The Generalized Mean and Error Analysis

The Generalized Mean and Error Analysis

The Affiliated High School of South China Normal University Author: Ketian Zhang

Faculty Adviser: Yiwen Huang

The Generalized Mean and Error Analysis

Abstract: In this paper, the definition of generalized mean 1 ( )] [

1

1

∑

=

= − ⁿ

i i

f f x

f n

x is

introduced. It summarizes the common characterizations and also expands the meaning of arithmetic, geometric and harmonic means. Secondly, the definition of average error is introduced, and the method of reducing error by averaging the values out is discussed. Based on the properties of convex function and the Jensen Inequality, the judgment of x_f ≤ x_g and the necessary and sufficient condition of x_f =x_g are discussed. Finally, the definition of effect coefficient which measures the effect of extreme values on a generalized mean is introduced. Some applications are contained in the paper as well.

Keywords: Mean; Error; Convex function; Jensen Inequality.

The aim and background of the research: What are the similarities of arithmetic, geometric and harmonic means? Whether the meaning of mean could be expanded? In scientific experiments, we often reduce errors by averaging the values. Are the effects different if we choose different means? In order to reduce the effect of extreme values, how to choose a kind of mean?

【 Main results ^】

The notations used in this paper are listed as follows.

I. Let and U be continuous random variables which satisfy (0， 1]-uniform distribution.

Un

U U¹, ²,...,

II. Denote by ⁱ

n

i

n U

A n

∑

=

=

1

1 ,

⎟

⎠

⎜ ⎞

⎝ ⎛∏

=

= n

i

G

Ui

n n

1

1

, ⁿ⁼

∑

H n

1

= i ₁Ui

.

III. For a continuous random variable X , let and denote the distribution function and the density function of

) (x

F_X p_X(x) X , respectively.

1. Generalized mean

Definition 1. If the inverse function of f(x) exists, then 1 ( )] [

1

1

∑

=

= − ⁿ

i i

f f x

f n

x is

called the f(x)-mean of x¹,x²,…,xⁿ.

Theorem 1. The arithmetic mean is x-mean, the geometric mean is -mean, and the harmonic mean is

lnx x

1-mean.

Proof. (1) Let a(x)=x. Then a⁻¹(x)=x

∑

=

=

⇒ ⁿ

i i

a x

x n

1

1 , that is, the arithmetic mean of x¹,x²,…,xⁿ.

(2) Let g(x)=lnx. Then g⁻¹(x)=e^x. For x¹,x²,…,xⁿ∈(0,+∞),

n n

i n i

n

i i n

i i n

i

g i x e x x

e n n x

e

x

∑ ∏ ∏ ∏

=

=

=

=

=

=

=

=

1 1

1 1

) (ln

^ ) 1ln

(

^ ) 1 ln

(

^ ,

which is the geometric mean of x¹,x²,…,xⁿ. (3) Let

x x

h 1

)

( = . Then x x

h 1

)

1( =

− . For x¹,x²,…,xn∈(0,+∞),

∑

∑

=

= _n

i i

n

i i

h

x n x n x

1 1

1 1

1

1 ,

which is the harmonic mean of x¹,x²,…,xⁿ. ■

Theorem 2. If the inverse function of f(x) exists, and g(x)=af(x)+b ( ), then

≠0 a

f

g x

x ≡ .

Proof. Note that the inverse function of g(x) is ¹( ) ¹( ) a

b f x

x

g⁻ = ⁻ − . Then

]}

) ( 1 [

{

1

1

∑

=

− +

= ⁿ

i

i

g af x b

g n x

] ) 1 (

[

1

1 f x b

an g

n

i

i +

=

∑

=

−

f n

i

i x

x n f

f =

=

∑

=

− 1 ( )]

[

1

1 ,

that is, xg =xf. ■

2. Average error

The variance and standard deviation usually used in mathematical statistics can characterize the data volatility, but the meaning of their specific value is not so straightforward. Therefore, the definition of average error is introduced. Moreover, we shall study some problems by using this definition.

Definition 2. For a random variable X , e^X =E(X −E(X)) is defined to be the average error of X. Moreover, e(x_f)=E(x_f −E(X)) is called the -average error of

) (x f X.

2.1. Basic properties of average error

Obviously, for a continuous random variable X , e^X =

∫

) (X e_X ≤σ .

