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

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{xThe 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