A MODEL FQR THE DISTRIBUTION OF INDIVIDUALS
BY SPECIES IN AN ENVIRONMENT

Thesis for the Degree of Ph. D.
MICHIGAN STATE UNIVERSITY
John W. McCIoskey
1965

 

L!
L [B R A R 1'
Michigan Stat:
University

  
 

THESIS

 

This is to certify that the

thesis entitled

I A MODEL FOR THE DISTRIBUTION
OF INDIVIDUALS BY SPECIES
IN AN ENVIRONMENT

presented by
John W. McCloskey
has been accepted towards fulfillment

of the requirements for

Major professowj

Date 27‘? /Z /T (5

0—169

 

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ABSTRACT

A MODEL FOR THE DISTRIBUTION
OF INDIVIDUALS BY SPECIES
IN AN ENVIRONMENT

by John W. McCloskey

The problem considered in this thesis is that
of developing a model for biological environments
so that, for samples of individuals obtained from
the environment, the number of species and the

number of individuals in the respective species can

be predicted, It is assumed that the number of

individuals in the environment, as well as the num-
ber of species, is countably infinite so that only
in environments where these quantities are very
large will the model be realistic.

In Chapter 1 the model is developed and in
Chapter 2 a procedure developed to obtain maximum
likelihood estimates of the parameters of the model
using a sample of data already gathered from the

environment. Since there are three parameters in

the model the estimates are obtained from the

 

John W. McCloskey

simultaneous solution of three equations which is
accomplished by means of an iterative Newton procedure.

As a means of studying the behavior of the model
a simulation procedure was developed in Chapter 4
which would choose a sample from the model for a
given set of parameters. This procedure uses random
variables having Binomial, Poisson, Hypergeometric,
Truncated Poisson and Exponential distributions.
Methods were thus developed in Chapter 3 to produce
random variables with these specified distributions
rapidly and with as few input random variables as
possible. The fundamental technique used in obtain-
ing these random variables is the acceptance-rejection
technique introduced by von Neumann.

Chapter 5 and Chapter 6 are devoted to the
analysis of data that was taken from actual biological
environments. The analysis is accomplished through
procedures developed in the previous chapters and the
Control Data 3600 computer used for the actual calcula-
tions. Several FORTRAN 60 programs were used for these

calculations which are tabulated in the appendix.

   

'."IIDI'IDHAL8 FY 5230133
‘m1di.-mflm‘ 7,

       
  
 
   
     
   

 

A THESIS
Submitted to
Michigan State University

in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY

Department of Statistics

 

 

ACKNOWLEDGMENTS

I would like to thank my academic advisor,
Professor Herman Rubin, for his many comments and
suggestions in the course of the research. His willing-
ness to discuss the problem and to exchange ideas in the
early stages of the research was especially helpful. Also,
the ideas expressed by Professor Philip Clark concerning
the presentation of the simulated data were very helpful
and greatly appreciated.

I am also grateful to the Department of Statistics,

Michigan State University, and the Office of Naval

Research for their financial support during the writing

of this thesis.

ii 3
1

I

 

i

TABLE OF CONTENTS

Page
Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Section 1: General Discussion of the Model. . . . . . . . 1
Section 2: Development of the Model . . . . . . . . . . . 7

Section 3: A Special Case of the Model. . . . . . . . . .10

Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . .13
Section 1: Maximum Likelihood Estimates of Parameters . .13
Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . .25

Section 1: Acceptance Rejection Procedures. . . . . . . .25

Section 2: Fitting Discrete Distributions nifltlarge

Means. . . . . . . . . . . . . . . .30

Section 3: Procedures for Discrete Distributions with
Small Means. . . . . . . . . . . . . . . . .38
Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . .42
Section 1: Simulation of the Model. . . . . . . . . . . .42
Section 2: Simulation in the Special Case . . . . . . . .51
Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . .52
Section 1: Data Analysis. .52
Chapter 6 . -70
Section 1: Investigation of Species per Genus Data. . . .70
APPENDIX. .77
BIBLIOGRAPHY. . . . . . . . . . . . . . . . . . . . . . . 116

iii

 

fi

LIST OF TABLES

Macrolepedoptera Data .
Theoretical Frequencies for Macrolepedoptera Data .
Goodness of Fit Test.

Simulated Test #1; a = 1, A = 40.2576 .

Simulated Test #2; a = 1, A = 40.2576 .

Simulated Test #3; a = 1, A = 40.2576 .

Simulation with a = 1, A = 40.2576 for increasing N .

Orthoptera Data .

Simulation Test for Orthoptera Data .

iv

Page

. 53
. 54
. 55
. 57
. 58

.‘59

62

. 75

76

  

T~lfiQnR€_thfi collection of all the individuals of a certain _

 
  

55.Gnr_e§ample butterflies, present in an environment. Consider
'*l_pp§rtition of the individuals into species designated by {31,sz,...}

Vhflre the species are arbitrarily named 81,8 ,... and suppose the

 

the number of species is very large and in shmpling from the
environment there is assumed to be a strictly positive probability
of finding a new species regardless of the number of species that

have already been found. Also define a probability pi for each
as
species 3. such that 2 p. = l.
' 1 i=1 1
Consider now the task of choosing a sample of N individuals

independently from the environment. Let these individuals be

designated by II’IZ" "IV' the individuals are chosen according
l
to the restriction
PEI. is from the species 3.] = p. for i = 1,2,- .N
1 J J j = 1,2,.

After tne individuals are chosen from the environment the
sample will contain say 3 species for which there are n1 species
with one individual, n2 species with two individuals and in general
“i sPecies with i individuals subject to the conditions
N N
Z n. = s and 2 i ni = N.
i=1 1 i=1
' Object of this report is to develop a model for natural

 

  

' V

fi

2.

Consider therefore the generalization of the probabilities
pi where for each species si in the environment it is assumed there
is an "intensity" xi proportional to pi. Let 2 = _;;xi where z is
defined to be the total intensity of the environment. Define an
intensity function f to be a non-negative integrable function on
I0,60 with the property that (i) for any 6 > 0, f€f(x)dx = + m

m w 0
and I f(x)dx < + m and (ii) I xf(x)dx < + m.

eThe model can now be stgted as follows: Given an intensity
function f for an environment, for any interval [a,b) with
O < a < b E +m the number of species present with intensity X1.- in
tge interval a 5 xi < b has a Poisson distribution with mean
f f(x)dx and for disjoint intervals the number of species with
intensities in the respective intervals have independent Poisson
distributions. Condition (i) on f is made so that the expected
number of species will be infinite and (ii) is made so that the
total intensity will be almost surely finite. Let Ui be a random
variable representing the number of individuals observed from
species 51.- for i = 1,2,.... Suppose U1, 2,... to be independent
Poisson random variables with means ksxi’ where ks is a positive
constant and xi the intensity of the respective species. Define
a sample to be an observation of the random vector U = (U1,U2,...)
and define Ym = (number of Ui = m) for m = 1,2»--

The development which follows in this section is an attempt
to give motivation for the actual development of the model in the
next section. Thus, let X = (X1,X2,..-) be a set 0f intensities

obtained from the process and define Z = f Xi and the species

‘..

 

3.

with intensity xi will be designated species si.
Then
x.

PEI1 is from species sj] = 22-for j = 1,2,.... Let s: be
the species of individual I1 and let V1 be the intensity of
this species. Choose a second individual 12 randomly from the
environment and examine its species. If it is different from
8:, let s; be its species and V2 the intensity of this species. 1
If however 12 is from the same species as I1 continue selecting
individuals independently from the environment until one is

found which has a different species than I1 and let the species

* -
of this individual be 32. Consider now the two random variables

_ l _ 2
211 - Z“ and W2 —Z-V

Theorem 1: Suppose that w1 and Wé are independent and identically

distributed according to a distribution H on [0,1]. If

1
)\+1

 

0 ( E(w1) < 1, then define A such that E(w1) = It then
follows that d H(u) = lCl—w)A-1dw.

Proof: Let YI. for i = 1,2, be the proportion of individuals in
the environmen: from the same species as individual II' The

individuals 11 and 12 are chosen independently from the same

environment so YI and YI are independent and identically
1

distributed. YI is defined as follows

and wl with probability w1

2 (1—w1)w2 with probability (l-Wl).

 

.“' —————----IIIIlllIIIIIIIIIIIIIIIIIIIIIII'IIIIIIIIIIIIIIIIIIIII

Let the rth moment of H be pr. Then for r > 0
r _ r _ r
gr 2041 ) - E<YI ) - E(YI )
1 2
r+l r r+l
12w1 + w2 (1-w1) J
+1
= ”r+l + H‘r E(1-W1)r

r+l
- k r+l
‘ ”r+l + “‘r [1:0 ('1) (k >“kJ

Solv1ng for “r+l »

r
k r+l
u- =u [1-E(-1)<>u]
+
r l r k=0 k k
1 + (_1)r+1 u
r
From this equation ”r+l is determined by p0,p1,...,ur unless
r is even and pr = 1. If however pr=l for r > 0 the distribution
is concentrated at one and all “k = 1. This distribution with uk= l
for all k indicates that all individuals are from the same species
which violates the assumption that the environment contain an infinite
number of species.
In order to determine the moments pr an equality must first be

established. Thus

1 l r11 _1
IE (l-x)r+1 (1-x)A-1dx =2! 1:0 (-1)k<r1:1> xk(l-x)A dx

r+l l
k r+l k A-1
= Z (-1) x (l-x) dx
k=0 Ck) g
r+l r+l .
k r+l k P+
a -1 [(+1 1" A = E (-1) k1 1112
REO ( ) k) I‘LL—LN k=0 (1‘5 raw-1+1)

k+1+l)
IIIIIIIIIiL.

 

____———-----IIIIIII

Thus

§HWC®MFA =lwfiflbonm(mNF
H, k 144me i EAL

l"(r+2+}\)

- (-1)r+l (r+l)! EL

1
r+l+1 I‘( r+2+).)

It is to be show now with the use of the above equation that

1
H = klr A+1 by induction. Obviously p. = ~ and assume
1‘ (k+1+x) 1 “1

RIF A+l

”k: I"(k+1+)\) for k= 0, 1,2,...r.

From the recursion formula for “r+l

_ rll‘ l+1 r k (r+l kJFgAH)
“r+l ‘ mam) l 1 ‘ 1:0 ('1) k) 1“(k+1+>.)
r+l I“ A+1
1+ (-1) r! T_Hr+l+l)
KL

r! F )\+1 1 _ _ + I
g I"(k+1+>\) I: 1 ' r+1+A ( DH 1(r I) F r+2+?\ ]
N

1+(-1)r+1 r! I‘ x+1

I"(r+l+)»)
r+l _ r+l gr+12 r! F A+l
r! l" A+l [ r+1+}\ + ( I) r+1+K F—(T+% ]
=l" k+l+K
( ) 1+ (-1)r+l r! ng+12
I“(r+1+A)

_Sr+l!!l"g}\+12
=I"(r+2+7\)

k-l
Consider the rth moment of the distribution X(1-X) for 0 5 x5 1 J

 

fi

0 otherwise

1
I xr 1(lcx)A-1dx = A F r+l F A = r1 F A+l .
o I‘(r+1+l) I"(r+1+>.)

This distribution has the desired moments and since its moment
generating function exists in a neighborhood of zero

dH(w) = A(l-w)A-1dw

  

'. 6.1"; f(v-‘n

 
  
 

if), fiiz halos-ant of the Mode! '
"W", be an, intensity function and
. ‘ “‘1 .

‘ is“: ”39$ intensities

snaps-avian section us i
. . .

