mathematical statistics 01

7/24/2019 Mathematical Statistics 01

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Mathematical Statistics

Text book: 1. Introduction to the Theory of Statistics by Mood Graybill et. al.2. Introduction to Mathematical Statistics by Robert V. Hogg & Allen T. Craig


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Unit1Random variables,

Distribution functions,and Expectation



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Definition: A random variable (r.v.) is a function, whichassigns to each sample element a real number.

Example: Consider the experiment of tossing a single coin.

So the sample space S={head, tail}. Let the

random variable X denote the number of heads.

So define X(c)=1 if c=head, and X(c)=0 if c =

tail.



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Definition: Discrete random variable: A random

variable X will be defined to be discrete if

the range ofXis countable.

Example:

Example:

{ }

)(6

1)( 6,,3,2,1 xIxfX L=

{ } )(2

1)( 1,0 xIxfX =



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Definition: If X is a discrete r.v. with distinct values

then the probability density(mass) function (p.d.f. or p.m.f.) is defined

by

Example: If then

,,,,, 21LL

nxxx

LL ,,,,),()( 21 nX xxxxxXPrxf ="==

{ }

),(6

1)( 6,,2,1 xIxfX L= =)1(Xf

6

1)1( ==XPr



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Theorem: A function is a p.d.f. of a r.v. X if and

only if (iff)

a.

b.

Example:1.Findksuch that

is ap.d.f.

2. Findksuch that

is ap.d.f.

)(xfX

,1)(0 xfX x"

=xX xf 1)(

{ }

)()2

1

()( ,2,1 xIkxfx

X L=

{ } 10),()( ,1,0


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Definition: The (cumulative) distribution function

(c.d.f.) of a random variable X, denoted by

is defined to be for

every real numberx.

),(XF )()( xXPrxFX =



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RemarkThe properties of :

1.

2. for

(i.e. is a nondecreasing function )

3.

4.

5. where is the left-

hand limit of at

6. is continuous from the right.

)(XF

,0)( =-XF 1)( =XF

)()( bFaF XX .ba


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Example: Let

(a) Find

(b) Find

(c) Find


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Example: Let

(a) Find

(b) Find


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Definition: Continuous random variableA r.v. X iscalled continuous if there exists a function

such that for

every real numberx.

Definition: The p.d.f. of a continuous r.v. X is the

function that satisfies

)(Xf -=x

XX duufxF )()(

.,)()( xduufxFx

X "= -



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Theorem: A function is a p.d.f. of a continuous

r.v. Xiff

a.

b.

Example: Findksuch that is a

p.d.f.

Example: Findksuch that is a

p.d.f.

)(xfX

xxfX " ,0)(

- =1)( dxxfX

)()( ),( xIkexfux

X ---

=

)()( ),0(2 xIekxxf xX

- =



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Remark: IfXis a continuousr.v., then

(a)

(b)

(c)

)()( xfxFdxd XX =

=


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Example: Let Define

Find thep.d.f.ofY.

Example: LetX~ Define

Find thep.d.f.ofY.

Example: Let Define Find

thep.d.f.ofY.

).()(~ )1,0( xIxfX X = .ln2 XY -=

).(100

1)( )100,0( xIxfX = XY 10=

).()(~ )1,0( xIxfX X = .XY =



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Definition: The expected value or mean of ar.v.

denoted by is

ifXis continuous

ifXis discrete

),(Xg

[ ],)(XgE

[ ]

,)()()(

-= dxxfxgXgE X

,)()(=x

X xfxg



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Example: LetX Find

Example: LetX~ Find

Example: LetX~ Find

Example: LetX~ Find

Example: Let Find

).(~ lpoisson ).(XE

).,( pnB ).(XE

).,(2

smN ).(XE

{ } ).(2

1)( ...,2,1 xIxf

x

X

= ).(XE

).(1

11)(~ ),(2 xIxxfX X -

+=p

).(XE



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Remark: Let X be a r.v. and let a, b be constants. Then

for any functions and whoseexpectations exist,

)(1 Xg )(2 Xg

bXgEaxbXga

XgExXg

XgbEXgaEXbgXagE

XgaEXagE

bbE

+=+

=

=

)]([then,allfor)(If.5

0)]([then,allfor0)(If.4

)]([)]([)]()([.3

)]([)]([.2

)(.1

11

11

2121

11



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Definition: A mode of a distribution of one r.v. X of

the continuous or discrete type is a value of

xthat maximizes thep.d.f.

