S TATIS TIKA UGM YOGYAKARTA

Introduction of Mathematical S tatistics 2

By :Indri Rivani Purwanti (10990)Gempur Safar (10877)Windu Pramana Putra Barus (10835)Adhiarsa Rakhman (11063)

Dosen :. . , . ., . .Prof Dr Sri Haryatmi Kartiko S Si M Sc

THE US E OFMATHEMATICAL

S TATIS TICS

Introduction to M athematical Statistics (IM S) can be applied for the whole statistics subject, such as:

Statistical M ethods I and II Introduction to Probability M odels M aximum Likelihood Estimation Waiting Times Theory Analysis of L ife-testing models Introduction to Reliability Nonparametric Statistical M ethods etc.

S TATIS TICAL METHODS

In Statistical M ethods, Introduction of M athematical Statistics are used to:

introduce and explain about the random variables , probability models and the suitable cases which can be solve by the right probability models.

How to determine mean (expected value), variance and covariance of some random variables,

Determining the convidence intervals of certain random variables

Etc.

Lee J. Bain & Max Engelhardt

Probability Models

M athematical Statistics also describing the probability model that being discussed by the staticians.

The IM S being used to make student easy in mastering how to decide the right probability models for certain random variables.

Lee J. Bain & Max Engelhardt

INTRODUCTION OF RELIAB ILITY

The most basic is the reliability function that corresponds to probability of failure after time t.

The reliability concepts:If a random variable X represents the lifetime of failure of a unit, then the reliability of the unit t is defined to be:

R (t) = P ( X > t ) = 1 – F x (t)

Lee J. Bain & Max Engelhardt

MAXIMUM LIKELIHOOD ES TIMATION

IM S is introduces us to the M LE,

Let L(0) = f (x1,....,xn:0), 0 Є Ω, be the joint pdf of X1,....,Xn. For a given set bof observatios, (x1,....,xn:0), a value in Ω at which L (0) is a maximum and called the maximum likelihood estimate of θ. That is , is a value of 0 that statifies f (x1,....,xn: ) = max f (x1,....,xn:0),

Lee J. Bain & Max Engelhardt

ANALYS IS OF LIFE-TES TING MODELS

M ost of the statistical analysis for parametric life-testing models have been developed for the exponential and weibull models.

The exponential model is generally easier to analyze because of the simplicity of the functional form.

Weibull model is more flexibel , and thus it provides a more realistic model in many applications , particularly those involving wearout and aging.

Lee J. Bain & Max Engelhardt

NONPARAMETRIC S TATIS TICAL METHODS

The IM S also introduce to us the nonparametrical methods of solving a statistical problem, such as:

one-sample sign testBinomial Test Two-sample sign test wilcoxon paired-sample signed-rank test wilcoxon and mann-whitney tests correlation tests-tests of independence wald-wolfowitz runs test etc.

Lee J. Bain & Max Engelhardt

K ETERK AITAN K ONVERGENSI

EXAMPLEWe consider the sequence of ”standardized” variables:

( ) ( )n n n

n n

Y np t nptZ Y

n

tM t e e Mσ σσ

− − = =

nn

Y npZnpq−=

With the simplified notation n npqσ =By using the series expansion

( ) ( )n n n nnnnpt t pt te pe q e pe qσ σ σ σ− − = + = +

21 2ue u u= + + + L

( )2 2 2

2 21 1 12 2

n

n n n n

pt p t t tp pσ σ σ σ

= − + + + + + + −

L L

( )2

12

nd nt

n n

= + +

Where d(n) → 0 as n →

∞

( ) 2 2limn

tZn

M t e→ ∞

=

( )0,1dnZ Z N∴ → :

APPROXIMATION FOR THE BINOMIAL DIS TR IBUTION

[ ] 0.5 0.5n

b np a npP a Y bnpq npq

+ − − −≤ ≤ = Φ − Φ

Example:A certain type of weapon has probability p of working successfully. We test n weapons, and the stockpile is replaced if the number of failures, X, is at least one. How large must n be to have P[X ≥ 1] = 0.99 when p = 0.95?Use normal approximation.

[ ]1 0.99P X ≥ =

[ ]1 0 0.99P X− ≤ =

0 0.5 0.051 0.990.05 0.95

nn

+ − − Φ = g g

0.5 0.05 0.010.218

nn

− Φ = 0.5 0.05 2.330.218

nn

− = −

20.25 0.05 0.0025 0.258n n n− + =20.0025 0.308 0.25 0n n− + =

2 24 0.308 0.308 4 0.0025 0.25 122 ( )2 2 0.0025

b b acna

− ± − ± −= = = +g gg

X : number of failuresp : probability of working successfully = 0.95q : probability of working failure = 0.05

ASYMPTOTIC NORMAL DISTRIBUTIONS

( )0.1dnn

Y mZ Z Nc n

−= → :

nX

If Y1, Y2, … is a sequence of random variables and m and c are constants such that

as , then Yn is said to have an asymptotic normal distribution with asymptotic mean m and asymptotic variance c2/n.