  

let X = (xl.x§,x3....)

obtained from the process described in

   

ng f as the intensityfunction. Let

 

z = 2 xi and let Z have density g. Define V1 to be a random
' i=1

 

X.
variable such that P(V1 = xilx) = z—1 for i = 1,2,... and define
Y 3 Z - V1.
Lemma 1:

If f is a continuous intensity function the joint
density h of V1 and Y can be expressed in the form

v1f(v1)
h(v1,y) = W $00
The proof of this result was obtained by Professor Herman Rubin
and is to be published in a paper by him.

Make the substitution 2 = v1 + y so that

v1f(v )

hv z(v1.2) = 2‘ g(z—v1). Now integrating with respect to
1’

V1 to obtain the density of the total intensity
2 z v1f(v1)
g(z) =f h.V Z(v1,z)dv1 = I z‘gh—VIMV1 .
0 1’ 0
V1
Define W = Z— . Then

hi‘,,z(w,z) = W = wzf(wz)g(z(l-w)).
Theorem 2: If hw'z(w,z) = wzf(wz)g(z(1-w)) for 0 S w 5 1 and
0 < z < a and if [Ra/'20,) é :p(w) and assuming f and g to be twice
differentiable then f(x) = c x-lekx for O < x < an and
8(2) = c'zfi'ekz for O < z < co .

hw,z(""z’_ wz f(wz) (z(l-w))
PM“ hWis”) = *3— ' gm

g(z

   

= (90¢)

 

 
 
  

“l’ ,

103 w + log 2 + log f(wz) + log g(z(1-w))
= 108 @(2) + log 3(2)
Let ¢1(wz) - log f(wz)
and ¢2(2(1-W)) = log g(2(1-W))
thus
log w + log 2 + W1(wz) + ¢2(z(1-w)) = log <p(w) + log g(z)

taking derivative with respect to w and then with respect to z

1 l . ' W

E + z *1 (wz) ‘ Z *2 (2(1-W)) - EYéil
ii (W2) + WZ Wi'(WZ) - Wé (z(l-w)) - z(l-w) W; (z(l-w)) = 0-
Thus

wz WE (wz) + Wi (wz) = z(l-w) $3 (z(l-w)) + W5 (z(l-w))

Since the above equation is valid for all values of z and w
the following must be true

wz WY (wz) + vi (wz) = k

and z(l-w) $3 (z(l-w) + W5 (z(l-w)) = k.
Solving then these two differential equations
H l =
u¢1<u)+¢1(u> k
u ¢i (u) = ku + H
H
‘4‘1' (U) =k+;
W1 (u) = ku + H log u + M

f(u) = e¢l(u) = c uH eku

Similarly

8(v) = e¢2(v) c c' vH' ekv

Finding now the particular solution

‘.

 

"-3-". _ w: E vs a l-w
' h by]. 3(2
H H esz c'z (1 -w)H' ekz(l-w)

.WSCWZ '
H'CZH ekz

 

l
c wn+1 zH+l(1_w)H

which implies that H - -1 yielding the final result f(u)-c u'leku .
From the above analysis and in an effort to make the model as

general as possible the form of the function f was decided to be
-cx

           

x
Obviously A > O and due to the restrictions of the model c > 0
since £ f(x)dx a 0 as N a m because the total intensity of the

largee species is almost surely finite. Also a 3 1 because if a < 1

 

 

*7 Ae‘c" A Ael'“
thenJ: f(x)dx = 1! dx 5 E —a dx = l-a < 0° controdicting the
x

restriction that the expected number of small species present be
infinite.

From the development in Chapter 2 it can easily be observed
that the transformation x 4 Ax, c a c/A, ks 4 k8 /A, A a All

In:
preserves the model so that only a, k s/(k 8+6) and A/(k S+C) are

identifiable. For this reason only the cases c = O and c - 1 need

be considered. The general form of the function f was taken to be

 

f(x) = xfor the work which immediately follows while in Chapter 6

X“
a generalized form of the case c = 0 is considered.

 

 

      
 

I J
V A \\ , 1
" 5'; 4‘ 3.5.5331. (has. 0: the Model

Consider now a special case of the model developed in t

I

he

 
 

-X
gfevious section where f(x) = Ae

-— .
X

 

Knowing 8(a) has the form

   

H' -z
g(l) = c's e and using the previously established equation for g(s),

   
   
  
   
   
    
  
    
 
   
  
  
 

 

 

z v1f(v1)
8(8) = ghvl z(V1.z)dv1 =f -——;—-— g(z-v1)dv1°
-v
2 —z+v
H' l
_1...' = I -——-—-—— c'(z-v ) e dv
‘ , O v12 1 1
-z z -z H'+l
_ Ac'e H' _ Ac'e z
- z {(z—vl) dvl - z(H'+l) '
This equation implies H' = A - 1 and since
” 1 1
I c' 2A- e_z dz = F(A), then c' = -—-
0

TCA)’
Therefore g(z) = f%X3 2A“1 e-z.

For j = 1,2,... define Vj to be a random variable such that

 

P(Vj = Xi I X) = 1 for all i except

those i's for which xi = Vk for k = 1,2,...j-1.

Let

    
   

 
 

Vi
wi = 1—1
— 2 V
i=1 3
z xf(x)
By repeated application of the formula g(z) = f— g(z -x)dx
1-1 0
which was zpreviously established 2-12 vj
i=1 i
f(v )2 vl v f(v ) v. f(v. )
8(2) = zv—1--—1 -—2-——2- i—ll—gu-svmvi. ..~
- a (z-vl) 1—1 j-l j

 

(1:1F.Yj)

1 i

ll.
i—l
z z-v1 ztglvj -
- v f(v) v f(v) v.f(v.) 1
= f I ... 1 1 2 2 ... 1 1 g(z- 2 v )dv ...dv
o o o z (z-v ) 1—1 j=1 J 1 1
1
(z— 2 v.)
i=1 3 ,
so the joint density for V1,V2, ...,Vi,Z where v0 = 0 becomes
TE]. vjf(v1.) i
h (v ,v ,...,v.,z)= g g(z—ZVJ.
V1,V2,...,Vi,Z 1 2 l J=1 j‘l J=1 J
z— E vk
k=O
-x

Theorem 3: In an environment where f(x) = A? and

_ 1 A-l —z . . .
g(z) — m)z e and where Vi' wi,Z and the JOlnt denSLty

V1,V2,...,Vi,z are defined as above, wi is distributed

according to the distribution

A—l

 

 

*
= _ , < 1.
h (wi) A(1 wi) for 0 5 W1 _
Proof: For i i 3 the joint density
7' i i
h (v ,v ,...,v.,z) = IT v f(v.) ] g(z- v.)
V1,V2,...,Vi,Z 1 2 l l. i=1 J]_ J \ i=1 J
(‘z— 2 v I)
k=0 k
i i
- L 1 — 2 v )A-le_z
— F(A) [ II; j—1 (2 ,=1 J
“— <2- ): vk>
k=0
1 l 1 1‘1 _2 V A_1 —Z
A [ T‘- 1 (z— E v )A ( l — ) e .
PM) J- 1‘1 j=1 1—1
2- E vk z— 2 v.
k=0 J

   

fi

12.

Make the substitution Wi = 1_i so that
z —2 V.
i=1 3

Vl’VZ’""Vi_1wi:z(v1’""vi-lwi’z) =

 

i i-l 1-1
A -
f"‘ [ ll -} ] (Z - E V.)A 1 (l—w.)A-1e_z.
(A) .=1 J 1 \ ,_ J l
J C 2- Z V / J_1
k=0 k

Integrating this density then

i—2
z- E v.
z j=l J
hwi’z(wi,z) = % ... g hV (v

,...,v. ,w.,z)dv.
1’V2""Vi-l’wi’z 1 1-1 1 1-

1.

.dv1

_ A A—l A-l -z
— ffx3 z (l—wi) e .

as

*

Integrating now with respect to 2, h (wi) = ghwi’z(wi,z)dz
.é.

A—l A—l -z
F(A) (l—wi) z e

u
OL—na

dz = A(l-wi)A—1. For i = 1,2, the

same procedure is followed with simplification in the integration.

   

  

"7.1.: mm Likelihood Estimates of Parameters

 

all. “he general form of the intensity function has been established
to be f(x) - -:2;- where A, a are parameters of the function. In
any sample that is taken from the model the number of individuals
in each species is assumed to be Poisson with mean proportional
to the intensity of the species; that is the number of individuals
in the sample from the 1th species is Poisson with mean ks xi where
ks is defined to be the intensity of the sample. This parameter
k8 is also to be estimated.

Suppose now that data is available from this model and it is

 

desired to estimate the above parameter. Let ym be the number of
Species with m individuals in the sample, I the number of individuals
and s the number of species. The ”following trivial equations are

to hold 2 ym s and 2 m ym I.
m=1 m=1

In accordance with the above notation the probability that
there will be m individuals in the sample from a species with
m
X -k x
intensity x is (k S!) e s and the expected number of species in
m

the sample with m individuals is

m
D k x as - - (ksx) ‘ksx
f‘S—e s f(x)dx=zj)““‘xa’exTe dx
0

 

m
Aka I‘gm-oH-l) (ks m-ar+1 1.111114%

m’ (k3+1)m‘°*1= (k 81"“)

ks+1rml

 

   
 

 

‘Eht total umber of species present in the sample has

 

I aifltribution,the ym are independent and have a Poisson
‘ hflitbution with mean B 11‘" L)

The density thus becomes

°’ l l
f(yl:Y2;y3,---;B;Tlfl) ‘ TT e m_m
m=1 y

ym

where ym is as previously

“I
defined and Am = 311‘“ 11%

the sample with :11 individuals.

, the expected number of species in

The logarithm of the density as a function of the three

parameters ignoring constants becomes
L<B,a,n> = 2 - Wan—11+

Now simplifying the first term

m§1 y m[log B+m log n+log F(mqr+l)- -logml].

Q Q mm
m P m-a+l _ _ B may -x
m§l - B“ m! _ m§1 m! g x e dx
com m I a
= f 21 - Bml xm qu e-xdx = -B g x'a e x (eTlx - l)dx
0m=

From Bierens DeHaan [1] table #90 equation #6

-qx_ e-rx dx _ _ _ p_ p .
g(e ) xP+1 - P P(1 p) (r q ) for p < l

 

so
Let P a o - 1. Then -B g ( e.(1'n)x - e x )xqa dx -

-B 0,—1 I‘(2-a)<1-(1-n)"'1) = -31" (1.0,) [(1—n)°"1 - 1 J
USing the above and simplifying the second term, the likelihood
function thus becomes L(B,o,n) - _. -
-BF(1-a)[(1-1])°"1-1 11+ 8 log B + I log 11+”. 8 1my logI‘(m-a+l)-;1ynlog ml

It was found that in taking the derivative of the sham; j , 4

 

  

i

15.

for T] near one and a near one. To eleviate this difficulty the
substitution (l-n) = e.q was made. The likelihood function
L(B,a,q) thus becomes
L(B,U,Q) - - BT(1-a) [e-q(a-1)-l] + 8 log B +Ilog (l-e-q)

Q as
+ mgl ym log T(m-a+1) - mglym 108 ml.

Taking the derivatives with respect to these parameters

-fl -q(°’-1) g
LB‘aB"F(1'a)[e '1]+B
-q
L = g‘L = - B 1"(1-cr) (l-a) e'q("'1) + I L
q q l-e-q

1_

= - B F(2-oz) e'qm'” + I
eq-l

and using the notation
a 1 _§
WK) = alog I1(X) = my ax I’(X)

so that

7 UK) = I1(X) WX)

then

La: % = BTU-a) “1-01) [e'q(°"1)-1J

+ qBF(lay) e-q(a-1) - Zlym W(m-a+1)
m:

In finding a solution for the equations La = Lq = LB= —O a
Newton approximation in three variables was first attempted but
abandoned since the matrix involved in using this method is almost

Singular causing instability in the procedure.