Proof. If X is a continuous random variable, then

∫

∫ ∫

∞

− X −E(X) p_X(x)dx] ≤ [X −E(X)] p_X(x)dx

[ ^{2 2} ，

that is, e_x² ≤Var(X). Hence e_X ≤σ(X).

If X is a discrete random variable, the result can be proved similarly. ■

Theorem 4. If the mathematical expectation and average error of a random variable X both exist, then for any constant ε> 0, we have

ε ^eε^X X

E X

P( − ( ) ≥ )≤ . Proof. If X is a continuous random variable, then

∫

∫

−

≤ −

=

≥

−

} ) (

; { }

) (

; {

) ) (

) ( ( )

) ( (

ε

ε

ε

ε

X E X x

X X

E X x

X

X E X p x dx

dx x p X

E X P

ε ε

X X

dx e x p X E

∫

≤ 1 ( ) ( )

.

If X is a discrete random variable, then the result can be proved similarly. ■

The result of Theorem 4 is similar to Chebyshev’s Inequality. They both give the upper bound of large deviation probability of occurrence. Since the probability of any events is no more than 1, we call “the above upper bound is less than 1” “meaningful”.

For Theorem 4, when ε >e_X , the upper bound is meaningful; for Chebyshev’s Inequality, when ( ) 1

2 <

ε X

Var (namely, ε >σ(X)), the upper bound is meaningful.

If follows from e_X ≤σ(X) that the meaningful scope of Theorem 4 is large than that of Chebyshev’s Inequality.

Example 1. The average error of U. Solution. Since

0（x≤0 or x>1）, ^pU(x)= 1 （_0<x≤1）,

and 2

) 1 (U =

E , hence

4 ) 1 2 ( 1 2 )

(1 2 1

1 ¹₂

0

1 2 1 1

0 − ⋅ = − + − =

=

∫

∫

∫

e_U .

Answer. The average error of U is 4

1.

Example 2. Let X ~N(μ,σ²). Then σ π

= 2

eX .

Proof. e x e dx

x X

2 2

2 ) (

2

1 _σ^μ

σ μ π ^{− −}

∞ +

∞

∫

=

x e dx

x

2 2

2

2

1 _σ

σ π

∞ − +

∞

∫

=

x e dx

∫

= 0

2 ²

2

2

2 1 ^σ

σ

π ^.

Let 2σ

t= x . Then

π ²σ

2

0

2 ⋅

=

∫

e_X ^t = σ

σ π π

2 2 2

2 ) ( 1

0

2 ⋅ =

−

+∞

e−t . ■

Theorem 5. Let X₁,X₂,⋅ ⋅⋅,X_n~N(μ,σ²) and a(x)=x. Then σ π X n

e _a 2

)

( = .

Proof. By the additivity of normal distribution,

∑

= n

i

Xi 1

) , (nμ nσ² N

X ~ ( , )

2

N μ σn ，and E(X)=E(X_i)（i=1,2,...,n）. Hence σ

π σ

π n n

X e X

e 2 2

)

( = = ⋅ = . ■

It can be seen that the average error of _Xi（i=1,2,...,n）reduced n-times by solving X .

Example 3. The accuracy of an analytical balance is 5mg. But the accuracy in a chemical analysis need to be 1mg. Suppose the measured value satisfies normal distribution. How many times do the parallel measures need and then obtain the mean of x satisfying the requirement?

Solution. Suppose we need n times the parallel measures. From the above example, the error reduce 5-times implies n=5, and it follows that n=25.

Theorem 6. Let be random variables with independent and identical distribution. Then

} {X_n

) 1 (

lim

1

i n

i

n Xi E X

n

∑

+∞ =

→ .

Proof. Let E(X_i)=μ and . Since , , … , are independent and identical distribution, by Lindberg-Levy Central Limit Theorem,

) 2

(X_i =σ

Var X₁ X₂ X_n

n n X

n

i i

n σ

μ

∑

= +∞

→

lim 1 ~N(0,1), and

∑

= → ⁿ

i

n Xi

X n

1

lim 1 ~ ( , )

2

N μ σn . Since ( )= lim ² =0

+∞

→ n

X Var

n

σ ,

μ

=

X ，that is, 1 ( ) lim

1

i n

i

n Xi E X

n

∑

+∞ =

→ . ■

It shows that the number of parallel measures tends to positive infinity, the arithmetic mean of tends to . For most surveying instruments with uniform scale, is the true value.

