Simulations in Statistical Physics: Moments, Variance, Covariance, and Correlation, Slides of Statistical Physics

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Chapter 2

Topic: pdf

Simulations in

Statistical Physics

Moments & Variance

The Central Moments

The Central moments of x are defined as

The second central moment has a particular meaning:

This is also called variance of x.

n i

n

p x x

g x x x



2 2

2 2

2 2 ( ) ( )





x x

p x x

x x p x x

i

i i

i

i i

i

2 2

var{ x }  x  x 

The standard deviation of x is    var{ x }

Covariance & Correlation

The covariance is a measure of the independence of two random variables x

and y:

cov{ x , y }  xy  x  y 

Zero covariance does not imply independence of random variables.

Another quantity related to covariance is the correlation coefficient:

It is equal to zero when x and y are independent. Also,

var{ }

cov{ , }

2 2

x

x x x x

var{ }var{ }

cov{ , } ( , ) x y

x y  x y 

Its value is in between -1 and +1. Monte Carlo calculations try to take

advantage of the negative correlation as a measure of reducing the

variance.

Consider two events E0 and E1 that are mutually exclusive and exhaustive:

Binomial Probability Distribution Function

( 0 } 0 , 0.

{ 1 } , 1 .,

 

 

P E x

P E p x

The expected values for the real number x and its square are

( ).

( ) 1 ( 1 ) 0 ,

2 E x p

E x p p p



     

The variance of x is

Suppose there are N independent samples of these events And each has either 0 or

1 outcome. Then probability of x Successes out of N is

( 1 ).

var{ } 2

2 2

p p p p

x x x

   

  

x N x x

N P X x C p p

 {  } ( 1  )

 x  Np

The variance of x is var{ x }  np ( 1  p )

The average or mean of x is

Binomial pdf:

Example: Suppose Probability that an entering college student will graduate

is 0.4. Determine that out of 5 students none will graduate. (b) at least one

will graduate.

Binomial Probability Distribution Function

 

  ^ 

Pr{ at least one will graduate } = 1 – Pr { none will graduate } = 0.

Pr{ all will graduate } = 0.

Pr{ none will graduate } =

 

  ^ 

Pr{ one will graduate } =

Fitting Data by pdf

Suppose we have a problem of toss of 5 fair coins and observing

number of heads. The following table is generated as a result of

experiments.

Number of Heads (X)

Observed frequency (fo)

0 38

1 144

2 342

3 287

4 164

5 25

Then the average number of heads is

=( 38×0 + 144×1 + 342×2 + 287×3 +

164×4 + 25×5 ) / (1000)

= 2470/1000 = 2.

p = /N = 2.47/5 = 0.

q = (1 – p) = 0.

The Poisson Distribution

A random variable x is said to follow Poisson distribution, when

Where, l is a parameter of this distribution. It is easy to find that

 x   l

This distribution is fundamental in the theory of probability and

stochastic processes. It is of great use in applications such as

radioactive decay, queuing service systems and similar systems.

 l

l

var{ x }



n

n

t

P X x e

n l t^ l

Example: ten percent of the tools produced in a factory are turning out to be

defective. Find the probability that in a sample of 10 tools chosen at random

exactly two will be defective using binomial and Poisson distributions.

Poisson Distribution Function

The probability of a defective tool = p = 0.

Pr{ 2 defective in 10 } = { 10! × p^2 × (1 – p)^8 } /{ 2! × 8! } = 0.

Mean value = λ = Np = 10 (0.1) = 1.

Pr{ 2 defective in 10 } = λX^ exp(-λ)/X! = { (1)2 exp(-1)/2! = 0.

In general, Poisson approximation is good if mean is less than 5 and p is

less than 0.1.