 

" a—————-----IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII'IIIIII

16.

Therefore the following modified Newton

method in two variables
was used:

1. Initial estimates &1 and al are given
2. Solve equation LB = 0 for B to get initial estimate bl
3. Step two makes LB(§1,q1,&1) = 0 so that

C ° >=<LBB WU")
-Lq(§1,<31,611) LqB qu q

can be solved for Aq as follows

Aq =L EB -L L
BB qq qB Bq

q2=q1+4q

UI

Solve LB — 0 using estimates &1 and a2 to obtain 32
6. As in step 3 find A! by the equation

M _ 'LBBLoz
“L L L L

BB (10' 08 Ba!
7. 02 = a1 + AU
8. Continue iterating until desired accuracy is reached.

This procedure gives likelihood estimates a, B and q

from which can be calculated the other two parameters

   

fi

17.

The second derivatives of the likelihood function necessary

for the above method are as follows:

2

LB 3 é-lL. = .;E
B BBB B2
= = .. -q(a'l)

LEI = RIB = -F(1-a) W(141) [I-e-q(a'l)] + qr(141) e'QCI-l)

q
= _ F 2_ -q(a'1) _ I L
qu B(a 1) ( a) e (eq ‘1)2

Bra—a) 92(1-01) [l-e'q("'1)]

[...
II

+ BTU-a) w'u-oz) [1_e-q(0’-1)]

- - -l
-2 qBT(l-a) ¢(1-a) e-qGY 1) - quF(l-a) e q(a )

as

+ 2 y ¢'(m-oz+1)
m=l m

For the calculation of W(x) and ¢'(x) Stirling's asymptotic
series is used for log T(x+l). Thus

108 I"(X-H) =(x + %) logx -x+% log 2n

1 1 1 - —1— +
J’fi ' T66}? + 1260x5 1680x7

 

V fi

18.

15.1" 1 1-44-; 1'
#(x+l) axlog (x+1) log x +—— 12x2 + 120x4 252x6 + EZBEB +....

2x
1 1 1 1 1 1
c . _ _ __ _ ___ .___ _ ___ +
1 (x+1) x 2x2 + 6x3 30x5 + 42x7 30x9 ' ""

For x 2 lg ¢(x) and W'(x) are calculated from the above

equations. However for Small x the recursion formula
F(x+1) = xf(x) is used.
log F(x+l) = log x + log F(x)
Differentiating both sides
w<x+1) =§+ we)
and ¢'(x+1)= - fiz+ w'(x)

In the calculation of the Newton process it is often necessary
to evaluate the expression F(l-a) [e-q(a-l)_1]- It is often the case
that a is near one which requires that this expression be evaluated
with care to avoid the loss of several significant digits. For this

reason make the following substitution:

1_e'Q(°"1)

1"(1-ar) [e'q(°"121] = 1"(2-a) 0H

2
=r(2-QI) e112E inhi. q wherez=av 1.

¥

 

_|_

 

 

 

19.
Now let
tanh w l
h w - l + w2 2
3 + w 2

5 + w 2

7 + w
9 + .

expressed as a continued fraction and

945 + 105w + w2 2
945 + 420w + 15w

expressed as a ratio of polynomials reduced from the first five

Using this and hyperbolic identies

terms of the continued fraction.

 

 

 

 

- z l-tanh SE 1 - hflE
e fi— = 4 = 4
l + tanh SE 1 + hfli
4 4
and
2 tanh ~25 133
sinh 3; = .__—_——_T
l -tanh2 %% 1 (225)
Using all of these equations then
1 -%§ sinh 52E
F(141) [e-q(a- )-1] = F(2-o) e -EE?-—— '
2
(1-113;) h 923
= F(2-a) - E:— =
2

(WISE) [141252]

= 1‘ 2-01) ig—
( [1332512

k.

 

  
    
  
 

1‘ ,xv~nerties of the model require that the paramstarda be
‘ than or at least equal to one. Since the system is fairly
Lfihfitabbg.it was found in actual calculation that the iterative
Newton procedure described previously would sometimes give an

estimate of a less than one. To avoid this difficulty a restriction

was placed on the procedure as follows: Given that 61 - l + 6 then

a A 6
. { ai+1 if a1+1 2 1 +.!$
”1+1 ‘ ~
Hg if a1“ < 1 +fi
2 2

If indeed a = 1 it would be hoped that &i ~ 1 from above but
this is not the case since the method blows up; that is for & less
than about 1.005 (depending on the data) the error in calculating
A0 is larger than & itself which reduces the iteration to nonsense.
What results then is that the estimate is out half way to one each
time until which time the error in Au causes a large positive jump.
The estimate again approaches one and the process repeated until
the computer is stopped by a programmed check which halts the Newton
process after 50 iterations if no solution is reached. If this
happens a is set equal to one in the original likelihood equation
and another method used to estimate the other parameters.

Consider therefore the likelihood function

4 ° mF w - I.
L(B,1,n) = m§l - Bn -é$l-+ m§1ym [log B + m log n + logF(m) log m ]

m a
s - ll. 1 + 10 P(m)- 103 ml]
Bm§1m+m§1ymnogs+mogn 3

Using the expansion

 

 

' fi

 

21.
x-l l x-l 2 l -l 3
logx: —x-+§(T) +§(§x—)+.... forx>15
x-l
set U = —;—
in - x-l
x(1-n) = l
1
x — (I-N) for 0 < n < 1 then 1 < x < ”
thus
°° 1
2 fl? = log x = log (l:fi) = - log (l-n). Using this and
m=l m

again making the substitution (l-n) = e-q

Q
L(B,1,q) = -Bq + E yIn [log B + m iog(1-e'q) + log 1"(m) - log ml].
m=l

Taking the derivatives with respect to these parameters

 

m
L = -q + E y /B = -q + E
B m=1 m B

a ‘q
L =-B+ 2y me_q=-B+I—1
q m=1 m l-e eq-1

s .

In finding a solution LB = Lq = 0 make the substitution B = a into

the second equation to get

 

Which reduces to eq - l - E q = 0-

To find a solution to this equation consider the following

iterative procedure. Given an initial estimate <10 and Where qr

 

k

V fi

22.
is the root of the equation
x x
qr - q0 + 1:;;:g;% - q0 + E?;) where x, a, b are to be determined

as follows:

qo
LetA=-andA0=e -1-1q0

X

‘1 (1+—
Set er-1-1qr=e°B(x)

 

-1-A(qo+ )30.

x
B(x)

Then
q0 x ’ x
B(x) [e e B(x) - l - Aqo - A B(x)] = 0.

 

 

Expanding this equation and calling it Q(x) then

2 x3 X4
Q(x)= B(x)e q0[1+B(x)+m+ m-t m+....]

-B(x) - 13(x)q0- ix

4
* q0 q0 x2 x3 x
=B(X) AO+(e -l)x+e [m)+m+m)+.ml

 

1
Now expand the expressions Bk(x) <: 1+ax+bx

into polynomials

1‘ 3
< 1 A) 2
___—___ = a + a x + a x + a x + ..
k2 k3
1+ax+bx k0 k1

= 1,2,3 the expansion

After finding these polynomials for k

then becomes Q(X) =

2
= B(X)A*o + (eqo- k)x + eq OE-xz (l - ax<+ (a2 - b)x + ...)

 

I ..

fi

23.

+%x3 (1-2ax+ ...) +§x4 (1 + ....)J +

‘1o

2 3 1 1
+e x [3‘28]

q
* q 0
+ B(x) A0 + (e 0- X)x + 23— x
qox 4
+e “[271"+—(32 -b)]+....

Choose a and b such that the coefficient of x3 and x4 are zero.

Thus

and
l a 1 2
E-§+§(a-b)—O
1-11.1443“, _> b 1_
24 9 2 9 36
therefore
q
1 e02 5
Q(x)= A0 (1 + 3x - —-x 2) + (eqO - A)x + —E— x + asx + .
* q *

* 5
=AO+(eCl O-l+—O)x+(—-3—O)x2 +615x +--

As an approximation to Q(x) = 0 set the equation
*
* qo 3 e(10 A 2
- -— --- —— x = 0
A0 + (e l + 3)x + ( 2 36)
For the general quadratic

and solve for x as follows:

0x2 + Bx + 6 = 0

 

h;

 

52 -4a6) -a-\Is-4aa

Since B is positive in the neighborhood of qr the positive
root is taken to obtain the root of the quadratic nearer zero and
the last form is used in calculating x to avoid round off. The
procedure for finding the root of the equation eq - l - Xq = 0
is as follows:

1. Make initial estimate q0 = log(l+k log 1)

* .
2. Evaluate Ai = e 1 - l - lqi

3. Solve the equation* q. *
* qii A1 +e1 fi)2=Ofrx
Ai+(e - +3—)x (2 36x 0 .
x
4. q. = q.+'———"‘—‘
1+1 1 l l 2
1+3x'36x

5. Return to step 2 until desired accuracy is reached.

The method was designed for rapid convergence and in fact it
was found in actual compitation that five digit accuracy was
obtained in only two iterations.

USing q as found from the above procedure and from the original
equations remembering that A = B for the case in question where
Q = l the estimates obtained are

A s
A=€

and

 

 
 

  

25;

 
   

lienerating Random Variables for the Simulation

  
    
 
 

   

731-1»: Adaptation Rejection Procedures

W‘Zh the process of simulating the established model on the

high:.speed computer it is necessary to generate random variables

  
  

with certain specified distributions. Since the actual computa-

   
   
 

tion is to be done on the computer and the procedures used many
thousands of times it is necessary they be effecient and use the
minimum of input random variables. With these goals in mind it
i was decided that for discrete random variables an acceptance

' rejection procedure would be used. This method of generating
random variables with Specified distributions is discussed by

Rubin [5] and will be used in this problem in the following way:

P(ni)

—————— Q(ni)

-§__

 

 

 

 

 

 

 

 

 

 

 

 

 

I l

I l

l I I
l I I
I I l
I l I
I I l
l I I
l I I
I I l
I I I
I I I
I I I
I I I
l I I
I I I
l I I
I I I
I I l
I l I
l l l

I .

e I O n1-1 n1 “1+1 n1+2 - D O

 

 

 

     

 

  
 
 

.1

   
 

a. random variable with the distribution P(n 1.3 is

  
 
  

". Construct a frequency distribution Q(n

i) which dominates
n43... Obtain

an observation from the distribution Q(n i) and

  
 

    
 

decspt this observation x1 with probability P(xl) . If x1 is
c'(Tl)

rejected obtain a second observation from Q(n i) and repeat the

     
  

process until an observation is accepted. If the first accepted

  
  

observation is designated as xthen it has distribution P(ni ).

