TOPIC 2TOPIC 2

Random Variables Random Variables

Random VariablesRandom Variables

Random Variables

Discrete Random Variables

Continuous Random Variables

Discrete Random Variables Discrete Random Variables

• Random Variable :

A numerical outcome of an experiment

Example: the sum of two fair dice, count number of tails from tossing 2 coins

• Discrete Random Variable :

Whole number (0, 1, 2, 3, etc.)

Obtained by counting

Usually a finite number of values :

Poisson random variable is exception (∞)

Probability Mass Function (PMF)Probability Mass Function (PMF)

• A set of probability values

• List of all possible [x, P (x)] pairs

x = value of random variable X (outcome)

P (x) = probability associated with value

• Mutually exclusive (no overlap)

• Collectively exhaustive (nothing left out)

• 0 ≤ P (x) ≤ 1 for all x

•

• Also referred as discrete random probability distribution 1xP

PMF ExamplePMF Example

Experiment: Toss 2 coins. Count number of tails (as a random variable of X).

Discrete Probability Distribution

Random Var, x Probabilities, P(x)

0 1/4 = 0.25

1 2/4 = 0.50

2 1/4 = 0.25

All Possible Occurence

Visualization of PMF ExampleVisualization of PMF Example

Listing Table

Formula

P (x)n

x!(n – x)!!

= px(1 – p)n - x

Graph

.00

.25

.50

0 1 2x

P(x)

{ (0, .25), (1, .50), (2, .25) }

Note: p = trial probability = ½

n = sample size = 2

# Tails f (x)Frequency

P (x)

0 1 .252 .501 .25

(x)

1

2

Cumulative Distribution FunctionsCumulative Distribution Functions

• An alternative way of specifying the probabilistic properties of a random variable X is through the function

• This function is known as the cumulative distribution function of a discrete random variable

xXPxF

ExampleExample

• Suppose X has a probability mass function given by

• then the cumulative distribution function F of X is given by

6

13,

3

12,

2

11 XPXPXP

6

5

3

1

2

12122 PPXPF

2

1111 PXPF

16

1

3

1

2

132133 PPPXPF

Summary MeasuresSummary Measures

• Expected value or expectation of a discrete random variable (Mean of probability distribution) :

average value of the random variables (all possible values)

• Variance : Measures the spread or variability in the values taken by

the random variable

• Standard Deviation The positive square root of the variance

i

ii xPxXE

iii

iii xPxxPXEx

XEXE

XEXEXVar

22

22

22

2

ExampleExample

Experiment : You toss 2 coins. You’re interested in the number of tails. What are the expected value, variance, and standard deviation of this random variable, number of tails?

0 .25 -1.00 1.00

1 .50 0 0

2 .25 1.00 1.00

0

.50

.50

= 1.0

x P(x) x P(x) x – (x – ) 2 (x – ) 2 P(x)

.25

0

.25

2 = .50

= .71

ExercisesExercises

1) An office has four copying machines, and the random variable x measures how many of them are in use at a particular moment in time. Suppose that P(X=0) = 0.08, P(X=1) = 0.11, P(X=2) = 0.27, and P(X=3) = 0.33.

a) What is P(X=4) ?

b) What is P(X≤2) ? (Cumulative probability distribution)

2) Four cards are labeled $1, $2, $3, and $6. A player pays $4 selects two cards at random, and then receives the sum of the winnings indicated on the two cards. Calculate the probability mass function and the cumulative distribution functions of the net winnings.

3) A consultant has six appointment times that are open, three on Monday and three on Tuesday. Suppose that when making an appointment a client randomly chooses one of its remaining open times, with each of those open times equally likely to be chosen. Let the random variable X be the total number of appointment that have already been made over both days at the moment when Monday’s schedule has just been completely filled

a) What is the state space of the random variable of X

b) Calculate the probability mass function of X

c) What is the expected value and standard deviation of the total number of appointments that have already been made over both days at the moment when Monday’s schedule has just been completely filled?

