Consider the following simple example, namely the tossing of two dice, resulting in the following possible values
$$
\begin{equation*}
\{2,3,4,5,6,7,8,9,10,11,12\}.
\end{equation*}
$$
These values are called the domain.
To this domain we have the corresponding probabilities
$$
\begin{equation*}
\{1/36,2/36/,3/36,4/36,5/36,6/36,5/36,4/36,3/36,2/36,1/36\}.
\end{equation*}
$$
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
$$
\begin{equation*}
\{2,3,4,5,6,7,8,9,10,11,12\},
\end{equation*}
$$
appear in a random order. After 11 throws the results may look like
$$
\begin{equation*}
\{10,8,6,3,6,9,11,8,12,4,5\}.
\end{equation*}
$$
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).
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
$$
\begin{equation*}
p(x) = \mathrm{Prob}(X=x).
\end{equation*}
$$
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
$$
\begin{equation*}
\mathrm{Prob}(a\leq X\leq b) = \int_a^b p(x)dx.
\end{equation*}
$$
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.
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 \( x \)
$$
\begin{equation*}
P(x)=\mathrm{Prob(}X\leq x\mathrm{)} =
\int_{-\infty}^x p(x^{\prime})dx^{\prime}.
\end{equation*}
$$
The relation between a CDF and its corresponding PDF is then
$$
\begin{equation*}
p(x) = \frac{d}{dx}P(x).
\end{equation*}
$$
There are two properties that all PDFs must satisfy. The first one is positivity (assuming that the PDF is normalized)
$$
\begin{equation*}
0 \leq p(x) \leq 1.
\end{equation*}
$$
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
$$
\begin{align*}
\sum_{x_i\in\mathbb D} p(x_i) & = 1,\\
\int_{x\in\mathbb D} p(x)\,dx & = 1.
\end{align*}
$$
The first one is the most basic PDF; namely the uniform distribution
$$
\begin{equation}
p(x) = \frac{1}{b-a}\theta(x-a)\theta(b-x).
\tag{1}
\end{equation}
$$
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.
The second one is the Gaussian Distribution
$$
\begin{equation*}
p(x) = \frac{1}{\sigma\sqrt{2\pi}} \exp{(-\frac{(x-\mu)^2}{2\sigma^2})},
\end{equation*}
$$
with mean value \( \mu \) and standard deviation \( \sigma \). If \( \mu=0 \) and \( \sigma=1 \), it is normally called the standard normal distribution
$$
\begin{equation*}
p(x) = \frac{1}{\sqrt{2\pi}} \exp{(-\frac{x^2}{2})},
\end{equation*}
$$
The following simple Python code plots the above distribution for different values of \( \mu \) and \( \sigma \).
Another important distribution in science is the exponential distribution
$$
\begin{equation*}
p(x) = \alpha\exp{-(\alpha x)}.
\end{equation*}
$$
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
$$
\begin{equation}
\langle h \rangle_X \equiv \int\! h(x)p(x)\,dx
\tag{2}
\end{equation}
$$
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
$$
\begin{equation*}
\langle x^n \rangle \equiv \int\! x^n p(x)\,dx
\end{equation*}
$$
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 \( \mu \)
$$
\begin{equation*}
\langle x\rangle = \mu \equiv \int x p(x)dx,
\end{equation*}
$$
for a continuous distribution and
$$
\begin{equation*}
\langle x\rangle = \mu \equiv \sum_{i=1}^N x_i p(x_i),
\end{equation*}
$$
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 \( p(x) \).
A special version of the moments is the set of central moments, the n-th central moment defined as
$$
\begin{equation*}
\langle (x-\langle x\rangle )^n\rangle \equiv \int\! (x-\langle x\rangle)^n p(x)\,dx
\end{equation*}
$$
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.
