Chapter 2 Random variables 2.1 Random variables Definition. Suppose that S={e} is the sampling space of random trial, if X is a real-valued function.

Chapter 2 Random variablesChapter 2 Random variables

2.1 Random variables

Definition. Suppose that S={e} is the sampling

space of random trial, if X is a real-valued fun

ction with domain S, i.e. for each eS ， there

exists an unique X=X(e), then it is called that

X a Random vector.

Usually , we denote random variable by notati

on X, Y, Z or , , etc..

For notation convenience, From now on, we denote random variable by r.v.

2.2 Discrete random variables

Definition Suppose that r.v. X assume value x1, x2, …, xn, … with probability p1, p2, …, pn, …respectively, then it is said that r.v. X is a discrete r.v. and name

P{X=xk}=pk, (k=1, 2, … )

the distribution law of X. The distirbution law of X can be represented by

X ～ P{X=xk}=pk, (k=1, 2, … ) ，

or

~XX x1 x2 … xK …

Pk p1 p2 … pk …

(1) pk 0, k ＝ 1, 2, … ;

(2) 1

.1k

kp＝

.}{35

332

C

CCkXP

kk

＝＝

Example 1 Suppose that there are 5 balls in a bag, 2 of them are w

hite and the others are black, now pick 3 ball from the bag without

putting back, try to determine the distribution law of r.v. X, where X

is the number of whithe ball among the 3 picked ball.

In fact, X assumes value 0 ， 1 ， 2 and

2. Characteristics of distribution law

Several Important Discrete

R.V.

(0-1) distribution

let X denote the number that event A appeared in a trail,

then X has the following distribution law

X ～ P{X ＝ k} ＝ pk(1 － p)1 － k, (0<p<1) k ＝ 0 ， 1

or X

kp

1 0

p p1

and X is said to follow a (0-1) distribution

Let X denote the numbers that event A appeared in a n-repeated Bernoulli experiment, then X is said to follow a binomial distribution with parameters n,p and represent it by XB （ n,p). The distribution law of X is given as :

),...,1,0(,)1(}{ nkppkXP knkk

nC

Example A soldier try to shot a bomber with probability 0.02 that he can hit the target, suppose the he independently give the target 400 shots, try to determine the probability that he hit the target at least for twice.

Poisson theorem If Xn~B(n, p), (n ＝ 0, 1, 2,…) and n is large e

nough, p is very small, denote =np ， then

,...2,1,0,!

}{ kek

kXPk

Answer Let X represent the number that hit the target in 400 shots Then X ～ B(400, 0.02), thus

P{X2} ＝ 1 － P{X ＝ 0} － P {X ＝ 1} ＝ 1 － 0.98400-400)(0.02)(0.98399)=…

Now, lets try to solve the aforementioned problem by putting

=np ＝ (400)(0.02) ＝ 8, then approximately we have

P{X2} ＝ 1 － P{X ＝ 0} － P {X ＝ 1}

＝ 1 － (1 ＋ 8)e － 8 ＝ 0.996981.

Poisson distribution

X ～ P{X ＝ k} ＝ , k ＝ 0, 1, 2, … (0)

e

!k

k

Poisson theorem indicates that Poisson distribution is the

limit distribution of binomial distribution, when n is large

enough and p is very small, then we can approximate

binomial distribution by putting =np.

Random variable Discrete r.v.

Distribution lawSeveral important r.v.s

0-1 distribution

Bionomial distributionPoisson distribution

2.3 Distribution function of r.v.

Definition Suppose that X is a r.v., for any real number

x ， Define the probability of event {Xx}, i.e. P {Xx} the distribution function of r.v. X, denote it by F(x), i.e

.

F(x) ＝ P {Xx}.

xX

}{lim}{0

0 xXPxXPxx

}0{ 0 xF

It is easy to find that for a, b (a<b),

P {a<Xb} ＝ P{Xb} － P{Xa} ＝ F(b) －F(a).

)0()(}{ xFxFxXP

For notation convenience, we usually denote distribution fucntion by d.f.

Characteristics of d.f.

1. If x1<x2, then F(x1)F(x2);

2. for all x ， 0F(x)1 ， and

;1)x(Flim)(F,0)x(Flim)(Fxx

).x(F)x(Flim)0x(F 0xx

00

3. right continuous ： for any x ，

Conversely, any function satisfying the above

three characteristics must be a d.f. of a r.v.

