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Random Variables. an important concept in probability. A random variable , X, is a numerical quantity whose value is determined be a random experiment. Examples Two dice are rolled and X is the sum of the two upward faces.

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### Random Variables

A point is selected at random from a square whose sides are of length 1. X is the distance of the point from the lower left hand corner.

an important concept in probability

A random variable , X, is a numerical quantity whose value is determined be a random experiment

Examples

- Two dice are rolled and X is the sum of the two upward faces.
- A coin is tossed n = 3 times and X is the number of times that a head occurs.
- We count the number of earthquakes, X, that occur in the San Francisco region from 2000 A. D, to 2050A. D.
- Today the TSX composite index is 11,050.00, X is the value of the index in thirty days

Examples – R.V.’s - continued

- A point is selected at random from a square whose sides are of length 1. X is the distance of the point from the lower left hand corner.

point

X

- A chord is selected at random from a circle. X is the length of the chord.

chord

X

Definition – The probability function, p(x), of a random variable, X.

For any random variable, X, and any real number, x, we define

where {X = x} = the set of all outcomes (event) with X = x.

Definition – The cumulative distribution function, F(x), of a random variable, X.

For any random variable, X, and any real number, x, we define

where {X≤x} = the set of all outcomes (event) with X ≤x.

- Two dice are rolled and X is the sum of the two upward faces. S , sample space is shown below with the value of X for each outcome

The cumulative distribution function, F(x)

For any random variable, X, and any real number, x, we define

where {X≤x} = the set of all outcomes (event) with X ≤x.

Note {X≤x} =f if x < 2. Thus F(x) = 0.

{X≤x} ={(1,1)} if 2 ≤ x < 3. Thus F(x) = 1/36

{X≤x} ={(1,1) ,(1,2),(1,2)} if 3 ≤ x < 4. Thus F(x) = 3/36

Continuing we find

F(x) is a step function

- A coin is tossed n = 3 times and X is the number of times that a head occurs.

The sample Space S = {HHH (3), HHT (2), HTH (2), THH (2), HTT (1), THT (1), TTH (1), TTT (0)}

for each outcome X is shown in brackets

Examples – R.V.’s - continued

- A point is selected at random from a square whose sides are of length 1. X is the distance of the point from the lower left hand corner.

point

X

- A chord is selected at random from a circle. X is the length of the chord.

chord

X

Examples – R.V.’s - continued

- A point is selected at random from a square whose sides are of length 1. X is the distance of the point from the lower left hand corner.

point

X

S

An event, E, is any subset of the square, S.

P[E] = (area of E)/(Area of S) = area of E

The probability function

Thus p(x) = 0 for all values of x. The probability function for this example is not very informative

The Cumulative distribution function

The probability density function, f(x), of a continuous random variable

Suppose that X is a random variable.

Let f(x) denote a function define for -∞ < x < ∞ with the following properties:

- f(x) ≥ 0

Then f(x) is called the probability density function of X.

The random, X, is called continuous.

Thus if X is a continuous random variable with probability density function, f(x) then the cumulative distribution function of X is given by:

Also because of the fundamental theorem of calculus.

point

X

and

Graph of f(x)

p(x) = P[X = x] = the probability function of X.

This can be defined for any random variable X.

For a continuous random variable

p(x) = 0 for all values of X.

Let SX ={x| p(x) > 0}. This set is countable (i. e. it can be put into a 1-1 correspondence with the integers}

SX ={x| p(x) > 0}= {x1, x2, x3, x4, …}

Thus let

Proof: (thatthe set SX ={x| p(x) > 0} is countable)

(i. e. can be put into a 1-1 correspondence with the integers}

SX = S1 S2 S3 S3 …

where

i. e.

Thus the elements of SX = S1 S2 S3 S3 …

can be arranged {x1, x2, x3, x4, … }

by choosing the first elements to be the elements of S1 ,

the next elements to be the elements of S2 ,

the next elements to be the elements of S3 ,

the next elements to be the elements of S4 ,

etc

This allows us to write

A Discrete Random Variable

A random variable X is called discrete if

That is all the probability is accounted for by values, x, such that p(x) > 0.

Discrete Random Variables

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

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

Graph: Continuous Random Variableprobability density function, f(x)

A Probability distribution is similar to a distribution ofmass.

A Discrete distribution is similar to a pointdistribution ofmass.

Positive amounts of mass are put at discrete points.

p(x4)

p(x2)

p(x1)

p(x3)

x4

x1

x2

x3

A Continuous distribution is similar to a continuousdistribution ofmass.

The total mass of 1 is spread over a continuum. The mass assigned to any point is zero but has a non-zero density

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.

Since {X ≤ - ∞} = f and {X ≤ ∞} = S

then F(- ∞) = 0 and F(∞) = 1.

- F(x) is non-decreasing (i. e. if x1 < x2 then F(x1) ≤F(x2) )

If x1 < x2 then {X ≤ x2} = {X ≤ x1} {x1 < X ≤ x2}

Thus P[X ≤ x2] = P[X ≤ x1] + P[x1 < X ≤ x2]

or F(x2) = F(x1) + P[x1 < X ≤ x2]

Since P[x1 < X ≤ x2] ≥ 0 then F(x2) ≥F(x1).

- F(b) – F(a) = P[a < X ≤ b].

If a < bthen using the argument above

F(b) = F(a) + P[a < X ≤ b]

Thus F(b) – F(a) = P[a < X ≤ b].

- 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.

A function F is continuous if

One can show that

Thus p(x) = 0 implies that

For Continuous Random Variables Variables

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

f(x) slope

F(x)

x

- 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

The probability distribution with probability function

is called the Bernoulli distribution

p

q = 1- p

- Success (S)
- Failure (F)

Suppose that we have a experiment that has two outcomes (A Bernoulli trial)

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

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

Now assume that the Bernoulli trial is repeated independently n times.

Let

Note: the possible valuesof X are {0, 1, 2, …, n}

For n = 5 the outcomes together with the values of X and the probabilities of each outcome are given in the table below:

For n = 5 the following table gives the different possible values of X, x, and p(x) = P[X = x]

For general n, the outcome of the sequence of n Bernoulli trails is a sequence of S’s and F’s of length n.

SSFSFFSFFF…FSSSFFSFSFFS

- The value of X for such a sequence is k = the number of S’s in the sequence.
- The probability of such a sequence is pkqn – k( a p for each S and a q for each F)
- There are such sequences containing exactly k S’s
- is the number of ways of selecting the k positions for the S’s. (the remaining n – k positions are for the F’s

These are the terms in the expansion of (p + q)n using the Binomial Theorem

For this reason the probability function

is called the probability function for the Binomial distribution

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)

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

p(x)

x

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

p(x)

x

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