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.
| Discrete PDF | Continuous PDF | |
|---|---|---|
| Domain | \( \left\{x_1, x_2, x_3, \dots, x_N\right\} \) | \( [a,b] \) |
| Probability | \( p(x_i) \) | \( p(x)dx \) |
| Cumulative | \( P_i=\sum_{l=1}^ip(x_l) \) | \( P(x)=\int_a^xp(t)dt \) |
| Positivity | \( 0 \le p(x_i) \le 1 \) | \( p(x) \ge 0 \) |
| Positivity | \( 0 \le P_i \le 1 \) | \( 0 \le P(x) \le 1 \) |
| Monotonic | \( P_i \ge P_j \) if \( x_i \ge x_j \) | \( P(x_i) \ge P(x_j) \) if \( x_i \ge x_j \) |
| Normalization | \( P_N=1 \) | \( P(b)=1 \) |
With a PDF we can compute expectation values of selected quantities such as
$$
\begin{equation*}
\langle x^k\rangle=\sum_{i=1}^{N}x_i^kp(x_i),
\end{equation*}
$$
if we have a discrete PDF or
$$
\begin{equation*}
\langle x^k\rangle=\int_a^b x^kp(x)dx,
\end{equation*}
$$
in the case of a continuous PDF. We have already defined the mean value \( \mu \)
and the variance \( \sigma^2 \).
There are at least three PDFs which one may encounter. These are the
Uniform distribution
$$
\begin{equation*}
p(x)=\frac{1}{b-a}\Theta(x-a)\Theta(b-x),
\end{equation*}
$$
yielding probabilities different from zero in the interval \( [a,b] \).
The exponential distribution
$$
\begin{equation*}
p(x)=\alpha \exp{(-\alpha x)},
\end{equation*}
$$
yielding probabilities different from zero in the interval \( [0,\infty) \) and with mean value
$$
\begin{equation*}
\mu = \int_0^{\infty}xp(x)dx=\int_0^{\infty}x\alpha \exp{(-\alpha x)}dx=\frac{1}{\alpha},
\end{equation*}
$$
$$
\begin{equation*}
\sigma^2=\int_0^{\infty}x^2p(x)dx-\mu^2 = \frac{1}{\alpha^2}.
\end{equation*}
$$
Finally, we have the so-called univariate normal distribution, or just the normal distribution
$$
\begin{equation*}
p(x)=\frac{1}{b\sqrt{2\pi}}\exp{\left(-\frac{(x-a)^2}{2b^2}\right)}
\end{equation*}
$$
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
$$
\begin{equation*}
\mu = \int_0^{\infty}xp(x)dx=\frac{1}{b\sqrt{2\pi}}\int_{-\infty}^{\infty}x \exp{\left(-\frac{(x-a)^2}{2b^2}\right)}dx,
\end{equation*}
$$
which becomes with a suitable change of variables
$$
\begin{equation*}
\mu =\frac{1}{b\sqrt{2\pi}}\int_{-\infty}^{\infty}b\sqrt{2}(a+b\sqrt{2}y)\exp{-y^2}dy=a.
\end{equation*}
$$
Similarly, the variance becomes
$$
\begin{equation*}
\sigma^2 = \frac{1}{b\sqrt{2\pi}}\int_{-\infty}^{\infty}(x-\mu)^2 \exp{\left(-\frac{(x-a)^2}{2b^2}\right)}dx,
\end{equation*}
$$
and inserting the mean value and performing a variable change we obtain
$$
\begin{equation*}
\sigma^2 = \frac{1}{b\sqrt{2\pi}}\int_{-\infty}^{\infty}b\sqrt{2}(b\sqrt{2}y)^2\exp{\left(-y^2\right)}dy=
\frac{2b^2}{\sqrt{\pi}}\int_{-\infty}^{\infty}y^2\exp{\left(-y^2\right)}dy,
\end{equation*}
$$
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
$$
\begin{equation}
p(x)=\frac{1}{\sqrt{2\pi}}\exp{\left(-\frac{x^2}{2}\right)}.
\tag{3}
\end{equation}
$$
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
$$
\begin{equation*}
P(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^x\exp{\left(-\frac{t^2}{2}\right)}dt,
\end{equation*}
$$
which is difficult to evaluate in a quick way.
