x$ we receive. The correlation term, though, has to be
calculated at the end of the experiment since we need all the
measurements to calculate the cross terms. Therefore, all measurements
have to be stored throughout the experiment.




## Definition of Correlation Functions and Standard Deviation

Let us analyze the problem by splitting up the correlation term into
partial sums of the form

$$
f_d = \frac{1}{nm}\sum_{\alpha=1}^m\sum_{k=1}^{n-d}(x_{\alpha,k}-\langle X_m \rangle)(x_{\alpha,k+d}-\langle X_m \rangle),
$$

The correlation term of the total variance can now be rewritten in terms of
$f_d$

$$
\frac{2}{mn^2}\sum_{\alpha=1}^m\sum_{k<l}^n (x_{\alpha,k}-\langle X_m \rangle)(x_{\alpha,l}-\langle X_m \rangle)=
\frac{2}{n}\sum_{d=1}^{n-1} f_d
$$

## Definition of Correlation Functions and Standard Deviation
The value of $f_d$ reflects the correlation between measurements
separated by the distance $d$ in the samples.  Notice that for
$d=0$, $f$ is just the sample variance, $\sigma^2$. If we divide $f_d$
by $\sigma^2$, we arrive at the so called **autocorrelation function**

<!-- Equation labels as ordinary links -->
<div id="eq:autocorrelformal"></div>

$$
\begin{equation}
\kappa_d = \frac{f_d}{\sigma^2}
\label{eq:autocorrelformal} \tag{9}
\end{equation}
$$

which gives us a useful measure of the correlation pair correlation
starting always at $1$ for $d=0$.




## Definition of Correlation Functions and Standard Deviation, sample variance

The sample variance of the $mn$ experiments can now be
written in terms of the autocorrelation function

<!-- Equation labels as ordinary links -->
<div id="eq:error_estimate_corr_time"></div>

$$
\begin{equation}
\sigma_m^2=\frac{\sigma^2}{n}+\frac{2}{n}\cdot\sigma^2\sum_{d=1}^{n-1}
\frac{f_d}{\sigma^2}=\left(1+2\sum_{d=1}^{n-1}\kappa_d\right)\frac{1}{n}\sigma^2=\frac{\tau}{n}\cdot\sigma^2
\label{eq:error_estimate_corr_time} \tag{10}
\end{equation}
$$

and we see that $\sigma_m$ can be expressed in terms of the
uncorrelated sample variance times a correction factor $\tau$ which
accounts for the correlation between measurements. We call this
correction factor the *autocorrelation time*

<!-- Equation labels as ordinary links -->
<div id="eq:autocorrelation_time"></div>

$$
\begin{equation}
\tau = 1+2\sum_{d=1}^{n-1}\kappa_d
\label{eq:autocorrelation_time} \tag{11}
\end{equation}
$$

<!-- It is closely related to the area under the graph of the -->
<!-- autocorrelation function. -->
For a correlation free experiment, $\tau$
equals 1.






## Definition of Correlation Functions and Standard Deviation
From the point of view of
Eq. ([10](#eq:error_estimate_corr_time)) we can interpret a sequential
correlation as an effective reduction of the number of measurements by
a factor $\tau$. The effective number of measurements becomes

$$
n_\mathrm{eff} = \frac{n}{\tau}
$$

To neglect the autocorrelation time $\tau$ will always cause our
simple uncorrelated estimate of $\sigma_m^2\approx \sigma^2/n$ to
be less than the true sample error. The estimate of the error will be
too "good". On the other hand, the calculation of the full
autocorrelation time poses an efficiency problem if the set of
measurements is very large.  The solution to this problem is given by 
more practically oriented methods like the blocking technique.
<!-- add ref here to flybjerg -->





## Code to compute the Covariance matrix and the Covariance


```python
# Importing various packages
from math import exp, sqrt
from random import random, seed
import numpy as np
import matplotlib.pyplot as plt

# Sample covariance, note the factor 1/(n-1)
def covariance(x, y, n):
    sum = 0.0
    mean_x = np.mean(x)
    mean_y = np.mean(y)
    for i in range(0, n):
        sum += (x[(i)]-mean_x)*(y[i]-mean_y)
    return  sum/(n-1.)

n = 100
x = np.random.normal(size=n)
print(np.mean(x))
y = 4+3*x+np.random.normal(size=n)
print(np.mean(y))
z = x**3+np.random.normal(size=n)
print(np.mean(z))
covxx = covariance(x,x,n)
covyy = covariance(y,y,n)
covzz = covariance(z,z,n)
covxy = covariance(x,y,n)
covxz = covariance(x,z,n)
covyz = covariance(y,z,n)
print(covxx,covyy, covzz)
print(covxy,covxz, covyz)
w = np.vstack((x, y, z))
#print(w)
c = np.cov(w)
print(c)
#eigen = np.zeros(n)
Eigvals, Eigvecs = np.linalg.eig(c)
print(Eigvals)
```