Example:The random sample involve n = 40 lifetimes of electrical parts, Xi ~ EXP(100). By the CLT,

has an asymptotic normal distribution with mean m = 100 and variance c2/n = 1002/ 40 = 250.

n → ∞

100100

40

n nn

X XZn

µσ

− −= =2 2

(100)( ) 100

( ) 100

iX EXPE XVar X

θθ

= == =

:

AS YMPTOTIC DIS TRIBUTION OF CENTRAL ORDER S TATIS TICS

TheoremLet X1, …, Xn be a random sample from a continuous distribution with a pdf f(x) that is continuous and nonzero at the pth percentile, xp, for 0 < p < 1. If k/n → p(with k – np bounded), then the sequence of kth order statistics, Xk:n, is asymptotically normal with mean xp and variance c2/n, where

22

(1 )

( )p

p pcf x

−=

• ExampleLet X1, …, Xn be a random sample from an exponential distribution, Xi ~ EXP(1), so that f(x) = e-x and F(x) = 1 – e-x; x > 0. For odd n, let k = (n+1)/2, so that Yk = Xk:n is the sample median. If p = 0.5, then the median is x0.5 = - ln (0.5) = ln 2 and

[ ]2

2 2

0.5(1 0.5) 0.25 1(0.5)(ln 2)

cf

−= = =

Thus, Xk:n is asymptotically normal with asymptotic mean x0.5 = ln 2 and asymptotic variance c2/n = 1/n.

( ) 0.50.5 0.50.5 1 xx F x e−= = = − 0.5

0.50.5 ln 0.5xe x−⇔ = ⇔ − =

1

0.51ln 0.5 ln ln 22

x−

⇔ = − = =

THEOREMIf p

nY m →then

( ) 2

( )( ) 1 nn n

Var YP Y E Y εε

− < ≥ −

( ) (0,1)dn nZ n Y m c Z N= − → :

( ) nn n n

cZZ n Y m c Y mn

= − ⇔ = +

( ) ( ) .0nn n

c ccZE Y E m E Z m m mn n n

= + = + = + =

( ) ( )2 2 2

.1nn n

c c ccZVar Y Var m Var Zn n n n

= + = = =

( ) 2

( )( ) 1 nn n

Var YP Y E Y εε

− < ≥ −

( )2

2lim lim 1 1nn n

cP Y mn

εε→ ∞ → ∞

− < ≥ − =

Proof

( )2

21ncP Y m

nε

ε⇒ − < ≥ −

pnY m∴ →

THEOREM

pnY Y →

For a sequence of random variables, if

thend

nY Y →

For the special case For the special case Y = c, the limiting distribution is the degenerate distribution P[Y = c] = 1. this was the condition we initially used to define stochastic convergence.

pnY c →

( ) ( )png Y g c →

, then for any function g(y) that is continuous at c,If

THEOREM p

nX c →If Xn and Yn are two sequences of random variables such that and pnY d →

then:

[ ][ ]

1. .

2. .

3. c 1, for 0.

4. 1 1 if 0 1 for all , c 0.

5. if 0 1 for all .

pn n

pn n

pn

pn n

pn n

aX bY ac bdX Y cdX c

X c P X n

X c P X n

+ → +

→

→ ≠

→ ≠ = ≠

→ ≥ =

ExampleSuppose that Y~BIN(n, p).

( ) 2

ˆ( )ˆ ˆ( ) 1 Var pP p E p εε

− < ≥ −

( )ˆ( )E p E Y n np n p= = = 2ˆ( ) ( )Var p Var Y n npq n pq n= = =

( ) 2ˆ 1 pqP p pn

εε

− < ≥ − ( ) 2ˆlim lim 1 1

n n

pqP p pn

εε→ ∞ → ∞

− < ≥ − =

Thus it follows that ( ) ( )ˆ ˆ1 1pp p p p− → −ˆ pp Y n p= →

Theorem Slutsky’s Theorem If Xn and Yn are two sequences of random variables such that

pnX c → and , then:d

nY Y →

1. .

2. .

3. , for 0.

dn n

dn n

dn n

X Y c YX Y cYY X Y c c

+ → +

→

→ ≠

Note that as a special case Xn could be an ordinary numerical sequence such as Xn = n/(n-1).

nIf Y ,d Y → then for any continuous function g(y), ( ) ( )ng Y .d g Y →

( ) (0.1),dnIf n Y m c Z N− → : and if g(y) has a nonzero derivative at y = m, ( )' 0, theng m ≠

( ) ( )( )

(0.1)'

n dn g Y g mZ N

cg m− → :

Any Question ? ? ?