Probability Theory

School of Mathematical Science and

Computing Technology in CSU

Course groups of Probability and Statistics

§2.4 Figure characteristics of random variables

Distribution list is able to describe the statistical characteristics

of random variables completely ， However, in some practical p

roblems ， only need to know some characteristics of random v

ariables and thus do not need to derive a result of its distribution

function .For example ：

Assessment of the viability of an enterprise, only need to

know the level of per capita profit of the enterprises ；

Study the merits of rice varieties, we are concerned abou

t the average rice grains and the average weight of each piec

e costs ；

Test the quality of cotton, they should not only pay attenti

on to the average length of fiber, but also pay attention to th

e deviate degree between the length of fiber and the average

length, the longer the average length and the smaller the dev

iate degree, the better the quality.

Study the level of one shooter, we not only depend on his av

erage ring number whether high or not, but also depend on hi

s scope of impacts whether small or not, that is, whether the f

luctuations in data small.

From the above example we can see ， some values relating to random variables. Although we ca

n not completely describe the random variable ， but we

can clearly describe the important feature of random vari

ables in some respects. Characteristics of these figures h

ave great significance both in theory and practice.

One aspect of probability characteristics of

random variable are available to describe

by figures.

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The average values of random varia

bles ——Mathematical expectation The situation of random variable val

ues are deviate from the mean value on a

verage—— Square

The contentof this section

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Definition: Suppose discrete distribution of random variable X as

,2,1,)( kpxXP kk

If the infinite series

1kkk px

absolute convergence ,then called which the sum is random variable X as mathematical expectation ,recorded as

1k

kk pxEX

1.The definition of mathematical expectation Section I Mathematical expectation

n

k

knkkn ppkCEX

0

)1(

n

k

knk ppknk

nk

1

)1()!(!

!

n

k

knk ppknk

nnp

1

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

)!1(

1

0

)1(1 )1(

n

k

knkkn ppCnp

np

EXpnBX Calculated),,(~Known

Answer

Example 1

EXPX Calculated),0(),(~Known

Example 2

Answer

EX

e

kk

k

k

0 !

1

1

)!1(k

k

ke

Common mathematical expectation of random variable Distribution Expectation

pXP

pXP

1)0(

)1(p

B(n,p)nk

ppCkXP knkkn

,,2,1,0

)1()(

np

P()

,2,1,0!

)(

kk

ekXP

k

Probability distribution

Parameters for the 0-1 distribution of p

2. The nature of mathematical expectation

following)

in then expectatio almathematicexist meet we

variablerandom theconstant

is ,, variablerandom is, uppose

suppose We（ S cbaYX ，

cbEYaEXcbYaXE )()1(

0then ,0 uppose)2( EXXS

bEXabXa hen , uppose)3( tS

n

i

n

iii

i

EXXE

niX

EXEYXYE

YX

1 1

is then thereothereach of

t independen are21If:Promotion

)(Then variablesrandom

t independenmutually twoare Suppose)4(

）（，

），，（，

，

Prove ： Only prove Nature (4) at n=2

,2,1

ason distributi its record And

and of valuespossible all as recorded

and usely respectivevariable

random discrete a is that assumeWe

21

21

2121

jipbXaXP

XX

bbaa

X

ijji

i

，），（

，，，，，，

may

:known weassumptiont independen theFrom

)()( 21 jiij bXPaXPp

21

21

,21

,21

21

21

)()(

)()()(

,

EXEX

bXPbaXPa

bXPaXPbapbaXXE

baXX

bXaX

jjj

iii

jijiji

jiijji

ji

ji

so

， is there

, whenBecause

Answer Import the random variables

Example 7 A Civil Aviation bus contains 20 visitors leave th

e airport ， the visitors can get off at 10 stations ， if one sta

tion has no passengers to get off the bus will not stop ， tak

e X as the number of stops ， Calculate EX （ Suppose each

passenger get off at various stations have the same possibilit

y, and suppose whether the passengers get off or not are ind

ependent of each other ）

istation at offget someone1

istation at offget nobody ,0

，iX

10,,2,1 i

Then there is

1021 XXXX

10

9 is istation at offget not didpassenger

oneany ofy probabilit the titleby the Intended ,

20)10

9( is istation at off

getnot do passengers 20 ofy probabilit theSo

20)10

9(1 is i station

at offget someone of yprobabilit The

2020 )10

9(1)1(,)

