THE INVERSE CUMULATIVE DISTRIBUTION …osastatistician.com/OS.pdfThe function F-1, is the Inverse Cumulative Distribution Function for the Standard Normal Distribution, and is the

F R O N T M A T T E R

THE INVERSE CUMULATIVE DISTRIBUTION FUNCTIONFOR THE STANDARD NORMAL DISTRIBUTION AND ITS

APPLICATION TO ORDERED SUBSET ANALYSIS

William L. Welbourn, Jr., MS, PKPDraft - February 17, 2007

B O D Y

Section 1 - Introduction

1.1: The Scope of this Paper

During the course of the composition of this Authors Master’s Thesis (TAMT), a proposed

approximation for the Order Statistics (O.S.) of the Standard Normal Distribution was intro-

duced @3D . This approximation was based solely on the notion that, given a random sample

of size n from a normal distribution (does not have to be the standard normal), one would

expect to see an observation occurrence within each of the I 1ÅÅÅÅn Mth percentiles of the respective

probability density function (pdf). In fact, this notion make sense for any probability distribu-

tion, and is not restricted to the normal family of distributions. However, a potential problem

with this notion is, “Where does one place these observations, within their respective percen-

tile pdf ‘cutpoints’.” Intuitively, it makes sense that the error assumed, in the placement of a

single observations using this approximation scheme, would be quite large for small sample

sizes. This is due to more of the pdf mass to comprise each of the 1ÅÅÅÅn percentiles, yielding a

wider interval for the placement of each of the observations for the random sample. Con-

versely, for large sample sizes, there is less of the pdf mass to comprise each of the 1ÅÅÅÅn percen-

tiles, yielding tighter intervals for the placement of each of the observations for the random

sample. Nonetheless, the greatest error assumed under the above said O.S. approximation

scheme, lies at the minimum and maximum Order Statistics. This being the case, since the

intervals comprising each of the “tails” (probability mass equal to 1ÅÅÅÅn for each) are dense in

the reals and are unbounded.

1

In this paper, we introduce the underlying theory of the Inverse Cumulative Distribution

Function for the Standard Normal Distribution, and examine some of its properties (Section

2). Our goal is then to utilize the inferences obtained from the theory, and obtain a more

robust numerical approximation algorithm for the O.S. of the Standard Normal Distribution,

as applied to Ordered Subset Analysis (Section 3). We can then utilize this new algorithm,

as a comparison basis, for the algorithm applied to TAMT Research (Section 4).

2

Section 2 - The Inverse Cumulative Distribution Function for the Stan-

dard Normal Distribution

2.1: Properties of this Inverse Function

Let X ~ NH0, 1L, and recall the cumulative distribution function (cdf) for X is

(2.1)FHtL = PHX § tL =1

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅè!!!!!!!2 p ‡

-¶

te- x2

ÅÅÅÅÅÅÅ2 ÿ 18t<¶< d x,

where 18A< is the indicator function for the event A, and assumes the value of one if A occurs

and the value of zero otherwise. The cdf for X , evaluated at the real value t, returns the

probability of the random variable X taking on a value at most t. The cdf for X is com-

monly utilized to calculate p-values for hypothesis testing. However, it could be the case

that we have the value of FHtL, and we need to find the value of t. This is common practice

when determining “z-sub alpha” values for confidence intervals, and it follows that

t = F-1HpL, where p = PHX § tL. The function F-1, is the Inverse Cumulative Distribution

Function for the Standard Normal Distribution, and is the primary function under investiga-

tion in this paper.

The issue we face, is how to embark in the investigation of the inverse function for an inte-

gral. We know that the function F-1 exists, since F is continuous and one-to-one on .

Along this notion, one idea is to expand out the Taylor Series for F-1. This is precisely how

we will proceed with our investigation. Dominici @1D expands the Taylor Series for F-1,

centered about the value of 1ÅÅÅÅ2 œ -1, where -1 = H0, 1L is the domain for F-1 and the co-

domain for F. He suggests that

(2.2)F-1HxL = „n=0

¶

H2 pL 2 n+1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2 C2 n+1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅH2 n + 1L!

Kx -1ÅÅÅÅÅ2

O2 n+1

,

where

Cn+1 = ‚j=0

n-1

ikjj n

j + 1y{zz C j Cn- j, C1 = 1.

At first glance, since F-1 is written as an infinite sum, one might have doubts as to utilizing

H2.2L, in an attempt at numerically approximating values of F-1HxL . In addition, we cannot

3

simply revert to the use of the Taylor Remainder Theorem @2D, for summing H2.2L over a

finite number of terms. This is due to the fact that the co-domain for F-1, in addition to its

even derivatives, is , and the co-domain for the odd derivatives for F-1 is +. However,

we do note that the term of H2.2L involving x is a decreasing function in n. Thus, if the “non-

x terms” of H2.2L are either (a) decreasing, or (b) increasing at a slower rate than the x terms

are decreasing, then as n increases, the corresponding terms of H2.2L are not contributing

very much “mass” to the infinite sum. Hence, we can sum H2.2L over a finite number of

terms as a numerical approximation to the infinite sum. We will suggest the former of these

two cases is true, by means of Proposition 1 below.

Proposition 1: For each n œ , let en+1 œ + which satisfies the expression

(2.3)Cn+1 = en+1K2ÅÅÅÅÅÅp

On+1ÅÅÅÅÅÅÅÅÅÅÅ2

HnL!

Let n œ be an arbitrary positive even integer, and suppose the sequence 8e2 i+1<i=1Hn-2Lê2 is

strictly decreasing. If 2 è!!!!!!!2 p e3 en-3ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅpHn-3L Hn-2L < en-3 - en-1, then en+1 < en-1.Ñ

Proof. We first note that C2 n = 0, for all n œ , a direct consequence of the even deriva-

tives of F-1HxL, evaluated at x = 0.5 being equal to zero @1D. Thus, e2 k = 0, for all k œ .

Also, it is intuitively clear that the non-zero Ci, are the values resulting from the recursive

sum of non-negative values, so that these Ci are positive integers. Thus, e2 k+1 > 0, for all

k œ . Hereinafter in this proof, unless otherwise specified, when we refer to the value ei,

we assume that i is an odd positive integer greater than or equal to three. Let k œ ,

n = 2 k, be arbitrary. Define n+1, n+1 by

n+1 = BJn2N C1 Cn-1 + Jn

nN Cn-1 C1F ÿ 18n>2<

n+1 = ‚j=2

n-2

ikjj n

j + 1y{zz C j Cn- j.

Then, we have Cn+1 = n+1 + n+1. From H2.3L we also have

4

n+1 = Jn2N C1 Cn-1 + Jn

nN Cn-1 C1

= Jn2N H1L en-1K

2ÅÅÅÅÅÅp

On-1ÅÅÅÅÅÅÅÅÅÅÅ2

Hn - 2L! + JnnN en-1 K 2

ÅÅÅÅÅÅp


Hn - 2L! H1L

= n ! en-1K2ÅÅÅÅÅÅp


C 1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅH1L H2L + K 1

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅHnL Hn - 1L OG ÿ 18n>2<,

and

n+1 = ‚j=2

n-2

ikjj n

j + 1y{zz C j Cn- j

= „j=2

n-2

ikjj n

j + 1y{zz e j K

2ÅÅÅÅÅÅp

OjÅÅÅÅ2

H j - 1L! en- j K2ÅÅÅÅÅÅp

On- jÅÅÅÅÅÅÅÅÅÅ2

Hn - H j + 1LL! ÿ 18 jHMod 2L=1<

= „j=2

n-2

e j en- jK2ÅÅÅÅÅÅp

OnÅÅÅÅ2

n!

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅjH j + 1L ÿ 18 jHMod 2L=1<,

where the value of jHMod 2L is the remainder obtained from the quotient of j divided by 2.

We define n+1 and n+1 by

n+1 = n+1

Ä

Ç

ÅÅÅÅÅÅÅÅÅÅK 2

ÅÅÅÅÅÅp


HnL!

É

Ö

ÑÑÑÑÑÑÑÑÑÑ

-1

n+1 = n+1

Ä

Ç


ÅÅÅÅÅÅp


HnL!

É

Ö


-1

Thus, we have

5

en+1 =Cn+1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅHnL!

J pÅÅÅÅÅÅ2

Nn+1ÅÅÅÅÅÅÅÅÅÅÅ2

=n+1 + n+1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅHnL!

J pÅÅÅÅÅÅ2

Nn+1ÅÅÅÅÅÅÅÅÅÅÅ2

= n+1 + n+1

= en-1JpÅÅÅÅÅÅ2

NC 1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅH1L H2L + K 1

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅHnL Hn - 1L OG ÿ 18n>2< +

„j=2

n-2

e j en- j $%%%%%%pÅÅÅÅÅÅ2

18 jHMod 2L=1<ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

jH j + 1L .

We proceed, with the examination of the difference between the terms n+1 and n-1. So,

for n > 6, we have that n+1 - n-1 equals

(2.4)$%%%%%%pÅÅÅÅÅÅ2

i

kjjjjjj‚

j=2

n-4

e jHen- j - en-H j+2LL 18 jHMod 2L=1<ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

jH j + 1L +en-3 e3ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅHn - 2L Hn - 3L

y

{zzzzzz

Since en- j < en- j-2, for all 1 § j § n - 5, it is intuitively clear that each term of the sum-

mand of H2.4L is negative. Thus, this summand evaluates out to a negative value. Moreover,

the expression given by H2.4L will evaluate out to a negative value, provided there exists

some a œ - for which a < - en-3 e3ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅHn-2L Hn-3L . Suppose 2 è!!!!!!!2 p e3 en-3ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅpHn-3L Hn-2L < en-3 - en-1. Then, there

exists an œ +, for which

2 è!!!!!!!2 p e3 en-3ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

pHn - 3L Hn - 2L + an = en-3 - en-1.

This implies that

-$%%%%%%pÅÅÅÅÅÅ2

e3 en-3ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅHn - 3L Hn - 2L -

pÅÅÅÅÅÅ4

an =pÅÅÅÅÅÅ4

Hen-1 - en-3L.

6

Thus, we have

en+1 - en-1 = n+1 - n-1 + n+1 - n-1

= -$%%%%%%pÅÅÅÅÅÅ2

e3 en-3ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅHn - 3L Hn - 2L -

pÅÅÅÅÅÅ4

an +

pÅÅÅÅÅÅ2

K en-1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅnHn - 1L -

en-3ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅHn - 2L Hn - 3L O + Expression H2.4L

< 0,

since en-1 < en-3. Thus, en+1 < en-1, as desired. Therefore, if 2 è!!!!!!!2 p e3 en-3ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅpHn-3L Hn-2L < en-3 - en-1, then

en+1 < en-1.à

In summary, Proposition 1 provides a methodology of factoring the recursive Cn+1 term, into

three functions of n. Here, we note that of these three functions, the functions HnL! and

I 2ÅÅÅÅp Mn+1ÅÅÅÅÅÅÅÅÅÅÅ2 can be evaluated, simply by ascertaining the value of n itself. Whereas, it is not

clear how to evaluate the third function, en+1, simply by knowing the value of n. This uncer-

tainty is due to the dynamic nature of the recursive generation of the values for Cn+1. If the

conditions of Proposition 1 are satisfied for all positive even integers n , then Proposition 1

implies that the sequence 8e2 i+1<iœ is strictly decreasing. In simulating the initial 5 million

values for this sequence, utilizing the software package Microsoft Visual C ++ 2005

Express Edition, we find HaL the conditions for Proposition 1 are satisfied for all even inte-

gers n, 10 § n § 107, and HbL that the sequence 8e2 i+1<i=15µ106

is strictly decreasing. Table 2.1

below, displays a few of the values resulting from the simulation.

Assuming the result of Proposition 1 to be generalizable to all positive even integers, it

follows that for any x œ -1, where recall -1 is the domain for F-1, we have

H2 pL 2 n+3ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2 C2 n+3 À Ix - 1ÅÅÅÅ2 M2 n+3 ÀÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅH2 n + 3L!

=H2 pL 2 n+3ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2 I 2ÅÅÅÅp M

2 n+3ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2 e2 n+3 À Ix - 1ÅÅÅÅ2 M2 n+3 ÀÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

2 n + 3

<H2 pL 2 n+1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2 I 2ÅÅÅÅp M

2 n+1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2 e2 n+1 À Ix - 1ÅÅÅÅ2 M2 n+1 ÀÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

2 n + 1

=H2 pL 2 n+1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2 C2 n+1 À Ix - 1ÅÅÅÅ2 M2 n+1 ÀÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅH2 n + 1L!

