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Binomial distribution A binomial experiment has four properties: 1) it consists of a sequence of n identical trials; 2) two outcomes, success or failure, are possible on each trial; 3) the probability of success on any trial, denoted p, does not change from trial to trial; 4) the trials are independent. Let us conduct a binomial experiment with n trials, a probability p of success and let X be a random variable giving the number of successes in the binomial experiment. Then X is a discrete random variable with the range {0, 1, 2, ..., n} and the probability function p(i ) = Cni p i (1 − p ) n −i We say that X has a binomial distribution. Binomial distribution with n=10, p=0.5 0,3 0,25 0,2 0,15 p(x) 0,1 0,05 0 0 2 4 6 8 10 12 Binomial distribution with n=10, p=0.25 0,3 0,25 0,2 0,15 p(x) 0,1 0,05 0 0 2 4 6 8 10 12 Expectancy and variance of a binomial distribution For a random variable X with a binomial distribution Bi(n,p) we have E ( X ) = pn D( X ) = np(1 − p ) Using the identity iCni = i n! n (n − 1)! = = nCni −−11 i !(n − i )! (i − 1)!(n − i )! we can calculate n E ( X ) = ∑ iC p q i n i =0 i n −i n =∑ iC p q i =1 i n i n −i n n −1 i =1 i =0 n =∑ nCni −−11 p i q n −i = i =1 = p ∑ nCni −−11 p i −1q n −i = pn∑ Cni −1 p i q n −i = pn( p + q ) The same identity can be used to calculate D(X) n −1 = pn E ( X 2 ) = ∑ i 2Cni p i q n −i =np ∑ iCni −−11 p i −1q n −i = n n i =0 i =1 n i −1 i −1 n −i n = np ∑ Cn −1 p q + ∑ (i − 1)Cni −−11 p i −1q n −i = i =1 i =1 n n −1 i i n −i i − 2 i − 2 n −i = np ∑ Cn −1 p q + (n − 1) p ∑ Cn − 2 p q = i =0 i =2 n−2 n −1 = np ( p + q ) + (n − 1) p ∑ Cni − 2 p i q n −i − 2 = i =0 { = np 1 + (n − 1) p( p + q ) n−2 }= np{1 + (n − 1) p} = = np(1 − p + np ) = npq + (np ) 2 D( X ) = E (X 2 ) − [E ( X )] ⇒ D( X ) = npq 2 Poisson distribution Poisson distribution is defined for a discrete random variable X with a range {0, 1, 2, 3, ... } denoting the number of occurrences of an event A during a time interval (T1 ,T2 ) if the following conditions are fulfilled: 1) given any two occurrences A1 and A2, we have P( A2 | A1 ) = P( A2 ) 2) the probability of A occurring during an interval (t1 ,t2 ) with T1 ≤ t1 < t 2 ≤ T2 equals to c(t2 − t1 ) where c is fixed 3) denoting by P12 the probability that E occurs at least twice between t1 and t2, we have P12 lim =0 t1 →t 2 t − t 2 1 The Poisson distribution Po(λ) is given by the probability function p( x ) = e ∞ ∑e x =0 x λ −λ x λ −λ x! 2 3 λ λ λ = e − λ 1 + + + + L = e − λ e λ = 1 x! 1! 2! 3! Poisson law with E(X) = 5 0,2 0,15 0,1 p(x) 0,05 0 0 5 10 15 The expectancy and variance of a random variable X with Poisson distribution are given by the following formulas E( X ) = λ D( X ) = λ ∞ E( X ) = ∑ e −λ λx x! x =0 ∞ E (X 2 ) = ∑ e x λ −λ x! x =0 ∞ = e −λ λ ∑ x =0 ∞ x =∑ e −λ x =1 ∞ x 2 =∑ e x =1 λx x! x λ −λ x! λx −1 ∞ x =e λ ∑ −λ x =1 ( x − 1)! ∞ x 2 =e −λ λ ∑ x =1 ∞ λx x =0 x! == e λ ∑ λx −1 ( x − 1)! −λ =λ x= λx x x ∞ ∞ λ λ ( x + 1) = e −λ λ ∑ x + ∑ = e −λ λ (eλ λ + e λ ) = x! x = 0 x! x =0 x! = λ2 − λ D( X ) = E (X 2 ) − [E ( X )] = λ2 + λ − λ2 = λ 2 Relationship between Bi(n,p) and Po(λ) For large n, we can write Bi (n, p ) ≈ Po (np ) Comparing Bi(20,0.25) with Po(5) 0,25 0,2 0,15 0,1 0,05 Poisson Binomial 18 15 12 9 6 3 0 0 Normal distribution law A continuous random variable X has a normal distribution law with E(X) = µ and D(X) = σ if its probability density is given by the formula 1 f (x) = e 2π σ − ( x − µ )2 2σ 2 Standardized normal distribution A continuous random variable X has a standardized normal distribution if it has a normal distribution with E(X) = 0 and D(X) = 1 so that its probability density is given by the formula 1 f (x) = e 2π − x2 2 Standardized normal distribution law (Gaussian curve) There are several different ways of obtaining a random variable X with a normal distribution law. For large n, binomial and Poisson distributions asymptotically approximate normal distribution law If an activity is undertaken with the aim to achieve a value µ, but a large number of independent factors are influencing the performance of the activity, the random variable describing the outcome of the activity will have a normal distribution with the expectancy µ. Of all the random variables with a given expectancy µ and variance σ2, a random variable with the normal distribution N(µ,σ2) has the largest entropy.

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