Xi E(X_i) )

(X_i E

Let {X_n} be random variables with independent and identical distribution. Let Xf be the mean of X₁,X₂,⋅ ⋅⋅,X_n. By Theorem 4, lim _f ¹[ ( ( _i))]

n X f⁻ E f X

+∞

→ = .

For example, if satisfies (0，1]-uniform distribution, then . It follows that

Xi f(x)=lnx

) 1 ln (

^

lim ¹

0 x dx

e X_f

n→+∞ =

∫

Since ln ( ln ) 1 lim( ln ) 1

0 1

0 1

0 = − =− − ₊ − =−

∫

) 1 1 (

lim = ≠ =

+∞

→ f i

n E X

X e .

It can be seen that if the number of parallel measures tends to positive infinity, then the f(x)-mean of X_i is not always equal to the true value.

Theorem 7. If Y =aX +b, then e^Y= ae^X, where a and b are constants.

Proof. Since E(Y)=E(aX +b)=aE(X)+b，we have dx x p b x aE b aX

e^Y =

∫

a

∫

= ( ) ( )

= ae^X .

If the variation of f(x) over {x p_X(x)≠0} is small, then for the random variable Y = f(X), we have Y ≈ f^'(x₀)X +b（x₀∈{x p_X(x)≠0}, is a constant）. By Theorem 5，

b

X

Y f x e

e ≈ ^'( ₀) . ■

Example 4. Let the hydrogen ion concentration of a bottle of dilute hydrochloric acid be 0.056mol/L（the error is no larger than 0.001mol/L）. Question: How many decimal places should the PH value of such dilute hydrochloric acid retain?

Solution. Suppose the hydrogen ion concentration is x, then pH =−lgx and 10

ln

' 1

pH_x =−x . Since , then , the

variation of ^' is relatively

] 057 . 0 , 055 . 0

∈[

x −7.896≤ pH_x^'≤−7.619

pHx small. Hence e_pH ≈7.7e_x，and the pH value retains two decimal places, that is,

25 . 1 056 . 0

lg =

−

=

pH .

2.2. The average errors of

) 2 ( 2 )

( 1 2 2 1

2 U U x PU x U

F^A = + ≤ = ≤ − .

Establish the plane rectangular coordinate system with abscissa axis U1 and vertical axis U2. Note that locate in the field surrounded by ,

and abscissa axis, vertical axis (except the origin) is an equally likely event. Then it satisfies the condition of geometric probability model.

) ,

(U1U2 U₁ =1 U₂ =1

See Fig. 1. (a), when

2

0<x≤ 1, 2 1 ² 2 ²

2 ) 2 1 ( / )

2

( x x

S U x U

P ≤ − = Δ^ABC = = .

See Fig. 1. (b), when 1 2

1 <x< ,

2 2 1

2 1 2(1 )

2 ) 2 2 1 ( 1 / ) 1

( ) 2

( x x

S U

x U

P DEF − = − −

−

=

−

=

−

≤ ^Δ .

2=1 U

1=1 U

2=1 U

1

2 2x U

U = −

1

2 2x U

U = −

1=1 U

¹

（a） （b）

Figure 1.

Since 1 0<U¹+2U² ≤

, then 0 （_x≤0）; 2x² （

2

0<x≤1）; _1−2(1−x)² （ 1

2

1 <x< ）; 1 （x≥1）.

2 ) 1 2 (1 ))

(

( ₂

2 = _A =

A E U F

F .

0 （x≤0 or x≥1）;

p^A2(x)= 4x （

2

0<x≤1）;

4−4x （ 1 2

1 <x< ）.

= ) (

2 x F^A

e(A₂)=

∫

x xdx x )(4 4x)dx 2

( 1 4

2 )

(1 ¹

2 2 1

1

0 − ⋅ + − −

=

∫ ∫

6 ) 1 2 3 3

( 4 3 )

( 4

1

2 1 2

2 3 1

0 2

3+ + − + − =

−

= x x x x x .

2.2.2. The properties of G₂

) (

) (

) (

1 2 2 2

1

2

U U x P x U U P x

F^G = ≤ = ≤ .