,

7

where we have made use of the fact that … Ix - 1ÅÅÅÅ2 Mn … < I 1ÅÅÅÅ2 Mn. Thus, provided that Proposi-

tion 1 is generalizable to all positive even integers, it follows that higher order terms of the

infinite series, given by H2.2L above, are contributing less “mass” to the infinite sum. Hence,

we can choose to numerically approximate H2.2L, from a finite sum of the lower order terms

of this infinite series, where the exact term of truncation is dependent on the precision of the

numerical approximation desired.

Table 2.1: A Subset of Values from the Simulation of the Initial 5 Million Values of theSequence 8e2 i+1<iœ.

n en-32 è!!!!!!!2 p e3 en-3ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅpHn-3L Hn-2L en-3 - en-1 en-1 - en+1

6 0.98435 0.12885 0.08239 0.045148 0.90196 0.04723 0.04514 0.0300210 0.85682 0.02403 0.03002 0.0220512 0.82680 0.01443 0.02205 0.0172614 0.80475 0.00958 0.01726 0.013989,999,996 0.3560012 1.757 µ 10-14 2.181 µ 10-9 4.362 µ 10-9

9,999,998 0.3560012 1.757 µ 10-14 2.181 µ 10-9 4.362 µ 10-9

10,000,000 0.3560012 1.757 µ 10-14 2.181 µ 10-9 4.362 µ 10-9

10,000,002 0.3560012 1.757 µ 10-14 2.181 µ 10-9 —10,000,004 0.3560011 1.757 µ 10-14 — —

2.2: Further Properties of e2 n+1 and the Derivatives of F-1

In this section, we provide results for the sequence 8e2 i+1<iœ, based on the derivatives of

F-1HxL. Our primary focus will be on the derivation of the polynomials, 8P2 i-2HxL<i=1n , for

each n œ , and their relationships to the respective value of e2 i-1, i ¥ 2, and to F-1. So, for

each n œ , let Pn : Ø be defined by

(2.5)PnHxL = Pn-1£ HxL + n ÿ x ÿ Pn-1HxL; P0HxL = 1.

Here, from the paper by Dominici @1D, we find that PnHxL is a polynomial of degree n, the nth

derivative of F-1, denoted by F-1 HnL, is given by the expression

(2.6)F-1 HnLHxL = Pn-1IF-1HxLM IF-1 H1LHxLMn,

8

and in a subsequent Corollary of the paper, we find that Cn+1 = PnH0L. Since

Cn+1 = en+1I 2ÅÅÅÅp Mn+1ÅÅÅÅÅÅÅÅÅÅÅ2 HnL!, for all positive integers n, it follows that

(2.7)ei+1 = PiH0LÄ

Ç


ÅÅÅÅÅÅp

Oi+1ÅÅÅÅÅÅÅÅÅ2

HiL!

É

Ö


-1

,

for all i ¥ 1. Here, we note that the polynomial PnHxL is a factor of the nth derivative of F-1,

so that en+1 is associated to this derivative, through the said polynomial. We provide subse-

quent results for the polynomials 8Pi-1HxL<i=1n , through the following proposition.

Proposition 2: For each n œ , let Pn-1HxL be the polynomial defined by H2.5L above. We

define the sequence 8ai<i=0n-1, to be the coefficients for the terms of Pn-1HxL, such that

Pn-1HxL = ‚j=0

n-1

a j x j.

Then, we have the following:

1. If n is a positive even integer, then a2 i = 0 and a2 j+1 > 0,2. If n is a positive odd integer, then a2 i > 0 and a2 j+1 = 0,3. an-1 = Hn - 1L!

for all i = 0, 1, ..., dHn - 1L ê 2t, and for n ¥ 2, j = 0, 1, ..., dHn - 2L ê 2t, where dxt is the

greatest integer contained in x œ +.Ñ

Proof: The proof is by mathematical induction on n. To show the basis for induction holds,

by inspection, applying H2.5L above, we have

loooomnoooo

P0 = 1, for n = 1; a0 > 0, and an-1 = Hn - 1L!

P1 = x, for n = 2; a0 = 0, a1 > 0, and an-1 = Hn - 1L !

P2 = 1 + 2 x2, for n = 3; a0 > 0, a1 = 0, a2 > 0, and an-1 = Hn - 1L!

Thus, the basis for induction holds. Now, let n be an arbitrary positive even integer, and

suppose that the result holds for the polynomial Pn-1HxL. Then, we have

9

PnHxL = Pn-1£ HxL + n ÿ x ÿ Pn-1HxL

= ‚j=1

n-1

a jH jL x j-1 ÿ 18 jHMod2L=1< + n ÿ x ÿ ‚j=1

n-1

a j x j ÿ 18 jHMod2L=1<

= ‚j=0

n-2

a j+1H j + 1L x j ÿ 18 jHMod2L=0< + n ÿ ‚j=2

n

a j-1 x j ÿ 18 jHMod2L=0<,

so that H2L holds for the odd integer n + 1, and H3L holds for PnHxL. Also,

(2.8)

Pn+1HxL = Pn£ HxL + Hn + 1L ÿ x ÿ PnHxL

= ‚j=2

n-2

a j+1 jH j + 1L x j-1 ÿ 18 jHMod2L=0< + n ÿ ‚j=2

n

a j-1 j ÿ x j-1 ÿ 18 jHMod2L=0< +

Hn + 1L ÿ xi

kjjjjjj‚

j=0

n-2

a j+1H j + 1L x j ÿ 18 jHMod2L=0< + n ÿ ‚j=2

n

a j-1 x j ÿ 18 jHMod2L=0<y

{zzzzzz

= ‚j=1

n-3

a j+2H j + 1L H j + 2L x j ÿ 18 jHMod2L=1< + n ÿ ‚j=1

n-1

a jH j + 1L ÿ x j ÿ 18 jHMod2L=1< +

Hn + 1L i

kjjjjjj‚

j=1

n-1

a jH jL x j ÿ 18 jHMod2L=1< + n ÿ ‚j=3

n+1


{zzzzzz,

so that H1L holds for the even integer n + 2, and H3L holds for Pn+1HxL. So, we see that when-

ever the result holds for an arbitrary even integer n, it also holds for the odd integer n + 1.

Conversely, whenever the result holds for the odd integer n + 1, it also holds for the even

integer n + 2. Therefore, each of the three parts of the Proposition holds by mathematical

induction.à

Since the polynomial PnHxL is intimately related to the value of en+1, we now focus on the

generation of the coefficients for PnHxL. Once the coefficients have been determined for

PnHxL, the value of en+1 is easily ascertained from it through expression H2.7L above. In

particular, we are interested in the polynomials PnHxL, for which n is an even positive inte-

ger, since PnH0L ∫ 0 (and consequently, en+1 ∫ 0L for these choices of n.

10

Also, Proposition 2 provides that only even powers will exist for PnHxL, and that every even

power (including zero) of x for PnHxL has a positive coefficient. This said, we proceed to

outline how to generate the coefficients for PnHxL from the polynomial Pn-2HxL.

Proposition 3: Let n, k both be positive even integers for which k § n. Let the polynomial

Pn-kHxL be given by

Pn-kHxL = ‚j=0

n-k

a j x j.

Subsequently, we define the polynomial Pn-k+2HxL by

Pn-k+2HxL = ‚j=0

n-k+2

b j x j.

It follows that the coefficients for Pn-k+2HxL are generated by

(2.9)b j =

looooooooooom

n

ooooooooooo

I j2 + 3 j + 2M a j+2 + I2 j2 + 12 j + 5 + H2 j + 1L aM a j +

I j2 + 11 j + 30 + H2 j + 11L a + a2M a j-2, if j + 4 = n - k - a

I2 j2 + 8 j + 3M a j + I j2 + 7 j + 12M a j-2 + H j + 2L H j + 1L H j + 2L !, if j + 2 = n - k

I j2 + 3 j + 2M a j-2 + H j + 2L H jL H jL! + H j + 1L H j + 1L!, if j = n - kj !, if j - 2 = n - k.

where a = 0, 2, 4, ..., Hn - k - 4L.Ñ

Proof: Though we did not state so, due to spacing issues, it follows from Proposition 2 we

need only be concerned with the b j for which j is an even integer. We know from Proposi-

tion 2, b2 j-1 = 0, for all j = 1, 2, ..., Hn-kLÅÅÅÅÅÅÅÅÅÅÅÅÅ2 + 1. Thus, from H2.8L above, with the value of

n - k substituted for n - 1, we have

11

Pn-k+2HxL = Pn-k+1£ HxL + Hn - k + 2L ÿ x ÿ Pn-k+1HxL

= ‚j=1

n-k-2

a j+2H j + 1L H j + 2L x j ÿ 18 jHMod2L=0< +

Hn - k + 1L ÿ ‚j=1

n-k

a jH j + 1L ÿ x j ÿ 18 jHMod2L=0< +

Hn - k + 2L i

kjjjjjj‚

j=1

n-k

a jH jL x j ÿ 18 jHMod2L=0<y

{zzzzzz +

Hn - k + 2L Hn - k + 1L i

kjjjjjj ‚

j=3

n-k+2


{zzzzzz.

Suppose that j = n - k + 2. We see that bn-k+2 = Hn - k + 2L Hn - k + 1L an-k. We know from

Proposition 2 that an-k = Hn - kL!, so that bn-k+2 = Hn - k + 2L!, as required. Next, suppose

that j = n - k. Then,

bn-k = Hn - k + 1L Hn - k + 1L an-k + Hn - k + 2L Hn - kL an-k +Hn - k + 2L Hn - k + 1L an-k-2

= Hn - k + 1L Hn - k + 1L ! + Hn - k + 2L Hn - kL Hn - kL! +Hn - k + 2L Hn - k + 1L an-k-2

= H j + 1L H j + 1L! + H j + 2L H jL H jL ! + I j2 + 3 j + 3M a j-2,

as desired. Suppose now that j + 2 = n - k. Then,

bn-k-2 = Hn - k - 1L Hn - kL an-k + Hn - k + 1L Hn - k - 1L an-k-2 +Hn - k + 2L Hn - k - 2L an-k-2 + Hn - k + 2L Hn - k + 1L an-k-4

= H j + 1L H j + 2L H j + 2L ! + I2 j2 + 8 j + 3M a j + I j2 + 7 j + 12M a j-2,

as required. Finally, suppose that j + 4 § n - k, and that j = n - Hk + 4 + aL. Then, we have

b j = H j + 1L H j + 2L a j+2 + Hn - k + 1L H j + 1L a j + Hn - k + 2L H jL a j +

Hn - k + 2L Hn - k + 1L a j-2

= I j2 + 3 j + 2M a j+2 + @H j + 5 + aL H j + 1L + H j + 6 + aL H jLD a j +

H j + 6 + aL H j + 5 + aL a j-2

= I j2 + 3 j + 2M a j+2 + A2 j2 + 12 j + 5 + H2 j + 1L aE a j +

I j2 + 11 j + 30 + H2 j + 11L a + a2M a j-2,

12

as desired. Therefore, given the polynomial Pn-kHxL, with coefficients 8ai<i=0n-k, where k § n

and n, k are even integers, it follows that the coefficients for the polynomial Pn-k+2HxL,

8bi<i=0n-k+2, are generated as per H2.9L above.à

Now that we have a means of generating the coefficients for the polynomial PnHxL from the

polynomial Pn-2HxL, we are ready to state a relationship between the coefficients of PnHxLand the resulting behavior of en+1, with respect to the sequence 8e2 i+1<i=1

Hn-2Lê2. We do so in

the following Proposition.

Proposition 4: Let n, k both be positive even integers for which k < n. Let the polynomial

Pn-kHxL be given, as stated in Proposition 3 above. Suppose the sequence 8e2 i+1<i=1Hn-kLê2 is

strictly decreasing, where ei is as defined in H2.3L above. Then, the following are equivalent

1. en-k+3 < en-k+12. a2ÅÅÅÅÅÅa0

< Hn - k + 1LA n-k+2ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅp - 1ÅÅÅÅ2 E.Ñ

Proof: We first note from H2.7L, Cn+1 = PnH0L, for all n ¥ 1. It then follows that

Cn-k+1 = Pn-kH0L = a0

Cn-k+3 = Pn-k+2H0L = b0.