Some other PDFs which one encounters often in the natural sciences are the binomial distribution
$$
\begin{equation*}
p(x) = \left(\begin{array}{c} n \\ x\end{array}\right)y^x(1-y)^{n-x} \hspace{0.5cm}x=0,1,\dots,n,
\end{equation*}
$$
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
In order to compute the mean and variance we need to recall Newton's binomial formula
$$
\begin{equation*}
(a+b)^m=\sum_{n=0}^m \left(\begin{array}{c} m \\ n\end{array}\right)a^nb^{m-n},
\end{equation*}
$$
which can be used to show that
$$
\begin{equation*}
\sum_{x=0}^n\left(\begin{array}{c} n \\ x\end{array}\right)y^x(1-y)^{n-x} = (y+1-y)^n = 1,
\end{equation*}
$$
the PDF is normalized to one.
The mean value is
$$
\begin{equation*}
\mu = \sum_{x=0}^n x\left(\begin{array}{c} n \\ x\end{array}\right)y^x(1-y)^{n-x} =
\sum_{x=0}^n x\frac{n!}{x!(n-x)!}y^x(1-y)^{n-x},
\end{equation*}
$$
resulting in
$$
\begin{equation*}
\mu =
\sum_{x=0}^n x\frac{(n-1)!}{(x-1)!(n-1-(x-1))!}y^{x-1}(1-y)^{n-1-(x-1)},
\end{equation*}
$$
which we rewrite as
$$
\begin{equation*}
\mu=ny\sum_{\nu=0}^n\left(\begin{array}{c} n-1 \\ \nu\end{array}\right)y^{\nu}(1-y)^{n-1-\nu} =ny(y+1-y)^{n-1}=ny.
\end{equation*}
$$
Another important distribution with discrete stochastic variables \( x \) is the Poisson model, which resembles the exponential distribution and reads
$$
\begin{equation*}
p(x) = \frac{\lambda^x}{x!} e^{-\lambda} \hspace{0.5cm}x=0,1,\dots,;\lambda > 0.
\end{equation*}
$$
In this case both the mean value and the variance are easier to calculate,
$$
\begin{equation*}
\mu = \sum_{x=0}^{\infty} x \frac{\lambda^x}{x!} e^{-\lambda} = \lambda e^{-\lambda}\sum_{x=1}^{\infty}
\frac{\lambda^{x-1}}{(x-1)!}=\lambda,
\end{equation*}
$$
and the variance is \( \sigma^2=\lambda \).
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
$$
\begin{equation*}
\lim_{n\rightarrow \infty}\left(\begin{array}{c} n \\ x\end{array}\right)y^x(1-y)^{n-x} e^{-\lambda}=\sum_{x=1}^{\infty}\frac{\lambda^x}{x!} e^{-\lambda}.
\end{equation*}
$$
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
$$
\begin{align}
\mathrm{Cov}(X_i,\,X_j) & = \langle (x_i-\langle x_i\rangle)(x_j-\langle x_j\rangle)\rangle
\tag{4}\\
&=\int\cdots\int (x_i-\langle x_i\rangle)(x_j-\langle x_j\rangle)P(x_1,\dots,x_n)\,dx_1\dots dx_n,
\tag{5}
\end{align}
$$
with
$$
\begin{equation*}
\langle x_i\rangle =
\int\cdots\int x_i P(x_1,\dots,x_n)\,dx_1\dots dx_n.
\end{equation*}
$$
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.
# 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)
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
$$
Cov(X_i, X_j) = 0 \hspace{0.1cm} (i\neq j).
$$
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
$$
\begin{equation*}
\{x_1, x_2,\dots\,x_k,\dots\}.
\end{equation*}
$$
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 \( p_X^{\phantom X}(x) \).
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
$$
\begin{equation*}
\langle x_{\alpha} \rangle \equiv \frac{1}{n}\sum_{k=1}^n x_{\alpha,k}.
