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The concepts of moments, variance, covariance, and correlation in the context of statistical physics. It covers the definitions of central moments and variance, the relationship between covariance and correlation, and the binomial probability distribution function. Examples and calculations are provided to illustrate the concepts.

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2011/2012

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Download Simulations in Statistical Physics: Moments, Variance, Covariance, and Correlation and more Slides Statistical Physics in PDF only on Docsity! Chapter 2 Topic: pdf Simulations in Statistical Physics Docsity.com 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 xxp xxxg )( )()( 22 22 22 )()( xx xxp xxpxx i i i i i i i 22}var{ xxx }var{xThe standard deviation of x is Docsity.com Consider two events E0 and E1 that are mutually exclusive and exhaustive: Binomial Probability Distribution Function .0,0}0( .,1,}1{ xEP xpEP The expected values for the real number x and its square are .)( ,0)1(1)( 2 pxE pppxE 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 22 pppp xxx xNx x N ppCxXP )1(}{ Npx )1(}var{ pnpx The variance of x is The average or mean of x is Binomial pdf: Docsity.com Example: Probability of getting at least four heads in 6 tosses of a fair coin is Binomial Probability Distribution Function 4 5 66 4 6 5 6 6 6 6 6 1/ 2 (1/ 2) 1/ 2 (1/ 2) 1/ 2 (1/ 2) 4 5 6 15 6 1 22 11 64 64 64 64 32 In 100 tosses of a fair coin the mean number of head and variance are 100 1/ 2 50x Np var{ } (1 ) 100(1/ 2)(1 1/ 2) 25x Np p Then the standard deviation is 5. Docsity.com 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 0 5 5 0.4 (0.6) 0.07776 0.08 0 Pr{ at least one will graduate } = 1 – Pr { none will graduate } = 0.92 Pr{ all will graduate } = 0.01024 Pr{ none will graduate } = 1 4 5 0.4 (0.6) 0.2592 0.26 1 Pr{ one will graduate } = Docsity.com 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 l x 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 ,,3,2,1,0, ! )( }{ n n t exXP n t ll Docsity.com 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 = 0.1 Pr{ 2 defective in 10 } = { 10! × p2 × (1 – p)8 } /{ 2! × 8! } = 0.1937 Mean value = λ = Np = 10 (0.1) = 1. Pr{ 2 defective in 10 } = λX exp(-λ)/X! = { (1)2 exp(-1)/2! = 0.1839 In general, Poisson approximation is good if mean is less than 5 and p is less than 0.1. Docsity.com Example: If the probability that an individual suffers a bad reaction from injection of a given serum is 0.001. Find the probability that (a)out of 2000 individuals exactly 3 will suffer bad reaction. (b)Zero person will suffer. (c)One person will suffer. Poisson Distribution Function Mean value = λ = Np = 2000 (0.001) = 2. Pr{ 3 will suffer bad reaction } = λX exp(-λ)/X! = { (2)3 exp(-2)/3! = 0.180 Pr{ 0 will suffer bad reaction } = λX exp(-λ)/X! = { (2)0 exp(-2)/0! = 0.134 Pr{ 1 will suffer bad reaction } = λX exp(-λ)/X! = { (2)1 exp(-2)/1! = 2/e2 =0.268 Docsity.com