10

9()0( ii XPXP

20)10

9(1)1(1)0(0 iii XPXPEX

So

10

11021 )(

iiEXXXXEEX

784.8])10

9(1[10 20 （ time

s）

That is

time

Arrived

yProbabilit

50:930:910:9

50:830:810:8

6

2

6

3

6

1

Example 8 According to regulation ， one st

ation everyday 8 ： 00 ~ 9:00,9:00 ~ 10:00 bot

h happen to have a bus reach the station ， bu

t the reach time is random ， and the arrive ti

me are independent of each other , the law is

. time waitinghis of

n expectatio almathematic thecalculate

00,:8at station reach theA visitor )1(

Answer

minute）

by counted（ is visitor the

of time waitingthe Suppose

X

. time waitinghis of

n expectatio almathematic thecalculate

20,:8at station reach theA visitor )2(

is X ofon distributi The)1(

X

kp

503010

6

2

6

3

6

1

)points(33.33 EX

X

kp

9070503010

6

1

6

2

6

1

6

3

6

1

6

1

6

2

6

3

is X ofon distributi The)2(

example for，above tableOn the

6

3

6

1)()()()70( BPAPABPXP

，： "108at arrive busfirst The" case theisA

thoseAmong

，： "309at rrive bus second he" case theis aT B

)min(27.2236

290

36

370

36

150

6

230

6

310

EX

then,)(

if variablesrandom discreteFor

kk pxXP ,X

3. The mathematical expectation of random variable function

exists)]([)(

function continuousits variables

random a is thereSupposeTheorem

xgEXgY ，X,

)]([ XgE k

kk pxg )(

)function continuous is(,),(.,

variablerandom offunction theisSuppose

gYXgZYX

Z

is lawon distributi theY) (X,

variablerandom discrete ldimensiona- two theIf

,2,1,,),( jipyYxXP ijji

ijji pyxgYXgEEZ ),()],([

isthereThen

X 1 3

P 3/4 1/4

Y 0 1 2 3

P 1/8 3/8 3/8 1/8

X

1 0 3/8 3/8 0

3 1/8 0 0 1/8

Y 0 1 2 3

)(,,Calculate XYEEYEX

Answer

Example 9 Known the joint distribution of（ X,Y ） is

8

1)33(0)23(0)13(

8

1)03(

0)31(8

3)21(

8

3)11(0)01()(

XYE

4

9

2

3,

2

3 EYEX

For a particular disease survey , n indi

viduals need a blood test , blood tests

can be two ways:

(1)Tests separately for each person's bl

ood , need to test n times totally ；

Blood program selection

33 、、 Simple Simple applicationapplication of of mathematica mathematica

l expectationl expectation

Suppose the probability of tested positive is p i

n someone area ， and each is a positive person ar

e independent of each other. Try to select a method

which can reduce the number of tests. .

K individuals will be mixed with the blood te

sts ， if the test results become negative , then the

k individual blood tests only once ； If the results

become positive, then the k individuals will have b

lood test one by one to identify sick persons, then

k individual blood tests to be k + 1 times.