Example1 Suppose that X has distribution

law given by the table

)(xF

x0

1

1 2

}{)( xXPxF ＝

X 0 1 2

P 0.1 0.6 0.3Try to determine the d.f. of X

0 0

0.1 0 1

0.7 1 2

1 2

x

x

x

x

For discrete distributed r.v.,

X ～ P{X= xk} ＝ pk, k ＝ 1, 2, …

the distribution function of X is given by

xxkk

k

pxXPxF:

}{)(

Suppose that the d.f. of r.v. X is specified as follows,

2

212

1131

1

)(

xb

xx

x

xa

xF

Try to determine a,

b and }2{},2{},10{},0{ XPXPXPXP

Is there a more intuitive way to express the distribution

Law of a r.v. ？ Try to observe the following graph

a b

?}{ bXap

2.4 Continuous r.v.Probability density function

Definition Suppose that F(x) is the distribution function of r.v. X ， if there exists a nonnegative function f(x) ， (-<x<+)

， such that for any x ， we have

xduufxXPxF )()()( ＝＝

then it is said that X a continuous r.v. and f(x) the density function of X , i.e. X ～ f(x) , (-<x<+)

The geometric interpretation of density function

b

adu)u(f)bXa(P ＝

2. Characteristics of density function

(1 f(x)0 ， (-<x<) ；

(2) .1)( ＝

dxxf

(1) and (2) are the sufficient and necessary propertie

s of a density function

xaexf )(

Suppose the density function of r.v. X is

Try to determine the value of a.

(3) If x is the continuous points of f(x), then

)()(

xfdx

xdF

Suppose that the d.f. of r.v. X is specified as follows, try to determine the density function f(x)

02

11

02

1

)(xe

xexF

x

x

（ 4 ） For any b ， if X ～ f(x) ，

(-<x<) ， then P{X=b} ＝ 0 。

And

b

adxxfbXaP

bXaPbXaP

)(}{

}{}{

＝＝

＝

Example 1. Suppose that the density function of X is specified by

Try to determine 1)the d.f. F(x), 2)P{X(0.5,1.5)}

其他0

212

10

)( xx

xx

xf

Distribution function

Monotonicity

Standardized

Right continuous

Density function

F(x)…f(x)Nonnegative

P{a<X<b}

Suppose that the distribution function of X is specified by

0 1

( ) ln 1

1

x

F x x x e

x e

Try to determine

(1) P{X<2},P{0<X<3},P{2<X<e-0.1}.

(2)Density function f(x)

1. Uniformly distribution

if X ～ f(x) ＝ 1

,

0

a x bb a

， el se

。 。

0 a b

ab

cddxab

dxxfdXcPd

c

d

c

＝＝＝ 1

)(}{

)x(f

x

It is said that X are uniformly distributed in

interval (a, b) and denote it by X~U(a, b)

For any c, d (a<c<d<b) ， we have

Several Important continuous r.v.

2. Exponential distribution

If X ～

0x,0

0x,e)x(f

x

＝

It is said that X follows an exponential distribution with parameter >0, the d.f. of exponential distribution is

)x(f

x

0

0,0

0,1)(

x

xexF

x

＝

Example Suppose the age of a electronic instrument is X (year),

which follows an exponential distribution with parameter 0.5, try to

determine

(1)The probability that the age of the instrument is more than 2 years.

(2)If the instrument has already been used for 1 year and a half, then

try to determine the probability that it can be use 2 more years.

,00

05.0)(

x

xexf

0.5x

37.0)1(

1

2

0.5x edx0.5e2}P{X

}5.1|5.3{)2( XXP

37.0

1

1.5

0.5x

3.5

0.5x

e

dx0.5e

dx0.5e

}5.1{

}5.1,5.3{

XP

XXP

The normal distribution are one the most important

distribution in probability theory, which is widely applied

In management, statistics, finance and some other ereas.

3. Normal distribution

A

B

Suppose that the distance between A ， B is ，the observed value of X is X, then what is the density function of X ？

where is a constant and >0 ， then, X is said to follow

s a normal distribution with parameters and 2 and rep

resent it by X ～ N(, 2).