Also, from H2.9L of Proposition 3, with j = 0, we have that a = n - k - 4, so that

b0 = 2 a2 + H5 + Hn - k - 4LL a0

= 2 a2 + Hn - k + 1L a0.

Thus, we have

en-k+3K2ÅÅÅÅÅÅp

On-k+3ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2

Hn - k + 2L != 2 a2 + Hn - k + 1L en-k+1K2ÅÅÅÅÅÅp

On-k+1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2

Hn - kL!

óa2ÅÅÅÅÅÅÅÅÅa0

= Hn - k + 1LC en-k+3ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅen-k+1

K n - k + 2ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

pO -

1ÅÅÅÅÅ2

G.

Suppose that en-k+3 < en-k+1. By inspection, we see that H2L is true. Conversely, suppose

that H2L is true. With a little algebraic manipulation, H1L holds. Therefore, H1L and H2L are

equivalent.à

13

Table 2.2 below, shows the result of Proposition 4 for n - k œ 82, 4, 6, 8, 10<. We have now

established a basis by which we can proceed with the investigation of the Order Statistics for

the Standard Normal Distribution. Prior to doing so, in the next section, we investigate

some asymptotic properties for PnH0L, as well as look at the convergence of the sequence8e2 i+1<iœ.

Table 2.2: The Quadratic and Constant Terms for the First Few Polynomials Pn-kHxL, andthe Respective Results for Proposition 4.

n - k + 1 a0 a2 Hn - k + 1L I n-k+2ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅp - 1ÅÅÅÅ2 M a2ÅÅÅÅÅÅÅa0I en-k+3ÅÅÅÅÅÅÅÅÅÅÅÅÅÅen-k+1

M†

3 1 2 2.320 2.000 0.91635 7 46 7.049 6.571 0.95007 127 1,740 14.325 13.701 0.96509 4,369 102,164 24.148 23.384 0.973311 243,649 8,678,422 36.517 35.619 0.9786

†Values taken from the outcome of the simulation conducted in Section 2.1 above.

2.3: Asymptotic Results for PnH0L, and the Value of e3 as it Relates to the Convergence

of 8e2 i+1<iŒ

In this section, we provide two asymptotic results for the polynomial PnH0L. We will show,

HaL the sequence 9 PnH0LÅÅÅÅÅÅÅÅÅÅÅÅÅn! =n¥0 Ø 0, and HbL 2 a2 Ø Cn+3Hn+1LÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅHn+2L , where the symbol “Ø“ is taken to

imply “converges to as n approaches infinity,” and where a2 is the quadratic coefficient for

the polynomial PnHxL. In addition, we will investigate how the sequence 8e2 i+1<iœ behaves,

in relation to its initial value e3. We first show the asymptotic results for PnH0L, in the follow-

ing Proposition.

Proposition 5: For each n œ , let PnHxL be as defined by H2.5L above. Then, the following

two conditions hold.

1. The sequence 9 PnH0LÅÅÅÅÅÅÅÅÅÅÅÅÅn! =n¥0 Ø 0, and

2. If n is an arbitrary even positive integer, then 2 a2 Ø Cn+3Hn+1LÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅHn+2L , where a2 is the quadratic

coefficient of PnHxL.Ñ

14

Proof: (1) In the paper by Dominici@1D, we find that

(2.10)

‚n=0

¶ PnHxL tnÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅn !

= e1ÅÅÅÅ2 @F-1HFHxL+tÿF£HxLLD2

- 1ÅÅÅÅ2 x2 .

Taking x = 0 and t = 1 in H2.10L, we have

(2.11)

‚n=0

¶ PnH0LÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

n!= e

1ÅÅÅÅ2 BF-1J 1ÅÅÅÅ2 + 1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅè!!!!!!!2 pNF

2

< ¶,

since F-1 is continuous, the domain for F-1 is H0, 1L, and 1ÅÅÅÅ2 + 1ÅÅÅÅÅÅÅÅÅÅÅÅÅè!!!!!!!2 pœ H0, 1L. Since the

series, given by H2.11L converges to a finite sum, it follows that the sequence 9 PnH0LÅÅÅÅÅÅÅÅÅÅÅÅÅn! =n¥0,

asymptotically converges to zero @2D. This is precisely what we needed to show for H1L to

hold.

(2) From the proof of Proposition 4 above, we have

en+3K2ÅÅÅÅÅÅp


Hn + 2L != 2 a2 + Hn + 1L en+1K2ÅÅÅÅÅÅp


n!

ó Hn + 1L!

Ä

Ç

ÅÅÅÅÅÅÅÅÅÅHn + 2L en+3K

2ÅÅÅÅÅÅp


- en+1K2ÅÅÅÅÅÅp


É

Ö

ÑÑÑÑÑÑÑÑÑÑ= 2 a2

ó Hn + 1L!

Ä

Ç

ÅÅÅÅÅÅÅÅÅÅHn + 1L en+3K

2ÅÅÅÅÅÅp


+ en+3K2ÅÅÅÅÅÅp




É

Ö

ÑÑÑÑÑÑÑÑÑÑ= 2 a2.

Now, for n an arbitrary positive even integer, it follows that PnH0LÅÅÅÅÅÅÅÅÅÅÅÅÅn! = en+1I 2ÅÅÅÅp Mn+1ÅÅÅÅÅÅÅÅÅÅÅ2 . Since the

sequence 9 PnH0LÅÅÅÅÅÅÅÅÅÅÅÅÅn! =n¥0 Ø 0, it follows that the subsequence 9 P2 nH0LÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅH2 nL! =n¥0

Ø 0. Thus, the

sequence :en+1I 2ÅÅÅÅp Mn+1ÅÅÅÅÅÅÅÅÅÅÅ2 >

n=2 iØ 0, where i œ . It immediately follows that the sequence

:en+1I 2ÅÅÅÅp Mn+1ÅÅÅÅÅÅÅÅÅÅÅ2 >

n=2 i is a Cauchy sequence. Hence, the sequence

:en+3I 2ÅÅÅÅp Mn+3ÅÅÅÅÅÅÅÅÅÅÅ2 - en+1I 2ÅÅÅÅp M

n+1ÅÅÅÅÅÅÅÅÅÅÅ2 >n=2 i

Ø 0. Thus,

15

2 a2 = Hn + 1L!

Ä

Ç

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅHn + 1L en+3K

2ÅÅÅÅÅÅp


+ en+3K2ÅÅÅÅÅÅp




´̈ ¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨̈¨̈ ¨≠ Æ¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨̈¨̈approaches zero, as nØ¶

É

Ö

ÑÑÑÑÑÑÑÑÑÑÑÑÑÑÑÑ

ØHn + 2L! Hn + 1LÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

n + 2 en+3K

2ÅÅÅÅÅÅp


=Hn + 1LÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅn + 2

Cn+3,

as desired. Therefore, the sequence 9 PnH0LÅÅÅÅÅÅÅÅÅÅÅÅÅn! =n¥0 Ø 0, and if a2 is the quadratic coefficient for

the polynomial PnHxL, where n is an even positive integer, we have that 2 a2 Ø Cn+3Hn+1LÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅn+2 .à

The final thing that is desired to investigate in this section, is the behavior of the sequence

8e2 i+1<iœ, in relation to its intial value, e3. It can be shown that if there exists h œ + and

N œ for which en+1ÅÅÅÅÅÅÅÅÅÅen-1> 1 + hÅÅÅÅn , for all even integers n > N , then the sequence 8e2 i+1<iœ will

diverge to positive infinity. Similarly, if there exists h œ + and N œ for whichen+1ÅÅÅÅÅÅÅÅÅÅen-1

< 1 - hÅÅÅÅn , for all even integers n > N , then the sequence 8e2 i+1<iœ will converge to the

value of zero. So, suppose e3* ∫ e3, and let 8e2 i+1

* <iœ be the resulting sequence, applying the

dynamical system, en+1* = n+1 + n+1, where

n+1 = en-1* J p

ÅÅÅÅÅÅ2

NC 1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅH1L H2L + K 1

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅHnL Hn - 1L OG ÿ 18n>2<

n+1 = „j=2

n-2

e j* en- j

* $%%%%%%pÅÅÅÅÅÅ2

18 jHMod 2L=1<ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

jH j + 1L .

Table 2.3 below, shows the results for several choices of e3*, and Figure 2.1 below displays a

subset of these choices for e3*.

16

Table 2.3: Summary of the Sequence 8e2 i+1* <iœ for Strategic Choices of e3

*.

e3* h† IN†M e2 N+1 n‡ e2 n+1

e3 - 10-5 0.3 H68, 738LHaL 0.3710 106 0.0756e3 - 10-6 0.3 H721, 097LHaL 0.3378 106 0.3195e3 - 10-7 0.09 H632, 880LHaL 0.3770 106 0.3690e3 0.05 H1LHaL e3 5.0 µ 106 0.3560e3 + 10-7 0.05 H3, 515, 751LHbL 0.3807 4.0 µ 106 0.3822e3 + 10-6 0.1 H532, 704LHbL 0.4173 106 0.4400e3 + 10-5 1.0 H57, 154LHbL 0.4573 106 1.858e3 + 10-4 1.0 H33, 938LHbL 0.7345 106 > 3.35 µ 106

e3 + 10-3 1.0 H3, 456LHbL 0.8278 106 > 1.19 µ 1069

e3 + 10-2 1.0 H356Lb 0.9678 440, 828 > 1.69 µ 10305

†The suggested value (from the simulation conducted) for which the convergence/divergence criterion of the sequence 8e2 i+1* <iœ holds.

‡The number of elements simulated from the sequence 8e2 i+1* <iœ, such that the simulated subset of this sequence is 8e2 i+1

* <i=1n .

HaLThe simulated sequence 8e2 i+1* <i=1

n , suggests the sequence 8e2 i+1* <iœ converges to a finite non-negative real number.

HbLThe simulated sequence 8e2 i+1* <i=1

n , suggests the sequence 8e2 i+1* <iœ diverges to +¶.

Figure 2.1: The Initial 500,000 Values of the Sequence 8e2 i+1* <iœ for Five Choices of e3

*.

17

The results for these choices of e3*, suggests that the sequence 8e2 i+1

* <iœ converges when

e3* < e3, and diverges when e3

* > e3. It remains unclear whether the sequence 8e2 i+1<iœ

converges. Expressions H2.7L and H2.10L above imply

lim-

t Ø H pÅÅÅÅÅ2 L0.5‚n=0

¶ PnH0L tnÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

n != lim-

t Ø H pÅÅÅÅÅ2 L0.5„n=0

¶

en+1ikjjj 2 t2

ÅÅÅÅÅÅÅÅÅÅÅÅp

y{zzz

nÅÅÅÅ2

$%%%%%%2ÅÅÅÅÅÅp

= lim-

t Ø H pÅÅÅÅÅ2 L0.5 e

1ÅÅÅÅ2 BF-1J 1ÅÅÅÅ2 + tÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅè!!!!!!!2 pNF

2

= ¶,

where “lim- ” is the left hand limit. This in turn implies that the series ⁄n=0¶ en+1 diverges.

Thus, if the sequence 8e2 i+1<iœ converges, then HaL it converges slowly, HbL it converges to a

real number greater than zero, or both HaL and HbL. From the data presented in Table 2.1

above, it appears that HaL is indeed the case. In fact, we see that following the generation of

the first 5 million terms of the sequence 8e2 i+1<iœ, the value for e107+1 º 36.2 % e3. So, we

can only make the inference here that if the said denumerable sequence converges, it does so

at a “slow pace.”

18

Section 3 - The Order Statistics for the Standard Normal Distribution

3.1: Properties of the Expected Values of the Order Statistics

We begin by introducing some notation. Let X ~ NH0, 1L, and let fX HxL be the pdf for X ,

evaluated at x œ . Let n œ be arbitrary, and let 8Xi<i=1n be a random sample (R.S.) from

the distribution of X . For each i = 1, 2, ..., n, let XHiL be the ith Order Statistic for the R.S.

8Xi<i=1n . Recall, the density function (pdf) for XHiL, evaluated at x œ , fXHiL HxL, is given by

(3.1)fXHiL HxL = i ÿ JniN fX HxL HFHxLLi-1 H1 - FHxLLn-i,

where FHxL is the expression given by H2.1L of Section 2 above @3D. Now, in TAMT, the

investigation of the expected value for Xêêê was key in showing assymtotic properties for the

Ordered Subset Analysis (OSA), where n Xêêê= ⁄i=1

n Xi. In particular, the focus of that OSA

investigation, was on sorting the R.S. 8Xi<i=1n in descending order. Thus, in investigating the

expected value of Xêêê, for a given subset of the OSA, it was (and is currently) necessary to

focus our attention on the expected value of XêêêHiL, where Hn - i + 1L Xêêê

HiL = ⁄ j=in XH jL, i § n.