    0.09525604390305954
    4.231031843972674
    0.04124568972977915
    0.831883346005529 8.296268771337688 5.337726521457732
    2.4701318674515136 1.7159602029423742 5.229477584776403
    [[0.83188335 2.47013187 1.7159602 ]
     [2.47013187 8.29626877 5.22947758]
     [1.7159602  5.22947758 5.33772652]]
    [12.99430696  0.07864499  1.39292669]


## Random Numbers

Uniform deviates are just random numbers that lie within a specified range
(typically 0 to 1), with any one number in the range just as likely as any other. They
are, in other words, what you probably think random numbers are. However,
we want to distinguish uniform deviates from other sorts of random numbers, for
example numbers drawn from a normal (Gaussian) distribution of specified mean
and standard deviation. These other sorts of deviates are almost always generated by
performing appropriate operations on one or more uniform deviates, as we will see
in subsequent sections. So, a reliable source of random uniform deviates, the subject
of this section, is an essential building block for any sort of stochastic modeling
or Monte Carlo computer work.





## Random Numbers, better name: pseudo random numbers

A disclaimer is however appropriate. It should be fairly obvious that 
something as deterministic as a computer cannot generate purely random numbers.

Numbers generated by any of the standard algorithms are in reality pseudo random
numbers, hopefully abiding to the following criteria:

  * they produce a uniform distribution in the interval [0,1].

  * correlations between random numbers are negligible

  * the period before the same sequence of random numbers is repeated   is as large as possible and finally

  * the algorithm should be fast.





## Random number generator RNG
 The most common random number generators are based on so-called
Linear congruential relations of the type

$$
N_i=(aN_{i-1}+c) \mathrm{MOD} (M),
$$

which yield a number in the interval [0,1] through

$$
x_i=N_i/M
$$

The number 
$M$ is called the period and it should be as large as possible 
 and 
$N_0$ is the starting value, or seed. The function $\mathrm{MOD}$ means the remainder,
that is if we were to evaluate $(13)\mathrm{MOD}(9)$, the outcome is the remainder
of the division $13/9$, namely $4$.




## Random number generator RNG and periodic outputs

The problem with such generators is that their outputs are periodic;
they 
will start to repeat themselves with a period that is at most $M$. If however
the parameters $a$ and $c$ are badly chosen, the period may be even shorter.

Consider the following example

$$
N_i=(6N_{i-1}+7) \mathrm{MOD} (5),
$$

with a seed $N_0=2$. This generator produces the sequence
$4,1,3,0,2,4,1,3,0,2,...\dots$, i.e., a sequence with period $5$.
However, increasing $M$ may not guarantee a larger period as the following
example shows

$$
N_i=(27N_{i-1}+11) \mathrm{MOD} (54),
$$

which still, with $N_0=2$, results in $11,38,11,38,11,38,\dots$, a period of
just $2$.




## Random number generator RNG and its period
Typical periods for the random generators provided in the program library 
are of the order of $\sim 10^9$ or larger. Other random number generators which have
become increasingly popular are so-called shift-register generators.
In these generators each successive number depends on many preceding
values (rather than the last values as in the linear congruential
generator).
For example, you could make a shift register generator whose $l$th 
number is the sum of the $l-i$th and $l-j$th values with modulo $M$,

$$
N_l=(aN_{l-i}+cN_{l-j})\mathrm{MOD}(M).
$$

## Random number generator RNG, other examples
Such a generator again produces a sequence of pseudorandom numbers
but this time with a period much larger than $M$.
It is also possible to construct more elaborate algorithms by including
more than two past terms in the sum of each iteration.
One example is the generator of [Marsaglia and Zaman](http://dl.acm.org/citation.cfm?id=187154)
which consists of two congruential relations

<!-- Equation labels as ordinary links -->
<div id="eq:mz1"></div>

$$
\begin{equation}
   N_l=(N_{l-3}-N_{l-1})\mathrm{MOD}(2^{31}-69),
\label{eq:mz1} \tag{12}
\end{equation}
$$

followed by

<!-- Equation labels as ordinary links -->
<div id="eq:mz2"></div>

$$
\begin{equation}
   N_l=(69069N_{l-1}+1013904243)\mathrm{MOD}(2^{32}),
\label{eq:mz2} \tag{13}
\end{equation}
$$

which according to the authors has a period larger than $2^{94}$.