Suppose that the density fucntion of X is specified by

2

221~ ( )

2

x

X f x e x

(1) symmetry

the curve of density function is symmetry with respect to x= and

f() ＝ maxf(x) ＝ .2

1

Two important characteristics of Normal distribution

(2) influences the distribution

， the curve tends to be flat ，

， the curve tends to be sharp ，

4.Standard normal distribution A normal distribution with parameters ＝ 0 and 2 ＝ 1 is said to

follow standard normal distribution and represented by X~N(0, 1) 。

.,2

1)( 2

2

xexx

and the d.f. is given by

xdte

xXPx

xt

,

}{)(

221

2

the density function of normal distribution is

The value of (x) usually is not so easy to compute

directly, so how to use the normal distribution

table is important. The following two rules are

essential for attaining this purpose.

Z~N （ 0 ， 1 ） , （ 0.5)=0.6915,

P{1.32<Z<2.43}=(2.43)-(1.32)=0.9925-0.9066

注： (1) (x) ＝ 1 － ( － x) ；

(2) 若 X ～ N(, 2) ，则

).(}{)(

x

xXPxF

1 X~N(-1,22),P{-2.45<X<2.45}=?

2. XN(,2), P{-3<X<+3}?

EX2 tells us the important 3 rules, which are widely

applied in real world. Sometimes we take P{|X- |

≤3} ≈1 and ignore the probability of {|X- |>3}

Example The blood pressure of women at age 18 are

normally distributed with N(110,122).Now, choose a

women from the population, then try to determine (1)

P{X<105},P{100<X<120};(2)find the minimal x such that

P{X>x}<0.05

105 1101 { 105} 0.42 1 0.6628 0.3371

12P X

Answer:（）

120 110 100 110{100 120}

12 12

0.83 0.83 2 0.7967 1 0.5934

P X

{ } 0.05P X x (2)令

1101 0.05

12

x

1100.95

12

x

1101.645

12

x

129.74x

Distribution law of the function of discrete r.v.s

Distribution of the function of r.v.s

Suppose that X ～ P{X ＝ xk} ＝ pk, k ＝ 1, 2, …

and y ＝ g(x) is a real valued function, then Y ＝ g(

X) is also a r.v., try to determine the law of Y..

Example

X

Pk

-1 0 1

31

31

31

Determine the law of Y=X2

Y

Pk

1 0

31

32

or

Y ＝ g(X) ～ P{Y ＝ g(xk)} ＝ pk ， k ＝1, 2, …

GenerallyX

Pk

Y=g(X)

kxxx 21

kppp 21

)()()( 21 kxgxgxg

Density function of the function of continuous r.

v.

1. If Xf(x),-< x< +, Y=g(X) , then one can try to determine the density function, one can determine the d.f. of Y firstly

FY (y) ＝ P{Yy} ＝ P {g(X) y} ＝

yxgdxxf

)()(

dy

ydFyf Y

Y

)()(

and differentiate w.r.t. y yields the density funciton

Example Let XU(-1,1), tyr to determine the d.f. and density function of Y=X2

dxxfyF

xxgyxxf

yx

XY

X

2

2)(0

112

1

其它

ydxyFy

y

Y 2

1)(

其它0

102

1)(')(

yyyFyf YY

If y<0 0)( yFY

If 0≤y<1

If y≥1 1)( yFY

y y

Mathematical expectation

Definition 1. If X~P{X=xk}=pk, k=1,2,…n, define

n

kkk pxXE

1

)(

the mathematical expectation of r.v. X or mean of X.

1

||k

kk px

Definition 2. If X~P{X=xk}=pk, k=1,2,…, and

define

1

( ) k kk

E X x p

the mathematical expectation of r.v. X

Example 2 Toss an urn and denote the points by X, try to determine the mathematical expectation of X.

6

1

1 7( )

6 2k

E X k

Definition 3 Suppose that X~f(x), -<x<, and

.)()( dxxxfXE

dxxfx )(||

the mathematical expectation of r.v. X

then define

Example 3. Suppose that r.v. X follows Laplace distribution with density functio

n

Try to determine E(X).

x

xf exp2

1)(

dxxx

XE

exp

2)(

Let

exp | |2

xt

tt dt

dttexp0

Mathematical expectation of several important r.v.s

1. 0-1 distribution

ppP

X

1

01EX=p

2. Binomial distribution B(n, p)

n

k

knk ppknk

nkXE

1

)1()!(!