1 The density functions of the random variables corresponding to the vertices in the figure of the function may be equal to zero. Such vertices can not be attained, and the same below.

When 0<x<1, we establish the plane rectangular coordinate system with abscissa axis U1 and vertical axis U2 (see Fig. 2.). Note that locate in the field surrounded by ,

) , (U1U2 1=1

U U₂ =1 and abscissa axis, vertical axis (except the origin) is an equally likely event. Then it satisfies the condition of geometric probability model. Since

1 2

2 U

U ≤ x is the field surrounded by

1

2 U

U = x , , and

abscissa axis, vertical axis, then

1=1

U U₂ =1

1

2 U

U = x intersects with y=1 at the intersection point (x²,1). Hence

( ) ( 1 ²)/1

1 1 2 1

2

2 ₂ dU x

U x U

U x

P ≤ =

∫

1=1

x²lnU¹¹₂ x² U

x +

=

1 2 2

U U =x

2 2

2 x lnx

x −

= .

Since 0<G2≤1, then

Figure 2.

0 (x≤0);

x²−2x²lnx (0<x<1); 1 (x≥1);

and ln 2 0.5966 4

) 1 2 (1 ))

(

( ²

2 = ^G = + =

G E U F

F ,

0 (x≤0 or x≥1); −4xlnx (0<x<1);

F^G2(x)=

=

= ( )

)

( ^2'

2 x F x

p^{G G}

dx x x x

G

e ( 4 ln )

2 ) 1

( ¹

2 =

∫

Let )ln 2

3 ( 4 9 ) 4 (

3 2 2

3 x

x x x x x

g = + − − . It can be checked that

) ln 4 2)(

( 1 )

'(

x x x

x

g = − − .

Hence ) lim ( )] 0.1989 2

(1 [ 2)]

(1 ) 1 ( [ )

( 2 = − − − 0 =

→ μ

μ g

g g

g G

e .

2.2.3. The properties of H₂

2 ) ( 1 )

1 ( 2 ) (

1 1 2

2 1

2

x U U xU P x U U P x F^H

≤ −

=

≤ +

= .

When 0<x<1 , we establish the plane rectangular coordinate system with abscissa axis U1

and vertical axis U2 (see Fig. 3.). Note that locate in the field surrounded by

) , (U1U2 1=1

U , U₂ =1 and abscissa

axis, vertical axis (except the origin) is an equally likely event. Then it satisfies the condition of geometric probability model. Since

x U U xU

≤ −

1 2 1

2 is the field surrounded

by U x

U xU

= −

1 1

2 2 , U₁=1 , U₂ =1 and abscissa axis, vertical axis, then x

U U xU

= −

1 2 1

2 intersects with U²=1 at the intersection point ,1) (2

x x

− . Hence

2=1 U

1=1 U

x U U xU

= −

1 1

2 2

Figure 3.

x dU x

x U

xU x

U U xU P

x

x + −

= −

≤ −

∫

− 2 2

2 )

( ^{1 1}

2 1

1 1

1 2

x x x

x U xU

x

x + −

−

=

−

+ ln(4 2 )] 2

4 [2

1

2 1

2

1 2 1)

2 ln(

2 −

+

= x

x x .

Since 0<H²≤1， 0 (x≤0); 2 1)

2 ln(

2 −

+ x

x x (0<x<1);

1 (x≥1);

then 0.6373 8

3 ln 2 ) 1 2 (1 ))

(

( ²

2 = ^H = + =

H E U F

F ,

0 (x≤0 or x≥1);

) 2 2 1 ln(

1+ − + −

x x

x x (0<x<1).

= F^H2(x)

= ( )

)

( ^2'

2 x F x

pH H =

The average error is e H E x E x x p^H (x)dx 2

) 1 ) ( ( )

( ^{1 2}

2 ^{= − =}

∫

Let 3 6

) 2 2 3ln(

) 4 2 1 ln(

4 ) ( 3 )

( ^{3 2} x² x

x x x x x

f = − − − − + + . It can be checked that

187

) ( 2) ( 1 )

( ²

' x x p x

f = − ^H （_0<x_<1）.

Hence 0.2221

3 3 1 8 ln 2 21 ln 4 )]

( [ ) ( )

( ¹ ₀²¹

2

2 = f x 1 + −f x = − + =

H

e .