## Random number generator RNG, other examples
Instead of  using modular addition, we could use the bitwise
exclusive-OR ($\oplus$) operation so that

$$
N_l=(N_{l-i})\oplus (N_{l-j})
$$

where the bitwise action of $\oplus$ means that if $N_{l-i}=N_{l-j}$ the result is
$0$ whereas if $N_{l-i}\ne N_{l-j}$ the result is
$1$. As an example, consider the case where  $N_{l-i}=6$ and $N_{l-j}=11$. The first
one has a bit representation (using 4 bits only) which reads $0110$ whereas the 
second number is $1011$. Employing the $\oplus$ operator yields 
$1101$, or $2^3+2^2+2^0=13$.

In Fortran90, the bitwise $\oplus$ operation is coded through the intrinsic
function $\mathrm{IEOR}(m,n)$ where $m$ and $n$ are the input numbers, while in $C$
it is given by $m\wedge n$.





## Random number generator RNG, RAN0

We show here how the linear congruential algorithm can be implemented, namely

$$
N_i=(aN_{i-1}) \mathrm{MOD} (M).
$$

However, since $a$ and $N_{i-1}$ are integers and their multiplication 
could become greater than the standard 32 bit integer, there is a trick via 
Schrage's algorithm which approximates the multiplication
of large integers through the factorization

$$
M=aq+r,
$$

where we have defined

$$
q=[M/a],
$$

and

$$
r = M\hspace{0.1cm}\mathrm{MOD} \hspace{0.1cm}a.
$$

where the brackets denote integer division. In the code below the numbers 
$q$ and $r$ are chosen so that $r < q$.






## Random number generator RNG, RAN0

To see how this works we note first that

<!-- Equation labels as ordinary links -->
<div id="eq:rntrick1"></div>

$$
\begin{equation}
(aN_{i-1}) \mathrm{MOD} (M)= (aN_{i-1}-[N_{i-1}/q]M)\mathrm{MOD} (M),
\label{eq:rntrick1} \tag{14}
\end{equation}
$$

since we can add or subtract any integer multiple of $M$ from $aN_{i-1}$.
The last term $[N_{i-1}/q]M\mathrm{MOD}(M)$ is zero since the integer division 
$[N_{i-1}/q]$ just yields a constant which is multiplied with $M$.





## Random number generator RNG, RAN0
We can now rewrite Eq. ([14](#eq:rntrick1)) as

<!-- Equation labels as ordinary links -->
<div id="eq:rntrick2"></div>

$$
\begin{equation}
(aN_{i-1}) \mathrm{MOD} (M)= (aN_{i-1}-[N_{i-1}/q](aq+r))\mathrm{MOD} (M),
\label{eq:rntrick2} \tag{15}
\end{equation}
$$

which results
in

<!-- Equation labels as ordinary links -->
<div id="eq:rntrick3"></div>

$$
\begin{equation}
(aN_{i-1}) \mathrm{MOD} (M)= \left(a(N_{i-1}-[N_{i-1}/q]q)-[N_{i-1}/q]r)\right)\mathrm{MOD} (M),
\label{eq:rntrick3} \tag{16}
\end{equation}
$$

yielding

<!-- Equation labels as ordinary links -->
<div id="eq:rntrick4"></div>

$$
\begin{equation}
(aN_{i-1}) \mathrm{MOD} (M)= \left(a(N_{i-1}\mathrm{MOD} (q)) -[N_{i-1}/q]r)\right)\mathrm{MOD} (M).
\label{eq:rntrick4} \tag{17}
\end{equation}
$$

## Random number generator RNG, RAN0
The term $[N_{i-1}/q]r$ is always smaller or equal $N_{i-1}(r/q)$ and with $r < q$ we obtain always a 
number smaller than $N_{i-1}$, which is smaller than $M$. 
And since the number $N_{i-1}\mathrm{MOD} (q)$ is between zero and $q-1$ then
$a(N_{i-1}\mathrm{MOD} (q))< aq$. Combined with our definition of $q=[M/a]$ ensures that 
this term is also smaller than $M$ meaning that both terms fit into a
32-bit signed integer. None of these two terms can be negative, but their difference could.
The algorithm below adds $M$ if their difference is negative.
Note that the program uses the bitwise $\oplus$ operator to generate
the starting point for each generation of a random number. The period
of $ran0$ is $\sim 2.1\times 10^{9}$. A special feature of this
algorithm is that is should never be called with the initial seed 
set to $0$.





