Continuous random variable theory.pdf

7/24/2019 Continuous random variable theory.pdf

1/101

(Continuous Random Variable )

Continuous Densities:


2/101

X can assume all possible values 0x1

Since possible values of X are non countablein 0x1, then what happens to point

probabilities?


3/101

Definition:

A random variable is continuous if it can

assume any value in some interval

( or intervals ) of real numbers and the

probability that it assume any specific value

is 0 ( zero ).


4/101

CONTINUOUS PDF (Density Function)

Definition:Let X be a continuous randomvariable. A function f(x) is called continuousdensity ( probability density function i.e pdf ) if

1 . f ( x ) 0

2 . ( ) 1f x d x

3. For anyab we havec

ad xxf )(c)xP(a.3


5/101

(Thus P[aXc] is area under graph of y=f(x)

betweenx =a andx =c.)


6/101

Remark: It is a consequence of the definition

that for any specified value of X, say xo , wehave P[X=xo] = 0, since

[ ] ( ) 0

o

o

x

o

xP X x f x d x

Remark :f (x 0)P [X=x 0].

In factf is analogous to the density of mass

whereas probability is analogous to the mass

itself.


7/101

Remark. If X assumes values in some finiteinterval [a,c], we simply set f(x) = 0 for all

x[a,c].


8/101

Every conceivable point on the line segment

could be the outcome of the experiment.Since X is continuous r.v, we have

][

][

][][

cxaP

cxaP

cxaPcxaP


9/101

Let X be the continuous r.v. with density

f(x). The cumulative distribution function (cdf ) for X ,denoted by F(X) , is defined by

F(X) = P ( X x ) , all x

=

PROBABILITY by using cdf:F(x)

P( aXc ) = F(c) F(a).

x

ds)s(f


10/101

Example : If pdf of a random variable X is

otherwise0

3x11)-(xr

1x0)1(

)(

xr

xf

(i) Find r and graph of f(x) (ii) Find F(x)


11/101

3x1

3x15

)1(

5

1

1x0)2

(52

)(

2

2

x

xx

xF


12/101

Theorem: Let F be the continuous cdf of a

continuous r.v with pdf f, then

abledifferenti

isFwhichatxallfor

),()( xFd x

d

xf


13/101

(5) (Continuous uniform distribution) A

random variable X is said to be uniformlydistributed over an interval (a,c) if its

density is given by

cxaac

xf

,1)(

(a)Show that this is a density for a continuousrandom variable.

Sol: acf or

ac

Since

,01


14/101

Secondly,

11

d xac

c

a

(b) Sketch the graph of the uniform density.

X=a X=c

f(x)

x

1/(c-a)


15/101

( ii )Shade the area in the graph of part (b)

that represents P[Xa+c)/2].

X=a X=c

f(x)

x(a+c)/2


16/101

(c ) Find the probability pictured in part: ii

(e) Let (l,m) and (d,f) be subintervals of (a,c)

of equal length. What is the relationship

between P[lX m] and P[dX f]

Sol: Probability is same on equal

length of interval.

]c

c


17/101

(10) Find the general expression for the

cumulative uniform distribution for arandom variable X over (a,c)

cxaacax

d sac

xXPxFx

a

,

)(

1][)(


18/101

cx

cxaac

ax

ax

xF

,1

,

,0

)(


19/101

Section 4.2

Def: Let X be a continuous random variable

with pdf f. The expected value of X is defined

as

[ ] ( ) .E X x f x d x

Again E(x) exists if and only if

| | ( )x f x d x is f inite


20/101

For a random variable X which is a function,

say H(x), the definition takes the form:

[ ( )] ( ) ( ) .E H x H x f x d x

provided

| ( ) | ( )H x f x d x is f inite


21/101

Moment generating function :

( ) [ ]

( )

w here is densi ty o f .

tXX

tx

m t E e

e f x d x

f X


22/101

Variance :

Variance is shape parameter in the

sense that a random variable with

small variance will have a compact

density; one with a large variance will

have a density that is rather spread

out or flat.


23/101

MGF of uniform distribution random

variable on (a, c) :

ace

t

ac

d x

eeEta

c

a

txtx

tc

e1

)(


24/101

Either by using mgf or directly, mean and

variance can be found.

12)()(

2)(

2

acXVar

caXE


25/101

Section 4.3:

Definition : The Gamma function is the functiondefined by

1

0

1 1

1

( ) , 0.

1

lim lim0

exists for 0.

z

z z

r

z e d z

s

z e d z z e d zr s


26/101

4.3: Gamma Random variable

Gamma Function: which is an improper

integral allows us to define exponential

and chi- squared random variables.