Hence, we now investigate the expected value of XHiL, where i = 1, 2, ..., n, is arbitrary,

since the expected value of XêêêHiL is a function of XHiL. Let XHM L denote the median of the

R.S. 8Xi<i=1n . We proceed by showing that EHXH1+iLL = -EHXHn-iLL, where i is any non-nega-

tive integer for which i § M - 1, where EHAL denotes the expected value of the event A.

This result intuitively makes sense in light of the symmetry of the normal distribution, and

we show that this is indeed the case nonetheless.

Proposition 6: Let M be the median of the numbers of the sequence 81, 2, ..., n<, where n

is the sample size of the random sample 8Xi<i=1n , as previously defined. Then, we have

EHXH1+iLL = -EHXHn-iLL, for all 0 § i § M -3 - 18M œ<ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

2,

where 18A< is the indicator function for the event A as previously defined in H2.1L of Section

2.1 above. Further, if M œ , then EHXHM LL = 0.Ñ

Proof: We first note that the function … F-1HxL … is symmetric over -1, centered about the

value 1ÅÅÅÅ2 œ -1. To see this, let x1, x2 œ -1, for which 0 § x1 - 1ÅÅÅÅ2 = 1ÅÅÅÅ2 - x2. This implies

that

19

… F-1Hx1L … = F-1Hx1L ikjjjsince Kx1 -

1ÅÅÅÅÅ2

O2 n+1

¥ 0, for all n œ ‹ 80<y{zzz

= „n=0

¶

anKx1 -1ÅÅÅÅÅ2

O2 n+1

= „n=0

¶

anK1ÅÅÅÅÅ2

- x2O2 n+1

= -„n=0

¶

anKx2 -1ÅÅÅÅÅ2

O2 n+1

= -F-1Hx2L= … F-1Hx2L …,

where an = H2 pL 2 n+1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2 C2 n+1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅH2 n+1L! . Thus, … F-1HxL … is symmetric over -1, centered about the

value 1ÅÅÅÅ2 œ -1. Moreover, we see for any value x1 œ -1, that F-1Hx1L = -F-1H1 - x1L.Let i § n be arbitrary. Then, by definition, we have

(3.2)

EHXHiLL = ‡xœ

x ÿ fXHiL HxL d x

= ‡xœ

x ÿ i ÿ JniN fX HxL HFHxLLi-1 H1 - FHxLLn-i d x

= n ÿ Kn - 1i - 1

O ‡limxØ-¶ FHxL

limxØ¶ FHxL IF-1HuL ui-1H1-uLn-iM du

= n ÿ Kn - 1i - 1

O ‡0

1F-1HuL ui-1H1 - uLn-i

d u,

where we have made the substitution, u = FHxL. Let bi-1,n-i : @0, 1D Ø @0, 1D, be the function

defined by bi-1,n-iHxL = xi-1H1 - xLn-i. Here, we make the observation that

bi-1,n-iHxL = bn-i,i-1H1 - xL, for any x œ @0, 1D. Suppose 0 § i § M - 3-18M œ<ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2 . Then,

(3.3)EHXHi+1LL = n ÿ Kn - 1

iO ‡

0

1F-1HuL uiH1 - uLn-Hi+1L

d u

= -n ÿ Kn - 1i

O ‡0

1AF-1H1 - uLE H1 - uLn-Hi+1L HuLi d u

20

= n ÿ Kn - 1i

O ‡1

0AF-1HvLE HvLn-Hi+1L H1 - vLi d v Hv = 1 - uL

= -n ÿ Kn - 1i

O ‡0

1AF-1HvLE HvLn-Hi+1L H1 - vLi d v

= -n ÿ K n - 1n - Hi + 1L O ‡

0

1AF-1HvLE HvLn-Hi+1L H1 - vLi d v

= -EHXHn-iLL,

as desired. Now, suppose that M œ , so that with i = M - 1, H3.3L implies that

EHXHM LL = -EHXHM LL. But, this holds if and only if EHXHM LL = 0. Therefore,

EHXH1+iLL = -EHXHn-iLL, for all 0 § i § M -3 - 18M œ<ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

2,

and if M œ , it follows that EHXHM LL = 0.à

Though the result of Proposition 6 appears rather trivial and non-essential to the theme of

the Paper at hand, we will make use of the properties unfolded within the proof, in the subse-

quent Proposition which follows. We now turn our focus toward the trucation of terms, for

the infinite series given by H2.2L, and the effect this has on the numerical approximation for

EHXHiLL. In the Proposition which follows, we will show that each subsequent term of the

said series, contributes less “mass” to the value of EHXHiLL, than does any single earlier term

of the series, provided that M < i, where M is the median of the sequence of integers

81, 2, ..., n<, as previously defined.

Proposition 7: For each m = 0, 1, 2, ..., let am = H2 pL 2 m+1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2 C2 m+1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅH2 m+1L! , the coefficient of the m + 1st

term of the series given by H2.2L above. For each i = 1, 2, ..., n, we define EHXHiLmL, as

follows

EHXHiLmL = n ÿ Kn - 1i - 1

O ‡0

1amKx -

1ÅÅÅÅÅ2

O2 m+1

xi-1H1 - xLn-i d x.

Let M be as previously defined. Then the following are true:

1. EHXHiLL = ⁄m=0¶ EHXHiLmL, and

2. If M < i and e2 m+3 < e2 m+1, then EHXHiLm+1L < EHXHiLmL.3. If n = 109 and M < i, then EHXHiLmL < EHXHi+1LmL.Ñ

21

Proof: (1) Here, from H3.2L above, we have

EHXHiLLÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅnI n-1

i-1 M= ‡

0

1F-1HxL xi-1H1 - xLn-i

d x

= ‡0

1

„m=0

¶

amKx -1ÅÅÅÅÅ2

O2 m+1

xi-1H1 - xLn-i d x

= ·0

1

„m=0

¶

am „j=0

2 m+1

K2 m + 1j

O x2 m+1- jK-1ÅÅÅÅÅ2

Oj xi-1 ‚

p=0

n-i

Kn - ip

O H-xLn-i-p d x

= ·0

1

„m=0

¶

am „j=0

2 m+1

„p=0

n-i

K2 m + 1j

O Kn - ip

O H-1L j+n-i-p K 1ÅÅÅÅÅ2

Oj x2 m- j+n-p d x,

an integral of a sum of polynomials of degree 2 m + n. This implies that we may pass the

integral through the summand of m. Thus, we have

EHXHiLLÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅnI n-1

i-1 M= ·

0

1

„m=0

¶

am „j=0

2 m+1

„p=0

n-i

K2 m + 1j

O Kn - ip


Oj x2 m- j+n-p d x

= „m=0

¶

·0

1

am „j=0

2 m+1

„p=0

n-i

K2 m + 1j

O Kn - ip


Oj x2 m- j+n-p d x

= „m=0

¶

‡0

1amKx -

1ÅÅÅÅÅ2

O2 m+1

xi-1H1 - xLn-i d x

= „m=0

¶

EHXHiLmLÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅnI n-1

i-1 M,

which provides that EHXHiLL = ⁄m=0¶ EHXHiLmL, as desired.

(2) Here, suppose M < i, and let x1 œ I 1ÅÅÅÅ2 , 1M, m œ be arbitrary. It follows that

bi-1,n-iH1 - x1L < bi-1,n-iHx1L. To see this, we note that 1 < x1ÅÅÅÅÅÅÅÅÅÅÅÅ1-x1. Since M < i, and

M = n+1ÅÅÅÅÅÅÅÅÅÅ2 , it follows that n - i < i - 1. This implies that I x1ÅÅÅÅÅÅÅÅÅÅÅÅ1-x1Mn-i

< I x1ÅÅÅÅÅÅÅÅÅÅÅÅ1-x1Mi-1. Thus,

22

bi-1,n-iH1 - x1L = H1 - x1Li-1 Hx1Ln-i

< Hx1Li-1 H1 - x1Ln-i

= bi-1,n-iHx1L,

as desired. Now, in the midst of the proof of Proposition 6 above, we showed that

F-1Hx1L = -F-1H1 - x1L, where we applied the result that Ix1 - 1ÅÅÅÅ2 M2 m+1= I 1ÅÅÅÅ2 - x2M2 m+1.

Thus, it follows that

(3.4)H1 - 2 x2L2 m+1 bi-1,n-iHx2L < H2 x1 - 1L2 m+1 bi-1,n-iHx1L,

for all x1 œ I 1ÅÅÅÅ2 , 1M, where x2 = 1 - x1 œ I0, 1ÅÅÅÅ2 M. This implies that

‡0

0.5H1 - 2 x2L2 m+1 bi-1,n-iHx2L d x2 < ‡

0.5

1H2 x1 - 1L2 m+1 bi-1,n-iHx1L d x1.

Let hHm, KL = KH2 x1 - 1L2 m+1, where K œ +. Since K, H2 x1 - 1L œ +, then the natural

logarithm of hHm, KL exists and is differentiable. Differentiating the function LnHhHm, KLL,with respect to m, we obtain

dÅÅÅÅÅÅÅÅÅÅÅÅd m

LnHhHm, KLL =1

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅhHm, KL h£Hm, KL

=d

ÅÅÅÅÅÅÅÅÅÅÅÅd m

HLnHKL + H2 m + 1L LnH2 x1 - 1LL= 2 LnH2 x1 - 1L,

so that h£Hm, KL = 2 KH2 x1 - 1L2 m+1 LnH2 x1 - 1L < 0. Since bi-1,n-iHx2L < bi-1,n-iHx1L, it

follows that h£Hm, bi-1,n-iHx1LL < h£Hm, bi-1,n-iHx2L. This result, along with H3.4L above,

implies that

H1 - 2 x2L2 m+1 bi-1,n-iHx2L I1 - H1 - 2 x2L2M < H2 x1 - 1L2 m+1 bi-1,n-iHx1L I1 - H2 x1 - 1L2M,

which in turn implies that

H2 x1 - 1L2 m+3 bi-1,n-iHx1L - H1 - 2 x2L2 m+3 bi-1,n-iHx2L <

H2 x1 - 1L2 m+1 bi-1,n-iHx1L - H1 - 2 x2L2 m+1 bi-1,n-iHx2L.

Assuming the result of Proposition 1 to hold for all even positve integers, it follows that

23

I2 ÿ 2ÅÅÅÅp ÿ pM2 m+3ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2 e2 m+3

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2 n + 3

Ä

ÇÅÅÅÅÅÅÅÅÅKx1 -

1ÅÅÅÅÅ2

O2 m+3

bi-1,n-iHx1L - K 1ÅÅÅÅÅ2

- x2O2 m+3

bi-1,n-iHx2LÉ

ÖÑÑÑÑÑÑÑÑÑ

<

I2 ÿ 2ÅÅÅÅp ÿ pM2 m+1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2 e2 m+1

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2 n + 1

Ä

ÇÅÅÅÅÅÅÅÅÅKx1 -

1ÅÅÅÅÅ2

O2 m+1

bi-1,n-iHx1L - K 1ÅÅÅÅÅ2

- x2O2 m+1

bi-1,n-iHx2LÉ

ÖÑÑÑÑÑÑÑÑÑ,

which in turn implies that

‡0.5

1am+1Kx1 -

1ÅÅÅÅÅ2

O2 m+3

bi-1,n-iHx1L d x1 - ‡0

0.5am+1K

1ÅÅÅÅÅ2

- x2O2 m+3

bi-1,n-iHx2L d x2 <

‡0.5

1amKx1 -

1ÅÅÅÅÅ2

O2 m+1

bi-1,n-iHx1L d x1 - ‡0

0.5amK 1

ÅÅÅÅÅ2

- x2O2 m+1

bi-1,n-iHx2L d x2.