## Random number generator RNG, RAN0 code

            /*
             ** The function
             **           ran0()
             ** is an "Minimal" random number generator of Park and Miller
             ** Set or reset the input value
             ** idum to any integer value (except the unlikely value MASK)
             ** to initialize the sequence; idum must not be altered between
             ** calls for sucessive deviates in a sequence.
             ** The function returns a uniform deviate between 0.0 and 1.0.
             */
        double ran0(long &idum)
        {
           const int a = 16807, m = 2147483647, q = 127773;
           const int r = 2836, MASK = 123459876;
           const double am = 1./m;
           long     k;
           double   ans;
           idum ^= MASK;
           k = (*idum)/q;
           idum = a*(idum - k*q) - r*k;
           // add m if negative difference
           if(idum < 0) idum += m;
           ans=am*(idum);
           idum ^= MASK;
           return ans;
        } // End: function ran0() 


## Properties of Selected Random Number Generators

As mentioned previously, the underlying PDF for the generation of
random numbers is the uniform distribution, meaning that the 
probability for finding a number $x$ in the interval [0,1] is $p(x)=1$.

A random number generator should produce numbers which are uniformly distributed
in this interval. The table  shows the distribution of $N=10000$ random
numbers generated by the functions in the program library.
We note in this table that the number of points in the various
intervals $0.0-0.1$, $0.1-0.2$ etc are fairly close to $1000$, with some minor
deviations. 

Two additional measures are the standard deviation $\sigma$ and the mean
$\mu=\langle x\rangle$.




## Properties of Selected Random Number Generators
For the uniform distribution, the mean value $\mu$ is then

$$
\mu=\langle x\rangle=\frac{1}{2}
$$

while the standard deviation is

$$
\sigma=\sqrt{\langle x^2\rangle-\mu^2}=\frac{1}{\sqrt{12}}=0.2886.
$$

## Properties of Selected Random Number Generators
The various random number generators produce results which agree rather well with
these limiting values. 

<table border="1">
<thead>
<tr><th align="center">$x$-bin </th> <th align="center"> ran0 </th> <th align="center"> ran1 </th> <th align="center"> ran2 </th> <th align="center"> ran3 </th> </tr>
</thead>
<tbody>
<tr><td align="center">   0.0-0.1     </td> <td align="right">   1013      </td> <td align="right">   991       </td> <td align="right">   938       </td> <td align="right">   1047      </td> </tr>
<tr><td align="center">   0.1-0.2     </td> <td align="right">   1002      </td> <td align="right">   1009      </td> <td align="right">   1040      </td> <td align="right">   1030      </td> </tr>
<tr><td align="center">   0.2-0.3     </td> <td align="right">   989       </td> <td align="right">   999       </td> <td align="right">   1030      </td> <td align="right">   993       </td> </tr>
<tr><td align="center">   0.3-0.4     </td> <td align="right">   939       </td> <td align="right">   960       </td> <td align="right">   1023      </td> <td align="right">   937       </td> </tr>
<tr><td align="center">   0.4-0.5     </td> <td align="right">   1038      </td> <td align="right">   1001      </td> <td align="right">   1002      </td> <td align="right">   992       </td> </tr>
<tr><td align="center">   0.5-0.6     </td> <td align="right">   1037      </td> <td align="right">   1047      </td> <td align="right">   1009      </td> <td align="right">   1009      </td> </tr>
<tr><td align="center">   0.6-0.7     </td> <td align="right">   1005      </td> <td align="right">   989       </td> <td align="right">   1003      </td> <td align="right">   989       </td> </tr>
<tr><td align="center">   0.7-0.8     </td> <td align="right">   986       </td> <td align="right">   962       </td> <td align="right">   985       </td> <td align="right">   954       </td> </tr>
<tr><td align="center">   0.8-0.9     </td> <td align="right">   1000      </td> <td align="right">   1027      </td> <td align="right">   1009      </td> <td align="right">   1023      </td> </tr>
<tr><td align="center">   0.9-1.0     </td> <td align="right">   991       </td> <td align="right">   1015      </td> <td align="right">   961       </td> <td align="right">   1026      </td> </tr>
<tr><td align="center">   $\mu$       </td> <td align="right">   0.4997    </td> <td align="right">   0.5018    </td> <td align="right">   0.4992    </td> <td align="right">   0.4990    </td> </tr>
<tr><td align="center">   $\sigma$    </td> <td align="right">   0.2882    </td> <td align="right">   0.2892    </td> <td align="right">   0.2861    </td> <td align="right">   0.2915    </td> </tr>
</tbody>
</table>