 
 

This procedure is to be used for Binomial, Poisson and

 
  
  
   
  
  
   
     
    
 
   

Hypergeometric distributions and in these cases the distribution
Q(ni) will take the form of a uniform over the mode and discrete

exponential over each tail with parameter a1 over the right tail

and parameter a2 over the left tail. The parameters a and a

1 2
are determined by taking the ratio of two consecutive probabilities
of the distribution P(ni) and it was determined that the most
efficient place to calculate al and 02 was about /§—’standard
deviations on either side of the mode of the distributions being

 

considered.
Thus let M = mode of distribution P(ni)
Let N1=M+[/§op]
and NzaM-[fiopj_1 .
then determining 01

Q(Nl) = Q(Nl) .eal E P(Nl)
Q<N1+1) Q(Nl)‘e‘°’1 fire—1+1)

   

P(N )
so that a! - log 1
1 < P(N —+11))

    

 

27.
and similarly for 02
Q (N2+l) - Q(N2+l) a _ e02 _ P(N2+1)
Q(NZ) ‘ Q(N2 +1)e'2 ‘ ‘ P(Nz)

 

so that 02 = 1°8< P(N 2+”)
P(Nz)

The first term of the right exponential is Nl-k where

[ log P(M) - log P(N1)]
— \

 

and
£0 Q(Nl-k+1) = iEO P(Nl) e

kn!

ka/ °° -i0’ 1
=P(N)e 1,2 e 1= P(N1)e
1 i=0 -a
l-e 1

also the last term of the left exponential is

N2 + 1 + j where
log P(M) - 10g P(N2+1)
.- {NJ
“2

co co 02(j-i)

iEO Q(N2 + 1 +j - i) = iEO P(N2+1) e

jQ’
2 _ P(N2+1)e

jQ/2 co -
= P(N2+l) e iEOe _02

l-e
Q(i) being thus defined in the tails let

Q(i)=P(M) forN +l+j<i<N1-k

2
so that ja m1
2 P N e
ifom = P(N2 “)6 + P(M) (NI-NZ-k-j-n + L1.)—
‘27;* .-.-“1

1 -

 

 

V . —‘_

28.

For ease of computation make the substitutions
u - log P(M) - log P(Nl) - kal
v = log P(M) - log P(N2+l)-ja2

which reduces the sum to

 

 

E Q(il - P(M) [e-v + (N -N -k-j-2) + 9-“ J
i=_co 1_e_Q12 1 2 143—01 .

By letting T =i=§°° Q(i) /P(M) =

L ___eu
= [ be“)!2 + (N1 ' N2 ' k ' J ' 2) + l-e-a1]

and normalizing this quantity a random variable with the dis-
tribution P(i) can be found as follows:

1) Let U1 be a uniform random variable

 

2) If U < the observation is to be taken from
1 -a2
T(l-e ) 1
the left tail. Thus choose N0 = [- — log U11] where U11 is a

“2
uniform random variable and the brackets indicates the greatest
integer contained in the bracketed quantity. The observation

thus becomes N = N2 + l + j - N0. Then accept N with probability

Q(N) _V _u
3) If EL-——b S Ul S 1 - 2—‘-—=U the observation is to
T(l-e 2) T(l-e 1)
be taken from the uniform range as follows
e-v
U1 ' T 1 -a2
_ #4 - -k-'-2
R - _u -v (N1 N2 J )
1 e e

-T(l-e-al)- T(l-e‘32)

and let N0 = [R]

 

 

29.

so that the observation N is

N = N2 + j + 2 + No

and N is accepted with probability aéflg .

4) If U > 1 - -£L--- the observation is taken from the
1 T(l-e-al)

right tail using the same procedure as in step 2. That is choose
_ i J . . . .
N0 - [- a11°g U12 where U12 is again uniform only this time

let the observation be

N = N1 - k + N0 and accept N with probability §{%% .

5) If at step 2,3 or 4,N is rejected obtain a new uniform
U2 and repeat the process until an observation N is accepted. N
will then be distributed according to the distribution P(ni).

In steps 2,3,4 the acceptance rejection part of the procedure
is handled in comparison with an exponential random variable E0 in
the following way: Accept N if

E0 2 - log %%%% = log Q(N) - log P(N)

Where in the left tail

log Q(N) = log P(N2 + 1) + (j ' N0) a2

in the right tail
10g Q(N) = log P(Nl) + (k - NO) a1
and in the uniform range
108 Q(N) = log P(M).
This method of comparison is used so it is not necessary to
calculate Q(M), Q(N2 + l) and Q(Nl) using instead already calculated

quantities.

 

llllllliiisns—-——-———e~

   
 

‘ Wtfng Discrete Distributions with Large gleam

  
   

'k‘ Ebb'problsm at hand the Poisson, Binomial and Hypergso-

 

iifric distributions are used.

 
 

It is therefore necessary to

 
 

=dstarmine the distribution Q(n

 

1) as developed in the previous

 

section for these cases-but first it will be necessary to develop

 
 

some machinery for the calculation of logirl which is necessary

 
 

to.evaluate in all three of the above mentioned distributions

when calculating log P(ni).

, The first equation used is Stirling's asymptotic approxima-
tion to nI. From this
log n! = (n + 3) log n - n + k 108 2" + ¢(n)
l 1 1 l
"he” W“) ' '13:: ' 3—66n3 + 1260n5 ' l680n7 i

Consider now the product

22:: I“(n+l) P(mg) a n! 1.3.5,__(2n_3)(2n_1) g 22,,

IT F(2n+1).
Taking the logarithm of both sides

95 log 1T + log I“(2n+1) - 2n log 2 + log P(mi) + 108 n1

 

and using the more general form of Stirling's equation
log F(x+1) =- (XI-35) logx - x + 15 log 211 + <p(x)
Q
-m
where ¢(x) a Ego Cm x .
Thus
35 log 11 + log I‘(2n+1) - 2,103 n + log I‘(2n+2) - log (21.1.1)

- 95 log n + (Zn-(~35) log(2n+l) - (2n+1) + 35 log 2n +m§ oc n1"“(2m-1)

Also i log n + log F(2n+l) ‘

9.. 2n log 2 + log l"(n+3/2) - log (nu-1;) + 108 1“.

 

 

‘ ~—————-----————""‘TIIIIJIIIIIIIIIIIIIIIIIIIIIIIIII

31.

Collecting terms then and combining these two equations
log n! = $10gfl + (2n+%) log(2n+l) - (2n+§)-log 2 + % log 2

a: as C
- n 1og<n+ss> - (ma) -m§0cm<n+a>'"‘ +m§o #2 (warm
2

= 95108 21'T + (n+52) 108(n+%) - (M’s) - Watt)
where
m
‘1’(n+35) -m§0 Cm(n+%)-m (1-2'“‘>.

Note that this function W is not the logarithmic derivative
of the gamma function used in Chapter 2.

Since the original equation for log n! was an asymptotic
approximation, log n! and therefore Y(n+%) cannot be calculated
in this way for small n. To find Y(n+%) for n = 0,1,...10
calculate Y(11 + %) = Y(ll.5) from the already developed formula
and use a backwards recursion formula which is now to be derived.

103 n! = J5103 2TT + (n+3!) 10g (Mi) - (Ms) - ‘i’(n+35)

Aslog 2n + (ma) log (1+ 51;) + (ma) log n - (ma)
- Y<n+%)

also
log n! = log n + log (n—l)!

= log n + % log 2n + (n-%) log (1-%B)-+ (“-5) 108 n-(n—%)

 

-Y(n-%).
Combining these equations
1 .1.
Y(n-%) = Y(n+%) + 1 + (n—s) log(l- 3;) - (n+g) log (1+ 2“
111111_l__11_m_
=Y(n+%)+1+(“'%)['E'EZEZ'§8n3'416n4 533; J
1 1 1 1 1 i 1 i 1 — ...
‘(“5)[fi'§m+§m3'4m4+sfi5+ 3
1 1 1 1 +
=‘i’n+ + + + +... ___2k
( %) 22-2-3n2 24.4-5n4 26-6-7-n6 22k-2k(2k+1)n

 

 

‘T i

32.

The first seven terms of this expansion are used for n = 1,
2,...,10 but in calculating Y(%) an additional four terms are used.
Consider now a careful calculation of the expression
(L+x) log(l+x)-x which will be useful in calculating logP(ni). Make

the substitution l+x = 1?? so that

l
x=—21 and y=i
l-y 2+x

and under the assumption that x > -1 it follows that lyl < l.

_=L".‘£ <H1>-3Y—
Thus (L+x) 108(L+X) x l-y log 1-y l-y '

The evaluation of this expression will be broken into two cases
First if A; < |y| < 1 then
x
(L+x) log(L+x) -x = (1+x) log(1+x) -(2+x) (2 + 2x - x)
= xy + (L+x) [log(1+x) - 2y].

Secondly if lyl s % use the expansion

1+ 2 3 2 5
log<fi>=2y+3y +Ey +

Thus
(L+x) log(L+x) -x

2
= fix) 23 £5 - J—
<1_y [2y+3y +5y + J l-y

2
=3L it}: 23 §5+§7+39+...]
l-y+l-y[3y +5y 7y 9y

2 5 2 7 2 9
=xy+ (1+x) [%y3+§y +7y +'9'Y + ---J

With these equations consider now the calculation of Q(ni)
for desired distributions.
1) Poisson Distribution: let A be the parameter of the Poisson

distribution.

 

ms—

 

33.

Then obviously M = [A]

N1=M+[/§Y]

N2=M-[/§A]-1

where in each case the bracket indicates the greatest integer

contained in the bracket.

 

Also
Q(Nl) — a1 _ P(Nl) _ N1+ 1
Q(N1+l) ‘ e ’ P(N1+1) ' A

so that 01 = log(N1+l) - log A

 

Similarly
Q(N2+1) _eaz = P(N2+l) = A
Q(NZ) P(Nz) N2+1
so that 02 = log A - log (N2+l)
, A“ -A
and finally P(n) = ET e for n = 0,1,2;3:4:---

so logP(n) = -X + n log K - log n!

= - A + n log A - (ms) log (me) + (mg) - A; log 2n + Vows).

Make the substitution n = A + n. Then

log P(n) = ->» + (Mme) IogA - (A+p+%) log(A+u+%) - (A+p+%)
+ 15 log21'r - ‘1’(>~+u+%) - a log K

= -A[1 + if 3 1og[1 + #1 + MKS] - a 1og2nA + ‘Y(7\+u+»‘5)

= -)~{[1 + £52] log [1 + 5:5] - iii-35} - $5 logZTrA + ‘I’()\+p+%).

2) Binomial Distribution: Let p,n be the parameters of the
Binomial distribution
then M = [(n+l)p]
Nl-M+[/m]
stM-Um] - 1

 

k

34.

 

a N +1 or n-N
andasbeforee1=l—i:£andez= __‘3..P_
n-Nl p N2+l l-p
N +1 n-N
= __ hp = 2 _ l-p
so 01 log n-Nl + log p and 02 log N2+l log p
and finally P(k) = 41—- pk(l-p)n-k for k = o l n
k!(n-k)! ’ """

log P(k) = log n! + k log p + (n-k) log(l-p)
-log k! - log(n-k)! .
Using the derived formula for log x! then and the identity
log n! = log(n+1)1 - log(n+l)
log P(k) = (rt-+35) log(n+l) - (n+1) + % log 211 + cp(n+l)
- [(k+%) log(k+%)- (k+%) + % logZW-Y(k+%)3
- [(n-k+%) log(n-k+%) - (n-k+%) + tlog 2n - Y(n-k*%)]
+ k log p + (n-k) log(l-p).
Make the substitution k + 35 = (n+1)p+p into the above equation.
108 P(k) = (n+%) 108(n+1) - [(n+1)p+u] 108((n+1)P+u)
- [(n+1)q-u] log((n+1)q-n)-+ [(n+1)p+n] 10g p
+ [(n+1)q-n] log q - l5 log 2n + cp(n+1) + ‘1’(k+15) + Y(n-k+%)
- g log p - a 103 q

= - [mum mm + (73517)] - [(n+1)p+u] Iog<n+1>p

- [(n+1)q-u] 10311 - (311351 - [(n+1)q-u] log(n+1)q

+ [(n+1)q-u] 108 q + [(n+1)p+u] log 9 + (n+1) log(n+1)
- % log (n+1) 2npq + ¢(n+1) + Y(k+%) + Y(n-k+%)

= - (n+1) p{[1+ inf—up] log(l + THE—17p) -(;f1—)p}

_ (n+1)q {[1 - TEE—17¢” 108(1 ' (Tl-PETE.) - < (tr;1)q D}

- % log(n+1)2npq + m(n+l) + Y(k+%) + Y(n-k+%) .