Thus, we have

EHXHiLm+1LÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅnI n-1

i-1 M= ‡

0

1am+1 Kx -

1ÅÅÅÅÅ2

O2 m+3

bi-1,n-i HxL d x

= am+1 ikjjj‡

0.5

1Kx1 -

1ÅÅÅÅÅ2

O2 m+3

bi-1,n-i Hx1L d x1 - ‡0

0.5K 1

ÅÅÅÅÅ2

- x2O2 m+3

bi-1,n-i Hx2L d x2y{zzz

< am ikjjj‡

0.5

1Kx1 -

1ÅÅÅÅÅ2

O2 m+1

bi-1,n-i Hx1L d x1 - ‡0

0.5K 1

ÅÅÅÅÅ2

- x2O2 m+1

bi-1,n-i Hx2L d x2y{zzz

= ‡0

1am Kx -

1ÅÅÅÅÅ2

O2 m+1

bi-1,n-i HxL d x

=EHXHiLmLÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅnI n-1

i-1 M,

so that EHXHiLm+1L < EHXHiLmL, as required.

(3) For clarity is provided in Appendix A.

Therefore, H1L EHXHiLL = ⁄m=0¶ EHXHiLmL, for all i § n, H2L if M < i and e2 m+3 < e2 m+1, then

EHXHiLm+1L < EHXHiLmL, and H3L if n = 109 and M < i, then EHXHiLmL < EHXHi+1LmL.à

24

3.2: The Expected Value of the ith Order Statistic

In this section, we evaluate the integral given by H3.2L above, and suggest a reduction for-

mula for the expected value of XHiL, where M < i § n, and where M is as previously defined.

We proceed with the statement of the integral reduction formula for EHXHiLmL .

Proposition 8: For each i = 1, 2, ..., n, and m = 0, 1, 2, ..., let EHXHiLmL and am be as previ-

ously defined in Proposition 7 above. Then,

(3.5)EHXHiLmL = nKn - 1i - 1

O am „j=0

2 m+1

K2 m + 1j

OCK2 m + n - j + 1i

O ÿ i ÿ 2 jG-1

H-1L2 m+1- j.Ñ

Proof: From Proposition 7 above, we have

EHXHiLmLÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅn ÿ I n-1

i-1 M= ‡

0

1amKx -

1ÅÅÅÅÅ2

O2 m+1

xi-1H1 - xLn-i d x

= ‡0

1amKx - 1 +

1ÅÅÅÅÅ2

O2 m+1

xi-1H1 - xLn-i d x

= am ‡0

1 ‚

j=0

2 m+1

K2 m + 1j

O Hx - 1L2 m+1- j H2L- j xi-1H1 - xLn-i d x

= am ‡0

1 ‚

j=0

2 m+1

K2 m + 1j

O H-1L2 m+1- j H2L- j xi-1H1 - xL2 m+1- j+n-i d x

= am „j=0

2 m+1

K2 m + 1j

O H-1L2 m+1- j H2L- j ‡0

1xi-1H1 - xL2 m+1- j+n-i

d x

= am ‚j=0

2 m+1

K2 m + 1j

O H-1L2 m+1- j H2L- j ikjj GHiL ÿ GH2 m - j + n - i + 2L

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅGH2 m - j + n + 2L

y{zz

= am „j=0

2 m+1

K2 m + 1j

OCK2 m + n - j + 1i

O ÿ i ÿ 2 jG-1

ÿ H-1L2 m+1- j,

as desired, where we have applied the relation GHnL = Hn - 1L!. Therefore,

25

EHXHiLmL = nKn - 1i - 1

O am „j=0

2 m+1

K2 m + 1j

OCK2 m + n - j + 1i

O ÿ i ÿ 2 jG-1

H-1L2 m+1- j,

for all i = 1, 2, ..., n, and m = 1, 2, ...à

Since EHXHiLL = ⁄m=0¶ EHXHiLmL, it follows that

(3.6)EHXHiLL = n Kn - 1i - 1

O „m=0

¶

am „j=0

2 m+1

K2 m + 1j

OCK2 m + n - j + 1i

O ÿ i ÿ 2 jG-1

H-1L2 m+1- j.

Expression H3.6L can now be utilized to numerically approximate the expected value of the

order statistics for the random sample 8Xi<i=1n . These computed expected values will then be

used as a comparison basis for the “formulae approximated” expected values calculated in

this Author’s Master’s Thesis.

3.3: Further Properties of EHXHiLmL and the Function gH j, iL

So, expression H3.6L stated in the previous Section, provides a means by which to calculate

the expected value of the order statistics for the random sample 8Xi<i=1n . We stated and

proved under certain regulatory conditions (I.e., Proposition 1 holds for all even positive

integers, and M < i), that the sequence 8EHXHiLmL<m=0¶ is strictly decreasing. However, the

magnitude of the difference between consecutive elements of this sequence remains unclear.

It could be the case that a great many elements of this sequence must be utilized (summed),

to obtain a numerically precise approximation for the infinite sum of expression H3.6L. In

this Section, we introduce some of the properties for EHXHiLmL (where m is arbitrary), which

lead to improvement in the computational efficiency of this value.

Here, for a fixed value of m, we define the function g : H j, iL Ø + by

gH j, iL = K2 m + 1j

OCK2 m + n - j + 1i

O ÿ i ÿ 2 jG-1

,

the jth term of the sum over EHXHiLmL of expression H3.6L above, where

j œ 8k œ ‹ 80< : k § 2 m + 1< and i œ 8k œ : M < i § n<. We will investigate how g

behaves for changes in the value of j, and determine where g achieves its maximum value

26

on its domain, while holding the value of i fixed. Then, we will investigate how g behaves

for changes in the value of i, holding j fixed at the value which maximizes g. In particular,

we will address and answer the issue of how the maximum for g (with respect to j) behaves

for dynamic values of i.

So, for an arbitrary value of i, we consider maximizing g, allowing the value of j to be

dynamic. Now, an option to maximizing g, is to determine where the derivative of g (with

respect to j) is equal to the value of zero. The problem with this notion is that g entails

products and quotients of factorials, the derviatives of which are not trivially obtainable. To

avoid taking the derivative of g, we note that the Gamma Function HGHnL = Hn - 1L!L is

continuous, differentiable (XXX Reference here), and non-zero over the interval H0, ¶L,which implies that the domain for g is H0, ¶L µ H0, ¶L, and we are not restricted to integer

values for i and j. Thus, if we can find a value for j, such that gH j, iL = gH j + 1, iL, then the

Mean Value Theorem for Derivatives provides that there exists c œ H j, j + 1L for which

(3.7)d

ÅÅÅÅÅÅÅÅÅÅd j

gHc, iL =gH j + 1, iL - gH j, iLÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅH j + 1L - j

= 0.

This implies that c œ H j, j + 1L is a local extrema for g. Hence, we proceed to investigate

the function gH j, iL for consecutive values of j, and solve the equation gH j+1,iLÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅgH j,iL = 1. Taking

the natural logarithm of gH j, iL (we can do this, since 0 < x!, for all x œ H0, ¶L), we have

LnHgH j, iLL = Lnikjj H2 m + 1L! Hi - 1L! H2 m + 1 + n - j - iL!

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅj! H2 m + 1 - jL! H2 m + 1 + n - jL! 2 j

y{zz

= ‚k=1

2 m+1

LnHkL + ‚k=1

i-1

LnHkL + ‚k=1

2 m+1+n- j-i

LnHkL -

i

kjjjjj‚

k=1

j

LnHkL + ‚k=1

2 m+1- j

LnHkL + ‚k=1

2 m+1+n- j

LnHkL + j ÿ LnH2Ly

{zzzzz,

and

LnHgH j + 1, iLL = LnHgH j, iLL - LnH2 m + 1 + n - j - iL - LnH j + 1L +LnH2 m + 1 - jL + LnH2 m + 1 + n - jL - LnH2L.

Thus,

27

gH j + 1, iLÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

gH j, iL =H2 m + 1 - jL H2 m + 1 + n - jLÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2 H2 m + 1 + n - j - iL H j + 1L .

We denote this ratio as hH j, iL, and differentiating h, with respect to j, we have

(3.8)

dÅÅÅÅÅÅÅÅÅÅd j

hH j, iL =-H2 m + 1 - jL

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2 H j + 1L H2 m + 1 - j + n - iL ik

jj1 +2 m + 1 - j + nÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

j + 1y{zz -

2 m + 1 - j + nÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2 H j + 1L H2 m + 1 - j + n - iL ik

jj1 -2 m + 1 - j

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2 m + 1 - j + n - i

y{zz

< 0,

for all i § n and j § 2 m. This implies that g is an increasing function of j , for all j § jè, and

g is a decreasing function of j, for all j > jè, where jè is the value for which hI jè, iM = 1. That

is, H3.8L implies that there can be at most one local extrema II.e., dÅÅÅÅÅÅÅd j gH j, iL = 0M for g, when

j § 2 m + 1, and this extrema is a local maximum. Thus, if there does not exist a local

extrema for g, when j § 2 m + 1, then H3.8L provides that the maximum for g occurs when

j ª 2 m + 1. We summarize these maximization results for g in the following Proposition.

Proposition 9: Let hH j, iL be as defined above, where we allow j to be any non-negative

real value for the function g, and where i, M < i § n, is considered a fixed positive integer.

We define J = 8 j : hH j, iL = 1<. It then follows that

1. J Õ +,2. » J » = 2, where » J » denotes the number of elements of the set J ,

3. MaxH8gH j, iL : M < i § n; j œ , 0 § j § 2 m + 1<L = ; gHc, iL, if c < 2 m + 1gH2 m + 1, iL, otherwise

,

where c œ HMinHJL, MinHJ + 1LL.Ñ

Proof: For notational clarity, let m = 2 m + 1. Then, it can be shown that

gH j + 1, iLÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

gH j, iL = 1 ó Hm - 3 j - 2L Hm - j + nL + 2 iH j + 1L = 0

ó J =loomnoo

-è!!!!!!!!!!!!!!!!!!!!

2 - 12 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

6,

+è!!!!!!!!!!!!!!!!!!!!

2 - 12 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

6

|oo}~oo

,

where = 4 m + 3 n - 2 Hi + 1L and = m2 + Hn - 2L m - 2 n + 2 i. Let b£ = + 2 i,

c£ = - 2 i, and fk : 81, 2, ..., n< Ø , k = 1, 2, be defined by

28

fkHiL =Hb£ - 2 iL + H1 - 2 ÿ 18kHMod2L=0<L "#############################################Hb£ - 2 iL2 - 12 Hc£ + 2 iLÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

6.

Now, to show that J Õ and » J » = 2 simultaneously hold, we need to show that

2 - 12 > 0. Here, we note that i § n, so that 4 m + 12 n - 16 i + 8 > 0 and 24 Hn - iL > 0.

Thus,

0 < H4 m + 12 n - 16 i + 8L + 24 Hn - iL< 4 m2 + H12 n - 16 i + 8L m + 24 Hn - iL + H3 n - 2 Hi + 1LL2

= 2 - 12 ,

so that J Õ , and » J » = 2. Thus, H2L holds. To show that H1L holds, we need to show for

each k = 1, 2, that fkHiL > 0. It is clear that > 0. Thus, if f2HiL > 0, then

8f1HiL, f2HiL< Õ +, which would signify that H1L holds. So, we proceed to show that

f2HiL > 0. Here, for each i = 1, 2, ..., n, we define the symbol ' *i ', by *i œ 8 < , > , =<,for which the relation f2HiL *i 0 holds. Then,

f2HiL *i 0 óHb£ - 2 iL -

"#############################################Hb£ - 2 iL2 - 12 Hc£ + 2 iLÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

6*i 0

ó Hb£ - 2 iL *i"#############################################Hb£ - 2 iL2 - 12 Hc£ + 2 iL

ó 12 Hc£ + 2 iL *i 0.

Now, M < i, so that n ¥ 2 and > m2 + Hn - 2L Hm - 1L ¥ 1. This implies that *i = ' > ', so

that f2HiL > 0, for all i, such that M < i. Since 0 < f2HiL < f1HiL, for all M < i, it follows that

J Õ +, so that H1L holds. Finally, to show that H3L holds, we need to show that m < f1HiL,for all M < i § n. Equivalently, we need to show è!!!!!!!!!!!!!!!!!!!!

2 - 12 > 2 Hm - n + i - 1L - n.