## Simple demonstration of RNGs using python
The following simple Python code plots the distribution of the produced random numbers using the linear congruential RNG employed by Python. The trend displayed in the previous table is seen rather clearly.


```python
#!/usr/bin/env python
import numpy as np
import matplotlib.mlab as mlab
import matplotlib.pyplot as plt
import random

# initialize the rng with a seed
random.seed() 
counts = 10000
values = np.zeros(counts)   
for i in range (1, counts, 1):
    values[i] = random.random()

# the histogram of the data
n, bins, patches = plt.hist(values, 10, facecolor='green')

plt.xlabel('$x$')
plt.ylabel('Number of counts')
plt.title(r'Test of uniform distribution')
plt.axis([0, 1, 0, 1100])
plt.grid(True)
plt.show()
```

## Properties of Selected Random Number Generators
Since our random numbers, which are typically generated via a linear congruential algorithm,
are never fully independent, we can then define 
an important test which measures the degree of correlation, namely the  so-called  
auto-correlation function defined previously, see again Eq. ([9](#eq:autocorrelformal)).
We rewrite it here as

$$
C_k=\frac{f_d}
             {\sigma^2},
$$

with $C_0=1$. Recall that 
$\sigma^2=\langle x_i^2\rangle-\langle x_i\rangle^2$ and that

$$
f_d = \frac{1}{nm}\sum_{\alpha=1}^m\sum_{k=1}^{n-d}(x_{\alpha,k}-\langle X_m \rangle)(x_{\alpha,k+d}-\langle X_m \rangle),
$$

The non-vanishing of $C_k$ for $k\ne 0$ means that the random
numbers are not independent. The independence of the random numbers is crucial 
in the evaluation of other expectation values. If they are not independent, our
assumption for approximating $\sigma_N$ is no longer valid.





## Autocorrelation function
This program computes the autocorrelation function as discussed in the equation on the previous slide for random numbers generated with the normal distribution $N(0,1)$.


```python
# Importing various packages
from math import exp, sqrt
from random import random, seed
import numpy as np
import matplotlib.pyplot as plt

def autocovariance(x, n, k, mean_x):
    sum = 0.0
    for i in range(0, n-k):
        sum += (x[(i+k)]-mean_x)*(x[i]-mean_x)
    return  sum/n

n = 1000
x=np.random.normal(size=n)
autocor = np.zeros(n)
figaxis = np.zeros(n)
mean_x=np.mean(x)
var_x = np.var(x)
print(mean_x, var_x)
for i in range (0, n):
    figaxis[i] = i
    autocor[i]=(autocovariance(x, n, i, mean_x))/var_x    

plt.plot(figaxis, autocor, "r-")
plt.axis([0,n,-0.1, 1.0])
plt.xlabel(r'$i$')
plt.ylabel(r'$\gamma_i$')
plt.title(r'Autocorrelation function')
plt.show()
```

As can be seen from the plot, the first point gives back the variance and a value of one. 
For the remaining values we notice that there are still non-zero values for the auto-correlation function.

















## Correlation function and which random number generators should I use
The program here computes the correlation function for one of the standard functions included with the c++ compiler.

        //  This function computes the autocorrelation function for 
        //  the standard c++ random number generator
        
        #include <fstream>
        #include <iomanip>
        #include <iostream>
        #include <cmath>
        using namespace std;
        // output file as global variable
        ofstream ofile;  
        
        //     Main function begins here     
        int main(int argc, char* argv[])
        {
             int n;
             char *outfilename;
        
