Statistical Distributions

． Bernoulli Distribution

． Binomial Distribution

． Hypergeometric Distribution

． Multinomial Distribution

． Poisson Distribution

Discrete Type

． Normal Distribution

． Gamma Distribution

． Exponential Distribution

． Chi-square Distribution

． Student’s T Distribution

． F Distribution

Continuous Type

． Bernoulli Distribution

An experiment that can have one(X) of two outcomes: Success(S, x=1), Failure(F, x=0) Bernoulli experiment P(S) = P(X=1) = p , P(F) = P(X=0) = 1 - p = q

Probability distribution function

..,

,,)()()(

wo

xppxXPxf

xx

0

101 1

)(~ pBerX

pqppXEXEXVar 22222 )()()(

pXE )(

． Binomial Distribution

Repeated n times independent Bernoulli experiment

i.e. a Binomial experiment possesses the following properties:

10 the experiment consists of a fixed number n of trials 20 the result of each trial can be classified into one of two categories

30 the probability p of a success remains constant for each trial

40 each trial of the experiment is independent of the other trials

Binomial experiment

Let r.v. X be the number of successes in the n trials of a Binomialexperiment, then X is called the Binomial distribution, ),(~ pnBinX .

The probability distribution function is

..,

,,,,,)()()(

wo

nxppCxXPxf

xnxnx

0

2101

where)!(!

!xnx

nC n

x

npXE )(

)()()()( pnpXEXEXVar 12222

mean

variance

Let nXXX ,,, 21 be a random sample with Ber(p) and

nXXXX 21 , then

),(~ pnBinX and

np

XEXEXE

XXXEXE

n

n

)()()(

)()(

21

21

)(

)()()(

)()(

pnp

XVarXVarXVar

XXXVarXVar

n

n

121

212

． Hypergeometric Distribution

◆ Sampling with replacement (WR) v.s. Sampling without replacement (WTR)

R red balls

N balls n ballssampling

Population

Sample

Let X be the number of red balls in the sample, then the distributionof X is the hypergeometric distribution, ),,(~ NRnHypX

The probability distribution function (PDF) is

..,

},min{},max{,)()(

wo

RnxRNnC

CCxXPxf N

n

RNxn

Rx

0

0

meanN

RnXE )(

.

variance1

12

NnN

NR

NR

nXVar )(

Theorem:

If ),,(~ NRnHypX , then for each value nx ,,,, 210 , and as

N and R with pNR , a positive constant ,

xnxnxN

n

RNxn

Rx

NppC

C

CC

)(lim 1

Example:

Ten seeds are selected from a bin that contains 1000 flower seeds,

of which 400 are red flowering seeds, and the rest are of other colors.

10 P(exactly five red flowering seeds)=0.2013

mean = 4 variance = 2.378

20 P(five red flowering seeds by the Binomial approximation)=0.2007

mean = 4 variance = 2.40

Extension:

． Multinomial Distribution

If each trial has several different outcomes, label the different possible

types resulting from each trial by i where ,,,, ki 21

the probability of each type at each trial is pi , and the count of each

of the types in a sample of size n as Xi , then the probability of),,,( kXXXX 21 is

k

ii

k

ii

k

i

xik

ii

xk

xx

k

kkk

pnxpx

n

pppxxx

n

xXxXxXPxxxf

i

k

111

1

2121

221121

1

21

,,!

!

!!!

!

),,,(),,,(

),,,,(~ kpppnMULTX 21

H.W.

寫出 “ Extended Hypergeometric Distribution”

． Poisson Distribution

the number of cars that are red, out of every 10 cars that pass a

certain spot on a road

Binomial distribution

the number of red cars that pass the spot per hour, without specifying

how many cars in total there are

Poisson distribution

)(~ PoiXr.v.

..,

,,,,!)()(

wo

xx

exXPxf

x

0

210

mean = variance = λ

例 :

假設到達某醫院病患人數符合 Poisson 過程 , 且平均每小時 1人到達 , 則

10 P(1 小時內無病患到達 ) 3679001

00 101

.!

)()(

ee

fXP

20 P(1 小時內病患到達人數少於 4 人 )

981001

343

0

1

.!