2.2.4. Conclusions

In conclusion, e(A₂)<e(G₂)<e(H₂)<e_U . It can be seen that choosing different means, the effect of reducing the error may be different.

2.3.

-average error for small error numbers

The small error number is defined to be a number which is largely greater than its error. If a set of small error numbers satisfies uniform distribution in the error range, without loss of generality, suppose ( U)

b b a bU a

X_i = + = + . Let

b

n= a. Then . It is easy to see that n is very large for small error numbers. Now we discuss the properties of the means when

) (n U b X_i = +

+∞

→

n .

Lemma 1. If the inversion function of exists, and are both

derivable, then .

) (x

f f(x) f⁻¹(x) 1

) ( )]

(

[ ^'

'

1 ⋅ =

− f x f x

f

) (x f y=

Proof. See Fig. 4. is equal to the slope of the tangent line of at

the point , and is

equal to the slope of the tangent line

of at the point .

)

'( x

f k1 l f(x) ^y=x

)) ( , (x f x

A f^−1'[f(x)]

k2 m

)

1( x

f ⁻ B(f(x),x)

)

1( x f y= ⁻

The images of and

(resp. A and B) are symmetric with regard to the line

) (x f

y= y= f⁻¹(x)

x

y= . Then by the symmetry,

Figure 4.

I. If l//m，then l、m are both parallel to the liney=x. Hence 1

2

1=k =

k , and then k¹k²=1.

II. If intersects with , then , l m l m intersect with the line y=x at a point C. Draw two lines at the point C which are parallel to x-axis and y-axis, respectively.

Since ∠1=∠3+∠4+∠5，∠2=∠3，∠4=∠5,

then ∠1+∠2=2(∠3+∠4)=2×45°=90°, and k¹⋅k²=tan∠1⋅tan∠2=1. Consequently， f^−1'[f(x)]⋅ f^'(x)=1. ■

Lemma 2. Suppose the inversion function of exists, and are both derivable. If is an increasing (resp. decreasing) function over

) (x

f f(x) f⁻¹(x) )

(x

f I , then

is an increasing (resp. decreasing) function over )

1( x

f⁻ M ={f(x)x∈I}.

Proof. Suppose f(x) is an increasing (resp. decreasing) function over I ,

is an increasing (resp. decreasing) function over （ ）. Let . Then and have different signs. Then

)

1( x f⁻ N N ⊆M x0∈N )

( ⁰

' x

f f^−1'[f(x⁰)]

0 )]

( [ )

( ^{0 1' 0}

' x ⋅ f⁻ f x <

f ,

which is a contradiction to Lemma 1. This completes the proof. ■

Lemma 3. If the inverse function of exists, and are both derivable, then

) (x

f f(x) f⁻¹(x)

xf= ( )]

) 1 (

[ ¹

2 1

n x x f

n f f

n

i

i +

∑

− is monotonically increasing, where x¹ is a variable，andx², x³, ..., xⁿ are constantvalues.

Proof. Since f(x) is a continuous function and has the inverse function, then )

(x

f is monotonical.

I. f(x) is monotonically increasing.

Then f⁻¹(x) and

n x x f

n f

n

i

i ( )

)

1 ( ¹

2

∑

=

are monotonically increasing.

Therefore ( )]

) 1 (

[ ¹

2 1

n x x f

n f f

n

i

i +

∑

− is monotonically increasing, i.e., x^f is

monotonically increasing.

II. f(x) is monotonically decreasing.

189

Then f⁻¹(x) and

n x x f

n f

n

i

i ( )

)

1 ( ¹

2

∑

=

are monotonically decreasing.

Therefore ( )]

) 1 (

[ ¹

2 1

n x x f

n f f

n

i

i +

∑

− is monotonically increasing, i.e., x^f is

monotonically increasing.

The result follows. ■

Theorem 8. Suppose a function has the inverse function . If and are both derivable, then

) (x

f f⁻¹(x)

) (x

f f⁻¹(x) L lim(x^f n)

n −

= →+∞ and

2

2

1 U

U +

have the same distribution, where x1=n+U1，x2=n+U2, and x^f is the -average mean of and .

) (x f x1 x₂

Proof. Note that ]

2

) 1 ( ) 1 [ (

2 ] ) ( )

[ ( ¹

1 + < ≤ ₋ + + +

− f n f n

f n x

f n

f f ^f , i.e., n<x^f ≤n+1.