Clearly, if 2 Hm+i-1LÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ3 < n, this will indeed be the case. So, suppose that 2 Hm+i-1LÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ3 ¥ n. Similar

to that done above, for each i = 1, 2, .., n, let *i œ 8 < , > , =<, for which the relation

f1HiL *i m holds. Then,

f1HiL *i m óHb£ - 2 iL + "#############################################Hb£ - 2 iL2 - 12 Hc£ + 2 iLÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

6*i m

ó "#############################################Hb£ - 2 iL2 - 12 Hc£ + 2 iL *i 2 Hm - n + i + 1L - n ¥ 0ó H24 n - 24 iL m + 24 Hn - iL *i 0,

29

which implies that *i = ' > ', since i § n and 2 Hm+i-1LÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ3 ¥ n implies that m > 1. Suppose

2 m < f2HiL < 2 m + 1. Then, the Mean Value Theorem and expression H3.7L above, provide

that there exists c œ Hf2HiL, f2HiL + 1L, for which g£Hc, iL = 0. Thus, if c < 2 m + 1, then

MaxHgH j, iLL = gHc, iL. Otherwise, MaxHgH j, iL = gH2 m + 1, iL. Conversely, suppose that

f2HiL § 2 m, then f2HiL + 1 § 2 m + 1, so that the c œ Hf2HiL, f2HiL + 1L for which g£Hc, iL = 0,

lies within the domain of j. Hence, MaxHgH j, iL = gHc, iL. This is precisely what we needed

to show for H3L to hold true. Therefore, the three conditions of the Proposition are true.à

Now that we have established where the maximum of g occurs, for dynamic values of j and

i fixed, we now investigate the behavior of the “shift” in the value of f2HiL as i changes in

value. The simplicity of this notion is due to the fact that f2HiL is dependent only on the

value of i and is independent of j. We proceed to take the derivative of f2HiL, with respect

to i, as this will allow us to determine the behaviour of this function as the value of i

changes. Here,

f2£HiL =

-2 - 2-1 IHb£ - 2 iL2 - 12 Hc£ + 2 iLM- 1ÅÅÅÅ2 H2 Hb£ - 2 iL H-2L - 12 H2LLÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

6

= -2ÅÅÅÅÅ6

i

k

jjjjjjjj1 -Hb£ - 2 iL + 6

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ"#############################################Hb£ - 2 iL2 - 12 Hc£ + 2 iL

y

{

zzzzzzzz

> 0,

since Hb£-2 iL+6ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ"###########################################Hb£-2 iL2-12 Hc£+2 iL> 1. Thus, as i increases, the value of j for which the maximum of

g occurs also increases, provided that the maximum value of g occurs when j = f2HiL. Fig-

ure 3.1 below, shows the functions gI j, i = M + 1ÅÅÅÅ2 M, gH j, i = 750L and gH j, i = nL, denoted

by g1, g2, and g3, respectively, when m = 25 and n = 1000. From this Figure, we see that

when i ª n, the function g is strictly increasing across the domain of j. This suggests for

large choices of i (with respect to nL, that we might computationally evaluate the summand

of expression H3.5L, beginning with the upper limit value of j = 2 m + 1, rather than the

lower limit value. This is due to the fact that we will be utilizing the programming language

C ++ to evaluate the terms of this summand. With the application of this programming

language, we are limited to positive numerical values in the interval I1.7 E-307, 1.7 E307M.Suppose the value of the kth term (k chosen arbitrarily) of this summand is less than the

30

minimum positive real number recognized by the C ++ programming language (I.e., less

than 1.7 E-307 ), and that k < Min 8f2HiL, 2 m + 1< . Since g is strictly increasing for every

j § k, it follows that each term of the summand, prior to the kth term is also not recognized

by C ++ as a real number. Thus, at least k terms of the summand may be truncated from the

numerical calculation of EHXHiLmL. So, by beginning with the upper limit of the summand of

H3.5L, we are able to determine that these k terms are not recognized by C ++, through the

calculation of a single term... Namely, the kth term. This leads to improved computational

efficiency in the calculation of EHXHiLmL, and hence EHXHiLL also. In fact, we will show in

Proposition 11 below, that this observation is not limited to large values of i, and evaluating

expression H3.5L beginning with the upper limit potentially improves computational effi-

ciency for all M < i § n.

Figure 3.1: A Sample of Three Graphs of the Function gH j, iL for m = 25 and n = 1000.

5 10 15 20 25

1

2

3

4

j

gHj,iL

gHj, i=M+0.5L Htimes 3.2µ10303L

gHj, i=1000L Htimes 1.33µ1011LgHj, i=750L Htimes 1.0µ10250L

In reviewing expression H3.5L, we note that each of the terms for the summand, entails the

product of two combinatoricals. So, the final thing that we will look at, prior to leaving

Section 3.3, is an equivalent formula for H3.5L which involves the function h solely. This

31

will allow for a simpler computation of EHXHiLmL, and is summarized in Proposition 10 as

follows.

Proposition 10: For each M < i § n, m = 0, 1, 2, ..., and j § 2 m + 1, let EHXHiLmL, gH j, iL,and hH j, iL be as previously defined in this Section. We define bm = amH2L-H2 m+1L. It then

follows that

(3.9)EHXHiLmL = bm

i

k

jjjjjjjjj1 + „

j=0

2 m

H-1L j+1 ‰k=2 m- j

2 m

HhHk, iLL-1y

{

zzzzzzzzz.Ñ

Proof: We first note that nI n-1i-1 M am gH2 m + 1, iL = bm. To see this, we have

nKn - 1i - 1

O am gH2 m + 1, iL = nKn - 1i - 1

O 22 m+1 bm K2 m + 12 m + 1

OBJniN ÿ i ÿ 22 m+1F

-1

=n I n-1

i-1 M 22 m+1

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅI n

i M ÿ i ÿ 22 m+1 bm

= bm,

as desired. We also have that gH j, iL = @hH j, iLD-1 gH j + 1, iL, for all 0 § j § 2 m. Then, for

0 § k § j, we have

gH j - k, iL = gH j + 1, iL ‰r= j-k

j

@hHr, iLD-1,

so that

nKn - 1i - 1

O am gH2 m - k, iL = n Kn - 1i - 1

O am gH2 m + 1, iL ‰r=2 m-k

2 m

@hHr, iLD-1

= bm ‰r=2 m-k

2 m

@hHr, iLD-1.

32

Hence,

EHXHiLmL = nKn - 1i - 1

O am „j=0

2 m+1

K2 m + 1j

OCK2 m + n - j + 1i

O ÿ i ÿ 2 jG-1

H-1L2 m+1- j

= n Kn - 1i - 1

O am ‚j=0

2 m+1

H-1L j+1 gH j, iL

= bm „j=0

2 m

H-1L j+1 ‰r=2 m- j

2 m

@hHr, iLD-1 + bm

= bm

i

k

jjjjjjjjj1 + „

j=0

2 m

H-1L j+1 ‰r=2 m- j

2 m

@hHr, iLD-1y

{

zzzzzzzzz,

as required. Therefore, EHXHiLmL = bmI1 + ⁄ j=02 m H-1L j+1 ¤k=2 m- j

2 m HhHk, iLL-1M.à

With the result of Proposition 10, we proceed to suggest a methodology of increasing the

computational efficiency in obtaining the value of H3.9L.

Proposition 11: Let 8sn<n=0p be a strictly increasing sequence of positive real numbers,

where p ¥ 1. Then,

1. … ⁄n=0p H-1Ln+1 sn … < sp.Ñ

Proof: Here, we have

‚n=0

p

H-1Ln+1 sn = ‚j=0

dHp-1Lê2tHsp-2 j - sp-2 j-1L H-1Lp+1 - s0 ÿ 18pHmodL 2=0<.

Suppose that p is a positive even integer. Then,

33

‚n=0

p

H-1Ln+1 sn = ‚j=0

dHp-1Lê2tHsp-2 j-1 - sp-2 jL - s0

> ‚j=0

p-1

Hsp- j-1 - sp- jL - s0

= -sp.

Next, suppose that p is a positive odd integer. Then,

‚n=0

p

H-1Ln+1 sn = ‚j=0

dHp-1Lê2tHsp-2 j - sp-2 j-1L

< ‚j=0

p-1

Hsp- j - sp- j-1L

= sp - s0

< sp.

Since 0 < sp - s0, it immediately follows that … ⁄n=0p H-1Ln+1 sn … < sp, as desired.à

Now, in H3.8L above, we showed that dÅÅÅÅÅÅÅd j hH j, iL > 0. Also, hH j, iL is greater than zero and is

continuous for all i § n and j § 2 m, so that 1ÅÅÅÅÅÅÅÅÅÅÅÅhH j,iL is greater than zero and continuous for

these values of i and j. From the application of the chain rule, we find

dÅÅÅÅÅÅÅÅÅÅd j

ikjj 1

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅhH j, iL

y{zz = H-1L ÿ

1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ@hH j, iLD2 ik

jj dÅÅÅÅÅÅÅÅÅÅd j

hH j, iLy{zz > 0,

for all i § n and j § 2 m. Suppose there exists k œ , 0 < k < 2 m, for which

¤ j=k2 m @hH j, iLD-1 < 1. Since 1ÅÅÅÅÅÅÅÅÅÅÅÅhH j,iL is increasing for increasing j, it follows that

‰j=0

2 m

@hH j, iLD-1 < ‰j=1

2 m

@hH j, iLD-1 < … < ‰j=k-1

2 m

@hH j, iLD-1 < ‰j=k

2 m

@hH j, iLD-1

Thus, Proposition 11 provides that

ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ„

j=2 m-k

2 m

H-1L j+1 ‰p=2 m- j

2 m

@hHp, iLD-1

ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ< ‰

j=k

2 m

@hH j, iLD-1.

34

Now, it can be shown that bm = e2 m+1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2 m+1 , for all m ¥ 1, and b0 =è!!!!!!!2 pÅÅÅÅÅÅÅÅÅÅÅÅÅ2 . Thus, for all m ¥ 1,

we have

EHXHiLmL = bm i

k

jjjjjjjjj1 + „

j=0

2 m

H-1L j+1 ‰p=2 m- j

2 m

Hh Hp, iLL-1y

{

zzzzzzzzz

=e2 m+1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

2 m + 1 i

k

jjjjjjjjj1 + „

j=2 m-k

2 m

H-1L j+1 ‰p=2 m- j

2 m

HhHp, iLL-1 + „j=0

2 m-Hk+1L

H-1L j+1 ‰p=2 m- j

2 m

HhHp, iLL-1y

{

zzzzzzzzz.

Thus, EHXHiLmL œ I, where I = @g - d, g + dD, such that g and d are given by

(3.10)

g =e2 m+1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

2 m + 1 i

k

jjjjjjjjj1 + „

j=0

2 m-Hk+1L

H-1L j+1 ‰p=2 m- j

2 m

HhHp, iLL-1y

{

zzzzzzzzz

d =e2 m+1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

2 m + 1

ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ„

j=2 m-k

2 m

H-1L j+1 ‰p=2 m- j

2 m

@hHp, iLD-1

ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ.

Suppose there exists k œ for which ¤ j=k2 m @hH j, iLD-1 < 10-305. Then, assuming that

e2 m+1 < 2 m + 1, it follows (Proposition 11) that d < 10-305. Thus, EHXHiLmL º g, with an

error in this approximated value of less than 10-305. Hence, as we mentioned from the

observations made for Figure 3.1 above, summing expressions H3.5L ê H3.9L beginning with

the upper/lower limits of the respective summand, leads to improvements in the computa-

tional efficiency of the respective summands for these expressions.

3.4: Methods

In this Section, we outline the methodology to generate the Order Statistics for the Standard

Normal Distribution, utilizing the theory introduced in the subsequent sections of this paper.

Since the Order Statistics, generated by H3.6L, will be used as a comparison basis to those

generated by TAMT, for consistency purposes, we choose the value of n to be one billion.

In addition, due to extraordinary computational processing time, we limit the number of

elements of the sequence 8e2 i+1<iœ to the initial five million values. We will show in Propo-

sition 12 below, that more than 9ÅÅÅÅ2 Im2 - 3 m + 2M computations are required to generate the

35

sequence 8e2 i+1<i=1m . Taking m to be the value of five million, this implies that more than

1.1 µ 1014 calculations are required to generate the sequence 8e2 i+1<i=1m . For each M < i § n,

the value of m will be chosen such that

m = Min 95 µ 106, Min 9m£ œ : EHXHiLm£ L < EIXHnL5µ106 M==. H3.11L

That is, m can assume the value of five million as a maximum, thereby exhausting all ele-

ments of the sequence 8e2 i+1<i=15µ106

in the calculation of EHXHiLL. And, if the precision of

EHXHiLm£ L is less than that of the value of EIXHnL5µ106 M, for some m£ < 5 µ 106, then the proper

subset 8e2 i+1<i=1m£ of 8e2 i+1<i=1

5µ106 will be used for the calculation of EHXHiLL. Part H3L of Proposi-

tion 7 above, implies that m£ = 5 µ 106 iff i = n, and that m£ < 5 µ 106 for all M < i < n. We

limit the value of m to that given by H3.11L, for two reasons: H1L The computational time to

compute the value of EHXHiLL is greatly improved, and H2L it does not make logical sense to

improve the numerical precision of EHXHiLL, for i < n, beyond that of the precision of the

maximum O.S. Finally, for each m = 0, 1, ..., and M < i § n, we define the value of

EHXHiLmL to be

EHXHiLmL =loomnoo

g + d, if m = 0 or ¤ j=02 m @hH j, iLD-1 > 10-305

g, if m ¥ 1 and there exists k > 0 for which ¤ j=k2 m @hH j, iLD-1 < 10-305

,

where g and d are as defined in H3.10L above. Expression H3.9L will be utilized to generate

the value of EHXHiLL. A simulation, resembling that of Appendix D of TAMT, will then be

conducted on these generated values of EHXHiLL, so that a comparison may be made to the

results shown in said Appendix. We leave Section 3, with the following Proposition.