\end{equation*}
$$
The sample variance is:
$$
\begin{equation*}
\mathrm{Var}(x) \equiv \frac{1}{n}\sum_{k=1}^n (x_{\alpha,k} - \langle x_{\alpha} \rangle)^2,
\end{equation*}
$$
with its square root being the standard deviation of the sample.
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
$$
\begin{equation}
\langle X_m \rangle= \frac{1}{m}\sum_{\alpha=1}^mx_{\alpha}=\frac{1}{mn}\sum_{\alpha, k} x_{\alpha,k},
\tag{6}
\end{equation}
$$
where the last sums end at \( m \) and \( n \).
The total variance is
$$
\begin{equation*}
\sigma^2_m= \frac{1}{mn^2}\sum_{\alpha=1}^m(\langle x_{\alpha} \rangle-\langle X_m \rangle)^2,
\end{equation*}
$$
which we rewrite as
$$
\begin{equation}
\sigma^2_m=\frac{1}{m}\sum_{\alpha=1}^m\sum_{kl=1}^n (x_{\alpha,k}-\langle X_m \rangle)(x_{\alpha,l}-\langle X_m \rangle).
\tag{7}
\end{equation}
$$
We define also the sample variance \( \sigma^2 \) of all \( mn \) individual experiments as
$$
\begin{equation}
\sigma^2=\frac{1}{mn}\sum_{\alpha=1}^m\sum_{k=1}^n (x_{\alpha,k}-\langle X_m \rangle)^2.
\tag{8}
\end{equation}
$$
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 \( \mathrm{Cov}(X,Y) \).
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
$$
\begin{equation*}
\sigma_m=
\frac{\sigma}{\sqrt{m}}.
\end{equation*}
$$
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.
Our estimate of the true average \( \mu_{X} \) is the sample mean \( \langle X_m \rangle \)
$$
\begin{equation*}
\mu_{X}^{\phantom X} \approx X_m=\frac{1}{mn}\sum_{\alpha=1}^m\sum_{k=1}^n x_{\alpha,k}.
\end{equation*}
$$
We can then use Eq. (7)
$$
\begin{equation*}
\sigma^2_m=\frac{1}{mn^2}\sum_{\alpha=1}^m\sum_{kl=1}^n (x_{\alpha,k}-\langle X_m \rangle)(x_{\alpha,l}-\langle X_m \rangle),
\end{equation*}
$$
and rewrite it as
$$
\begin{equation*}
\sigma^2_m=\frac{\sigma^2}{n}+\frac{2}{mn^2}\sum_{\alpha=1}^m\sum_{k < l}^n (x_{\alpha,k}-\langle X_m \rangle)(x_{\alpha,l}-\langle X_m \rangle),
\end{equation*}
$$
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 \( k\ne l \).
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.
Let us analyze the problem by splitting up the correlation term into partial sums of the form
$$
\begin{equation*}
f_d = \frac{1}{nm}\sum_{\alpha=1}^m\sum_{k=1}^{n-d}(x_{\alpha,k}-\langle X_m \rangle)(x_{\alpha,k+d}-\langle X_m \rangle),
\end{equation*}
$$
The correlation term of the total variance can now be rewritten in terms of
\( f_d \)
$$
\begin{equation*}
\frac{2}{mn^2}\sum_{\alpha=1}^m\sum_{k < l}^n (x_{\alpha,k}-\langle X_m \rangle)(x_{\alpha,l}-\langle X_m \rangle)=
\frac{2}{n}\sum_{d=1}^{n-1} f_d
\end{equation*}
$$
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
$$
\begin{equation}
\kappa_d = \frac{f_d}{\sigma^2}
\tag{9}
\end{equation}
$$
which gives us a useful measure of the correlation pair correlation
starting always at \( 1 \) for \( d=0 \).