Then 0< lim( − )≤1

+∞

→ x^f n

n ,

and F^L(x)=0（x≤0），F^L(x)=1（x≥1）.

For 0<x<1，we have

] )

2 ) ( ) [ ( ( )

( ¹ f x¹ f x² n x

f P x

F^L = ⁻ + − ≤

=P(x²≤ f⁻¹[2f(n+x)− f(x¹)]) (By Lemma 3).

Let two dimensional Cartesian coordinate system consist of the abscissa axis x¹ and vertical axis x² . Note that (x1,x2) locate in the field surrounded by

n

x= ,y=n,x=n+1 andy=n+1 (except ) is an equally likely event. Then the conditions of geometric probability model are satisfied.

) , (n n

)]

( ) ( 2

[ ¹

2 f 1 f n x f x

x ≤ ⁻ + − is the region surrounded byx=n, , ,

and the below area of .

n

y= x=n+1 +1

=n

y f⁻¹[2f(n+x)− f(x¹)]

Then {f⁻¹[2f(n+x)− f(x¹)]}_x1^'=−f^−1'[2f(n+x)− f(x¹)]⋅ f^'(x¹).

When n→+∞, then f(n+x)= f(n), f(x1)= f(n), f^'(x¹)= f^'(n) and {f⁻¹[2f(n+x)− f(x¹)]}^' =−f^−1'[f(n)]⋅ f^'(n)=−1.

Hence f ⁻¹[2f(n+x)− f(x¹)] is the straight linel：

, where the line passes the point with slope -1.

1

2 (n x) (n x) x

x − + = + − l (n+x,n+x)

Then the line passes pointA(n,n+2x) and point E(n+1,n+2x−1).

When 2

0<x≤1（see Figure 5.（a））, we obtain

2 1 2

2 1 (2 ) 2

2 1 1 / )])

( ) ( 2 [

(x f f n x f x S x x

P ≤ ⁻ + − = ^ΔABC = = .

When 1

2

1 <x< （see Figure 5.（b））, we have

2 2

1 1

2 1 2(1 )

2 ) 2 2 1 ( 1 / ) 1

( )]) ( ) ( 2 [

( x x

S x

f x n f f x

P ≤ ⁻ + − = − ^ΔDEF = − − = − − ,

i.e., 0 （x≤0）;

2x² （

2

0<x≤1）;

1−2(1−x)² （ 1 2

1<x< ）; 1 （x≥1）.

Hence F^L(x)=F^A2(x). ■

It is shown that the distribution of lim(xf n)

n −

+∞

→ is independent from .

Combining Theorem 5, for any continuous function which has the inverse

function and ，

) (x f

) (x f

1

1 n U

x = + x²=n+U², we have

6 ) 1 ( ) (

lim = ₂ =

+∞

→ e x_f e A

n .

Example 5. LetR1 and R² be two same resistances labeled by “ （ ）”.

Suppose

Ω

2k ±0.1% R1 and R² are both uniform distribution within the error scope. Try to estimate the error of the resistance R raised from the parallel of R1 and R².

Solution. Since R1,R2∈(1998，2002] （Unit：Ω）, then

Figure 5.（b）

Figure 5.（a）

= ) (x F^L

+1

=n y

+1

=n x

l

n x=

n y=

+1

=n y +1

=n xl

n x=

n y=

) 5 . 499 ( 4 4

1998 1 1

1 U U

R = + = + .

Similarly, we obtain R²=1998+4U²=4(499.5+U²). Let f x 1x

)

( = and R_f be the ^f(x)-mean of R1 and R₂. Then

2

2 1

2

1 Rf

R R

R

R R =

= + .

Note that n=499.5 is relatively large. Therefore

5 2 . 2 499

2

1 U

U

R− ≈ + , i.e.,

. Hence 999

2 2+

≈ A

R ≈ = × = Ω

3 1 6 2 1 2 A²

R e

e .

3. The effect of

For four numbers 1, 1, 1, 4, the x-mean is 1.75 and the lnx-mean is 414

. 1

2= . Obviously, the effects of the two means raised by extreme value （4） are different. Extreme value（4） has more affect on the x-mean than the -mean.

