Continuous Random VariablesChapter 5

Nutan S. MishraDepartment of Mathematics and

StatisticsUniversity of South Alabama

Continuous Random Variable

When random variable X takes values on an interval

For example GPA of students X [0, 4]High day temperature in Mobile X ( 20,∞)Recall in case of discrete variables a simple event

was described as (X = k) and then we can compute P(X = k) which is called probability mass function

In case of continuous variable we make a change in the definition of an event.

Continuous Random VariableLet X [0,4], then there are infinite number of

values which x may take. If we assign probability to each value then

P(X=k) 0 for a continuous variableIn this case we define an event as (x-x ≤ X ≤ x+x ) where x is a very tiny

increment in x. And thus we assign the probability to this event

P(x-x ≤ X ≤ x+x ) = f(x) dxf(x) is called probability density function (pdf)

Properties of pdf

lim

lim

1)(

0)(upper

lower

dxxf

xf

(cumulative) Distribution Function

The cumulative distribution function of a continuous random variable is

Where f(x) is the probability density function of x.

a

itlower

dxxfaXPaFlim

)()()(

)()(

)()()(

aFbF

dxxfbxaPbxaPb

a

Relation between f(x) and F(x)

)()(

)()(

xfdx

xdF

dttfxFx

Mean and Variance

2lim

lim

22

lim

lim

22

lim

lim

)(

)()(

)(

upper

lower

upper

lower

upper

lower

dxxfx

dxxfx

dxxxf

Exercise 5.2To find the value of k

Thus f(x) = 4x3 for 0<x<1

P(1/4<x<3/4) = =

P(x>2/3) = =

4

1]4

1[]

4[...

1

10

41

0

3

1

0

3

k

kx

kxkSHL

kx

dxx4/3

4/1

34

dxx1

3/2

34

Exercise 5.7

)4()5()54(

9/59/41)3()3(

FFxP

FxP

Exercise 5.13

21

0

521

0

322

1

0

41

0

41

0

3

44*

444*

dxxdxxx

dxxdxxdxxx

Probability and density curves

• P (a<Y<b): P(100<Y<150)=0.42

Useful link: http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/cprob/cprob2.html

Normal DistributionX = normal random variate with parameters µ and

σ if its probability density function is given by

µ and σ are called parameters of the normal distribution

http://www.willamette.edu/~mjaneba/help/normalcurve.html

xxexf 22 2/)(2

1)(

Standard Normal Distribution

The distribution of a normal random variable with mean 0 and variance 1 is called a standard normal distribution.

-4 -3 -2 -1 0 1 2 3 4

Standard Normal Distribution

• The letter Z is traditionally used to represent a standard normal random variable.

• z is used to represent a particular value of Z.• The standard normal distribution has been

tabularized.

Standard Normal Distribution

Given a standard normal distribution, find the area under the curve

(a) to the left of z = -1.85

(b) to the left of z = 2.01

(c) to the right of z = –0.99

(d) to right of z = 1.50

(e) between z = -1.66 and z = 0.58

Standard Normal Distribution

Given a standard normal distribution, find the value of k such that

(a) P(Z < k) = .1271

(b) P(Z < k) = .9495

(c) P(Z > k) = .8186

(d) P(Z > k) = .0073

(e) P( 0.90 < Z < k) = .1806

(f) P( k < Z < 1.02) = .1464

Normal Distribution

• Any normal random variable, X, can be converted to a standard normal random variable:

z = (x – μx)/x

Useful link: (pictures of normal curves borrowed from:

http://www.stat.sc.edu/~lynch/509Spring03/25

Normal DistributionGiven a random Variable X having a normal

distribution with μx = 10 and x = 2, find the probability that X < 8.

-4 -3 -2 -1 0 1 2 3 4

4 6 8 10 12 14 16

z

x

Relationship between the Normal and Binomial Distributions

• The normal distribution is often a good approximation to a discrete distribution when the discrete distribution takes on a symmetric bell shape.

• Some distributions converge to the normal as their parameters approach certain limits.

• Theorem 6.2: If X is a binomial random variable with mean μ = np and variance 2 = npq, then the limiting form of the distribution of Z = (X – np)/(npq).5 as n , is the standard normal distribution, n(z;0,1).

