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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

var{ x }  x^2  x ^2

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

For a general function g(x), the variance is

Two random variables g(x) and h(x) are independent if they derive

from independent events. When two are independent then,

 g ( x ) h ( x )   g ( x )  h ( x ) 

For independent g and h variables,

2 2

2 2

2

( ) ( )

var{ ( )} [ ( ) ( ) ]

  













 

   

 

g g

p g x p g x

g x g x g x

j j

j j j

j

var{ ag ( x ) bh ( x )} a^2 var{ g ( x )} b^2 var{ h ( x )}

Moments & Variance

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{ , }^22

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 , E x^2 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 P { X  x } N^ C p ( 1  p ) 

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

The average or mean of x is

Binomial pdf:

Example: Probability of getting at least four heads in 6 tosses of a fair coin is

Binomial Probability Distribution Function

     

(^6 4 6 4 6 5 6 5 666 ) 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

 x  Np  100  (^) 1/ 2   50

var{ } x  Np (1  p )  100(1/ 2)(1 1/ 2)  25

Then the standard deviation is 5.

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

 

(^5 0 ) 0.4 (0.6) 0.07776 0. 0

    ^   

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

Pr{ all will graduate } = 0.

Pr{ none will graduate } =

 

(^5 1 ) 0.4 (0.6) 0.2592 0. 1

    ^   

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.

Fitting Data by pdf

Number of Heads (x)

Observed frequency (fo)

Pr { x heads} Frequency from theory (fe)

0 38 0.332 33

1 144 0.1619 162

2 342 0.3162 316

3 287 0.3087 309

4 164 0.1507 151

5 25 0.0294 29

Then we can compute theoretical values of frequency from binomial

distribution: p = 0.494 ; q = 0.506 x N x x N x

x

p q

x

 













 











( 0. 494 ) ( 0. 506 )

5 5

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 }

, 0 , 1 , 2 , 3 , ,

!

( )

{  }  



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.

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.

Pr{ 0 will suffer bad reaction } = λX^ exp(-λ)/X! = { (2)^0 exp(-2)/0! = 0.

Pr{ 1 will suffer bad reaction } = λX exp(-λ)/X! = { (2)^1 exp(-2)/1! = 2/e^2 =0.