35.

3) Hypergeometric Distribution: Let D,N,n be the parameters

of the distribution

 

 

 

 

 

 

_ (m1 D+l
“‘9" M - 1 T572]
N = M‘+ anN-D)(N-n)
. 1 m I) J
Nz-(N 1)
also
01 (Nr+l)(N-D- -n+N1+l) 02 (D-N2)(n-N2)
e = and e = .
(D- N 1) (n-N 1) (N2+1)(N-D~n+NZ+1)
Consider now the probability
D N-D
_ Q(n-x) _ D! LN—D)! nlgN-nzl
P(x) - -(D-x)!x! (n-x)!(n—D~n+x)! N!
C(N’ n ’D) for x = O,l,...,D.

=(D-x)! x! (n- x)! (N- D- n+x)!
Expand the factorials using the established formula and

make the substitution

1
Y = X'+ % - MO where M0 = (n;1+ 3+
Thus
103 x! = (we) log(wi) - (as) + 35 logzTr - Mme)
= (MO+Y) log(MO+y) - (Md+y) + % log2n - Y(Md+y)
=M 1 i’- 1 1+L)- 1— +ylogM -M +MologMO
01(+M0) °g( M0 MO] 0 o
+ % logZTT - “Moi-Y)
log(n-x)! = (n-x-I-Jfi) log(n-x‘l-lg) + 5. log21'r - Yul-yes)

- Y n+1-M -
(n+l-MO-y) log(n+l-M0-y) - (n+l-MO-y)-+ % logZW ( 0 y)

36.

= - - __1L_. '
("+1 o) [(1 n+l-M0) mg” ' 31%) 'nfifi] " y 10g(“H“Mo)
+ (n+l-M0) log(IH-l-MO) - (n+1-MO) + $5 log21T - ‘Y(n+1-MO-y).
Similarly as above
loD-x!=DI-1-M[ -—1—— -—L- ‘ J
0 0 O
- y log(D+l-M0) + (D+l-MO) log(D+l-MO) - (D+l-MO)
+ % logZfi - Y(D+1-MO'Y)
and finally

log(N-D-Mx)! = (N-D-n-I-MO) [(1 + W) log(l + W)
0 0

N-D-n+MO + y log(N D n+MO)

+ (N-D-n+MO) log(N-D-n+MO) - (N-D-n+MO)
+ % logZfl - Y(N-D-n+MO+y).

Combining these equations

*
logP(x) = c (N, ,D) - M (1+ L) log(1+ L) - y-
n (i M0 M0 M0]

= z _ _L
(n+1-Mb) [(1 - ;;{jfig) log(l ‘ “+I'MO) n+1-M0]

2 __ ._ _;2L__.
(D+l-MO) [(1 - B;%ffi;) 108(1 ' D+1-MO) D+l-M0]

_L_ - —X——
(N-D-MMO) [(1 431m) 108“ +N-D-n+MO) N-D-n+Mo]

MO(N-D-n+MO)
y 1° (n+l-MO) (D+1-MO)]

 

+ Y(M0+y) + ‘1’(n+l-M0-y) + ‘i’(D+l-MO-y) + ‘1’(N-D-n+MO+y).
M (N-D-n+M

)
' 0 O = l h' h eliminates this
A check Will show that (n+1'Mo)(D+1'M0) W 1C

term from consideration.

37.

Also for use in these acceptance rejection procedures the
*
constant term C (N,n,D) may be neglected since the procedure uses
only the ratios of the probabilities of the reSpective points

being considered.

38.

Section 3: Procedures for Discrete Distributions with Small Means.
The procedures in the previous section generate the desired
random variable using a small number of uniform random variables
but at the expense of considerable numerical calculations. When
the mean of the distributions under consideration is small, proce-
dures exist which use about the same number of uniform random
variables but which are much less involved. Such procedures used
in the simulation will now be considered.
1) Poisson: Let A be the mean of the Poisson distribution. Let
EI,E2,E3,... be independent exponential random variables which

are independent

are obtained by the equation E = -log Ui where U

i i

uniform random variables. Let J be the integer such that

J-l J
E E. < x s 2 E1
i=1 1 i=1

J
Then J-l has a Poisson distribution with mean X and igl Ei- X

is independent exponential. This result can be shown by directly

integrating the joint density of the Ei'
2) Truncated Poisson: This distribution is needed only in the

small mean case and its use will be shown later. Let X be the

mean of the distribution and as before let E1,E2,E3,... be

exponential random variables.

Let q be defined as the integer Such that
qk < E1 f (q+l)X.

Let J be the integer such that
J-l J
2 E < (q+l)k f 2 Ei'

i=1 1 i=1

39.

Then J-l has a truncated Poisson distribution with mean A
J

and 217-1- (q+l))» is independent exponential. This result can
i=11

also be shown by directly integrating the joint density of the Ei'

3) Binomial: Let N,p be the parameters of the Binomial distribution.
Define a = -log (l-p) and let g = Na. Divide the interval
(O,Na] into the N intervals I1 = ((i-l)a,fl1].

Let E1,E2,E3,...
Consider the points
i
xi = jgl Ej for i = l,2,3,...,k-l
where k is defined to be the first integer such that

k
= E E, > Na.
*1. H.

be independent exponential random variables.

Let NB = Number of intervals Ii which contain a point xi.
Then NB has a Binomial distribution with parameters N and p.
This can be shown directly by integrating the joint density of

the E,.
J

4) Hypergeometric: Let N,D and n be the parameters of the

distribution. Then

N- (N-D)!
P(x)= C—%%l>= (ID—L— (N __n—) IN! (N-D-n-i-x) ! (n-x)!

* x [(N-D-n)! 1- *Kx
(N-D)! CDCTEE') [Elmo-manna)! (jg—j ]

=(NTD-n) ! (N-n) 1N!

 

 

 

___E___
* N-D-n+l

where _£_*.= N D “n+1 and consequently P = L+ n
N-D-n+l

Let NIW B(D;p*) and accept N1 with probability

40.

N
n! (N-D-n) l < N-D-n+l 1
(N-D-nl-Nl)! (n-Nl)! n > '

*
If N is rejected let N q'B(D,p ) and repeat the

l 2

process. Let NH be the first accepted N Then NH has a Hyper-

i'
geometric distribution with parameters N,D,n. This procedure is
used for small mean Hypergeometric and it is to be noted that
for the case where x = 0,1 the acceptance probability is one so
that the acceptance rejection part of the procedure is ignored

when the Binomial random variable is zero or one.

Consider now a simplification of the factor,

_ n!(N-D-n)! N-D-n+l x
R(x) zN-D-n+x)l(n-x)! < n >'

Using the established formula for log x!
log n! = log n+log(n-l)! = (n+%)log n - n + % 1082TT + ¢(n)
log(n-X)! = (II-#35) log(n-x+%) - (n-x+%) + 3: 10g21T - Uri-#35)
Combining these with x log n
log n! - log(n-x) - x log n = - (n-x+%) log(l-Ԥ%fi)
- (fit-35) + <P(n) + wows).
Make the substitution u = x-%
= - (n-p) 1og(1-§) - u + cp(n) + Y(n-u).
Similarly as above
log(N-D-n)! - log(N-D-n+x)! + x log(N-D-n+l)

- - + Y N-D-n+l+p).
= - (N-D-n+L+u) log (1 + fi:%:;;1)‘+ u-+ ¢ (N D n+1) (

From this then

1

log R(n) = - n[(1 -‘%) 108(1 "E) "i? J
- (N-D-rH-l) [(1 + 3757.371) log (1 + fifty—+1) ' N-D-n-i-l]
+ (P(n) + Y(n_n) + (P(N-D'IH’I) + Y(N—D-n+l+l-1).

Let U be a uniform random variable. Accept the observation 1f

41.

U 5 R01) or equivalently if E = -log U 2 -log R01) which reduces

to E + log R01) 2 0.

Chapter 4
Section 1: Simulation of the Model.

Let Q be an environment. Recall that the Species in the
environment are to be such that for any interval [a,b) with

0 < a < b < ” the number of Species with intensities in this

b
interval has a Poisson distribution with mean I f(x)dx where
a
f(x) = A e-x.
a

X

Suppose that A and a are given and that a sample of N
individuals is to be taken from a computer Simulated environ-
ment. The problem reduces to first choosing the intensities
of the Species in the environment so that they satisfy the
above condition and then choosing the number of individuals
in each species Such that this number has a Poisson distribution
with mean proportional to the intensity of the reSpective Species.

This constant of proportionality will be designated by ks and will

be called the power of the sample.

Let x be the intensities of the Species that are to

1,x2,...

be selected and suppose a supply of independent exponential ran-

dom variables Ei’i = 1,2,... are available.

Noting that the waiting time for a Poisson process is ex-

ponential consider the following method of choosing the intensities.

Let

an
E = I f(x)dx and solve this equation for x1-

1 x
1
When this is done let
x
1 o
E = I f(x)dx and solve this equation for x2 and

2 x
2

42.

43.

continue finding intensities x1,x2,x3,x4,...
6
Notice that for c > O,I f(x)dx =‘+ " so that the method must
0
be modified for small x. The modification and the method of

determining a constant cs, which determines the intensity at which

the modification will be made, will be shown later.
-x

 

The function f(x) = Ae cannot be integrated directly between
x
two arbitrary positive numbers so that the solution of the equation

1
E1 = I f(x)dx for x is obtained through an acceptance-rejection

x1+ 1

r+l

procedure. No such procedure was found that was effecient over
the entire real line so that different procedures were used depend-
ing upon the portion of the real line that was being considered.
The following method for finding the intensities of the Species
was used:

1. Set y = x-+ a log x

3.”;

91: g:
so that l + x x

dx

2. Let y0 = x0 =+OD and set i = 1, set k = l and
set E0 = 0.
3. Set y: = xi_1 + a log xi_1 = yi_1 and in order
to determine xi let
y1 = xi-+ a log xi.
xi-l y: y

i
Ae-y __x— dy

Then I f(x)dx = I f(Y)‘;E§dY = I xH1
x. Y- yi

1 1

* 44.

yi

*
4. Set Ek = f Ae ydy .-. A(e‘Yi _ e’Yi)
yi
and solving for yi
*

-log (Ek + e i) + log A
k-l
-log (Ek + 1:; Ej) + log A

yi

k
-log ( 2 E ) + log A.
i=1 1

5. Solve the equation y1 = x1 + a log x1 for xi.

x
6. Acce t x. with b b ' i
p 1 pro a illtyxi+a

 

78) If x is rejected set * - ’
i yi — yi,1ncrease k by one and return
to step #4 provided x1 > 3.0.
b) If x1 is accepted increase k by one,increase i by one and
return to step #3 provided x1 > 3.0.
For intensities less than 3.0 a modification is made in the
procedure to obtain a higher degree of effeciency in choosing

the x

*
8. Let xi equal the last intensity calculated in step #5.

*
*Let k1 k: xN1= xi *
x* xi
1
Then I f(x)dx = flAe-XZ _ =xf1-A_e xCijxdx.
xi xi fa Zad xi 20
*
X.
1
9. Set Ek = 'I'A_ e-xdx and solving for x.
x. a 1
1 2
k
E E
x1=-log[j—klj +e 1]
A 2"“

10. If xi 2 2.0 accept x1 With probability' ';>F° If x1 is
i

45.

