Computing Interval Estimates for Components of Statistical Information

with Respect to Judgements on Probability Density

Functions

Victor G. Krymsky

Ufa State Aviation Technical University, Russia;

E-mail: kvg@mail.rb.ru

Kgs. Lyngby, Denmark, 2004

Imprecise Prevision Theory (IPT)

Starting Points: Fundamental Publications

[1], [2]

[1] Walley P., Statistical reasoning with imprecise probabilities, Chapman and Hall, New York, (1991);

[2] Kuznetsov V., Interval statistical models, Radio and Sviaz, Moscow, (1991) (in Russian).

Traditional Problem Formulation

in the Framework of IPT Constraints:

,0)( x

(1) .,...,2,1 ,)()(0

niadxxxfaT

iii

)()(inf0

)(

T

xdxxxg

It is necessary to find:

(2) )()(sup0)(T

xdxxxg

subject to constraints (1).

,1)(0 T

dxx and

as well as

Dual for the Initial Problem Statement

n

i

iiii

idiccadaccgM

10

,,0

)(sup

).()()(1

0 xgxfdccn

iiii

RdcRc ii , ,0subject to and for any x ≥ 0,

i=1,2,…,n:

n

iiiii

idiccadaccgM

10

,,0

)(inf)(

And

subject to RdcRc ii , ,0 and for any x ≥ 0,

i=1,2,…,n:

).()()(1

0 xgxfdccn

iiii

(3)

(4)

(5)

(6)

Let us find:

Important Conclusion Concerning Optimal

Solutions

(L. Utkin and I. Kozine, [3])

[3] Utkin L. and Kozine I. Different faces of the natural extension. In: Proceedings of the Second International Symposium on Imprecise Probabilities and Their Applications, ISIPTA '01, 2001, pp. 316-323.

Optimal solutions belong to a family of DEGENERATE distributions

(such probability densities are composed of δ-functions)

Distribution of Probabilistic Masses

Masses are concentrated in the fixed points:

∞

Δ x→0

x

Density

0

Use of Additional Judgements

Additional judgement can be

reflected by inequality:

(x)K=const,(7)

where is such that RK .1KT

T0 x

K

ρ(x)

Main Goal(Theorem)

If there is no any finite interval for which function g(x) can be represented in the form

(8)

where then function ρ(x) providing solution of optimization problem mentioned above, belongs to class of step-functions with minimum value equal to 0 and maximum value equal to K.

, ],,0[],[ Tx

n

iii xfhhxg

10 ),()(

,,...,, 10 Rhhh n

Some Comments

To provide (8) the system

)()(

.............................

),()(

),()(

),()(

)(

1

)(

1

1

10

xgxfh

xgxfh

xgxfh

xgxfhh

nn

i

nii

n

iii

n

iii

n

iii

must have at least one solution which is independent on x in some interval ].[α,βx

Applying Methodology of the Calculus of VariationsThe inequalities

(9)

should be excluded from direct consideration in order to allow operating in the open domain with the values of the function.

The requirement (x)≥0 can be replaced by denoting).()( 2 xzx The requirement (x)≤K can be reflected by equality

,)()( 22 Kxvxz where v(x) is newly introduced function.

0(x)K

(10)

(11)

Modified Formulation of the Problem

We would like to estimate

and

subject to

T

xzdxxzxg

0

2

)()()(inf

T

xzdxxzxg

0

2

)()()(sup

,)()( 22 Kxvxz

,1)(0

2 T

dxxz

,)()(0

2i

T

i adxxzxf .,...,2,1 ,)()(

0

2 niadxxzxf i

T

i

(12)

(13)

(14)

(15)

(16)

Lagrange Approach

n

iii xzxfxz

xvxzxxzxgvzF

1

220

222

)()()(

)()()()()(),(

n

niii xzxf

2

1

2 )()(

)()(),( 2 xzxgvzF

;0),(),(

z

vzF

dx

d

z

vzF

.0),(),(

v

vzF

dx

d

v

vzF

Equations of Euler – Lagrange:

(17)

The Necessary Conditions of Optimality

The equations look here as

follows:

;0)()()()()(1

0

n

iinii xfxxgxz

.0)()( xvx

. ],,0[],[ Tx

Let us fix any interval

Case 1. 0)( xz

inside the interval.Kxv )( .0)( xThen and

(18)

Case 2. ,0)( xz so0)( xv and .)( Kxz

Practical Implementation

Optimal probability density:

x x x x x x … x0 1 2 3 4 5

K

ρ(x)

,)(),(1

1

jx

jx

jj dxxgxxG

,...,n.,idxxfxx

jx

jx

ijji 21 ,)(),(

1

1

Denote:

(19)

(20)

0

Reformulation of the Problem Statement

We would like to estimate

,),(min0

122,...1,0

m

jjj

xxxxGK

m

jjj

xxxxGK

0122

,...1,0

),(max

m

jjj xxK

0212 ;1)(

.,...,2,1 ,),(0

122 niaxxKa i

m

jjjii

subject to

(21)

(22)

(23)

(24)

Example 1

The information concerning a continuous random variable X

is where K, T are fixed positive numbers. What are the bounds for the expectation M(X)?

* * *

Let us choose m=0.

Objective function:

,)()()( ],0[ xIKxx T

.2

)(1

0

20

21

0

x

x

Txx

KxdxKdxxxJ

Solution of Optimization Problem

Lower and upper bounds of J

interval:

.2

1min

KJ

.2

1

2

1)/1(max

KT

KKTJ

1/2K 1/K T

ρ(x)

K

(T-1/K) (T-1/2K) T

0

0

x

x

ρ(x)

K

Example 2

We add the constraint:

where is the indicator function. Here also any finite interval of x values for which

cannot be found, so the theorem can be applied. Further analysis shows, that m=1 is the best choice for such situation.

T

bdxxxf0

1 ,)()(

)()(],[1 xIxf

aa

)()(],[10 xIccxxg

aa

Example 2(Continuation 1)

To provide

we have to set:

(i) if

;afor 0

;for

;1

for 0

;1

0for

)(

TxK

bK

baxaK

axK

bK

bxK

x

minJ

:1

K

ba

Example 2(Continuation 2)

(ii) if :1

K

ba

.1

for 0

;1

afor

;for 0

;0for

)(

TxK

bKaa

K

bKaaxK

axK

ba

K

baxK

x

K

bKab

KJ

)1(

2

1min

As the result

.

)(1)(

2

1min

K

bKaabKaKa

KJ

or

Acknowledgements

• The research was initiated by Dr. Igor Kozine of Risø National Laboratory, Denmark, whose kind attention to this work is gratefully acknow-ledged.

• The work was partially supported by the grant T02-3.2-346 of Russian Ministry for Education which is also acknowledged.