Answer For the simple calculation ， Based n are multiples of k ， suppose divided into a total of n / k group, the number of tests for group i

required to be Xi

kp1 kp 11

Xi

P

1 k + 1

]11)[1(1 kki pkpEX

kpkk 1)1(

k

n

iiEXEX

1

kpkkk

n 1)1(

kpn k 1

)1(1

,01

)1(

kp kIf Then EX < n

Such as ，

10000110010

1999.0110000

,10,001.0,10000

10

EX

kpn

Section II SquareGuide example Test the quality of two groups of light bulbs, which were randomly selected 5, the measured lifetime (unit: hours) as follows:

A: 2000 1500 1000 500 1000

B: 1500 1500 1000 1000 1000

Let us compare the quality of these two groups of light bulbs

After calculated :Average life are :A:1200 B:1200

After Observated :A has large departurein useful life,B has small departurein useful life,so,B has better quality

Mathematical expectation Square

1. The definition of square

is that,,

called then,

,

DXX

EX)E(XEX)E(X

X22

as recordedof square theas

exist

if variablerandom a is Suppose:Definition

(X - EX)2 —— Random variable X the value

of deviation from the average of the situation

are a function of X is also a random variable E(X - EX)2 —— Random variable X the value

of the average deviation from the average

deviation from the degree - a number Note:

Variance reflects the random variable relative degree of its deviation from the mean.

2)(Square EXXEDX DXdeviation) (standard squareMean

,2,1,)( kpxXP kk

X is discrete random variables, probability distribution is:

k

kk pEXxDX 2

If X is continuous random variables, probability density is f (x)

dxxfEXxDX )(2

Commonly used formula for calculating the variance:

22 )(EXEXDX

2. The nature of the square ）Constant is(0)1( CDC

DXCCXD 2)()2(

DYDXYXD

YX

)( variablesrandom

t independenmutually twoare,Suppose)3(

then,

1)(isThat ,)Here(constant

one takingofy probabilit thetois0

toconditions sufficient andnecessary The)4(

CXPEXCC

XDX

)constantsarbitrary is(

)()()5( 22

a

aEXaXEDX

Example1 Suppose X ~ P (), Calculate DX.

Answer

0 !k

k

k

ekEX

1

1

)!1(k

k

ke

3. Square calculation

EXXXEEX )]1([2

!)1()]1([

0 k

ekkXXE

k

k

2

2

22

)!2(

k

k

ke

22EX

22 )(EXEXDX

Example 2 Suppose X ~ B( n , p) ， calculate DX Answer One follow the example above

calculate DX

Answer Two Import the random variables nXXX ,,, 21

happen matter times i of est

happenA matter times i of est

AT

TX i ,0

,1

nXXX ,,, 21 are independent of

ni ,,2,1 )1( ppDX i

n

iiXX

1

so,

)1(1

pnpDXDXn

ii

each other ，

Common random variable of variance

Distribution SquareProbability distribution

Parameters for the 0-1 distribution of p pXP

pXP

1)0(

)1(p(1-p)

B(n,p)nk

ppCkXP knkkn

,,2,1,0

)1()(

np(1-p)

P()

,2,1,0!

)(

kk

ekXP

k

Example 8 Suppose X express the requir

ed fire number of shooting independent

until hits the target n times. Known for e

ach target shooting in a probability of p ，calculate EX , DX

Answer X i express the required fire numbe

r of hit the target i - 1 times to hit the target

i times ， i = 1,2,…, n

1,2,1,)( 1 qpkpqkXP ki

1

1

1

1

k

k

k

ki kqpkpqEX

pqp

1

)1(

12

nXXX ,,, 21 are independent of

n

iiXX

1

each other , moreover

1

1

1

12 )1(k

k

k

ki kpqpqkkEX

pqkkpq

k

k 1)1(

2

2

px

dx

dpq

qxk

k 1

02

2

pxpq

qx

1

)1(

23

2

2

p

p

p

nEXEX

n

ii

1

Therefore,

21

)1(

p

pnDXDX

n

ii

Only know the expectations of random variables

and the variance can not determine their

distribution, such as:

222

112

p

p

pp

pDX i

X

P

-1 0 1

0.1 0.8 0.1

Y

P

-2 0 2

0.025 0.95 0.025

and

2.0,0 DXEX

2.0,0 DYEY

They have the same expectation and variance, but the distribution is different .