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Random Variables. Discrete Random Variables. For a discrete random variable X the probability distribution is described by the probability function p ( x ) = P [ X = x ], which has the following properties:. Continuous random variables.

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discrete random variables
Discrete Random Variables

For a discrete random variable X the probability distribution is described by the probability function p(x) = P[X = x], which has the following properties:

continuous random variables
Continuousrandom variables

For a continuous random variable X the probability distribution is described by the probability density function f(x), which has the following properties :

  • f(x) ≥ 0
the distribution function f x
The distribution function F(x)

This is defined for any random variable, X.

F(x) = P[X ≤ x]

Properties

  • F(-∞) = 0 and F(∞) = 1.
  • F(x) is non-decreasing (i. e. if x1 < x2 then F(x1) ≤F(x2) )
  • F(b) – F(a) = P[a < X ≤ b].
slide5

p(x) = P[X = x] =F(x) – F(x-)

Here

  • If p(x) = 0 for all x (i.e. X is continuous) then F(x) is continuous.
slide6

For Discrete Random Variables

F(x) is a non-decreasing step function with

F(x)

p(x)

slide7

f(x) slope

F(x)

x

  • For Continuous Random Variables Variables

F(x) is a non-decreasing continuous function with

To find the probability density function, f(x), one first finds F(x) then

slide10

Success (S)

  • Failure (F)

Suppose that we have a experiment that has two outcomes

These terms are used in reliability testing.

Suppose that p is the probability of success (S) and

q = 1 – p is the probability of failure (F)

This experiment is sometimes called a Bernoulli Trial

Let

Then

slide11
The probability distribution with probability function

is called the Bernoulli distribution

p

q = 1- p

slide13

We observe a Bernoulli trial (S,F)n times.

Let X denote the number of successes in the n trials.

Then X has a binomial distribution, i. e.

where

  • p = the probability of success (S), and
  • q = 1 – p = the probability of failure (F)
slide14

Example

A coin is tossed n= 7 times.

Let X denote the number of heads (H) in the n = 7 trials.

Then X has a binomial distribution, with p = ½ and n = 7.

Thus

slide16

Example

If a surgeon performs “eye surgery” the chance of “success” is 85%. Suppose that the surgery is perfomed n = 20 times

Let X denote the number of successful surgeries in the n = 20 trials.

Then X has a binomial distribution, with p = 0.85 and n = 20.

Thus

slide18
The probability that at least sixteen operations are successful

= P[X ≥ 16]

= p(16) + p(17) + p(18) + p(19) + p(20)

= 0.1821 + 0.2428 + 0.2293 + 0.1368 + 0.0388

= 0.8298

other discrete distributions

Other discrete distributions

Poisson distribution

Geometric distribution

Negative Binomial distribution

Hypergeometric distribution

the poisson distribution
The Poisson distribution
  • Suppose events are occurring randomly and uniformly in time.
  • Let X be the number of events occuring in a fixed period of time. Then X will have a Poisson distribution with parameter l.
slide22

is the probability function for the Binomial distribution with parameters n and p, and we allow n→ ∞ and p →0 such that np = a constant (= lsay) then

  • If
slide23

Suppose

Proof:

slide24

Now

Now using the classic limit

graphical illustration
Graphical Illustration

Suppose a time interval is divided into n equal parts and that one event may or may not occur in each subinterval.

n subintervals

time interval

- Event occurs

X = # of events is Bin(n,p)

- Event does not occur

As n→∞, events can occur over the continuous time interval.

X = # of events is Poisson(l)

example
Example

The number of Hurricanes over a period of a year in the Caribbean is known to have a Poisson distribution with l = 13.1

Determine the probability function of X.

Compute the probability that X is at most 8.

Compute the probability that X is at least 10.

Given that at least 10 hurricanes occur, what is the probability that X is at most 15?

solution
Solution
  • X will have a Poisson distribution with parameter l = 13.1, i.e.
the geometric distribution
The Geometric distribution

Suppose a Bernoulli trial (S,F) is repeated until a success occurs.

Let X = the trial on which the first success (S) occurs.

Find the probability distribution of X.

Note: the possible values of X are {1, 2, 3, 4, 5, … }

The sample space for the experiment (repeating a Bernoulli trial until a success occurs is:

S = {S, FS, FFS, FFFS, FFFFS, … , FFF…FFFS, …}

(x – 1) F’s

p(x) =P[X = x] = P[{FFF…FFFS}] = (1 – p)x – 1p

slide31

P[X = x] = p(x) = p(1 – p)x – 1 = pqx – 1

Thus the probability function of X is:

A random variable X that has this distribution is said to have the Geometric distribution.

Reason p(1) = p, p(2) = pq, p(3) = pq2 , p(4) = pq3 , …

forms a geometric series

the negative binomial distribution
The Negative Binomial distribution

Suppose a Bernoulli trial (S,F) is repeated until k successes occur.

Let X = the trial on which the kth success (S) occurs.

Find the probability distribution of X.

Note: the possible values of X are

{k, k + 1, k + 2, k + 3, 4, 5, … }

The sample space for the experiment (repeating a Bernoulli trial until k successes occurs) consists of sequences of S’s and F’s having the following properties:

  • each sequence will contain k S’s
  • The last outcome in the sequence will be an S.
slide33

A sequence of length x containing exactly k S’s

SFSFSFFFFS FFFSF … FFFFFFS

The last outcome is an S

The # of S’s in the first x – 1 trials is k – 1.

The # of ways of choosing from the first x – 1 trials, the positions for the first k – 1 S’s.

The probability of a sequence containing k S’s and x – k F’s.

the hypergeometric distribution
The Hypergeometric distribution

Suppose we have a population containing N objects.

Suppose the elements of the population are partitioned into two groups. Let a = the number of elements in group A and let b = the number of elements in the other group (group B). Note N = a+ b.

Now suppose that n elements are selected from the population at random. Let X denote the elements from group A. (n – X will be the number of elements from group B.)

Find the probability distribution of X.\

slide35

Population

GroupB(b elements)

Group A (a elements)

N - x

x

sample(n elements)

slide36

The number of ways x elements can be chosen Group A .

The number of ways n - x elements can be chosen Group B .

Thus the probability function of X is:

The total number of ways n elements can be chosen from N = a + b elements

A random variable X that has this distribution is said to have the Hypergeometric distribution.

The possible values of X are integer values that range from max(0,n – b) to min(n,a)