Naturally, a problem is raised that how to compare the effect of a kind of mean from extreme value conditions?

lnx

Theorem 9. Let f(x)be a continuous function and Uf the -mean of and . Then

) (x

f U1

U2 x² F (x) 1 (1 x)²

Uf ≤ − −

≤ .

Proof. When U¹≤x and U²≤x，then

x x f x f f

U f x f f

xf ≤ ⁻ + ≤ ⁻ + ]=

2 ) ( ) [ ( 2 ]

) ( )

[ ( ^{2 1}

1 and

2 2

1 ) ( )

( )

(U x PU x PU x x

P ^f ≤ ≥ ≤ ⋅ ≤ = .

When U1>x and U2>x , then f x f x x

U f f x f f

Uf > ⁻ + > ⁻ + ]= 2

) ( ) [ ( 2 ]

) ( )

[ ( ^{2 1}

1

and

P(U^f >x)=1−P(U^f ≤x)≥P(U¹>x)⋅P(U²>x)=(1−x)², i.e., P(U^f ≤x)≤1−(1−x)².

Hence x² F (x) 1 (1 x)²

Uf ≤ − −

≤ . ■

Especially, when x=E(U), we have

4 ) 3 4 (

1≤PUf ≤x ≤ . Let

4 )) 1 ( (U ≤E U =

P f and if one of U1 and U2 is larger thanE(U), then

Uf is larger thanE(U). It is shown that Uf has enormous implications by large numbers. Let

4 )) 3 ( (U ≤E U =

P f and if one of and is smaller than , then

U1 U2 E(U)

Uf is smaller than E(U). It is shown that Uf has enormous implications by small numbers.

Let two dimensional Cartesian coordinate system consist of the abscissa axis U1

and vertical axis U2. (See Figure 6.). Note that locate in the field

surrounded by 、 and the two axis (except the origin) is an equally likely event. Then the conditions of geometric probability model are satisfied.

Generally，let

) , (U1U2 1=1

U U₂ =1

)) ( (U E U P

d = ^f ≤ , U1≤E(U) and U2≥E(U), then the probability of U^f ≤E(U) is equal to

2 2 1 4

1 4) ( 1 2 1 SABCD

I − = −

=

= d

S d

p

Figure 6.

2

=1 Uf

and the probability of U^f >E(U) is 1−p.

Definition 3. The impact coefficient of the f(x) -mean is defined as

2 ) 1 2 (1

2 −

= _Uf

f F

p （ ]

2 ) ( )

[ ( ^{1 2}

1 f U f U

f

Uf = ⁻ + ）.

It is obvious that . If is more larger, then the -mean acted by small numbers has more impactions; if is more smaller, then the -mean acted by larger numbers has more impactions. If

1

0≤ p≤ p_f f(x)

pf f(x)

5 .

=0

pf , the -mean acted by larger numbers and small numbers has the same impactions.

) (x f

4.

The necessary and sufficient condition of comparing values and identical equality of means

4.1. Comparing values of means

Mean inequality shows the relationship of sizes between arithmetic mean, geometric mean and harmonic mean. For the generalized f(x) -mean and

) (x

g -mean, what are the relationship of their sizes?

Theorem 10. If is monotonically increasing and is a concave function，then

) (x

f f[g⁻¹(x)]

g

f x

x ≤ with equality holds if and only if x₁=x₂ =⋅ ⋅⋅=x_n. Proof. Let h(x)= f[g⁻¹(x)].

Since is monotonically increasing, then is monotonically increasing.

Then and are both monotonically increasing.

)

g(x g⁻¹(x)

) (x

h h⁻¹(x)

Note that h(x) is a concave function. By Jensen Inequality,

∑

∑

≥ ⁿ

i i n

i

i h y

y n h n

1 1

) 1 (

1 )

( , where the equality holds if and only if y₁ =y₂ =⋅ ⋅⋅=y_n.

Then 1 ( )]

1 [

1 1

1

∑

∑

−

=

≥ ⁿ

i i n

i

i h y

h n

n y .

Suppose xⁱ=g⁻¹(yⁱ)（i=1，2，...,n）, then yi=g(xi).

We have 1 ( )]

[ ) 1 (

1 1

1

∑

∑

−

=

≥ ⁿ

i i n

i

i f x

h n x

n g .