)()(

x

x

x

eXPXP

Hypergeometric distribution

n/N 0.05≦

Binomial distribution

Poisson distribution

n large, p small(rare event)

例 :

假設某種疾病治癒率為 2% , 若今有 100 位病患接受治療 , 試求最多三人被治癒之機率。

<Sol. Binomial>

令 r.v. X 表治癒人數 , 則 ).,(~ 020100BinX

859002010203 1003

0

100 ...)(

xx

xxCXP

<Sol. Poisson>

)(~)(., 22020100 PoiXnpsmallpn

85702

33

0

2

.!

)(

x

x

xe

XP

． Normal Distribution

X

Xμ

f(x)

r.v. X ~ N(μ, σ2)

02

1 2

2

2

,,,)( Rxexfx

the pdf for X is

2 )(,)( XVarXE

μ-σ μ+σ

X ~ N(μ, σ2)

μ-2σ μ+2σ

9970333

9540222

6830

.)()(

.)()(

.)()(

XPXP

XPXP

XPXP

μ

X ~ N(μ, σ2)

μa b

??)()()()(

b

a

xba dxedxxfaFbFbXaP

2

2

2

21

● X ~ N(μ, σ2) normalized Z ~ N( 0 , 1)

),(~ 10NX

Z

Z0

ψ(z)

● the pdf for X is

z

ezz

,)( 2

2

21

Z ~ N( 0 , 1)

● ??)()()(

z

zz dzedzzzzZP 2

2

21

z

( 查表 )

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0.00.10.20.30.40.50.60.70.80.91.01.11.21.31.41.51.61.71.81.92.02.1

0.50000.53980.57930.61790.65540.69150.72570.75800.78810.81590.84130.86430.88490.90320.91920.93320.94520.95540.96410.97130.97720.9821

0.5040               

0.9463  

0.97190.9778

 

0.5080               

0.9474  

0.97260.9783

 

0.5120               

0.9484  

0.97320.9788

 

0.5160               

0.9495  

0.97380.9793

 

0.51990.55960.59870.63680.67360.70880.74220.77340.80230.82890.85310.87490.89440.91150.92650.93940.95050.95990.96780.97440.97980.9842

0.5239               

0.9515  

0.9750  

0.5279               

0.9525  

0.9756  

0.5319               

0.9535  

0.9761 

0.5359               

0.9545  

0.9767  

Standard Normal Cumulative Distribution Function Φ(z)

● Properties : Z ~ N( 0 , 1)

50000 .)()( ZP

Z0

ψ(z)Z ~ N( 0 , 1)

z-z

10

20

)(

)(

)(

)()(

z

zZP

zZP

zZPz

1

1

ab

bXaPaFbFbXaP )()()(

30 X ~ N(μ, σ2)

例 : 1. P(– 0.15 Z 1.60) = Φ(1.60) – Φ(– 0.15)≦ ≦ = Φ(1.60) – [1 – Φ(0.15)] = 0.9452 – [1 – 0.5596] = 0.9452 – 0.4404 = 0.5048

2. P(Z – 1.9 or Z 2.1) ≦ ≧ = P(Z – 1.9) + P( Z 2.1) ≦ ≧ = Φ(– 1.9) + [1 – Φ(2.1)] = [1 – Φ(1.9)] + [1 – Φ(2.1)] = [1 – 0.9713] + [1 – 0.9821] = 0.0287 + 0.0179 = 0.0466

0 z

Z ~ N( 0 , 1)

● )()( zZPzZP

z

α α

● 11 )()()()( zzzz

6451961 0500250 .,. .. zz例 :

例 : 假設某一族群男性體重 (X) 呈常態分配 , 平均體重與標準差分 別是 80 及 5 公斤 , 則體重介於 65 和 75 公斤之間的比例有

15740

01035

8075

5

80

5

80657565

.

)..(

)()(

ZP

XPXP

H.W. 體重超過 85 公斤的比例有多少 ?

●

),(~)(

..

))(,(~

)(,),(~..

101

1

515

Npnp

npXZei

pnpnpNX

pnnpandpnBinXvr

10

)(.

pnpnpx

ZPxXP1

50

)(.

pnpnpx

ZPxXP1

5020

例 : Consider sample allele proportions for the ABO blood group system. For a sample of size 16 alleles from a population in which allele A has proportion 0.50, the numbers of A are X.

ondistributiBinomialbyXP 227206 .10

20 ionapproximatNormalbyZPXP 22660505016

50501666 .)