The sample variance of the \( mn \) experiments can now be written in terms of the autocorrelation function
$$
\begin{equation}
\sigma_m^2=\frac{\sigma^2}{n}+\frac{2}{n}\cdot\sigma^2\sum_{d=1}^{n-1}
\frac{f_d}{\sigma^2}=\left(1+2\sum_{d=1}^{n-1}\kappa_d\right)\frac{1}{n}\sigma^2=\frac{\tau}{n}\cdot\sigma^2
\tag{10}
\end{equation}
$$
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
$$
\begin{equation}
\tau = 1+2\sum_{d=1}^{n-1}\kappa_d
\tag{11}
\end{equation}
$$
For a correlation free experiment, \( \tau \)
equals 1.
From the point of view of Eq. (10) 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
$$
\begin{equation*}
n_\mathrm{eff} = \frac{n}{\tau}
\end{equation*}
$$
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.
# 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)
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.
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:
The most common random number generators are based on so-called Linear congruential relations of the type
$$
\begin{equation*}
N_i=(aN_{i-1}+c) \mathrm{MOD} (M),
\end{equation*}
$$
which yield a number in the interval [0,1] through
$$
\begin{equation*}
x_i=N_i/M
\end{equation*}
$$
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 \( 4 \).
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
$$
\begin{equation*}
N_i=(6N_{i-1}+7) \mathrm{MOD} (5),
\end{equation*}
$$
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
$$
\begin{equation*}
N_i=(27N_{i-1}+11) \mathrm{MOD} (54),
\end{equation*}
$$
which still, with \( N_0=2 \), results in \( 11,38,11,38,11,38,\dots \), a period of
just \( 2 \).
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 \( M \),
$$
\begin{equation*}
N_l=(aN_{l-i}+cN_{l-j})\mathrm{MOD}(M).
\end{equation*}
$$
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 which consists of two congruential relations
$$
\begin{equation}
N_l=(N_{l-3}-N_{l-1})\mathrm{MOD}(2^{31}-69),
\tag{12}
\end{equation}
$$
followed by
$$
\begin{equation}
N_l=(69069N_{l-1}+1013904243)\mathrm{MOD}(2^{32}),
\tag{13}
\end{equation}
$$
which according to the authors has a period larger than \( 2^{94} \).
Instead of using modular addition, we could use the bitwise exclusive-OR (\( \oplus \)) operation so that
$$
\begin{equation*}
N_l=(N_{l-i})\oplus (N_{l-j})
\end{equation*}
$$
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 \( m\wedge n \).
We show here how the linear congruential algorithm can be implemented, namely
$$
\begin{equation*}
N_i=(aN_{i-1}) \mathrm{MOD} (M).
\end{equation*}
$$
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
$$
\begin{equation*}
M=aq+r,
\end{equation*}
$$
where we have defined
$$
\begin{equation*}
q=[M/a],
\end{equation*}
$$
and
$$
\begin{equation*}
r = M\hspace{0.1cm}\mathrm{MOD} \hspace{0.1cm}a.
\end{equation*}
$$
where the brackets denote integer division. In the code below the numbers
\( q \) and \( r \) are chosen so that \( r < q \).
To see how this works we note first that
$$
\begin{equation}
(aN_{i-1}) \mathrm{MOD} (M)= (aN_{i-1}-[N_{i-1}/q]M)\mathrm{MOD} (M),
\tag{14}
\end{equation}
$$
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 \( M \).
We can now rewrite Eq. (14) as
$$
\begin{equation}
(aN_{i-1}) \mathrm{MOD} (M)= (aN_{i-1}-[N_{i-1}/q](aq+r))\mathrm{MOD} (M),
\tag{15}
\end{equation}
$$
which results
in
$$
\begin{equation}
(aN_{i-1}) \mathrm{MOD} (M)= \left(a(N_{i-1}-[N_{i-1}/q]q)-[N_{i-1}/q]r)\right)\mathrm{MOD} (M),
\tag{16}
\end{equation}
$$
yielding
$$
\begin{equation}
(aN_{i-1}) \mathrm{MOD} (M)= \left(a(N_{i-1}\mathrm{MOD} (q)) -[N_{i-1}/q]r)\right)\mathrm{MOD} (M).