             cin >> n;
             double MCint = 0.;      double MCintsqr2=0.;
             double invers_period = 1./RAND_MAX; // initialise the random number generator
             srand(time(NULL));  // This produces the so-called seed in MC jargon
             // Compute the variance and the mean value of the uniform distribution
             // Compute also the specific values x for each cycle in order to be able to
             // the covariance and the correlation function  
             // Read in output file, abort if there are too few command-line arguments
             if( argc <= 2 ){
               cout << "Bad Usage: " << argv[0] << 
        	 " read also output file and number of cycles on same line" << endl;
               exit(1);
             }
             else{
               outfilename=argv[1];
             }
             ofile.open(outfilename); 
             // Get  the number of Monte-Carlo samples
             n = atoi(argv[2]);
             double *X;  
             X = new double[n];
             for (int i = 0;  i < n; i++){
                   double x = double(rand())*invers_period; 
                   X[i] = x;
                   MCint += x;
                   MCintsqr2 += x*x;
             }
             double Mean = MCint/((double) n );
             MCintsqr2 = MCintsqr2/((double) n );
             double STDev = sqrt(MCintsqr2-Mean*Mean);
             double Variance = MCintsqr2-Mean*Mean;
        //   Write mean value and standard deviation 
             cout << " Standard deviation= " << STDev << " Integral = " << Mean << endl;
        
             // Now we compute the autocorrelation function
             double *autocor;  autocor = new double[n];
             for (int j = 0; j < n; j++){
               double sum = 0.0;
               for (int k = 0; k < (n-j); k++){
        	 sum  += (X[k]-Mean)*(X[k+j]-Mean); 
               }
               autocor[j] = sum/Variance/((double) n );
               ofile << setiosflags(ios::showpoint | ios::uppercase);
               ofile << setw(15) << setprecision(8) << j;
               ofile << setw(15) << setprecision(8) << autocor[j] << endl;
             }
             ofile.close();  // close output file
             return 0;
        }  // end of main program 


## Which RNG should I use?
* C++ has a class called **random**. The [random class](http://www.cplusplus.com/reference/random/) contains a large selection of RNGs and is highly recommended. Some of these RNGs have very large periods making it thereby very safe to use these RNGs in case one is performing large calculations. In particular, the [Mersenne twister random number engine](http://www.cplusplus.com/reference/random/mersenne_twister_engine/) has a period of $2^{19937}$. 

* Add RNGs in Python





## How to use the Mersenne generator
The following part of a c++ code (from project 4) sets up the uniform distribution for $x\in [0,1]$.

        /*
        
        //  You need this 
        #include <random>
        
        // Initialize the seed and call the Mersienne algo
        std::random_device rd;
        std::mt19937_64 gen(rd());
        // Set up the uniform distribution for x \in [[0, 1]
        std::uniform_real_distribution<double> RandomNumberGenerator(0.0,1.0);
        
        // Now use the RNG
        int ix = (int) (RandomNumberGenerator(gen)*NSpins);


## Why blocking?
**Statistical analysis.**

    * Monte Carlo simulations can be treated as *computer experiments*

    * The results can be analysed with the same statistical tools as we would use analysing experimental data.

    * As in all experiments, we are looking for expectation values and an estimate of how accurate they are, i.e., possible sources for errors.

A very good article which explains blocking is H. Flyvbjerg and H. G. Petersen, *Error estimates on averages of correlated data*,  [Journal of Chemical Physics 91, 461-466 (1989)](http://scitation.aip.org/content/aip/journal/jcp/91/1/10.1063/1.457480).

    




## Why blocking?
**Statistical analysis.**

    * As in other experiments, Monte Carlo experiments have two classes of errors:

      * Statistical errors

      * Systematical errors


    * Statistical errors can be estimated using standard tools from statistics

    * Systematical errors are method specific and must be treated differently from case to case. (In VMC a common source is the step length or time step in importance sampling)

    



## Code to demonstrate the calculation of the autocorrelation function
The following code computes the autocorrelation function, the covariance and the standard deviation
for standard RNG. 
The [following  file](https://github.com/CompPhysics/ComputationalPhysics2/tree/gh-pages/doc/Programs/LecturePrograms/programs/Blocking/autocorrelation.cpp) gives the code.

        //  This function computes the autocorrelation function for 
        //  the Mersenne random number generator with a uniform distribution
        #include <iostream>
        #include <fstream>
        #include <iomanip>
        #include <cstdlib>
        #include <random>
        #include <armadillo>
        #include <string>
        #include <cmath>
        using namespace  std;
        using namespace arma;
        // output file
        ofstream ofile;
        
        //     Main function begins here     
        int main(int argc, char* argv[])
        {
          int MonteCarloCycles;
          string filename;
          if (argc > 1) {
            filename=argv[1];
            MonteCarloCycles = atoi(argv[2]);
            string fileout = filename;
            string argument = to_string(MonteCarloCycles);
            fileout.append(argument);
            ofile.open(fileout);
          }
        
