Discrete Random Variables and Distributions The Bernoulli ...ms.mcmaster.ca/courses/20092010/term1/stats2d03/Lecture Notes- 05.… · Discrete Random Variables and Distributions ⇒

Discrete Random Variables and Distributions ⇒ The Bernoulli & Binomial Random Variables

♦ The Bernoulli & Binomial Random Variables

The Bernoulli Random Variable

Suppose the outcome of a random experiment can be classifiedas either a success or a failure.

The sample space in such an experiment is

S = {success , failure }

We then define a discrete random variable X as follows,

X (success) = 1

X (failure) = 0

The support of a Bernoulli random variable is

X = {0,1}.

Qihao Xie Introduction to Probability and Basic Statistical Inference



Suppose further the probability of success outcome is p for somep ∈ (0,1).

Then, X is is said to be a Bernoulli random variable with theprobability mass function

p(1) = P{X = 1}= p;

p(0) = P{X = 0}= 1−p.

Alternative, the probability mass function of a Bernoulli randomvariable can be written as

p(x) = P{X = x}= px (1−p)1−x , x = 0,1.




Proposition 3.2If X is a Bernoulli random variable, then the r th moment of X

E(X r ) = p, r = 1,2, . . . .

Proof:

Corollary 3.3The mean and variance of a Bernoulli random variable X are

E(X ) = p and Var(X ) = p(1−p).

Proof:




The Binomial Random Variable

Suppose we perform the same Bernoulli trial n independenttimes,

Each of which results in a success with probability p, and a failurewith probability 1−p.

Let X denote the number of successes that occur in the n trials,then X is said to be a Binomial random variable with parameter(n,p), and is denoted by

X ∼ Bin(n,p).




The support of a Binomial random variable is

X = {0,1,2, . . . ,n}.

The probability mass function of a Binomial random variable is

p(x) = P{X = x}=(

nx

)px (1−p)1−x , x = 0,12, . . . ,n.

Note 3.7: A Bernoulli random variable is a Binomial randomvariable with parameter (1,p).




Proposition 3.3If X is a Binomial random variable, then the r th moment of X

E(X r ) = npE[(Y +1)r−1]

, r = 1,2, . . . ,

where Y ∼ Bin(n−1,p).Proof:

Corollary 3.4The mean and variance of a Binomial random variable X are

E(X ) = np and Var(X ) = np(1−p).

Proof:




Proposition 3.4If X ∼ Bin(n,p), p ∈ (0,1), then as x goes from 0 to n, P{X = x} first increases

monotonically and then decreases monotonically, reaching its largest value when

x is the largest integer less than or equal to (n +1)p.

Proof:




Proposition 3.5If X ∼ Bin(n,p), p ∈ (0,1), then the cumulative distribution function of X is givenby

F (x) = P{X ≤ x}=x

∑k=0

(nk

)pk (1−p)n−k .

Corollary 3.5If X ∼ Bin(n,p), p ∈ (0,1), then

P{X = x +1}=p

1−p· n−x

x +1·P{X = x}, x = 0,1,2, . . . ,n−1.

Proof:




Example 3.10Find the cumulative distribution function of X ∼ Bin(3, 1

4 ).Solution:


Discrete Random Variables and Distributions ⇒ The Poisson Random Variable

♦ The Poisson Random Variable

DefinitionA random variable X is said to be a Poisson random variable withparameter λ if and only if the probability mass function of X is

p(x) = P{X = x}= e−λ λ x

x !, x = 0,1,2,3, . . . .

The support of a Poisson random variable is

X = {0,1,2,3, . . .}.

We denote a Poisson random variable with parameter λ asX ∼ Poi(λ ).

Note 3.8:∞

∑x=0

e−λ · λ x

x != 1.




Note 3.9: The Poisson random variable may be used toapproximate Binomial random variable when n is large and p issmall enough so that λ = np is of moderate size.




Example 3.11Suppose the probability of an individual car having an accident in a busy traffic

intersection is 0.0001. Between 5:00pm and 7:00pm, 1000 cards pass through the

intersection. What is the probability that there will be more than 1 accident in this

time period?

Solution:




Corollary 3.6The mean and variance of a Poisson random variable X are

E(X ) = λ = Var(X ).

Proof:




Proposition 3.6If X ∼ Poi(λ ), λ > 0, then the cumulative distribution function of X is given by

F (x) = P{X ≤ x}=x

∑k=0

e−λ · λ k

k !, x = 0,1,2, . . . .