Example: Find the mode of each of the following

distributions

).(xfX

{ }


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Definition: A (100p)th percentile of the distribution of a

r.v. X is a valueksuch thatand

Remark:1. A median of a r.v. X is a 50th percentile.

2. If X is a continuous r.v., then

(a) the (100p)th percentile of X is k satisfying

and

(b) the median ofXismsatisfying

pkXPr


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Example: Find the median of the following distributions:

zero elsewhere

ExampleFind the 20th percentile of

)()1(3)()( )1,0(2 xIxxfa X -=

,42or10,3

1)()(


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Homework:

1. Find the median of the distribution

where

2. Find the mode of the distribution

3. If then find the median.

4. Let X be a continuous nonnegative r.v. and

where and are

constants, Find

),(1

)( ),(

)(

xIexf

x

X qlq

l

q

--

= .,0 l,a m

.0,0,10 >>


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Definition: For each integer n the nth moment of X, is

defined as and the nth centralmoment ofX, is defined as

Definition: The variance of a r.v. X is its second centralmoment, The positive

square root of is the standard

deviation, ofX.

)(n

XE.)(

nEXXE -

.)()( 2EXXEXVar -=2)( s=XVar

,s



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Remark:1.

2. where a is a constant.

3. where a , b are

constants.

22 )()( EXEXXVar -=

,0)( =aVar

),()( 2 XVarabaXVar =



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Example: Find the variance of the following distributions:

{ } )(2

1)()(

)()(

),()(

),()(

,2,1 xIxfd

poissonc

pnBb

Gammaa

x

X L

=

l

ba



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Definition: Let X be a r.v. with c.d.f. The

moment-generating function (m.g.f.) of X,denoted by is

provided that expectation exists for

Theorem: IfXhasm.g.f. then

).(XF

),(tMX ),()(tX

X eEtM =

.0, >


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Example: Find for each of the following p.d.f.:)(tMX

{ } )(2

1)()(

)()(

),()(

)()()(

,2,1

),0(

xIxfd

poissonc

pnBb

xIexfa

x

X

xX

L

=

= -

l

l l



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Example: Let and define Find

thep.d.f.ofY, and

Example: Let and define Findthe p.d.f. of Y. (Please use c.d.f. method and

m.g.f.method)

)1,0(~UX .ln2 XY -=

)(YE ).(YVar

),(~

2

smNX .s

m-=

X

Y



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Example: LetXbe ar.v.such that

and

Find thep.d.f.ofX.

Example: Find the moments of the distribution that

hasm.g.f.

,!2

)!2()( 2

m

mXE

m

m

=

,0)( 12 =-mXEK,3,2,1=m

.1,)1()( 3


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Definition: If X is a r.v., the rth factorial moment of X

is defined as

Definition: The factorial moment generating function

of a r.v. X is defined (if it exists) as

)0( >r

[ ].)1()1( +-- rXXXE K

).()( XX tEtm =



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Remark: where we define

Example: Let Find

Example: Let Find

[ ]

),1()1()1( )(rXmrXXXE =+-- K

.|)()1( 1)( == tXr

r

rX tm

dtdm

).(~ lpoissonX [ ].)2)(1( -- XXXE

).,(~ pnBX ).(tmX



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Theorem: Let X be a r.v. and a nonnegative

function with domain the real line then

for every

Corollary: Chebyshev inequality. If X is a r.v. with

finite variance, then

for every or

)(g

[ ][ ]

,)(

)(k

XgEkXgPr .0>k

[ ]2

1

rrXPr - sm 0>r

[ ] .112r

rXPr -


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Example: IfXis ar.v.such that and

use the Chebyshev inequality to determine alower bound for

Example: Does there exist a r.v. X for which

3)( =XE ,13)( 2 =XE

[ ].82


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Homework:

1. Let Prove

2.LetXbe ar.v.withm.g.f. Prove that

and

).,(~ baGammaX .2

)2(

a

ab

e

XPr

.),( hthtMX

mathematical statistics 01

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