Then

∑ ∑

=

=

− ≥ ⁿ

i i n

i

i f x

x n n g

g f

1 1

1 1 ( )

)]

1 ( [

{ , and 1 ( )]

[ )]

1 ( [

1 1 1

1

∑ ∑

=

−

=

− ≥ ⁿ

i i n

i

i f x

f n x n g

g , i.e.,

g

f x

x ≤ with equality holds if and only if x₁=x₂ =⋅ ⋅⋅=x_n. ■

If g(x) is monotonically decreasing，then the same conclusion is also obtained.

Similarly, it is easy to obtain the following:

If f(x)is monotonically decreasing and f[g⁻¹(x)]is a concave function, then

g

f x

x ≥ .

If f(x) is monotonically increasing and f[g⁻¹(x)] is a convex function, then

g

f x

x ≥ .

If f(x) is monotonically decreasing and f[g⁻¹(x)] is a convex function, then x

x ≤ .

(The equality holds if and only if x₁ =x₂ =⋅ ⋅⋅=x_n.)

Example 5. （Mean inequality）

∏ ∑

∑

=

=

=

≥

≥ _n

i i

n n

i i n

i i

x x n

n x

1 1

1 ( 1 ) 1

1 （x_i >0）.

Proof. Let a(x)=x,g(x)=lnx and x x

h 1

)

( = . By Theorem 1, it only needs to prove x_a ≥x_g ≥x_h.

Since a^''[g⁻¹(x)]=(e^x)^'' =e^x >0, then a[g⁻¹(x)] is a convex function.

Note that a(x) is monotonically increasing. Then we have x_a ≥x_g .

By 1 0

1) ( 1) (ln )]

(

[ ^{1 '' '} ₂

'' ⁻ = = − = >

x x x x

h

g , g[h⁻¹(x)] is a convex function.

And g(x) is monotonically increasing. Then x_g ≥x_h holds.

From the above discussions, we have x_a ≥x_g ≥x_h. The proof is completed. ■

4.2. The relationship of sizes and impact coefficients of means

Theorem 11. If x^f ≤x^g holds, then p^f ≥ p^g. Proof. If U1 and U2 satisfy

2

≤1

Ug , thenU1 and U2 shall satisfy 2

≤1 Uf . Then )

2 (1 2)

(1 ^g

f U

U F

F ≥ .

Hence p^f ≥ p^g. ■

By Theorem 8 and its generalized version and Theorem 9, we obtain the following conclusion:

Corollary 1. Let . If , then ; if

, then .

)]

( [ )

(x f g ¹ x

h = ⁻ f^'(x)⋅h^''(x)<0 p_f ≥ p_g 0

) ( ) ( ^''

' x ⋅h x >

f p_f ≤ p_g

Example 6. Let ⁿ（ ， ）. Then is monotonically decreasing on .

x x

f( )= x>0 n>0 p_f n

Proof. Let n₂ >n₁ >0, f₁(x)=xⁿ¹and f₂(x)=xⁿ². Then

1 2

)]

( [ )

( _{2 1} ^{1 n}

n

x x f f x

h = ⁻ = .

Therefore, ²

1 2 1

'' 2 1

2

) 1 ( )

( = − ^{n −}

n

n x n n x n

h .

Since 1

1 2 >

n

n and x>0, then h^''(x)>0 holds.

Note that f₂^'(x)=n₂xⁿ²⁻¹ >0. Then f^'(x)⋅h^''(x)>0. We have p_f2 ≤ p_f1.

Hence _pf is monotonically decreasing on . _n ■

For f(x)=xⁿ（n>1）and p_f = p( is a constant), the value of may be searched by the following program（Pas Language）:

p n

{$N+}

var i,j,d:longint;

n,a,b,p,dx,s:extended;

function f(x,n:extended):extended;

begin

if x=0 then f:=exp(ln(0.5)*(n-1)/n)

else f:=exp(ln(exp(ln(0.5)*(n-1))-exp(ln(x)*n))/n);

end;

begin

readln(p,a,b,d);

repeat

n:=(a+b)/2;

dx:=f(0,n)/d;

s:=0;

for j:=1 to d-1 do begin

s:=s+f(j*dx,n)*dx;

end;

s:=2*s-0.5;

if s=p then begin

writeln(n);

break;

end;

if s>p then a:=(a+b)/2