*
rejected increase k by one, set xi = xi, return to step #9.
*
f =
I x1 is accepted increase k by one, set x1+1 xi, increase
i by one, return to step #9.
For the case xi < 2.0, x1 is rejected and the procedure

again modified.

*
11. Let x. = 2.0, increase k by one and set k = k.
xi: 1 x* x* 2
i i 1
Then i f(x)dx a i A e-x e-ldx = IA 8-1 61-x dx.
i 1 0' --1 xi CY
x e x
X? -a
12. Set Ek = i A x dx
i e
and solving for X1
1 (1 k l
-O’ -0' e _-
= - 2 15]] 1-01 .
x1 [2 A j=k2 J
l-xi
13. If xi‘z 1.0 accept xi with probability e . If xi
is rejected increase k by one, set x: = xi, return to Step #12.
If x1 is accepted increase k by one, set x:+1 = xi,increase i by

one, return to step #12. If xi < 1.0, reject xi since the

procedure breaks down at one.

6
AS was pointed out earlier, ‘f f(x)dx = a for e > 0 so

0
that procedures of the type used for large intensities are

impractical for very small intensities. Note that when choosing

a sample from the simulated environment the number of individuals

in each species has a Poisson distribution with mean ksxi' Here

k is unknown but it can be estimated and from this estimate a
S

method devised for choosing Species with small intensities which

have a high probability of appearing in the sample while over-
looking many which do not appear in the sample of N individuals.

46.

The expected number of individuals is represented by the equation

1 . 1 1

 

1' -x i-l
kaf(x)dx+ z kx =fk Ae dx
3 x + 2 k x
0 j=1 SJ 08 xa j=1 SJ
1
2 -1
l-a 1
gksUAx (l-xi-ilfmx-t- 2: x,].kiA(_1___L+_1_,
0 j=1 J 2'0 3-0 4(4-0)
1-1
2
”=14.

Setting this equal to N, the number of individuals to be taken

from the simulated environment, the estimate of k is
s

 

R3 = N i
1 1 1 '1
A ___.- ___. ______.
(Z-a 3-o/ + 4(4-a)) + 331*,

Using this estimate of k3 continue finding the intensities
of the Species in the simulated environment.

0.8
14. Set 88 = E;— , x1" = 1.0,

increase k by one and set k3 = k. Then
x* x?

i 1
‘f f(x)dx = ‘f -A— e-xdx.
x1.- x. #1
x? 1
1A
15. = --
SetEk J” adx
x1 x

and solving for x,

1
k __1_
xi =[1 " (if!) [3'ng Ej]] 1.0 '
-x,

16. Accept x with probability e 1. If x1 is rejected,

1
increase k by one, set x: = x1 and return to step #15

i 2 es. If x1 is accepted, increase k by one, set

x:+1 = x,, increase 1 by one and return to step #15 provided

provided x

x 2
i 6s'

47.

For xi < as then the probability that a Species with intensity

xi will have an individual present in the sample with power ks is

e-k 3x1. Therefore instead of solving the equation
1 - x*

13k = f f(x)dx
X.
1
for x1 and letting the number of individuals present inthe sample

from this species be

111 where niQITruncated Poisson with parameter
{ 18x1 with probability 1 - e' Sxi

0 with probability e-ksxi
an equivalent method for determining the individuals in the small

species is to solve the equation
*

x1

=‘r ksxf(x)dx for X1 and let the number of individuals

x1

present in the sample from this species be

ni where n,m Truncated Poissonawith parameter
1

 

. l-e-ksxi
{ k x. with probability
8 1 -R Sxi
ksxi-L+e Sxi
0 with prob. k x

This modification has the effect of skipping over some

Species which are in the environment but which do not appear

in the sample.

17. Set N* = 1 and x = xi, k4 = k
Xi x1 1 l a x

. A A -x = . - - d
f ksx f(x)dx =.f ksx ~75 e dx I ksA x e x
x x x x

i 1 i

48.

x1

18. Set Ek =‘I ksA xl-adx and solving for X1
x
i 1

(2'0) [jgk4Ej ] Z-a

.. 4.2-0 - >
1 E

 

12A
8
-x
19. If x1 > 0 accept xi with probability e
*
If x1 is rejected, increase k by one, set xi = x1 and return
*

to step #18. If x1 is accepted, increase k by one, set x1+1 = xi,

increase i by one and return to step #18.
The procedure is continued until a negative intensity is

reached. Let sN be the number of Species obtained. Consider now

the problem of finding the sample of N individuals and let n1 for

i = 1,2,...,sN be the number of individuals chosen from the Species

with intensity xi. Thus

 

 

n. is chosen from a Poisson distribution with
1 parameter ksxi for i = 1,2,...,N -l.
n. chosen from a truncated Poisson distribution
1 - x
. -e s i
with parameter k x, with probability . ,
s 1 k x,
A ”fisxi S 1
ksx,-1+e
' ‘ ' " 1 for i = N* ... s .
0 With probability ) , N
s k x,
N S 1
Let NT = i§l ni .
If NT = N then the sample is as chosen. If NT > N then NT - N

individuals must be independently rejected from the chosen sample.

This is accomplished be means of the Hypergeometric distribution

. I
where the number to be eliminated in the first SPeCieS is “1

N -N,n

which is distributed Hypergeometric With parameters NT’ T 1

th . .
and in general the number to be eliminated in the k Spec1es 18

49.

k-l
n' which is distributed Hypergeometric with parameter N - Z n
k k_1 1‘ i=1 i ’
NT-N-Z ni, nk. This is continued until all NT - N individuals

i=1

have been eliminated.

The number of individuals in each gpecies is
n
and E n: = N.

n: = n - n; for i = 1,2,...,s
i=1

i N

If however NT‘< N then N - NT more individuals must be chosen
from the model.
fijflg . The factor two is added to make the

N
probability of fallingTshort again vary small since it is better

Let Ak a 2k
3 s

to over estimate ks. The intensity of the sample is now

k; = k. + Ak so let n2 be the number of individuals that are to
s s

be added to the already selected Species where

n? N Poisson (Aksxi) for i = 1,2,...,SN.

Since some Species were skipped in the interval [0,x6) the
possibility that some of these may now appear in the enlarged
Sample must be considered. Let €* = Ei— . If 5* > x6 select the
new Species using the method described in steps #17-19 replacing
Rs by 4&8 and continue finding intensities until zero is reached.

The number of individuals present in the sample from these Species is

n? where If; N Truncated Poisson with parameter £1ka1 With
1

1 6-13ka1
probability -——Zfif;——'
S i

AA
0 with probability 8 i
' ' ber

for i = SN + l,...,sfi where SN 13 now the num

so that it is
of species present. The intensity x6 is chosen

50.

*
very unlikely that e < x.e but if this should happen the situation

can be corrected by decreasing the upper value of x6 from say-19;;§
s

. .6
to p0881b1Y‘2E- and rerunning the experiment.
s

I
SN
1 = 11
let NT i=1 ni
If Ni = N - NT then no individuals need be deleted. If Ni > N-NT
then Ni - N‘+ NT individuals must be eliminated from the Ni new

individuals chosen.

This is accomplished again using the Hypergeometric distribu-
tion by letting ni be the number of individuals eliminated from
the first species Where ni is distributed Hypergeometric with

T’ Ni - N‘+ NT’ n? and for the kth species
k-l ' k-l

n; is Hypergeometric (Nf - i§l n2, Ni - Ni+ NT - i§1 Hi; nk)'

parameters N'

The number of individuals in the respective Species is then

n? = n, + n; - n1 for i = 1,2,...,sN

1 1
l
*= "- ' .=S +1 cons
n1 n1 n1 for 1 N , , N
S!
N
e E n? = N.
wher i=1 1

If Ni < N - NT repeat the process for selecting new individuals
from the Species. Because of the method for determining Aks

however it is extremely unlikely that the adding of new individuals

Will be necessary more than once.

Chapter 4
Section 2: Simulation in the Special case.

The simulation of the model for the Special case developed
in Chapter 1 Section 3 is greatly simplified over the general
case. In taking the sample of size N in this case consider the
following procedure. Define W1 as before to be the proportion
of individuals of the ith sampled Species present in the environ-
ment neglecting the i-l Species already sampled.

Choose w1“'h(w) = A(1-W)A-1 where A is a parameter of the
model which is to be estimated. Choose mlm Binomial (N-l,w1)
where 1111 represents the number of times that this Species repeats
in selecting the remaining N-l individuals. Then n1 = m1-+ 1
represents the number of individuals from the first Species in
the random sample of N individuals. Now choose w23'h(w) and
again select

m “'Binomial (N-n1-1,W2) and let 112 = m2-+ l.

2 i-l
In general select wi“'h(WL select miB'Binomial (N- ZInJ-l,wi)
J:
and let n. = m.‘+ l.
1 1

Continue this process until a sample of N individuals has

been chosen.

51.

Chapter 5
Section 1: Data Analysis

Using the procedures developed in the previous chapters a
set of data will now be analized to indicate the fit of the
model in an actual environment. The data used for this purpose
was taken from Williams [4] and is reproduced in table 5.1.

From the maximum likelihood methods developed in Chapter 2
an estimate of the parameters for this set of data was found to be

6: = 1.0000 21 = 40.453 13 = 392.7.

Since & 8 l the estimation of the other two parameters reduces
to the Special case Where a is set equal to one in the likelihood
equations and an estimate of the other parameters obtained by the
procedure developed in Chapter 2 Section 2. This maximum likeli-
hood estimate was found to be

21 = 40.2576 13 = 387.2

According to the model each sample is such that the number
of individuals in the sample from a species with intensity x is
distributed Poisson with mean ksx. Using ks as an estimate 0f

kS then, the expected number of Species in the sample with m

individuals is

 

” c m -k x w A m -k x A -x
k x) S Ae
.(_s.2‘_). S d = (s e ——dx
1]; ml e f(x) x '(I). ‘111'1— x
“ m .. 1+1». 1 i m
AkS m-l -( s d _ S F m
= m! g x e 3 x — m! ( l)m
fc
_ e S m l _ 40.2576 99743)m.
- A ( Eé+l ) m — ‘m (°

52.

53.

Table 5.1
Macrolepedoptera Data
Observed captures of Macrolepedoptera in a light trap at Rothamstad
Journal of Animal Ecology Volume 12-13, pp.45-46.
Distribution of Species according to number of individuals present

in the sample.

 

l 2 3 4 5 6 7 8 9 10

0 35 ll 15 14 10 ll 5 6 4 4

10 2 2 5 2 4 3 3 3 3 4
20 l 3 3 l 3 l l 3 2 0
30 0 l 0 2 0 3 2 0 0 0
40 0 0 2 2 l 0 0 0 3 0
50 4 l l 2 0 0 l 2 O 3

 

also at 61,64,67,73,76(2),78,84,89,96,99,109,112,120,122,129,
135,141,148,149,151,154,177,181,187,190,199,211,221,226,235,239,
244,246,282,305,306,333,464,560,572,589,604,743,823,2349

TOTAL NUMBER OF INDIVIDUALS 15,609

TOTAL NUMBER OF SPECIES 240

54.

