Lektsii (1) / Lecture 10

ICEF, 2012/2013 STATISTICS 1 year LECTURES

BERNOULLI SCHEME. BINOMIAL RANDOM VARIABLES

Suppose there is a trial with two random outcomes, that we will call success or a failure. Examples: 1) a coin is tossed one time, success = head, failure = tail; 2) a detail is selected from a big lot, success = selected detail is not defective, failure = selected detail is defective; 3) a chip is tested, a chip works without breakdowns at least one month, failure = less than one month, etc. Now let’s assume that we independently make n such trials and the probability of success in each trial is π, 0 ≤π ≤1 independently on the number of a trial. This series of trials is called

Bernoulli scheme.

Definition. The total number of successes in Bernoulli scheme with n trials and probability of success π is called the binomial random variable with parameters (n,π) .

Let’s denote by 1 the appearance of success and by 0 the appearance of failure. Then the result of any Bernoulli scheme with parameters (n,π) can be represented as a sequence

s = (ε1,ε2 ,...,εn ) ,

It is clear that the total number of successes X is equal to ∑n εi .

i=1

Let’s assign to each s = (ε1,ε2 ,...,εn ) the probability Pr(s) = pk qn−k

EXPECTATION AND VARIANCE OF A DISCRETE RANDOM VARIABLE Let X be some discrete random variable,

Definition. Number E(X ) = µX = ∑xi pi is called the expectation or mean value

i

(математическое ожидание) of a random variable X.

Properties of expectation

1. If X ≥0 then E(X ) ≥0 (monotonicity).

2.If X, Y are two random variables and a, b are constants then E(aX +bY ) = aE(X ) +bE(Y ) (linearity of expectation)

Definition. Number V (X ) =σX2 = ∑(xi −µX )2 pi the variance (дисперсия) of a random

i

variable X. Number V (X ) =σX is called the standard deviation (стандартное отклонение) of a random variable X.

Properties of variance

1.V (X ) ≥ 0 for any random variable X, and V (X ) = 0 if and only if random variable is constant (i.e. is not random).

2.If a and b are two constants then V (aX +b) = a2V (X )

3.V (X ) = E(X 2 ) −[E(X )]2 = E(X 2 ) −µX2 .

Let X be a binomial random variable with parameters (n,π) (recall that n is the number of trials and π is a probability of success in each trial. According to the definition

Using a routine arithmetic one can prove that

E(X ) = nπ, V (X ) = nπ(1−π)

Example. The proportion of defective chips is 60%. Ten chips are selected at random from the large lot.

10

1) Probability that exactly half of chips is defective: P(X =5) = 0.650.45 = 0.20 .

5

2)The expectation of defective chips is E(X ) =10*0.6 = 6 .

3)The variance: V (X ) =σ2 =10*0.6*0.4 = 2.4, σ =1.55 .