Proposition 12: For each j ¥ 3, let c2 j+1 represent the number of calculations required to

obtain the value of e2 j+1. Then, the number of calculations required to obtain the values for

the sequence 8e2 i+1<i=3m , namely ⁄ j=3

m c2 j+1, is greater (approximately equal to) than9 Hm-2L Hm-1LÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2 .Ñ

Proof: Here, for any j ¥ 2, recall that e2 j+1 = 2 j+1 + 2 j+1, where

36

2 j+1 = e2 j-1JpÅÅÅÅÅÅ2

N ikjj 1

ÅÅÅÅÅ2

+1

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2 jH2 j - 1L

y{zz

2 j+1 = „k=2

2 j-2

e j e2 j-k $%%%%%%pÅÅÅÅÅÅ2

18kHMod 2L=1<ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

kHk + 1L .

For each j ¥ 2, let c2 j+1 and c2 j+1 be the number of calculations required to obtain the

values of 2 j+1 and 2 j+1, respectively. By inspection, we find that c2 j+1 = c2 j-1 , for all

j ¥ 3. Moreover, as far as the number of calculations are concerned, the terms 2 j+1 and

2 j-1, differ by the quotient e3 e2 j-3 è!!!!!!!2 pÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2 H2 j-3L H2 j-2L , which we denote by the symbol q. Now, we

assume that the value of "#####pÅÅÅÅ2 has been determined a-priori, as a constant, so that the num-

ber of calculations required to obtain the value of q is eight. This is broken down into five

calculations for the denominator and two calculations for the numerator. Each of the terms

2 j - x, where x œ 82, 3<, entails two calculations, namely H1L the product of the number two

and j, and H2L the difference between the result of this product and x. Since the product of

the terms H2 j - 3L and H2 j - 2L, entails a calculation, the aggregate number of calculations

for the denomiator of q is five. The numerator entails the products of the two epsilon terms,

e3 and e2 j-3, and the constant "#####pÅÅÅÅ2 . Thus, the aggregate number of calculations for the

numerator is two. Thus, to obtain the numerator and denomiator values for q, a total of

seven calculations are required. Finally, to obtain the value of q, the division of the result-

ing numerator and denomiator values is necessary. Thus, a total of eight calculations are

required to obtain the value of q. Moreover, the value of q is part of the summand for 2 j+1,

so that the “addition” of q is necessary to obtain the value of 2 j+1. Therefore,

c2 j+1b = c2 j-1b + 9, so that c2 j+1 = c2 j-1 + 9. Thus, for j ¥ 3, we have

c2 j+1 = c2 j-1 + 9= c2 j-3 + H2L H9L=ª

= c2 j-H2 j-5L + H j - 2L H9Lt

large j9 H j - 2L,

since the value of c5 is minute in comparison to large values of j. Given any m œ , m ¥ 3,

we have

37

‚j=3

m

c2 j+1 t‚j=3

m

9 H j - 2L

=9 Hm - 2L Hm - 1LÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

2.

But, the summand ⁄ j=3m c2 j+1 is precisely the number of calculations required to generate the

values of the sequence 8e2 j+1< j=3m . Therefore, the number of calculations required to gener-

ate the values for the sequence 8e2 j+1< j=3m is greater than 9 Hm-2L Hm-1LÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2 , but approximately

equal to this value.à

Table 3.1 below, shows how the summand ⁄ j=3m c2 j+1 increases as m increases. Here, we

note that to generate the one million terms for the sequence 8e2 j+1< j=16µ106

, in excess of the

initial five million terms, more than 4 µ 1013 calculations are required. This is almost half of

the number of calculations required to generate the initial five million terms for the said

sequence. Thus, we choose to terminate the generation of the sequence 8e2 j+3< jœ, with j

equal to the value of five million. We proceed to examine how the above methodology,

compares to that of Appendices C and D of this Author’s Master’s Thesis, as applied to

Ordered Subset Analysis.

Table 3.1: The Approximate Number of Calculations Required to Generate the Sequence8e2 j+1< j=1

m .

j m j I⁄k=3m j c2 k+1M† ⁄k=mj-1

mj c2 k+1

1 1,000,000 4.50 µ 1012 —2 2,000,000 1.80 µ 1013 1.35 µ 1013

3 3,000,000 4.05 µ 1013 2.25 µ 1013

4 4,000,000 7.20 µ 1013 3.15 µ 1013

5 5,000,000HaL 1.12 µ 1014 4.05 µ 1013

6 6,000,000 1.62 µ 1014 4.95 µ 1013

7 7,000,000 2.20 µ 1014 5.85 µ 1013

8 8,000,000 2.88 µ 1014 6.75 µ 1013

9 9,000,000 3.64 µ 1014 7.65 µ 1013

10 10,000,000 4.45 µ 1014 8.55 µ 1013

†Approximate number of calculations required to generate the sequence of values 8e2 k+1<k=0m j .

HaLApproximately 72 hours of computational time was required to generate the sequence 8e2 j+1< j=15µ106

, on an Intel P4 3.4GHZ platform with1GB of DDR2 533MHZ RAM, utilizing the C ++ programming language.

38

Section 4 - Comparison of the Two Methodologies to Generate the Order

Statistics for the Standard Normal Distribution, as Applied to Ordered

Subset Analysis

4.1: The Order Statistics

To begin Section 4, we clarify what is meant by the term “Two Methodologies.” This expres-

sion is equivalent to, “The Methodology utilized in this Author’s Master’s Thesis (Appendi-

cies C and D of said Paper) to generate the Order Statistics for the Standard Normal Distribu-

tion, as applied to the OSA, and the Methodology introduced in the current Paper to generate

the Order Statistics for the Standard Normal Distribution, as applied to the OSA.” We

denote the shorthand notation for these two methodologies by Thesis and InvF, respec-

tively. Prior to comparing how Thesis and InvF resemble/differ, as applied to the OSA,

we first compare the Order Statistics generated under each of these two methodologies.

Here, let n be as previously defined - the sample size of the random sample under investiga-

tion - where, as mentioned in Section 3 above, to be consistent with Appendix D of Thesis,

we choose the value of n to be one billion. Now, In vF has been outlined in detail in Sec-

tion 3.4 above. To this end, we have not described the details for Thesis. In summary,

Thesis defines the value of XHiL to be

(4.1)XHiL = F-1K1 -n - i + 1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

n + 1O,

for each i œ , for which j § i § n, where j œ is defined to be such that XH j-1L § 0 § XH jL.

The value for XHiL, given by expression H4.1L above, is then taken to be an approximated

value for the true value of EHXHiLL. The values(s) of j which satisfy the inequality,

XH j-1L § 0 § XH jL, is(are) such that j œ @M , M + 1D, where M is the median of the sequence

of values 81, 2, ..., n<, as previously defined in Section 3 above. Thus, we see that the order

statistics analyzed under the two methodologies, Thesis and InvF, are essentially the

same, differing only when M œ . And, when M œ , Thesis includes the investigation of

the O.S. XHM L, whereas InvF does not. For our investigation, M = 109+1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2 – . Thus, the

two methodologies investigate the same sequence of O.S., 8XHiL<i=M +1n . Table 4.1 below,

summarizes a few of the O.S., generated under each of the two methodologies.

39

Appendix A - Proofs

A.1: Proposition 7, Part 3

The proof for Part 3 of Proposition 7 relies on the results of the following four Lemma’s, inaddition to a computational analysis. So, we abstain from restating and proving this portionof Proposition 7, until we have stated and proved the results of the Lemma’s, as well assummarized the results from the numerical analysis.

Lemma A1: Let a, b, n œ be arbitrary, for which a + b = n and b < a. Then for anarbitrary element C œ , it follows that

‡xœ

xaH1 - xLb d x = „i=0

b¤ j=0

i-1 Hb - jLÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ¤ j=1

i+1 Ha + jL xa+i+1H1 - xLb-i

+ C.Ñ

Proof: The proof is by tabular integration (repetitious use of integration by parts). Let

gHxL = xa and f HxL = H1 - xLb. Then, the setup for the tabular integration of the function

f HxL ÿ gHxL is

f HxL and itsderivatives

gHxL and itsintegrals

H1 - xLb xa

- bH1 - xLb-1 xa+1Ha + 1L-1

ª ª

H-1Li ¤ j=0i-1 Hb - jL H1 - xLb-i I¤ j=1

i Ha + jLM-1 xa+i

ª ª

H-1Lb ¤ j=0b-1 Hb - jL I¤ j=1

b Ha + jLM-1 xa+b

0 I¤ j=1b+1 Ha + jLM-1

xa+b+1

It follows that

‡xœ

xaH1 - xLb d x = „i=0

b

H-1Li H-1Li Âj=0

i-1

Hb - jL H1 - xLb-i i

kjjjjjj‰

j=1

i+1

Ha + jLy

{zzzzzz

-1

xa+i+1 + C

= „i=0

b¤ j=0


i+1 Ha + jL xa+i+1H1 - xLb-i

+ C,

40

for any C œ . This is precisely what we needed to show.à

Lemma A2: Let a, b, n œ be arbitrary, for which a + b = n and b < a. Then,

(A1)‡1ÅÅÅÅ2

1Kx -

1ÅÅÅÅÅ2

O Kga,b-1HxL -aÅÅÅÅÅÅb

ga-1,bHxLO d x =GHbL GHa + 1LÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

GHn + 2L + K 1ÅÅÅÅÅ2

On+1

Hc + dL,

where gi, jHxL = xiH1 - xL j, is as defined in Proposition 6 of Section 3.1 above, and where

c =b - Ha + 2L

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅbHa + 1L Ha + 2L -

Hb + 1L GHbL GHa + 1LÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

GHn + 2L

d = „i=1

b-1¤ j=1


i+2 Ha + jL@Hi + 1L Hb - a - 2 i - 2LD.Ñ

Proof: Here, we can re-write the integrand of HA1L as the expression FHxL - GHX L, where

FHxL =aÅÅÅÅÅÅb

‡1ÅÅÅÅ2

1K 1

ÅÅÅÅÅ2

ga-1,bHxL - ga,bHxLO d x

GHxL = ‡1ÅÅÅÅ2

1K 1

ÅÅÅÅÅ2

ga,b-1HxL - ga+1,b-1HxLO d x.

Then, Lemma A1 above provides that

aÅÅÅÅÅÅÅÅÅÅÅ2 b

‡1ÅÅÅÅ2

1ga-1,bHxL d x =

aÅÅÅÅÅÅÅÅÅÅÅ2 b

Ä

Ç

ÅÅÅÅÅÅÅÅÅÅÅÅÅ

b ! Ha - 1L!ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅHa + bL !

- K 1ÅÅÅÅÅ2

Oa+b

„i=0

b¤ j=0

i-1 Hb - jLÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ¤ j=1

i+1 Ha - 1 + jL

É

Ö

ÑÑÑÑÑÑÑÑÑÑÑÑÑ

and

aÅÅÅÅÅÅb

‡1ÅÅÅÅ2

1ga,bHxL d x =

aÅÅÅÅÅÅb

Ä

Ç

ÅÅÅÅÅÅÅÅÅÅÅÅÅ

b ! a !ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅHa + b + 1L !

- K 1ÅÅÅÅÅ2

Oa+b+1

„i=0

b¤ j=0


i+1 Ha + jL

É

Ö

ÑÑÑÑÑÑÑÑÑÑÑÑÑ.