Table 5.2
Theoretical Frequencies for Macrolepedoptera Data
Distribution of the expected number of Species present in the

sample with parameters a = 1.0000, A = 40.2576, k8 = 387.2

1 2 3 4 5 6 7 8 9 10
0 40.15 20.03 13.31 9.96 7.96 6.61 5.64 4.93 4.37 3.92

10 3.55 3.25 2.99 2.77 2.58 2.41 2.26 2.13 2.02 1.91
20 1.81 1.73 1.65 1.58 1.51 1.45 1.39 1.34 1.29 1.24

30 1.20 1.16 1.12 1.08 1.05 1.02 .99 .96 .93 .91

 

 

40 .88 .86 .84 .81 .80 .78 .76 .74 .72 .71
50 .69 .68 .66 .65 .64 .62 .61 .60 .59 .57
also

61 - 70 5.14 151 - 200 7.31

71 - 85 6.34 201 - 300 8.57

86 -110 7.96 301 - 500 7.43

111 ~150 8.84 500 6.07

EXPECTED NUMBER OF INDIVIDUALS 15,587.7

EXPECTED NUMBER OF SPECIES 239.99

# of individuals
in Species

°°\‘O‘U14>uN.-a

9-10
ll-12
13-14
15-16
l7-l9
20-22
23-25
26-30
31-36
37-45
46-55
56-70
71-85
86-110

111-150
151-200
201-300
301-500
500 --

2

Goodness of Fit Test

Observed

35
ll
15
14
10
ll

5
6
8
4
7
7
9
8
7
7
6
7
11
9
5
4
8
7
8
4
__7_

240

X 095(24) = 36.42

55.

Table 5.3

40.
20
13.

9.

7
6.

5
4
8
6
5
5
6
5
4
6
6.
7
7
8
6
7
8
7
8
7
6

 

239.

Theoretical
frequency Frequency

15

.03

31
96

.95

61

.64
.93
.29
.80
.77
.00
.40
.44
.71
.68

63

.98
.02
.18
.34
.96
.84
.31
.57
.43
.07

99

(f

 

- f

fth

ob

.66
4.07
.21
1.64
.53
2.91
.07
.23
.01
1.15
.26
.80
1.06
1.20
1.11
.02
.06
.12
2.26
.08
.28
1.97
.08
.01
.04
1.58
.14
22.55

th)

2

56.

These values were calculated and are presented in table 5.2. It is

interesting to note that in this Special case the expected number of

Species present in the sample can be easly calculated by the formula

 

 

°° -f<sx °° -ksx Ae-x
f(l-e ) f(x)dx = f(l-e ) dx
0 0
a _ -(ké+l)x
A e - e = A A = A.A =
— Ag x dx Alog(ks+1) A 239.99.

Using these theoretical values a x2 goodness of fit test is
applied to the data in table 5.1 and the theoretical values in

table 5.2. The number of degrees of freedom for this test is j-3

where j is the number of categories. Here three degrees of freedom

are lost because of the estimation of the three parameters of the

model. The test is as shown in table 5.3 and is not significant

at the 5% level.

As an aid in studying the behavior of the model a simulation
procedure has been developed in the previous chapters. Three
independent samples of 15,609 individuals have been taken from the
model using the parameters estimated from the data in table 5.1
and the procedure developed in Chapter 4 Section 2. These three
sets of simulated data are reproduced in tables 5.4-5.6 and
should give the reader a good indication of the stability of the
model. Note in particular that the total number of SPECIES’pre-
sent in each of the simulated samples are very close and that
while the number of Species present in the samples with a given

number of individuals may have a large variation among samples

nevertheless the number of large, moderate and small species

57.

Table 5.4

Simulated Test #1

Distribution of Species according to number of individuals

present in the sample with parameters a'= 1.0000, A = 40.2576

 

l 2 3 4 5 6 7 8 9 10

0 41 27 8 6 8 9 7 8 4 5

10 3 3 2 3 5 l 4 3 2 0
20 0 2 l 0 0 l l 2 0 0
30 3 3 l O 4 0 0 0 2 3
40 0 0 1 l l 0 O 1 0 0
50 l l 1 0 0 0 l 0 0 O

 

also at 62,63,66,67,79,80,83,85(2),88,89,9l(2),93,94,96,97,105,
107,109(2),136,155,159(2),162,165,166,169,180,187,188,189,217,
222,246,247,255,260,273,277,287,324,325,345,350,405,408,440,
464,485,582,606,1385,1399.

TOTAL NUMBER OF INDIVIDUALS 15,609

NOTAL NUMBER OF SPECIES 235

58.

Table 5.5
Simulated Test #2
Distribution of species according to number of individuals

present in the sample with parameters a = 1.0000, A = 40.2576

 

l 2 3 4 5 6 7 8 9 10

0 40 14 15 10 8 7 3 8 3

10 4 3 l l 2 5 3 l 2 O
20 3 0 l 3 l 0 3 2 0 2
30 l l l 2 l 0 2 0 l 0
40 l 0 0 3 l 2 0 0 2 0
50 O 0 0 l l 0 l 3 0 0

 

also at 61(4),64,66,67,69,71(2),72(3),73,74,7S,93,94(2),97,100(2),
101,112,120,122,124(2),125,130,135,136,140,143,148,155,157,161,
l75(3),177,187,191,192,193,196,205,206,237,291,295(2),299,302,
305,325,348,349,394,405,426,573,808,819,1079.

TOTAL NUMBER OF INDIVIDUALS 15,609

TOTAL NUMBER OF SPECIES 243

59.

Table 5.6
Simulated Test #3
Distribution of species according to number of individuals

present in the sample With parametersa= 1.0000 A = 40.2576

 

l 2 3 4 5 6 8 9 10
O 45 18 12 10 7 4 8 3 4
10 6 3 l 2 l l 2 3 2 2
20 3 O l l 2 2 2 0 l 0
30 2 l 3 l 4 l 3 0 l 0
40 O 0 1 0 2 O 0 0 O l
50 4 0 2 0 0 l l l l 0

 

also at 61,62,67,68,70(2),71,75,77,79,80(2),89,92(2),102,104,
105,106,107,113,115,119,121,125(2),138,152,168,192,l96,208,218,

223(2),248,249,286,301,305,375,384,410,451,531,616,630,762,768,

1186,1711
TOTAL NUMBER OF INDIVIDUALS 15,609
TOTAL NUMBER OF SPECIES 233

60.

present remains quite stable among samples.

vWith the use of the simulated tests the question of the
accuracy of the estimates of the parameters when using the model
can be considered. The simulated data is now considered as the
original data to find the maximum likelihood estimates of the
parameters, again using the procedures developed in Chapter 2.
These estimates for the three simulated tests can be compared to

the values of the parameters used in obtaining the simulated

data as shown in the table below:

a A k
8

Values of parameters 1.000 40.2576 387.2

Estimates for simulated test #1 1.000 39.2429 397.7

Estimates for simulated test #2 1.000 40.8523 382.1

Estimates for simulated test #3 1.000 38.8425 401.8

Another point of interest is to consider the behavior of the
data as the number of individuals increases in the sample. Taking
a = l and A = 40.2576 table 5.7 shows the behavior of the data
where a sample of size 50 is first taken and then the sample in-
creased in small steps up to 15,609. It is to be remembered that
this collection of data only illustrates the behavior as n increases
in one sample but should serve as a guide for other samples. It
is to be noted for example that the number of Species with one
individual in the sample has already stabilized by the time 200
individuals are sampled.

In order to compare the simulated data to the theoretical

distribution for arbitrary N it is necessary to obtain an estimate

61.

of the parameter ks. Noting that the number of individuals
present in a sample from a species with intensity x1 is dis-
tributed Poisson with mean kai so that the expected number is

ksx., an estimate of this parameter for arbitrary N is obtained

(9
by setting the equation I ksx f(x)dx equal to N. Thus
0

 

- m
fk x d — k A f e xdx = k A
s s s
0 0
C O A N
and the estimate 18 k = -.
s A

Using this estimate the expected number of Species present

in the sample with m individuals for a sample with parameter
-x

 

 

 

 

 

h and where f(x) - Ae is
S
N
m (ES x)m "ksN E: A m m_1 '(l‘sN“1)X
N e f(x)dx = N f x e dx
0 m! m! 0
A fism T(m) is m
g N _ A <: N :>
I A - — A
m‘ (kS +1)m m ks +1
N N

.A<_N_>m
m N+A '

For given values of N and A this can easily be tabulated and in

particular compared with the data in table 5.7 for A = 40.2576.

Distribution of species according to number of individuals

Table 5.7

62.

 

 

 

 

 

present in the sample with parameterscr= 1.0000, A = 40.2576
for increasing N
N = 50 Number of Species = 28
1 3 4 5 9 10
0 15 8 2 1 O 0 0 0
N = 100 Number of Species = 41
. 1 3 4 5 6 8 9 10
0 20 8 3 2 5 l 0 2 0 0
N = 200 Number of species = 63
1 2 3 4 5 6 7 8 9 10
0 29 10 3 2 2 l 0
also at 12(2),21
N = 500 Number of Species = 94
1 2 3 4 5 6 7 8 9 10
0 31 19 8 9 4 1 5 2 0
10 3 1 l 0 1 0 0 O 2
20 0 2 l 1 0 0 0 0 0
also at 33,49
N = 1000 Number of species 119
1 2 3 4 5 6 7 8 9 10
0 32 19 12 10 6 8 2 3 l 2
10 2 2 l l l 2 l 0
20 0 0 l 0 1 l 0 1

also at 33,42,43,46,47,49,67,97

63.

N = 2000 Number of Species = 143

 

 

1 2 3 4 5 6 7 8 9 10
0 36 14 9 10 7 10 5 5 4 2
10 1 3 5 0 l 1 0 l
20 1 l 1 1 O 2 0 3
also at 34,35,40,47,49,54,56,59,80,83,93,95,97,144,203

N = 3000 Number of Species = 165
1 2 3 4 5 6 7 8 9 10
0 43 16 ll 8 8 7 4 3 6 6
10 4 2 3 2 0 1 2 2 3
20 3 2 0 1 0 2 0 0 1 0

also at 39(2),41,42,44(2),47,50(2),52(2),67,79,80,81,92,124,

125,130,136,142,208,319

N = 4000 Number of Species = 176

 

1 2 3 4 5 6 7 8 9 10

o 41 22 5 11 5 5 7 6 3 4

10 4 2 4 5 1 4 1 2 2 1
20 2 2 1 1 o 1 2 o 1 0
3o 2 2 o 2 o o o 1 1 o
40 o o o o o o o o 1 2
50 o o 1 o o o 1 0 0 0

 

also at 61,63,66,67,68,71,86,93,106,110,114,157,l61,l70,173,

190,295,434

64.

N = 5000 Number of Species ' 190

 

1 2 3 4 5 6 7 8 9 10

O 43 28 5 8 5 5 7 3 5 5

10 6 0 5 3 5 1 2 2 2 0
20 2 2 3 2 1 0 0 2 O 2
30 0 O 3 3 0 1 0 l O 1
40 0 1 1 0 0 O 1 0 0 0
50 O 0 1 0 0 0 O 0 l 0

 

a136 at 61,64,65,71,74,79,82,88(2),89,98,ll8,130,133,138,l95,

200,212,227,237,378,529

N = 6000 Number of species = 198

 

 

 

1 2 3 4 5 6 7 8 9 1o
0 4o 31 8 7 7 5 6 o 5 5
1o 2 5 5 2 1 4 7 1 1 o
20 1 o 3 2 3 1 o 1 1 1
3o 1 1 1 1 1 1 1 o 1 1
40 o 2 3 o 1 o 1 o o o
50 o o 1 2 o 1 o o o 0
also at 70(2),77,81,87,91,93,96,101,107,120(2),142,149,150,
171,218,236,248,275,285,449,633
N = 7000 Number of Species = 200
~ 1 2 3 4 5 6 7 8 9 10
0 35 29 9 8 11 2 7 4 1 5
1o 4 1 4 6 2 3 2 1 1 5
20 0 1 1 1 2 2 2 2 1 o
30 1 o 2 1 1 2 1 2 o 0
4o 0 2 o o 1 0 0 2 1 0
50 3 1 o o 0 1 0 1 0 1

 

also at 61,67,81,83,86,92,102,105,109(2),115,124,136,141,172,

173,175,197,256,277,288,332(2),531,742

65.