Gamma function:

0

1 0,)( d zez z


27/101

Theorem: (Properties of Gamma function)

1),1()1()(.2

1)1(.1

allf orProof:by definition of Gamma function, we have

0 0

zz0

1dzedzez)1(

by integration by parts, we have


28/101

)1()1(

)1(|

0,)(

0

1)1(

0

1

0

1

d zzeze

d zez

zz

z

Hint: by repeated use of L hospital rule,

we shall have


29/101

)!1()1()2()1()2()2)(1(

)1()1()(

,

)!1()(

nnnnnn

nnn

Since

nn

Thus, Gamma function is a generalization of the

Factorial notation


30/101

provedbenot to

)21(

0

2/1

d zez z


31/101

GAMMA RANDOM VARIABLE

A random variable X with density function

is said to have a Gamma Distribution with

parameters and,for x>0,>0,>0.

wiseother

xexxf

x

,0

0,0,0,)(

1

)(

/1


32/101

To check the necessary and sufficient condition

of pdf: 0xallfor0)x(f

0

11

0

/1

)(

1

,

)(

1)(,

d zez

zxandd zd xzx

Let

d xexd xxfFurther

z

x


33/101

p d faisxfH ence

ez z

)(

1

)( 0

1

Theorem: Let X be a gamma random variable

with parameters and. Then m.g.f for X isgiven by:

2

x

)x(Var.3

]X[E.2

1t,)t1()t(m.1


34/101

Proof:

)1(

)1(

)1(

)(1

)(

1

][)(,

0

)1

(1

/1

0

t

d zd xand

t

zx

xtzlet

d xex

d xexe

eEtmd efby

xt

xtx

tx

x


35/101

1,)1()(

)()1()(

1

)1()(

1)1(

)

1

(

)(

1

0

1

1

0

tttm

t

d zezt

t

d ze

t

z

x

z

z


36/101

0t

1

0tx

/)()t1(

/)t(m

dt

d]X[E.2

2

02

22

)1(

/))((][

tx tmd td

XE


37/101

0

2 12

2

00

2

2 2 2

( )2 ) [ ]

( ) (1 )3 ) [ ]

( 1)

T h u s [ ] ( 1) ( ) .

X

t

X

tt

d m tE Xd t

d m t d tE X

d t d t

V a r X

wiseother

xex

xf

x

,0

0,0,0,

)(

1

)(

/1


38/101

For >1 maximumValue of density is at

x=(-1)

Gamma(2 ,3) Gamma(2 0,0.5)

Gamma(1,4)


39/101

For >1 maximum

Value of gamma densityis at

x=(-1)

2

)(.

][.

1,)1()(.

xVariii

XEii

tttmi x


40/101

Exponential distribution : exponential

random variable is Gamma random

variable with=1.The density is

> 0 is the parameter of this exponentialdistribution. E[X]=, Var[X]=2 .

otherwise0

00, x1)(

x

exf


41/101

The c.d.f. of exponential distribution with

Parameter is given by

o x1

111)(

0

0

0

xxs

x

s

x s

ee

ed sexF


42/101

/1 , 0

( ) 0, otherwise.

xe x

F x


43/101

Moment generating function, Mean and

Variance of exponential distribution

Note: Put =1 in the gamma distribution

we get the required results.

2

1

)(][

)1()(

XVarmeanXE

ttmX


44/101

Poisson Process and Exponential dist :

For a Poisson process with parameter ,

the

w iting time

W is the time in the

given interval before the1

st

success.

Theorem :W has an exponential

distribution with parameter =1/.


45/101

Proof: This theorem is distribution of

waiting time

The distribution function F for W is given by

]wW[P1]wW[P)w(F

Here, we have that the first occurrence of the

event will take place after time w only if number

of occurrences in the time interval [0,w] is zero

Let X be the number of occurrences of

the event in this time interval [0,w].


46/101

X is a poisson random variable with parameter

w.

w

0w

e!0

)w(e

]0X[P]wW[P,Thus

w

ewWPwF

1][1)(


47/101

Since, in the continuous case, the derivative

of the cumulative distribution function isthe density

1with

varibalerandomlexponentia

anfordensityexactlyisThis

)()(

wewfwF


48/101

GAMMA RANDOM VARIABLE

A random variable X with density function

is said to have a Gamma Distribution with

parameters and,for x>0,>0,>0.

wiseother

xexxf

x

,0

0,0,0,)(

1

)(

/1


49/101

Chi-square distribution : If a random variable

X has a gamma distribution with parameters=2 and=/2 , then X is said to have achi-square ( 2 ) distribution with degrees ofFreedom and denoted by X2

, is a positive

Integer.