..

..(

． Gamma Distribution

),(~.. GamXvr

0001 1

,,,)(

)( xexxfpdfx

)( XE

2)( XVar

． Exponential Distribution

In the Gamma distribution ,),(~.. GamXvr

if

01

xexfpdfEXPXvr

x

,)(,)(~..

2 )(,)( XVarXE

1 then

Theorem: no-memory property

)()()(~ tXPaXtaXPExpX

． Chi-square Distribution

In the Gamma distribution ,),(~.. GamXvr

if then2

2 ,

02

1 21

2

2 2

2

xexxfpdfXvr

x

,)(

)(,)(~..

2 )(,)( XVarXE

例 :

)(~)(

),.(~,,,..

111 2

2

122

22

221

nXXSn

NXXXsr

n

ii

n

22

2

22 1

1

)(

)()(

SE

nSn

EE

12

112

121

4

2

222

2

22

nnnSVar

nSn

VarVar

)(

)()()(

)()(

)(

． Student’s T Distribution

)(~)(~),(~

tVZ

TthentindependenareVandNZif 210

Rxx

xfpdf

,)(2

12

11

2

21

20

)(,)( XVarXE

． F Distribution

),(~)(~)(~ 21

2

2

1

1

22

212

1

FV

V

XthentindependenareVandVif

01

22

2 2

2

1122

2

1

21

21 21

1

1

xxxxfpdf ,)(

)(

22

221

2122

22

2

442

22

22

,)()(

)()(

,)(

XVar

XE

例 :

),(~

)(

)(

)(

)(

)(

)(

)(~)(

)(~)(

),.(~,,,..),.(~,,,..

11

1

1

1

1

1

1

111

111

2

2

2

2

2

2

2

2

2

2

22

122

22

22

122

22

221

221

yxy

x

x

y

y

y

yy

x

x

xx

y

y

x

x

y

n

ii

yy

yyy

x

n

ii

xx

xxx

ynxn

nnFS

S

n

Sn

n

Sn

n

nF

tindependenarenYYSn

andnXXSn

tindependenare

NYYYsrandNXXXsr

y

x

yx

Basic Sampling

Distribution Theory

● Statistic is a function of a random sample.

e.q.

n

iiX

nX

1

1

● Statistic is a random variable. Sampling distribution

e.q.

),(~

),(~,,,

n

n

ii

n

NXn

X

NXXXsamplerandom

2

1

221

1

Inferential statistics

◆ Estimation of parameter

◆ Testing of statistical hypothesis

估計 (Estimation): 由母體抽出樣本 , 依據樣本統計量的 抽樣分配 , 推估母體參數真實值。

點估計 (point estimation)

區間估計 (interval estimation)

點估計 (point estimation)

e.q.

2S

Sample mean X Population meanEst.

Sample variance2Est. Population variance

Sample proportion pn

xp ˆ Est. Population proportion

i.e. ̂ Est. (Statistic) (Parameter)

‧ 不偏性 (unbiased)

假設 T 為母體未知參數 θ 之一估計量 , 若 E(T) = θ, 則 T 為 θ 的不偏估計量 (unbiased estimator)

參數的估計是否只存在唯一的估計量 ( 統計量 )?

‧ 有效性 (efficient)

假設 T1 及 T2 均為母體未知參數 θ 之不偏估計量 , 且 Var(T1) < Var(T2), 則 T1 較 T2 具有效性

‧ 一致性 (consistant)

假設 T 為母體未知參數 θ 之一估計量 , 若 , 則 T 為 θ 的一致估計量 (consistant estimator)

0

)(lim TVarn

例 : r.s. 321 XXX ,, 具期望值 μ , 變異數 σ2

令

則3

32

3

2

3321

321

2321

1XXX

TXX

TXXX

T

,,

33

3211

XXXETE )(

)( 2TE

23 )(TE

9

3

33

2

2

222321

1

XXX

VarTVar )(

9

5 2

2)(TVar

Unbiased estimator

例 : r.s. ),(~,,, 2

21 NXXX n

),(~ n

n

ii NX

nX

2

1

1

02

nXVar

XE

nn

limlim Consistant estimator

Unbiased estimator