          // Compute the variance and the mean value of the uniform distribution
          // Compute also the specific values x for each cycle in order to be able to
          // compute the covariance and the correlation function  
        
          vec X  = zeros<vec>(MonteCarloCycles);
          double MCint = 0.;      double MCintsqr2=0.;
          std::random_device rd;
          std::mt19937_64 gen(rd());
          // Set up the uniform distribution for x \in [[0, 1]
          std::uniform_real_distribution<double> RandomNumberGenerator(0.0,1.0);
          for (int i = 0;  i < MonteCarloCycles; i++){
            double x =   RandomNumberGenerator(gen); 
            X(i) = x;
            MCint += x;
            MCintsqr2 += x*x;
          }
          double Mean = MCint/((double) MonteCarloCycles );
          MCintsqr2 = MCintsqr2/((double) MonteCarloCycles );
          double STDev = sqrt(MCintsqr2-Mean*Mean);
          double Variance = MCintsqr2-Mean*Mean;
          //   Write mean value and variance
          cout << " Sample variance= " << Variance  << " Mean value = " << Mean << endl;
          // Now we compute the autocorrelation function
          vec autocorrelation = zeros<vec>(MonteCarloCycles);
          for (int j = 0; j < MonteCarloCycles; j++){
            double sum = 0.0;
            for (int k = 0; k < (MonteCarloCycles-j); k++){
              sum  += (X(k)-Mean)*(X(k+j)-Mean); 
            }
            autocorrelation(j) = sum/Variance/((double) MonteCarloCycles );
            ofile << setiosflags(ios::showpoint | ios::uppercase);
            ofile << setw(15) << setprecision(8) << j;
            ofile << setw(15) << setprecision(8) << autocorrelation(j) << endl;
          }
          // Now compute the exact covariance using the autocorrelation function
          double Covariance = 0.0;
          for (int j = 0; j < MonteCarloCycles; j++){
            Covariance  += autocorrelation(j);
          }
          Covariance *=  2.0/((double) MonteCarloCycles);
          // Compute now the total variance, including the covariance, and obtain the standard deviation
          double TotalVariance = (Variance/((double) MonteCarloCycles ))+Covariance;
          cout << "Covariance =" << Covariance << "Totalvariance= " << TotalVariance << "Sample Variance/n= " << (Variance/((double) MonteCarloCycles )) << endl;
          cout << " STD from sample variance= " << sqrt(Variance/((double) MonteCarloCycles )) << " STD with covariance = " << sqrt(TotalVariance) << endl;
        
          ofile.close();  // close output file
          return 0;
        }  // end of main program 
        
        


## What is blocking?
**Blocking.**

    * Say that we have a set of samples from a Monte Carlo experiment

    * Assuming (wrongly) that our samples are uncorrelated our best estimate of the standard deviation of the mean $\langle \mathbf{M}\rangle$ is given by

$$
\sigma=\sqrt{\frac{1}{n}\left(\langle \mathbf{M}^2\rangle-\langle \mathbf{M}\rangle^2\right)}
$$

* If the samples are correlated we can rewrite our results to show  that

$$
\sigma=\sqrt{\frac{1+2\tau/\Delta t}{n}\left(\langle \mathbf{M}^2\rangle-\langle \mathbf{M}\rangle^2\right)}
$$

where $\tau$ is the correlation time (the time between a sample and the next uncorrelated sample) and $\Delta t$ is time between each sample

    


## What is blocking?
**Blocking.**

    * If $\Delta t\gg\tau$ our first estimate of $\sigma$ still holds

    * Much more common that $\Delta t<\tau$

    * In the method of data blocking we divide the sequence of samples into blocks

    * We then take the mean $\langle \mathbf{M}_i\rangle$ of block $i=1\ldots n_{blocks}$ to calculate the total mean and variance

    * The size of each block must be so large that sample $j$ of block $i$ is not correlated with sample $j$ of block $i+1$

    * The correlation time $\tau$ would be a good choice




## What is blocking?
**Blocking.**

    * Problem: We don't know $\tau$ or it is too expensive to compute

    * Solution: Make a plot of std. dev. as a function of blocksize

    * The estimate of std. dev. of correlated data is too low $\to$ the error will increase with increasing block size until the blocks are uncorrelated, where we reach a plateau

    * When the std. dev. stops increasing the blocks are uncorrelated




## Implementation
    * Do a Monte Carlo simulation, storing all samples to file

    * Do the statistical analysis on this file, independently of your Monte Carlo program

    * Read the file into an array

    * Loop over various block sizes

    * For each block size $n_b$, loop over the array in steps of $n_b$ taking the mean of elements $i n_b,\ldots,(i+1) n_b$