\tag{17}
\end{equation}
$$
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 \( 0 \).
/*
** 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 =<PHONE_NUMBER>, 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()
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 \( \mu=\langle x\rangle \).
For the uniform distribution, the mean value \( \mu \) is then
$$
\begin{equation*}
\mu=\langle x\rangle=\frac{1}{2}
\end{equation*}
$$
while the standard deviation is
$$
\begin{equation*}
\sigma=\sqrt{\langle x^2\rangle-\mu^2}=\frac{1}{\sqrt{12}}=0.2886.
\end{equation*}
$$
The various random number generators produce results which agree rather well with these limiting values.
| \( x \)-bin | ran0 | ran1 | ran2 | ran3 |
|---|---|---|---|---|
| 0.0-0.1 | 1013 | 991 | 938 | 1047 |
| 0.1-0.2 | 1002 | 1009 | 1040 | 1030 |
| 0.2-0.3 | 989 | 999 | 1030 | 993 |
| 0.3-0.4 | 939 | 960 | 1023 | 937 |
| 0.4-0.5 | 1038 | 1001 | 1002 | 992 |
| 0.5-0.6 | 1037 | 1047 | 1009 | 1009 |
| 0.6-0.7 | 1005 | 989 | 1003 | 989 |
| 0.7-0.8 | 986 | 962 | 985 | 954 |
| 0.8-0.9 | 1000 | 1027 | 1009 | 1023 |
| 0.9-1.0 | 991 | 1015 | 961 | 1026 |
| \( \mu \) | 0.4997 | 0.5018 | 0.4992 | 0.4990 |
| \( \sigma \) | 0.2882 | 0.2892 | 0.2861 | 0.2915 |
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.
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). We rewrite it here as
$$
\begin{equation*}
C_k=\frac{f_d}
{\sigma^2},
\end{equation*}
$$
with \( C_0=1 \). Recall that
\( \sigma^2=\langle x_i^2\rangle-\langle x_i\rangle^2 \) and that
$$
\begin{equation*}
f_d = \frac{1}{nm}\sum_{\alpha=1}^m\sum_{k=1}^{n-d}(x_{\alpha,k}-\langle X_m \rangle)(x_{\alpha,k+d}-\langle X_m \rangle),
\end{equation*}
$$
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.
# 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.
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
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);
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).
// 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
$$
\sigma=\sqrt{\frac{1}{n}\left(\langle \mathbf{M}^2\rangle-\langle \mathbf{M}\rangle^2\right)}
$$
$$
\sigma=\sqrt{\frac{1+2\tau/\Delta t}{n}\left(\langle \mathbf{M}^2\rangle-\langle \mathbf{M}\rangle^2\right)}
$$
where \( \tau \) is the correlation time (the time between a sample and the next uncorrelated sample) and \( \Delta t \) is time between each sample
# 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 resampling method is also very popular. It is very simple:
It can be shown, see the article by Efron that it produces the correct standard deviation.
This method is very useful for small ensembles of data points.
$$
\tilde{\theta} = \frac{1}{K} \sum_{k=1}^K \theta_k,
$$
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 \( \angle\theta\rangle = \tilde{\theta} \pm \sigma_{\tilde{\theta}} \) .
# 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
@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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--- layout: post authors: ["Greg Wilson"] title: "Lab Meeting Reminder" date: 2014-05-21 time: "08:30:00" tags: ["Community", "Lab Meeting"] ---
Our monthly lab meeting will take place on Thursday, May 22 at 10:00 am and again at 7:00 pm, Eastern time. There's a lot to discuss, so please take a moment beforehand to go through the lab meeting Etherpad and see if there are any pull requests or enhancement proposals that you are particularly interested in.
We look forward to seeing you online.


