Discrete Random Variables-Part I

8/13/2019 Discrete Random Variables-Part I

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Discrete Random Variables

Distribution, Expectation, Variance


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Random Variable

A random variable is a functionwhose domain is the sample spaceandwhose range is a subset of the set of real numbers.

A random variable is a function that associates a unique numerical valuewith every outcome of an experiment. The value of the random variable willvary from trial to trial as the experiment is repeated. *

Experiment : Toss a coin twice

Sample Space : {HH, HT, TH, TT}

X = number of heads

X(HH) = 2, X(HT) = 1, X(TH) = 1, X(TT) = 0

X : {0,1,2} is a function.

Hence X is a random variable

(*http://www.stats.gla.ac.uk/steps/glossary/probability_distributions.html#randvar)
http://www.stats.gla.ac.uk/steps/glossary/probability_distributions.htmlhttp://www.stats.gla.ac.uk/steps/glossary/probability_distributions.htmlhttp://www.stats.gla.ac.uk/steps/glossary/probability_distributions.htmlhttp://www.stats.gla.ac.uk/steps/glossary/probability_distributions.html


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Experiment : Draw three balls at random from an urncontaining two redand two greenballs.

Sample Space = {RRG, RGG}

X = number of red ballsX(RRG) = 2, X(RGG) = 1

X : {1,2} is a function. Hence X is a random variable.

Experiment : Toss two dice.Sample Space = {(1,1), (1,2),, (1,6),, (6,1),(6,6)}

X = sum of the scores obtained in the two tosses

X((1,1)) = 2, X((1,2)) = 3, , X((6,6)) = 12

X : {2, ,12} is a function.X is a random variable


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Discrete Random Variable

A random variable is said to be discrete if the range of values takenby the random variable is eitherfinite or is countably infinite.

All the examples of random variables given till now are discreterandom variables

An example of a random variable which is not discrete is givenbelow :

Experiment : Choose three students at random from PGP-1

Sample Space : {{a, b, c}: a, b, c are students of PGP-1}

X = total sum of weights of the three students in fitness center

X can take any positive value.

Hence X is not a discrete random variable.


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Distribution of a Discrete Random

Variable

Let the random variable X take the values x1, x2, xn,

P(X = xi) = P({: X() = xi})

Let pi= P(X = xi). Then pi 0and pi= 1 The list of values of X along with their

corresponding probabilities of occurrence iscalled the probability distributionof the random

variable XX : x1 x2 xn

P(X = x) : p1 p2 pn


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Example :

Let X be the number of tails in two tosses of a fair

coin. Then X takes the values 0,1,and 2

P(X =0) = P({HH}) =

P(X=1) = P({HT, TH}) =

P(X=2) = P({TT}) =

Therefore the probability distribution of the random

variable X isX : 0 1 2

P(X=x): 1/4


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Example : Toss a die with one side

marked 1, two sides marked 2 and three

sides marked 3 twice. Let X be the sum ofscores in the two tosses. Now note that X

can take the values 2, 3, 4, 5 and 6. The

probability distribution of X is

X : 2 3 4 5 6

P(X=x): 1/36 1/9 5/18 1/3 1/4


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Three Coins are tossed.

Identify the Sample Space.

Let us define the random variable

X= No of HeadsNumber of Tails

Find the Probability Distribution of X


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Sample

Space

X = No of

(Heads

Tails)

P (X=x)

HHH 3 1/8

HHT 1 1/8

HTH 1 1/8HTT -1 1/8

THH 1 1/8

THT -1 1/8

TTH -1 1/8

TTT -3 1/8


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X = No of (Heads

Tails)

P (X=x)

3 1/8

1 3/8

-1 3/8

-3 1/8


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Expectation of a Discrete Random

Variable

Let X be a discrete random variable with

distribution

X : x1

x2

xn

P(X = x) : p1 p2 pn

Then the Expectation of X denoted as E(X)

is defined asE(X) = xi pi = xP(X=x)

= (value x probability)


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Example : Let X be the number of heads in two tosses of afair coin.

The probability distribution of the random variable X is

X : 0 1 2P(X=x): 1/4

Thus E(X) = 0. + 1. + 2. = 1

Example : In the case when the random variable X hasdistribution

X : 2 3 4 5 6

P(X=x): 1/36 1/9 5/18 1/3 1/4

E(X) = 2. 1/36 + 3.1/9 + 4. 5/18 + 5. 1/3 + 6.

= 14/3


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Expectation as Theoretical Average-

An Intuitive View

Let X be a random variable which takes valuesx1, x2, , xkwith probabilities p1, p2, , pkrespectively.

Suppose the experiment is repeated a largenumber, say n times, and each time the value ofX is recorded. Suppose in these n values,x1occurs f1times, x2occurs f2times and so on.

The average value of X is then (x1 f1+ + xk fk)/n As n, fi/npitherefore

(x1 f1+ + xk fk)/nx1 p1+ + xk pk = E(X)


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Properties of Expectation

Let X be a random variable and a, b are

real numbers then

E[a X + b] = a E[X] + b

If P(X 0) = 1 then E(X) 0


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Properties of Expectation

Suppose g(X) is a real valued function of X

Then we define the expectation of functionof random variable as

E[g(X)] = g(xi)P(X = xi) = g(xi) pi

Let g (X) = X2

Thus E(X2) = g(xi)P(X = xi) = xi2 pi


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VARIANCE

If a random variable X has the following probabilitydistribution (t > 0)

X : -t t

P(X=x) :

Then E(X) = 0 for any t but as t increases then the spreadof the distribution increases.

Variance is a measure of spread of a probabilitydistribution.

Variance of a random variable X is defined as

Var(X) = E(XE(X))2= E(X2)(E(X))2 (Proof - exercise)

For the above random variable Var(X) = t2. So as tincreases Var(X) increases.


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STANDARD DEVIATION (S.D.)

The positive square root of Var(X) is called

the standard deviation of X or s.d.(X).

The main advantage of using standard

deviation is that the standard deviation is

measured in the same unit as the original

data whereas the variance is measured in

squared units.


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Properties of Variance and

Standard Deviation

Let X be a random variable and a, b are

real numbers

Var(a X + b) = a2Var(X)

s.d.(a X + b) = |a| s.d.(X)

Var(X) = 0 if and only is there is a real

number t such that P(X = t) = 1.

Since Var(X) 0, E(X2) (E(X))2


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Markovs Inequality

If X is a non-negative random variableand a is a real number > 0 then

P(X a) E(X) / a

Thus P(X > 2E(X))

P(X > 4E(X))

P(X > 100 E(X)) .01


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Chebyshevs Inequality

If X is a random variable with E(X) = andstandard deviation and if a is a real number > 0then

P(|X-| k) 2 / k2

Put X = (X-)2 and a =k2 in Markovs inequalityand you get Chebyshevs Inequality

Thus P(|X-| > 2) (k = 2)

P(|X-| > 3) 1/9 (k = 3)

P(|X-| > 10) .01 (k = 10)


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Example of Markov and

Chebyshevs Inequality Let us suppose that production of a factory is random

variable with mean 50

a) what is the probability that the weeks production willexceed 75

b) what is the probability that the weeks production willexceed 100

c) If the variance is 25, what can be said about theprobability that production will be between 40 and 60

P(X 75) E(X) / 75 = 50/75 = 2/3 P(X 100) E(X) / 75 = 50/100 =

P(|X-| k) 2 / k2


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P(|X-| k) 2 / k2

Given = 50 and 2= 100

P(|X-50| 10) 25/ 100 =

P(|X-50| < 10) 1- =

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Discrete Random Variables-Part I