Corollary 3.7If X ∼ Poi(λ ), λ > 0, then

P{X = x +1}=λ

x +1·P{X = x}, x = 0,1,2, . . . .

Proof:


Discrete Random Variables and Distributions ⇒ The Geometric Random Variable

♦ The Geometric Random Variable

DefinitionA random variable X is said to be a Geometric random variable with parameter pif and only if the probability mass function of X is

p(x) = P{X = x}= (1−p)x−1p, x = 1,2,3, . . . .

The support of a Geometric random variable is

X = {1,2,3, . . .}.

We denote a Geometric random variable with parameter p asX ∼ Geo(p).

Note 3.10:∞

∑x=1

(1−p)x−1p = 1.




Note 3.11: Consider n independent trials, each with probabilityp (0 < p < 1) of being a success, which are performed until asuccess occurs. Then, the number of trials X required until asuccess has the Geometric distribution with parameter p.

Example 3.12Toss a fair coin with a head is obtained. What is the probability ofthe 1st head appears on the 3rd toss?Solution:




Corollary 3.8The mean and variance of a Geometric random variable X with successprobability p are given by

E(X ) =1p

and Var(X ) =1−p

p2 ,

respectively.

Proof:




Proposition 3.7If X ∼ Geo(p), 0 < p < 1, then the cumulative distribution function of X is given by

F (x) = P{X ≤ x}=x

∑k=1

(1−p)k−1p, x = 1,2,3, . . . .


Discrete Random Variables and Distributions ⇒ The Negative Binomial Random Variable

♦ The Negative Binomial Random Variable

DefinitionA random variable X is said to be a Negative Binomial randomvariable with parameters r and p if and only if the probabilitymass function of X is

p(x) = P{X = x}=(

x−1r −1

)pr (1−p)x−r , x = r , r +1, r +2, . . . .

The support of a Negative Binomial random variable is

X = {r , r +1, r +2, r +3, . . .}.

We denote a Negative Binomial random variable with the numberof successes r observed and the success probability p on eachtrial as X ∼ NB(r ,p).




Note 3.12:∞

∑x=r

(x −1r −1

)pr (1−p)x−r = 1.

Note 3.13: Consider n independent trials, each with probabilityp (0 < p < 1) of being a success, which are performed until a totalof r successes has been accumulated. Then, the number of trialsX required until the r th success being obtained has the NegativeBinomial distribution with parameters n and p.

Note 3.14: A Geometric random variable with parameter p is justa Negative Binomial random variable with parameters r = 1 andp.




Example 3.13Toss a fair coin with a head is obtained. What is the probability of getting the 4st

head appears on the 10th toss?

Solution:





E(X ) =rp

and Var(X ) =r(1−p)

p2 ,

respectively.Proof:




Proposition 3.8If X ∼ NB(r ,p), 0 < p < 1, then the cumulative distribution function of X is givenby

F (x) = P{X ≤ x}=x

∑k=r

(x−1k −1

)pk (1−p)k−1, x = 1,2,3, . . . .


Discrete Random Variables and Distributions ⇒ The Hypergeometric Random Variable

♦ The Hypergeometric Random Variable

DefinitionA random variable X is said to be a Hypergeometric randomvariable with parameters n, N and m if and only if the probabilitymass function of X is

p(x) = P{X = x}=

(mx

)(N−mn−x

)(Nn

) , x = 0,1,2, . . . ,n.

The support of a Hypergeometric random variable is

X = {0,1,2, . . . ,n}.

We denote a Hypergeometric random variable with parametersn, N and m as X ∼ HG(n,N,m).




Note 3.15: The hypergeometric P(X = x) = 0 unless x satisfies

n− (N−m)≤ x ≤min(n,m).

Note 3.15:n

∑x=0

(mx

)(N−mn−x

)(Nn

) = 1.

Note 3.17: Consider a sample of size n is chosen randomly(without replacement) from an urn contained N balls, of which mare white and N−m are black. If X is the number of white ballsselected, then X has the Hypergeometric distribution withparameters n, N and m.




Note 3.18: For X ∼ HG(n,N,m) when m and N are large inrelation to n, the pmf of X is approximately equal to the pmf of aBin(n,p) random variable, where p = m

N .Proof:


E(X ) =nmN

and Var(X ) =N−nN−1

np(1−p),

respectively.

Proof:


Discrete Random Variables and Distributions The Bernoulli ...ms.mcmaster.ca/courses/20092010/term1/stats2d03/Lecture Notes- 05.… · Discrete Random Variables and Distributions ⇒

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