Answers to ExercisesAnswers to Exercises

1) a)

b)

2)

Calculation:

21.033.027.011.008.014

143210

1

XP

XPXPXPXPXP

xPi

i

46.027.011.008.0

2102

XPXPXPXP

6156,$3$

6146,$2$

6136,$1$

6113,$2$

6103,$1$

6112,$1$

XP

XP

XP

XP

XP

XP

155

6544

6433

6311

6200

6111

XPF

XPF

XPF

XPF

XPF

XPF

.,0431

,1421

etcX

X

612

34

!24!2

!442

C

Answers to ExercisesAnswers to Exercises

3) a)

b) Probability Mass Function Cumulative Distribution Function

c)

20123

456

!36!3

!663

C

50.020

10

20

10206

30.020

6

20

4105

15.020

3

20

144

05.020

13

63

53

63

63

43

53

63

33

43

63

33

C

CCXP

C

CCXP

C

CCXP

C

CXP

150.030.015.005.06

50.030.015.005.05

20.015.005.04

05.03

XP

XP

XP

XP

8874.07875.0

7875.05.025.563.025.55

15.025.5405.025.53

25.55.063.0515.0405.03

2

22

2222

iii

iii

xPxXVar

xPxXE

Continuous Random Variables Continuous Random Variables

• Continuous Random Variable

A numerical outcome of an experiment

Whole or fractional number

Obtained by measuring

Weight of a student (e.g., 115, 156.8, etc.)

Infinite number of values in interval

Too many to list like a discrete random variable

ExampleExample

Measure Time

Between Arrivals

Inter-Arrival

Time

0, 1.3, 2.78, ...

Experiment RandomVariable

PossibleValues

Weigh 100 People Weight 45.1, 78, ...

Measure Part Life Hours 900, 875.9, ...

Amount spent on food $ amount 54.12, 42, ...

• Defines the probabilistic properties of a continuous random variable

• Shows all values, x, and frequencies, f (x)

f (x) is a Probability Density Function (Not Probability Random Variable)

• Properties

Probability Density FunctionsProbability Density Functions

Value

(Value, Frequency)

Frequency

f(x)

a bx(Area Under

Curve)f x dx( )

All x 1

f x( ) a x b 0,

Continuous Random Variable Probability Continuous Random Variable Probability

Probability is Area Under Curve!

P a x b f x dxa

b

( ) ( )

f(x)

xa b

xall

dxxfxallP 1

Continuous Random Variable Probability Continuous Random Variable Probability

f(x)

xa

a

a

dxxfaXP 0 • This is in contrast to discrete random variables, which can have non zero probabilities of taking specific values.

• Continuous random variables can have nonzero probabilities of falling within certain continuous region (e.g. a ≤ x ≤ b)

Cumulative Distribution FunctionCumulative Distribution Function

• The cumulative distribution function of a continuous random variable X is defined as

• The cumulative distribution function F(x) is a continuous non-decreasing function that takes the value 0 prior to and at the beginning of the state space and increases to a value of 1 at the end

x

dyyfxXPxFendpointlower

Summary MeasuresSummary Measures

• Expected Value or expectation (Mean of random variable) :

Weighted average of all possible values

If the probability density function f(x) is symmetric then the expectation of the random variable x is equal to the point of symmetry

• Variance : Weighted average of squared deviation about mean

• Standard Deviation :

• Median :

xall

dxxfxXE

xall

dxxfxxE 222

2

5.0xF

Variances/Standard DeviationsVariances/Standard Deviations

• Variance shows the spread or variability in the values taken by the random variable

• Standard deviation is often used in place of the variance to describe the spread of the distribution

ExampleExample

Suppose that the diameter of a metal cylinder has a probability density function

f(x) = 1.5 – 6(x – 50)2

for 49.5 ≤ x ≤ 50.5• Is this a valid probability density function?• Is the probability density function symmetric? What is the

point of symmetry?• What is the probability that the metal cylinder has a

diameter between 49.8 and 50.1 mm?• What is the cumulative distribution function of the metal

cylinder diameter?• What is the expected diameter of the metal cylinder?• What is the variance and standard deviation of the metal

cylinder diameters?

Answer to the ExampleAnswer to the Example

• Is this a valid probability density function? Yes.

• What is the probability that the metal cylinder has a diameter between 49.8 and 50.1 mm?

• What is the cumulative distribution function of the metal cylinder diameter?

15.745.75505.4925.495.1505.5025.505.1

5025.15065.1

1

33

5.50

5.493

5.50

5.49

2

xxdxx

dxxfxall

432.0716.74148.75

5025.15065.11.508.491.50

8.493

1.50

8.49

2

xxdxxxP

5.745025.1

5025.15065.1

3

8.493

5.49

2

xx

yydyyxXPxFx

x

• What is the expected diameter of the metal cylinder?

• Is the probability density function symmetric? What is the point of symmetry? Yes. μ = 50 is the point of symmetry

• What is the variance and standard deviation of the metal cylinder diameters?