    * Take the mean and variance of the resulting array

    * Write the results for each block size to file for later
      analysis








## Actual implementation with code, main function
When the file gets large, it can be useful to write your data in binary mode instead of ascii characters.
The [following python file](https://github.com/CompPhysics/MachineLearning/blob/master/doc/Programs/Sampling/analysis.py)   reads data from file with the output from every Monte Carlo cycle.


```python
# Blocking
    @timeFunction
    def blocking(self, blockSizeMax = 500):
        blockSizeMin = 1

        self.blockSizes = []
        self.meanVec = []
        self.varVec = []

        for i in range(blockSizeMin, blockSizeMax):
            if(len(self.data) % i != 0):
                pass#continue
            blockSize = i
            meanTempVec = []
            varTempVec = []
            startPoint = 0
            endPoint = blockSize

            while endPoint <= len(self.data):
                meanTempVec.append(np.average(self.data[startPoint:endPoint]))
                startPoint = endPoint
                endPoint += blockSize
            mean, var = np.average(meanTempVec), np.var(meanTempVec)/len(meanTempVec)
            self.meanVec.append(mean)
            self.varVec.append(var)
            self.blockSizes.append(blockSize)

        self.blockingAvg = np.average(self.meanVec[-200:])
        self.blockingVar = (np.average(self.varVec[-200:]))
        self.blockingStd = np.sqrt(self.blockingVar)
```

## The Bootstrap method

The Bootstrap  resampling method is also very popular. It is very simple:

1. Start with your sample of measurements and compute the sample variance and the mean values

2. Then start again but pick in a random way the numbers in the sample and recalculate the mean and the sample variance.

3. Repeat this $K$ times.

It can be shown, see the article by [Efron](https://projecteuclid.org/download/pdf_1/euclid.aos/1176344552)
that it produces the correct standard deviation.

This method is very useful for small ensembles of data points.  


## Bootstrapping
Given a set of $N$ data, assume that we are interested in some 
observable $\theta$ which may be estimated from that set. This observable can also be for example the result of a fit based on all $N$ raw data. 
Let us call the value of the observable obtained from the original 
data set $\hat{\theta}$. One recreates from the sample repeatedly 
other samples by choosing randomly $N$ data out of the original set. 
This costs essentially nothing, since we just recycle the original data set for the building of new sets. 


## Bootstrapping, recipe
Let us assume we have done this $K$ times and thus have $K$ sets of $N$ 
data values each. 
Of course some values will enter more than once in the new sets. For each of these sets one computes the observable $\theta$ resulting in values $\theta_k$ with $k = 1,...,K$. Then one determines

$$
\tilde{\theta} = \frac{1}{K} \sum_{k=1}^K \theta_k,
$$

and

$$
sigma^2_{\tilde{\theta}} = \frac{1}{K} \sum_{k=1}^K \left(\theta_k-\tilde{\theta}\right)^2.
$$

These are estimators for $\angle\theta\rangle$ and its variance. They are not unbiased and therefore 
$\tilde{\theta}\neq\hat{\theta}$  for finite K. 

The difference is called bias and gives an idea on how far away the result may be from 
the true $\angle\theta\rangle$. As final result for the observable one quotes $\angle\theta\rangle = \tilde{\theta} \pm \sigma_{\tilde{\theta}}$ .



## Bootstrapping, [code](https://github.com/CompPhysics/MachineLearning/blob/master/doc/Programs/Sampling/analysis.py)

        # Bootstrap
            @timeFunction
            def bootstrap(self, nBoots = 1000):
                bootVec = np.zeros(nBoots)
                for k in range(0,nBoots):
                    bootVec[k] = np.average(np.random.choice(self.data, len(self.data)))
                self.bootAvg = np.average(bootVec)
                self.bootVar = np.var(bootVec)
                self.bootStd = np.std(bootVec)


## Jackknife, [code](https://github.com/CompPhysics/MachineLearning/blob/master/doc/Programs/Sampling/analysis.py)

        # Jackknife
            @timeFunction
            def jackknife(self):
                jackknVec = np.zeros(len(self.data))
                for k in range(0,len(self.data)):
                    jackknVec[k] = np.average(np.delete(self.data, k))
                self.jackknAvg = self.avg - (len(self.data) - 1) * (np.average(jackknVec) - self.avg)
                self.jackknVar = float(len(self.data) - 1) * np.var(jackknVec)
                self.jackknStd = np.sqrt(self.jackknVar)


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