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BRASSED OFF A film review by Mark R. Leeper
Capsule: This film is by turns a comedy, a serious drama, and an anti-Tory political tract. But Mark Herman who wrote and directed certainly knows how to create characters in whom the audience can place an emotional investment. This film about a century-old brass band in a dying Yorkshire mining town is predictable but has a lot of affecting human drama. Rating: low +2 (-4 to +4) 7 (0 to 10) New York Critics: 5 positive, 4 negative, 4 mixed
Since the 1950s the British have certainly known how to make modest but affecting films in ways that American filmmakers rarely attempt. Some simple comedies from the post-war era gave us eloquent and loving pictures of British village life still memorable today. That tradition is strong in BRASSED OFF, but mostly in the first half of the film. Director/writer Mark Herman hooks the audience with a light comic view of the mining village of Grimely, but increasingly the replaces the comedy with serious drama. We see the village having its most difficult times with the pit closings of the Margaret Thatcher administration.  Not that it is a total surprise with the opening of the film making some angry remarks about closing mines to replace with nuclear power plants.
The heart of the Yorkshire town of Grimely is its colliery. ('Ere. of Grimely--well, most of it anyway--is the Grimely Colliery Brass Band. The band has been around for a hundred years. But these are hard times for Grimely. The company looks like it might close the colliery and if there really is a pit closing, the whole town might just as well dry up and blow away. Already there are those in the band who are so depressed about what is happening to the town that they are ready to quit the band. Band leader Danny (Pete Postlethwaite, the currently best thing about THE LOST WORLD: JURASSIC PARK) cannot believe that people in the town will let the band die, even if the mine goes under. Just as the band is starting to founder, new life is breathed into it by a new member. Gloria (the radiant Tara Fitzgerald), granddaughter of a former great band member, returns to the town of her birth with her grandfather's flugelhorn. The first woman ever in the Grimely Colliery Brass Band plays the flugelhorn as well as her grandfather did. Suddenly the band starts looking and sounding better to the band members. While the future of the town is souring, the men are distracted for a few hours a week by music and a little flirting. And of course some of the wives are jealous. But things grow grimmer in Grimely as the company offers a job buyout. While some band members are sacrificing food for music, most of the town is looking at whether they want to mortgage the future of the town in the buyout or hope the mines are not closed. Mark Herman has a feel for the humanity of the people, no doubt based on his youth in Yorkshire.
Pete Postlethwaite is one of the British actors who does a great job and nobody seems to make much of a fuss about. He had major roles in films such as IN THE NAME OF THE FATHER, or bit parts in films such as THE USUAL SUSPECTS. His bicycle-riding band leader in denial about how serious the town's problems are is entirely different from roles he has done before. Tara Fitzgerald is a decent actress, and though I have not seen her in any really demanding roles, she is certainly a joy to watch. Ewan McGregor, like Postlethwaite, is also in two current films, this and THE PILLOW BOOK. And with his Scottish name it goes without saying that he was in TRAINSPOTTING. Veteran character actor Jim Carter (THE ADVOCATE, BLACK BEAUTY, and RICHARD III), looking like a heavy-set Leonard Rossiter, usually can be counted on for a bit a color.
Americans will be at a slight disadvantage in seeing BRASSED OFF. They will be informed at the outset what a colliery is, but the Yorkshire accent takes a little getting used to. What very likely were some funny lines will go past viewers not quick enough to pick up what is said. Curiously the anti-Tory sentiments may hit home from similarities in policy between the Conservative Tories and the Republican Party. But the political arguments will lose a little impact since coal-mining is shown to be a dangerous profession which shortens lives. The viewer may decide that it is just as well that the next generation is saved from going into the pits and instead is forced to find other work.
Not surprisingly, the score by Trevor Jones is big and brassy with some nice brass renditions of popular light classical themes. The score is played by the Grimethorpe Colliery Band who also supplied extras for the story's band. No doubt Grimely is based in no small part on Grimethorpe. This is a film that has more than a few moving moments and it worth looking for. There is more to this film than the trailers would lead one to think. I rate it a low +2 on the -4 to +4 scale.
Mark R. Leeper<EMAIL_ADDRESS>Copyright 1997 Mark R. Leeper
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