5065625.183765625.19125.745.495.755.50

505.075.05025.15065.15.50

5.49425.50

5.493

5.50

5.49

2

xxxxxdxxx

dxxfxXExall

22 65.165.1 xxxfxf

224.005.005.0025.0025.0

502.1505.05065.1505.50

5.4953

5.50

5.49

222

22

and

xxdxxx

dxxfxXVarxall

Answer to the ExampleAnswer to the Example

Answer to the ExampleAnswer to the Example

• Graphs of the example

49.5 50.550

μ = E(X)

f (x)

f (x) = 1.5 – 6(x – 50)2

Probability density function

49.5 50.5

F(x)

F (x) = 1.5x – 2(x – 50)3 – 74.5

0

1

Cumulative distribution function

σ = 0.224

0.5

50Median

Mean

Chebyshev’s InequalityChebyshev’s Inequality

• If a random variable has a mean µ and a variance σ2, then

11

12

cforc

cXcP

• For example, taking c = 2 and 3 gives

%8989.03

1133

%7575.02

1122

2

2

XP

XP

σσ σσσ σ

E (x) = µ

27

• Alternative ways of describing spread of data include determining the location of values that divide a set of observations into equal parts.

• The pth quantile or 100pth percentile of a random variable X with a cumulative distribution function F(x) is defined to be the value of x for which

p = 0.25 is called 25th percentile or lower quartile (Q1) p = 0.50 is called 50th percentile or median (Q2) p = 0.75 is called 75th percentile or upper quartile (Q3)

• Interquartile Range (IQR) = Q3 – Q1

Quantiles of Random Variables Quantiles of Random Variables

pxF

28

Quantiles of Random Variables Quantiles of Random Variables

Upper Quartile

MedianLower Quartile

Area = 0.25f(x)

Interquartile Range

ExampleExample

A random variable X has a probability density function

a) What is the value of A?

b) What is the median of X?

c) What is the lower quartile of X?

d) What is the upper quartile of X?

e) What is the interquartile range?

a)

43 xforx

Axf

866.1134212

11

4

3

4

3

AAxA

dxx

Adxxf

xall

Answer to the ExampleAnswer to the Example

48.35.03732.35.0732.3

5.0866.1

5.0

3

3

xxy

dyy

yF

x

x

b)

c)

d)

e)

24.325.03732.325.0732.3

25.0866.1

25.0

3

3

xxy

dyy

yF

x

x

74.375.03732.375.0732.3

75.0866.1

75.0

3

3

xxy

dyy

yF

x

x

5.024.374.3 IQR

Jointly Discrete Random VariablesJointly Discrete Random Variables

• Joint Probability Distribution

Y random values

Y1 Y2 Yn

XrandomValues

X1 p1,1 p1,2 p1,n

X2 p2,1 p2,2 p2,n

Xm pm,1 Pm,2 pm,n

n

j

m

iji YXP

1 1

1,

n

j

m

iijp

1 1

1or

• Joint Cumulative Distribution Function

y

j

x

iijpyYxXPyxF ,,

Marginal Probability DistributionMarginal Probability Distribution

Y random values

Y1 Y2 Yn

XrandomValues

X1 p1,1 p1,2 p1,n

X2 p2,1 p2,2 p2,n

Xm pm,1 Pm,2 pm,n

n

jjpxP

111

n

jjpxP

122

n

jmjm pxP

1

m

iipyP

111

m

iipyP

122

m

iinn pyP

1

Marginal distribution of x

Marginal distribution of y

Marginal Probability DistributionMarginal Probability Distribution

j

n

jji

m

ii yPYEyYxPXExX

1

22

1

22

YYXX 22

Variance:

Standard Deviation:

n

jjj

m

iii yPyYExPxXE

11

Expectation (Mean):

n

j

m

iijji pyxXYE

1 1

Conditional Probability DistributionConditional Probability Distribution

m

iij

ij

j

jijiyYi

p

p

yYP

yYxXPyYxXPp

j

,||

If two discrete random variables X and Y are jointly distributed, then the conditional distribution of random variable X conditional on the event Y = yj consists the probability values

What is this next equation about?

n

jij

ij

i

jiijxXj

p

p

xXP

yYxXPxXyYPp

i

,||

Covariance and CorrelationCovariance and Correlation

Covariance:

YX

YX

YX

YXYX

.Cov

)()(

,Cov,Corr

22

The correlation takes values between -1 and 1, and the discrete random variables x and y are• independent if Corr (X,Y) = 0 [or Cov(X,Y) = 0]• strongly dependent if Corr (X,Y) = -1 or 1 (negatively or positively)

To indicate the strength of the dependence of two random variables

YEXEXYE

YEYEXEXEYX

,Cov

In practice, the most convenient way to asses the strength of the dependence between two random variable is through their Correlation

ExampleExample

• A company that services air conditioner (AC) units in residence and office blocks is interested in how to schedule its technicians in the most efficient manner. Specifically the company is interested in how long a technician takes on a visit to a particular location, and the company recognizes that this mainly depends on the manner of AC units at the location that need to be serviced

Service Time (hours)

1 2 3 4

Number of AC units

1 0.12 0.08 0.07 0.05

2 0.08 0.15 0.21 0.13

3 0.01 0.01 0.02 0.07

n

j

m

iijp

1 1

1• Check that

107.02.01.01.

13.21.15.08.05.07.08.12.1 1

n

j

m

iijp

ExampleExample

• What is the probability that a location has no more than two AC units that take no more than 2 hours to service? (Joint cumulative probability function)

4301508081222 22211211 .....pppp,F

• What are the expected number, variance and standard deviation of AC? of the service time?

Service Time (hours)

1 2 3 4

Number of AC units

1 0.12 0.08 0.07 0.05

2 0.08 0.15 0.21 0.13

3 0.01 0.01 0.02 0.07

sum

0.32

0.57

0.11

sum 0.21 0.24 0.30 0.25

ExampleExample

62.0

386.011.079.1357.079.1232.079.11

79.111.0357.0232.01

2

2222

XX

X

XE

08.1

162.125.59.2430.59.2324.59.2221.59.21

59.225.430.324.221.1

2

22222

YY

Y

YE

• Is there any correlation between the number of ACs and the service hours?

34.0

08.162.0

224.0,,

224.079.159.286.4,

86.407.4308.2112.111 1

YX

n

j

m

iijji

YXCovYXCorr

YEXEXYEYXCov

PyxXYE

Positively correlated!

ExampleExample

• Suppose that a technician is visiting a location that is known to have three air conditioner units, what is the probability that the service time is four hours?

64.011.0

07.0

07.002.001.001.0

07.0

,|

4

3

34|4

|

3

jj

xX

n

jij

ij

i

jiijxj

p

pp

p

p

xXP

yYxXPxXyYPp

i

Linear Function of a Random VariableLinear Function of a Random Variable

If X and Y are two random variables, and

baXY

For some a, b that are real number, the expectation, the variance and the standard deviation of the random variable Y are

bXEaYE

XVaraYVar 2

XaYY

Linear Combination of Random VariablesLinear Combination of Random Variables

If X1 and X2 are two random variables, and

2121 XEXEXXE

and the variance

2122

12

212 ,Cov2 XXXXXX

If X1 and X2 are independent random variables so that Cov(X1,X2 ) = 0, then

The standard deviation

212

21 XXXX

22

12

212 XXXX

ExampleExample

1) Use the answers of the previous example (E(X), E(Y), E(XY), Var(X) and Var(Y)) and assume that X and Y are independent variables. Find the expectation and variance of the following random variables• 2X+6Y• 5X-9Y+8

From the previous example:

59.279.1 YEXE 162.1386.0 22 YX

12.1959.2679.126262 YEXEYXE

36.6859.2979.15895895 YEXEYXE

376.43162.136386.046262 22 YVarXVarYXVar

Then,

77.103162.181386.02595895 22 YVarXVarYXVar

Averaging Independent Random VariablesAveraging Independent Random Variables

Suppose that X1, X2 , ..…, Xn is a sequence of independent random variables each with an expectation μ and a variance σ2, and with an average

Then

n

XXXX n

21

n

n

n

XEXEXEXE n21

nn

n

n

XVarXVarXVarXVar n

2

2

2

221

ExampleExample

1) The weight of a certain type of brick has an expectation of 1.12 kg with a standard deviation of 0.03 kga) What are the expectation and variance of the average weight of

25 bricks randomly selected?b) How many bricks need to be selected so that their average

weight has a standard deviation of no more than 0.005 kg?

12.1

252521

XEXEXE

XE

000036.0

25

03.0 22

221

nn

XVarXVarXVarXVar n

a) Since independent variable, then

b)

12.12521 XEXEXE

36005.003.0

005.0005.0005.02

nnn

XVarX