Exercise 5.19

-4 -3 -2 -1 0 1 2 3 4

5.1

2/1 2

)5.1( dtexP t

Uniform distribution

The uniform distribution with parameters α and β has the density function

elsewhere 0

1

)(

xforxf

Exponential Distribution: Basic Facts

• Density

• CDF

• Mean

• Variance

, 0, 0

0, 0

xe xf x

x

1 , 0

0, 0

xe xF x

x

1E X

2

1Var X

Key Property: Memorylessness

• Reliability: Amount of time a component has been in service has no effect on the amount of time until it fails

• Inter-event times: Amount of time since the last event contains no information about the amount of time until the next event

• Service times: Amount of remaining service time is independent of the amount of service time elapsed so far

for all , 0P X s t X t P X s s t

Exponential Distribution

The exponential distribution is a very commonly used distribution in reliability engineering. Due to its simplicity, it has been widely employed even in cases to which it does not apply. The exponential distribution is used to describe units that have a constant failure rate. The single-parameter exponential pdf is given by:

where: · λ      = constant failure rate, in failures per unit of measurement, e.g. failures per hour, per

cycle, etc.·    λ   = .1/m·        m = mean time between failures, or to a failure.·        T = operating time, life or age, in hours, cycles, miles, actuations, etc. This distribution requires the estimation of only one parameter, , for its application.

Joint probabilities

For discrete joint probability density function (joint pdf) of a k-dimensional discrete random variable X = (X1, X2, …,Xk) is

defined to be f(x1,x2,…,xk) = P(X1 = x1, X2 = x2 , …,Xk = xk) for all possible values x = (x1,x2,…,xk) in X.

Let (X, Y) have the joint probability function specified in the following table

x/y 2 3 4 5

0 1/24 3/24 1/24 1/24 6/24

1 2/24 2/24 6/24 2/24 12/24

2 2/24 1/24 2/24 1/24 6/24

5/24 6/24 9/24 4/24 24/24

Joint distribution

Consider 042.24/1)2,0( f

243

242

241)2,1( yxP

2411

246

245)3,2( yxP

2410

246

242

242)4,1( yxP

244

242

242)2,0( yxP

122)1|2( xyP

122)1|3( xyP

126)1|4( xyP

122)2|4( xyP

25.024/6)4,1( f

Joint probability distribution

Joint Probability Distribution Function f(x,y) > 0

Marginal pdf of x & y

here is an example

x =1, 2, 3 y = 1, 2

x y

yxf 1),(

Ayx

yxfAYXP),(

),(]),[(

y

X yxfxf ),()(

x

Y yxfyf ),()(

21),(

yxyxf

Marginal pdf of x & y

Consider the following example

x = 1,2,3

y = 1,2

x xY

yxyYPyxfyf

3

1 21)(),()(

21

32

21

2

21

1

xxx

y yX

yxxXPyxfxf

2

1 21)(),()(

21

36

21

3

21

2

21

1 yyyy

Independent Random Variables

If

dependentff 052.57630

245

246)2(*)0( 21

tindependen)()()( yYPxXPyYxXP

)()(),( yfxfyxf YX

241

6

1*

4

1)5(*)0( 21 ff

Properties of expectations

for a discrete pdf, f(x), The expected value of the function u(x), E[u(X)] =

Mean = = E[X] =

Variance = Var(X) = 2 = x2=E[(X-)2] = E[X2] - 2

For a continuous pdf, f(x)

E(X) = Mean of X =

E[(X-)2] = E(X2) -[E(X)]2 = Variance of X =

S

dxxfx )(

Sxxfxu )()(

Sxxfx )(

S

dxxfx )()(2

Properties of expectationsE(aX+b) = aE(X) + b

Var (aX+b) = a2var(X)

Mean and variance of Z

x

Z Is called standardized variable

baxxx

Z

1

E(Z) = 0 and var(Z) = 1

Linear combination of two independent variables

Let x1 and x2 be two Independent random variables then their linear combination y = ax1+bx2 is a random variable.

E(y) = aE(x1)+bE(x2)

Var(y) = a2var(x1)+b2var(x2)

Mean and variance of the sample meanx1, x2,…xn are independent identically distributed

random variables (i.e. a sample coming from a population) with common mean µ and common variance σ2

The sample mean is a linear combination of these i.i.d. variables and hence itself is a random variable

nn x

nxn

xnn

xxxx

1...

11...21

21

nx

xE x

2

)var(

by denoted )(