Also, for each i = 0, 1, ..., b, we have

41

¤ j=0i-1 Hb - jL

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ¤ j=1

i+1 Ha + jL-

¤ j=0i-1 Hb - jL

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ¤ j=1

i+1 Ha - 1 + jL=

¤ j=0i-1 Hb - jL@Ha - Ha + i + 1LD

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ¤ j=0

i+1 Ha + jL

= -Hi + 1L ¤ j=0i-1 Hb - jL


i+1 Ha + jL,

so that

FHxL = GHbL K a !ÅÅÅÅÅÅÅÅÅÅÅÅÅ2 n!

-aa !

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅHn + 1L!O - K 1

ÅÅÅÅÅ2

On+1

i

k

jjjjjjjj„i=1

b

Hi + 1L ¤ j=1i-1 Hb - jL


i+1 Ha + jL+

1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅbHa + 1L

y

{

zzzzzzzz

For the integrand GHxL, Lemma A1 above provides that

1ÅÅÅÅÅ2

‡1ÅÅÅÅ2

1ga,b-1HxL d x =

Hb - 1L! a !ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2 Ha + bL!

- K 1ÅÅÅÅÅ2

Oa+b+1

„i=0

b-1¤ j=0

i-1 Hb - 1 - jLÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

¤ j=1i+1 Ha + jL

and

‡1ÅÅÅÅ2

1ga+1,b-1HxL d x =

Hb - 1L! Ha + 1L!ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅHa + b + 1L!

- K 1ÅÅÅÅÅ2

Oa+b+1

„i=0

b-1¤ j=0

i-1 Hb - 1 - jLÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ¤ j=1

i+1 Ha + 1 + jL.

Now, for each i = 0, 1, ..., b - 1, we have

¤ j=0i-1 Hb - 1 - jL


i+1 Ha + 1 + jL-

¤ j=0i-1 Hb - 1 - jL


i+1 Ha + jL=

¤ j=0i-1 Hb - 1 - jL@Ha + 1 - Ha + i + 2LD

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ¤ j=1

i+2 Ha + jL

= -Hi + 1L ¤ j=0i-1 Hb - 1 - jL


i+2 Ha + jL,

so that GHxL is equal to the expression

GHbL K a !ÅÅÅÅÅÅÅÅÅÅÅÅÅ2 n!

-Ha + 1L!ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅHn + 1L!

O - K 1ÅÅÅÅÅ2

On+1

i

k

jjjjjjjj„i=1

b-1 Hi + 1L ¤ j=0i-1 Hb - 1 - jL

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ¤ j=1

i+2 Ha + jL+

1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅHa + 1L Ha + 2L

y

{

zzzzzzzz

For each i = 1, 2, ..., b - 1, we also have

42

¤ j=0i-1 Hb - 1 - jL


i+2 Ha + jL-

¤ j=1i-1 Hb - jL


i+1 Ha + jL=

¤ j=1i-1 Hb - jL@Hb - iL - Ha + i + 2LD

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ¤ j=1

i+2 Ha + jL

=¤ j=1

i-1 Hb - jL Hb - a - 2 i - 2LÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

¤ j=1i+2 Ha + jL

.

Thus, we have

FHxL - GHxL =GHbL a !

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅGHn + 2L + K 1

ÅÅÅÅÅ2

On+1

Kd -Hb + 1L GHbL a !ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

GHn + 2L -1

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅbHa + 1L +

1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅHa + 1L Ha + 2L O

=GHbL GHa + 1LÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ


On+1

Hc + dL,

as desired.à

Lemma A3: Let a, b, n œ be arbitrary, for which a + b = n and b < a. Letf : @1, b - 1D Ø - be defined by

fHxL =GHbL GHa + 1L

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅGHb - x + 1L GHa + x + 3L Hx + 1L Hb - a - 2 Hx + 1LL,

and for each x = 1, 2, ..., b - 2, let fdHxL be the resulting function, when the differencefHx + 1L - fHxL is set equal to zero. Then,

1. The solutions to fd£ HxL = 0 are real, and Min 8x œ : fd

£ HxL = 0< < 0.2. Let S Õ be the solution set of real values to the equation fdHxL = 0. Then, S ∫ «. Subse-

quently, let m = Max 8x œ : fdHxL = 0<. Then,

‚i=1

b-1

fHiL >loomnoo

⁄i=1dµt+1 fHiL + Hb - 1 - Hdmt + 1LL fHdmt + 2L, if m œ H0, b - 2L and fd

£ HmL > 0

⁄i=1h fHiL + Hb - 1 - hL fHh + 1L, if m § 0, and fd

£ H0L > 0 ,

where dmt is the greatest integer contained in the real number m, and h œ for whichh < b - 1.Ñ

Proof: (1) By inspection, we find that fdHxL is a cubic polynomial, with coefficients ai for

the power of xi, i = 0, 1, 2, 3, given by

loooooom

noooooo

a0 = a2 + aH5 - 3 bL - 11 b + 2 b2 + 6a1 = a2 - 2 aHb - 5L - 14 b + b2 + 22a2 = 4 Ha - bL + 18a3 = 4

.

43

Thus, fd£ HxL is a parabola, with coefficients ai for the power of xi, i = 0, 1, 2 given by

looomnoooo

a0 = a2 - 2 aHb - 5L - 14 b + b2 + 22a1 = 8 Ha - bL + 36a2 = 12

.

Let 8x1, x2< be the solution set to the equation fd£ HxL = 0. Utilizing the quadratic formula, we

find that

x1 =2 Hb - aL - 9 - "####################################Ha - bL2 + 6 n + 15ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

6, x2 =

2 Hb - aL - 9 + "####################################Ha - bL2 + 6 n + 15ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

6,

and clearly, each of x1 and x2 are real numbers, for which x1 < x2. Also, since b < a, it

follows that x1 < 0. Thus, part H1L of the Lemma holds.

(2) Since fdHxL is a cubic polynomial, then the solution set to the equation fdHxL = 0 con-

tains either, (a) one real value and two conjugate imaginary values, or (b) three real values.

Thus, S ∫ «. So, let m = Max 8x œ : fdHxL = 0<. Suppose that m œ H0, b - 2L and

fd£ HmL > 0. Since fd

£ HxL is a convex parabola and x1 < 0, it follows that m > x2, and subse-

quently fd£ HxL > 0, for all x > m. Thus, fdHxL > 0, for all x > m. Hence, fHx + 1L - fHxL > 0,

for all x > m, so that

fHdmt + 1L < fHdmt + 2L < … < fHb - 1L < 0.

Thus,

‚i=dmt+2

b-1

fHiL > Hb - 1 - Hdmt + 1L fHdmt + 2L,

which implies that

‚i=1

b-1

fHiL = ‚i=1

dmt+1

fHiL + ‚i=dmt+2

b-1

fHiL

> ‚i=1

dmt+1

fHiL + Hb - 1 - Hdmt + 1L fHdmt + 2L,

44

as desired. Next, suppose that m § 0 and fd£ H0L > 0. Since fd

£ H0L > 0, x1 < 0, and fd£ HxL is a

convex parabola, it follows that fd£ HxL > 0 for all x ¥ 0. Thus, fHx + 1L - fHxL > 0, for all

x ¥ 0, so that

fH1L < fH2L < … < fHb - 1L < 0.

Thus,

‚i=h+1

b-1

fHiL > Hb - 1 - hL fHh + 1L,

for any h œ , for which h < b - 1. This implies that

‚i=1

b-1

fHiL = ‚i=1

h

fHiL + ‚i=h+1

b-1

fHiL

> ‚i=1

h

fHiL + Hb - 1 - hL fHh + 1L,

which establishes part H2L of the Lemma. Therefore, the two conditions of the Lemma hold.à

Lemma A4: Let a, b, n œ be arbitrary, for which a + b = n and b < a. Letg : @0, 1D Ø \ - be a convex and symmetric parabola on its domain, for which gI 1ÅÅÅÅ2 M = 0.Further, let f : @0, 1D Ø be a continuous function satisfying

f HxL =

looooooooom

n

ooooooooo

< 0, if x œ I 1ÅÅÅÅ2 , aÅÅÅÅÅn M= 0, if x œ 90, 1ÅÅÅÅ2 , aÅÅÅÅÅn = ‹ 818a<n-1<<

> 0, if x œloomnoo

I0, 1ÅÅÅÅ2 M ‹ I aÅÅÅÅÅn , 1M ÿ 18a<n-1<

I0, 1ÅÅÅÅ2 M ‹ I aÅÅÅÅÅn , 1E ÿ 18a=n-1<

.

If 0 < ‡1ÅÅÅÅ2

1f HxL d x, then 0 < ‡

1ÅÅÅÅ2

1gHxL f HxL d x.Ñ

Proof: Suppose 0 < Ÿ 1ÅÅÅÅ2

1 f HxL d x. For real values ai, i = 0, 1, 2, a2 > 0, let gHxL = ⁄i=02 ai x2.

Since g is a convex, symmetric parabola on @0, 1D, it follows that g ' I 1ÅÅÅÅ2 M = 0. Thus,

a1 = -a2. Also,

45

gK 1ÅÅÅÅÅ2

O = a2K1ÅÅÅÅÅ4

O - a2K1ÅÅÅÅÅ2

O + a0 = 0,

which provides that gHxL = a2Ix - 1ÅÅÅÅ2 M2. Hence, g and f are clearly continuous functions on

the interval @0, 1D. Thus, the product of these two functions is also continuous on this inter-

val, and hence integrable on this interval. Since g is increasing on the interval I 1ÅÅÅÅ2 , 1M , it

follows that

gJ aÅÅÅÅÅÅn

N f HxL § gHxL f HxL´̈ ¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨̈ ¨̈ ¨̈≠ Æ¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨̈̈¨¨̈ ¨̈

xœI 1ÅÅÅÅ2 , aÅÅÅÅÅn M

ï gJ aÅÅÅÅÅÅn

N ‡1ÅÅÅÅ2

aÅÅÅÅÅn

f HxL d x § ‡1ÅÅÅÅ2

aÅÅÅÅÅn

gHxL f HxL d x

gJ aÅÅÅÅÅÅn

N f HxL § gHxL f HxL´̈ ¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨̈ ¨̈ ¨̈≠ Æ¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨̈ ¨̈

xœH aÅÅÅÅÅn ,1L

ï gJ aÅÅÅÅÅÅn

N ‡aÅÅÅÅÅn

1f HxL d x § ‡

aÅÅÅÅÅn

1gHxL f HxL d x,

so that

0 < gJ aÅÅÅÅÅÅn

N ‡1ÅÅÅÅ2

1f HxL d x § ‡

1ÅÅÅÅ2

1gHxL f HxL d x,

as required.à

Computational Analysis A1: Let a, b œ be arbitrary, for which a + b = 109 and b < a.We will show computationally, via the C++ programming language, that

‡1ÅÅÅÅ2

1Kx -

1ÅÅÅÅÅ2

O Kga,b-1HxL -aÅÅÅÅÅÅb

ga-1,bHxLO d x =GHbL GHa + 1LÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ


On+1

Hc + dL > 0,

where c and d are as defined in Lemma A2 above. Clearly, each of the values for c and dare less than zero. Thus, to show the desired result, it suffices to show

GHbL GHa + 1LÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

GHn + 2L > K 1ÅÅÅÅÅ2

On+1 ƒƒƒƒƒƒƒƒƒƒ

Hc + dLƒƒƒƒƒƒƒƒƒƒ.

Here, we note that d = ⁄i=1b-1 fHiL, where fHiL is as defined in Lemma A3 above. It can be

readily shown that

fHb - 1L <-Hb + 1L GHbL GHa + 1LÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

GHn + 2L ,

so that Lemma A3 provides

46

c + d >b - Ha + 2L

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅbHa + 1L Ha + 2L +

loomnoo

⁄i=1dµt+1 fHiL + Hb - Hdmt + 1LL fHdmt + 2L ÿ 18mœH0,b-2L and fd

£ HmL>0<

⁄i=1h fHiL + Hb - hL fHh + 1L ÿ 18m§0,and fd

£ H0L>0<.

We denote the terms of the right hand side of the above inequality by a and b, respectively.Since c + d < 0, it follows that » c + d » = -Hc + dL. Thus,

» c + d » < » a + b » § » a » + » b »,

so that if we show

(A2)GHbL GHa + 1LÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ

GHn + 2L > K 1ÅÅÅÅÅ2

On+1

H » a » + » b »L,

then we have established the desired result. Since a and b are each to assume very largeinteger values, we will show HA2L, utilizing the natural logarithm function.

47

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