50/101

0x,ex2)

2

(1)x(f 2/x

1

2

2/

E[X2 ]=, Var[X2 ]=2

=2 and=/2 ,


51/101


52/101


53/101


54/101

We do not have explicit formula for CDF F

of X

2

. In stead values are tabulated onp. 695-696 as below (F occurs in margin here,

and related value of r.v. inside the table):


55/101

If F is CDF for Chi square random variable

Having 5 degrees of freedomF(1.61)= .1 , F(4.35)= .5

P[X2 < t]

F 0.100 0.2 50 0.500

5 1.61 2 .67 4.35

6 2 .2 0 3.45 5.35

7 2 .83 4.2 5 6.35


56/101

38. Consider a chi squared random variable

with 15 degrees of freedom.(i) What is the mean of ?

2

15X

Sol: Chi square distribution is Gamma with

=/2 &=2 , Mean ===15 and2 =2 = 30


57/101

otherwise0

0,2)2/15(

1)(

,2/,2var

2/1)2/15(2/15

2

xexxf

hencewithiablerand omisX

x

Sol:

(ii) What is the expression for the density for

?2

15X


58/101

( iii) What is the expression for the moment

Generating function for 215X

Sol:

2/1,)21()1()(2/15

ttttm x


59/101

P[X2

< t]

F 0.005 0.010 0.900

15 4.60 5.2 3 2 2 .3

6 2 .2 0 3.45 5.35

7 2 .83 4.2 5 6.35

.02 5

6.2 6

10.0900.01

]3.2 2[1]3.2 2[ 215215

XPXP

( iv) Find ]3.2 2[ 215

XP


60/101

875.002 5.0900.0)2 6.6()3.2 2(

]3.2 22 6.6[ 215

FF

XP

P[X2 < t]

F 0.005 0.010 0.900

15 4.60 5.2 3 2 2 .3

6 2 .2 0 3.45 5.35

7 2 .83 4.2 5 6.35

.02 5

6.2 6

]3.2 22 6.6[Find(V) 215

XP


61/101

Sol. P[X2 15 >2 r]=r.

freedomofdegree150.2 5205.0

f or

P[X2 15 >2 0.05]=0.05.

P[X2 < t]

F 0.005 0.010 0.950

15 4.60 5.2 3 2 5..0

.02 5

6.2 6

V Find 2 0.05, &2 0.01 for 15

degree of freedom


62/101

,6.30201.0

P[X2 15 >2 r]=r.

P[X215

>20.01

]=0.01.

P[X2 < t]

F 0.005 0.010 0.990

15 4.60 5.2 3 30.6

.02 5

6.2 6


63/101

4.4 The Normal Distribution

A random variable X with density f(x)

is said to have normal distribution with

parameters and > 0, where f(x) isgiven by:

.0);,(,,2

1)(

2

2

2

xexf

x

)(0)()( fi


64/101

- 1f(x)dx(ii)

),(-x0)()( xfi

-

2

1

-

2

1

2

-

2

1

2

2

2

2

2

1

2

1

let

12

1

proveto

d zed xe

d zd xzx

d xe

zx

x


65/101

0

2

1

-

2

1 22

2

12

2

1d zed ze

zz

0

)(2

1

0

0

2

1

0

2

1

0

2

1

0

2

1

22

22

22

d x d ye

d yed xeII

Id zed ze

yx

yx

zz


66/101

2

0

2/

0

0

)(2

12/

0

2

dd we

rd rde

w

r

0

)(2

1

0

22

d x d ye

yx


67/101

2I..

ei

12

2

2

12

2

1

0

2

1

-

2

1 22

d zed ze

zz


68/101

Standard Normal Distribution

If =0 =1 then normal randomvariable is called standard normal

variable symbol for normal r.v is Z.


69/101

The probability density function isgiven by

).,(,2

1)(

2

2

zezf

z

d ze

d zzfzZPzF

z z

z

Z

2

2

2

1

)()()(


70/101

Standard Normal distribution

Theorem: Let X be normal with meanand standard deviation. The variable

is standard normal. Z has mean 0 and

standard deviation 1.