!)(

nkppCkXP knkkn ,...1.0)1(}{

knkn

k

ppknk

n

)1()!()!1(

!

1

)1(11

1

)1()!()!1(

)!1(

knk

n

k

ppknk

nnp

np

lnln

l

ln ppCnpkl

1

1

01 )1(1令

3.Poisson distribution

...,2,1,0,!

}{~ kek

kXPXk

4. Uniform distribution U(a, b)

1, ,

~ ( )0, ,

a x bX f x b a

el se

0 1

1

;)!1(!

)(k k

kk

kee

kkXE

b

a

badxab

xXE ;

2)(

5.Exponential distribution

00

0)(

x

xexf

x

1

dxexXE x

0

)(

0

xxde

dxexe xx

0

0

6. Normal distribution N(, 2)

x,e21)x(f~X

2

2

2

)x(

dxex

XEx

2

2

2

)(

2)(

2

2 ;2

tx tt e dt

令

EX 1 Suppose that the distribution law of X

Try to determine the mathematical expectation of Y=X2

X

Pk

-1 0 1

31

31

31

Y

Pk

1 0

31

32 3

2

3

10

3

21)( YE

Mathematical expectation of the functions of r.v.s

Theorem 1 let X~P{X=xk}=pk, k=1,2,…,then the mat

hematical expectation of Y=g(X) is given by the follo

wing equation and denoted by E(g(X))

.)()]([)(1

kk

k pxgXgEYE

Answer Y=ax is strictly monotonic with respect to x with inverse function

a

byyh

)(

Thus the density function of Y is given by

aa

byfyf XY

1)()(

EX2 ： Suppose that r.v. X follows standardized distribution

Try to determine the mathematical expectation of Y=aX+b

dya

ey

YEa

by

1

2)( 2

2

dxe

bax x

2

2

2

b

ae

a

by

1

2

12

2

Theorem 2 If X~f(x), -<x<, the mathematical expectation of Y=g(X) is specified as

.)()()]([)( dxxfxgXgEYE

Suppose that X follows N(0 ， 1) distribution, try

to determine E(X2), E(X3) and E(X4)

2

2

2

1)(

x

exf

dxex

XEx

22

2

2

2)(

2

2

2

x

dex

dxex

2

2

2

1

1

dxex

XEx

23

3

2

2)(

0

dxex

XEx

24

4

2

2)(

23 2

2

x

dex

dxex x

22 2

23

3

1. E(c)=c, where c is a constant;

2 。 E(cX)=cE(X), c is a constant.

Properties of mathematical expectation

Proof. Let X~f(x), then

dxxcxfcXE )()(

)()( XcEdxxxfc

Example 2. Some disease will occur with probability 1% ， investigate 1000 people now, it is necessary to check the blood. The method is clarified these people into ten group with each group 100 and check the mixed blood sample. If the result is negative, it is not need to do any test any more, if it is positive, then ,it is necessary to test each blood sample respectively, try to determine the average times needed for the test.

Let Xj is the number to be taken of jth group, and X the number to be taken in 1000 people, then

10,...1j

Xj

Pj

1 101

100%)99( 100%)99(1

]99.0100

11[1000 100

)()()(10

1

10

1

j

jj

j XEXEXE

)]99.01)(101(99.0[10 100100

644

)99.01)(101(99.0 100100 jEX

Variance

Definition 1 Suppose that X is a r.v. with EX2<∞, define E (X-EX)2 the variance of r.v. X and denote it by DX

E (X-EX)2= E [X2-2EXEX +E2X ]=EX2- E2X

The variance of several important r.v.s

(0-1) distribution with P(X=0)=p, then DX=pq

Binomial distribution B(n,p), then DX=npq

Poisson distribution with parameter

then, DX=EX=

The variance of several important r.v.s

Uniform distribution: Suppose that r.v. X is uniformly distributed on interval [a,b], then

Normal distribution: then

Exponential distribution: then

21( )

12DX b a

( )X E

2

1DX

2( , )X N 2DX

Definition of mathematical expectation

Properties of mathematical expectation

Mathematical expectation of the functions of r.v.

Several important expectation