XW

X


71/101

)()()( aX

PaWPaFW

,2

1)(

2

2

2 d seaXP

a s

d zzzs ds,s,


72/101

,2

1)()(

2

2

2 d seaXPaF

a s

W

d zzz

s

ds,s,

,2

1)()( 2/

2

d zeaXPaF

a

z

W

=FZ(a)


73/101

Moment Generating Function

Let Z be normally distributed withparameters=0 and=1 , then the momentgenerating function for Z is given by

,

2/2)( tZ etm


74/101

Rajiv

-

2

1

-

2

1

2

2

2

1

2

1

)()(

d xe

d zeeeEtm

ztz

ztztz

Z

22

22

2

2

1

22

2222

tzt

tttzzztz


75/101

d wd z

d zeetm

tz

tZ

wt)-(zlet

21)(

-

22/

2

2

-

22/

2

2

21)( d weetm

w

tZ


76/101

Moment Generating Function

Let X be normally distributed withparameters and, then themoment generating function for X isgiven by

, E(X2) =

2

22

)(tt

X etm


77/101

)()()( )( zttxX eEeEtm

)( ztt eeE )( ztt eEe

2/

22

tt ee


78/101

Mean and Standard deviation for

Normal distribution

Theorem : Let X be a normal randomvariable with parameters and. Then is the mean of X and is its standard

deviation.


79/101

aF

cF ZZc)XP(a

),N(isX(i)If

d zed zzfzZPzFz zz

Z

22

2

1)()()(

)(1)(F)(Z

zFziiZ

3.5zif1)(F

-3.5zif0)(F)(

Z

Z

z

ziii


80/101


81/101


82/101

Correction for continuity for a discrete random

variable approximated with continuous random

variable: Half Unit correction

i. P(aX b)=P(a-0.5 Xb+0.5)

ii. P(Xb)= P(Xb)= P(Xb+0.5)

= P(Xb+0.5)

iii. P(X a)= P(a-0.5 X)=P( X a-.5)

iv. P(X= a)=P (a- 0.5


83/101

Normal Approximation to the

Poisson Distribution

Let X be Poisson with parameters.

Then for large values ofs, X isapproximately normal with mean

,.......3,2,1,0,!

)(

xx

sexf

xs


84/101


85/101

Chebyshevs Inequality

We are interested to find a lower bound

on probability that X is inside a interval

symmetric about mean or upper bound

on probability that X is outside a interval

symmetric about mean .


86/101

Chebyshevs Inequality

Let X be a random variable with variance

2 and mean , then for any positivenumberm ,

P[ |X-|m ]1/m 2

Chebyshevs Inequality for upper

bound


87/101

P f f Ch b h I lit


88/101

Proof of Chebyshev Inequality

for upper bound for continous X

IIIIIId xxfx

d xxfxd xxfx

let

c

c

c

c

)()(

)()()()(

mc

2

222

22


89/101

)0(

)()()()(

222

IIas

d xxfxd xxfx c

c

)xcrc-x-if)((

)()(

2

2

ocxas

d xxfcd xxfcc

c


90/101

)()(2 cXcPcXcP

)()(2 cXcPcXcP

2

1)(

)(

mmXP

mXcP


91/101

Log-Normal Distribution

The positive random variable Y is

said to have a log normal

distribution, if logeY is normally

distributed. i.e. logeYN(,)=X

Point : & are not mean & standarddeviations of Log normal randomvariable.


92/101


93/101

Problem 45/ p 146

Let X be normal with mean andvariance2. Let G denote thecumulative distribution for Y = eX

and let F denote the cumulativedistribution for X.

Show that (i) G(y) = F(lny),

y>0

(ii) Find density of Y


94/101

G(y) = P(Yy) = P(eX y) = P(X lny)

Now X ~ N(,). Therefore,y>0

0y,(lny)F

,2

1

)ln()(

X

ln

2 2

2

d xeyXPyG

y x

.0);,(, x

1

ln2

y x


95/101

0y0

0y,(lny)F

,2

1)ln()(

X

2 2

d xeyXPyG

y

0y0

0,1

dy

dF

dy

dG X

yy

2


96/101

otherwise0

0y,1

2

1 22

2ln

y

eyd y

d G

,2

1

)(ln

ln

2 2

2

d xeyF

y x

Hence the density for Y is given by


97/101

Hence, the density for Y is given by

.other wise0

0,);,(,

2

1)(

2

2

2

ln

ye

y

yg

y


98/101

Example

From a usual pack of 52 cards, cards are

drawn randomly with replacement till the

red card appears. If X denotes the number

of card drawn, using Chebyshevsinequality, find a lower bound for P[ | X-2|

Continuous random variable theory.pdf

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Transcript of Continuous random variable theory.pdf