N = 8000 Number of Species = 205

 

1 2 3 4 5 6 7 8 9 10

0 35 23 16 7 14 3 3 3 5 1

10 4 1 5 1 6 3 3 1 1 2
20 0 3 2 1 2 2 3 0 3 O
30 1 1 2 1 0 1 1 0 2 1
40 0 2 0 1 0 0 l 2 0 1
50 0 0 1 3 0 1 0 0 0 1

 

also at 61,64,66,67,72,75,94,95,97,108,111,116,122,123,129,149,

150,167,192,l95,204,227,284,314,325,382,393,600,860

N = 9000 Number of Species = 207

 

1 2 3 4 5 6 7 8 9 10
0 35 22 13 8 9 9 0 7 2 4
10 5 2 2 1 6 3 3 1 1 3
20 1 O 1 2 2 2 3 1 2 1
30 1 1 0 2 0 2 1 0 1 1
40 0 1 1 1 2 1 1 0 0 1
50 0 O 0 1 1 0 1 O 3 0

 

also at 61,64,65(2),68,71,72,77,81,84,99,105(2).118,120,121,
131,140,147,163,173,182,213,224,234,265,324,360,365,433,445,

676,961

66.

N = 10,000 Number of Species = 210

 

1 2 3 4 5 6 7 8 9 10

0 32 25 13 7 9 6 4 6 5 4

10 1 3 3 3 2 2 2 3 3 2
20 2 2 0 2 1 0 1 1 1 4
3O 3 O 1 0 1 4 0 0 1 0
40 0 1 1 2 0 1 0 3 1 0
50 1 1 1 0 0 0 0 0 1 0

 

also at 61,62,65(2),71(2),72(2),74(2),76,80,86,88,93,112,113,
123,130,132,138,141,154,160,191,192,199,236,244,274,293,355,391,

397,479,492,751,1093

N = 11,000 Number of Species = 215

 

1 2 3 4 5 6 7 8 9 1o
0 34 25 12 8 9 6 4 7 4 2
10 4 3 o 4 1 3 4 3 1 o
20 5 1 1 1 1 3 1 1 0 1
30 1 1 1 2 2 1 1 3 o 1
40 1 o o o 2 1 1 o o 1
50 1 1 2 o 1 1 o o 2 o

 

also at 64,66,69,7o,71,77(3),79,80,81,85(2),95,96,105,121,126,
135,145,148,154,155,173,175,207,213(2),256,261,298,316,385,440,

442,529,533,335,1215

67.

N = 12,000 Number of species = 220

 

1 2 3 4 5 6 7 8 9 10

0 37 23 12 8 7 11 2 6 5 2

10 3 4 2 2 1 2 2 6 1 1
20 2 1 1 2 2 2 2 2 2 0
3O 1 0 1 0 1 2 2 1 0 3
40 0 1 1 1 0 1 0 1 2 0
50 0 2 0 1 O 1 1 1 O 2

 

also at 62,63,67,70,73,80,81,85(2),86(2),88,89,93,94,101,105,
121,128,144,148,157,162,169,171,187,l89,223,227,230,276,279,322,

348,423,474,484,575,592,916,1327

N = 13,000 Number of Species = 221

 

1 2 3 4 5 6 7 8 9 10
o 38 20 10 11 7 9 5 1 9 2
10 1 2 5 4 1 2 2 2 2 3
20 1 1 o 1 3 1 3 1 3 2
3o 1 1 1 1 o o 1 0 3 1
40 o 1 1 o 2 2 o 1 1 o
50 1 1 o 1 1 0 o 1 2 1

 

also at 64(2),65,66,67,71,74,77,84,90,92(2),94,96,98,99,101,
102,106,114,13o,138,161,165,177,181,189,201,207,239,247,252,

305,311,342,374,452,515,518,628,646,989,1415

68.

N = 14,000 Number of Species = 224

 

1 2 3 4 5 6 7 8 9 1o

0 39 20 1o 10 8 8 5 2 8 5

10 1 2 1 3 2 3 4 1 2 3
20 1 1 2 o 1 1 1 2 2 0
3o 1 5 4 o o 1 o o 1 0
4o 0 1 3 o o 1 2 1 1 o
50 1 1 1 1 o o 1 1 1 o

 

also at 62,63(2),64,69,70,71,72(2),78,84(2),90,96,97(2),99,
103,106(2),107,109,116,123,136,148,171,179,194,195,197,2o3,221,

223,258,269(2),333,339,400,477,556,559,679,684,1063,1528

N = 15,000 Number of Species = 230

 

1 2 3 4 5 6 7 8 9 10

o 43 17 13 6 1o 9 5 4 4 5

1o 5 2 1 2 3 o 2 3 3 2
20 2 1 2 1 3 o o o 1 1
30‘ 2 2 2 3 5 o o 1 o 0
40 o 1 o 1 1 1 o o 3 2
50 o 1 1 o o 1 2 o o 1

 

 

also at 62,64,65,66,68(2),73,75(2),77,79,83,87(2),99,1o1,102,
103,106,108,111,115,117,119,123,133,144,164,184,190,205,206,213,

216,235,238,275,287,294,356,369,399,43o,508,588,602,728,741,1134,

I

1647

69.

N = 15,609 Number of species = 233

 

1 2 3 4 5 6 7 8 9 10
0 45 18 12 6 10 7 4 8 3 4
10 6 3 1 2 1 1 2 3 2 2
20 3 0 1 1 ' 2 2 2 0 1 0
30 2 1 3 1 4 1 3 0 1 0
40 0 0 1 0 2 0 0 0 0 1
50 4 0 2 O 0 1 1 1 1 0

 

also at 61,62,67,68,70(2),71,75,77,79,80(2),89,92(2), 102,104,
105,106,107,113,115,119,121,125(2),138,152,168,192,196,208,218,
223(2),248,249,286,301,305,375,384,410,451,531,616,630,762,768,
1186,1711.

Chapter 6
Section 1: Investigation of Species per Genus Data

In an effort to determine the different types of environments for
which the model holds data from Williams[6] on Orth0ptera was investigated.
It is realized that the data is in the form of Species per genus which
is quite a different concept from the individuals per Species data that
had previously been considered but this data seemed to Show some of the
same properties as the other data and it was heped that this biological
Situation could also be explained by the model. Applying therefore the

methods of the previous chapters the maximum likelihood estimate of the

parameters was

A

a=1. 1056 A=2 31.065 ks=16. 3

In comparing the actual data , reproduced in table 6.1, to the
theoretical expected values obtained using the above estimates of the
parameters it was determined that the model fit rather well for the
small and moderate genera but that the theoretical values for the
larger genera were too small. This conclusion was reinforced when
three samples of 4112 species were taken using a=1.1056 and A=231.065
and it was found that the largest genus among the three samples con-
tained only 80 Species, far below the number that was actually
encountered.

With this result in mind it was decided that an adequate fit
might be obtained if the form of the function f(x) was altered to
accommodate this new situation . It was decided that the term e"'x in
the numerator made the function f(x) decrease too rapidly. For large
intensities it was decided to try the form f(x)= "$5 where q is a

parameter with the restriction ZSq<m.
70.

71.

In determining q for the Orthoptera data make the definition

SPCEa,b) = total number of Species in the genera which have ni Species
with a s n1'< b..Adjusting so that k8 = 1 the eXpected number of Species
is defined by the equation.r: x f(x) dx for the interval [a,b).

Using SPCEa,b) as an estimate of the number of Species in the

sample which are in genera having an intensity in the interval [a,b)

consider the following equations

C C
I x f(x) dx =f 11 x1‘9 dx = SPc[30,c)
30 30

Q Q
I x f(x) dx = IA xl-q dx = SPCCC:°°)
c c

for 50 < c < 100
and q >2

Integrating and eliminating A from the two above equations the
solution for q is seen to be

16g SPCI: 30,1») - log SPC[c,co)
q = 2- log 30 - log c

From the graph of q as a function of c for 50 <c< 100 a good
choice for q in this case seems to be q = 3. A130 for small genera
the function f(x) appears to take the general form similar to f(flwg—

so that the expected number of genera with m species is

 

C» Q
f xm e'x f(x) dx =f A xm‘l e"x dx = A Pgmz = _A
0 m1 0 m1 m! m

Combining these two characteristics it was decided that the

function f(x) should take the form f(x) = '-T;$;$z where A and a

x
are positive constants. An attempt at finding a maximum.likelihood
estimate of these parameters became very messy so that an estimate

was obtained from a simultaneous solution of the equations

72.

co 00
A
xfx dx= dx=SPC30w=904
I... H I... W [1)
a: Q

x e'x

—-—- f(x) dx =I -—-—2—A e-x dx = 320
1! 0 (7&3) .

ID

The estimates obtained were
; = 10 A = 37,000

The simulation procedure used in the case where f(x) = Fifi-5?-
is quite similar to the procedure deve10ped in Chapter 4 Section 1
except that some changes are needed in finding the intensities due to
the different form of the function f(x). As before let Ek, k=1,2,3,...
be an infinite supply of exponential random variables and consider the
following procedure for producing a sample of size N with known

parameters a and A.

1. Set k=1, set i=1, set x: = + co

 

 

 

x: x3}: x}? A ‘
A — — C x >2
2. Ix]: f(x) dx—IXimdx- Ix]: x3 x+a dx
5!:
xi A .
3. Set Ek = I :{3 dx and SOIVIng for Xi
xi
.. A 1
xi " 2 ‘12—"
\ >3 E3
l j=1
X’ 2
4. Accept xi with probability (Ii—2" .
5. If xi is rejected, set x: = xi, increase k by one, and return
to step #3 provided Xi > _a___ If xi is accepted, increase k by

/2-1.

one, set x241 == xi, increase 1 by one and return to step #3

. a
provided Xi > 2 _ 1 ,

is the point where the acceptance

 

The point xi=

f2-1

73.

probability< :5?) is equal to one half so that it becomes
desirable to modify the procedure at this point to increase the

effeciency.

6. Set k1= k. Also
2" 9.‘

x1 x1 A :>2 d
Ix]: f(X) dx =1 *fx1 x(x+a)2dx =f:* 2x _3{_x_ x+a x
xi A 1
7. Set Ek='fx1Ԥ -3dx and solving for xi

 

1
1i-[231r‘=\| Krl
\)22 Ej+ 2 E.
PM 3‘1 J

 

8. Accept xi

with probability (ii)?

9. If xi is rejected, set xii: = xi, increase k by one, and
return to step #7 provided xi > a. If xi is accepted, set x§+1= xi,
increase k by one, increase i by one and return to step #7 provided
xi > a,

At the point x = a another modification is to be made to increase

the effeciency.

10. Set k2=k, set le= x?

 

x* X? x? A a 2

i 1 A — < >dx
f f(x) dx=f , mde-fx. 2:52 x+a

xi x* X1

11. Set Ek= f:i —-zdx and solving for Xi

1 , -gZL £131.13]

x1=x~N1e
2

12. Accept xi with probability (xi-+3 > '

 

. > .
k by one, increase i by one and return to step #11 provided x:L es

74.

The constant e is determined similar to the procedure used
s
before.
The expected sample Size is

Q m Q
A 1 k A
g sx (X) x g ksx x(x+a)2dx ksA g (;:332 dx a .

Setting this equal to N to obtain an estimate for kS

* aN
ks - 7r -
S t N* ‘ *
e = = =
14. 1. and k4 k, Xe xi.

For the small intensities the modification which skips over

some of the genera which do not appear in the sample is again employed.
* * x*
X.

 

x145 lA xikSA
f ksx f(x)dx=f ksx —--—2 =f -;S-2- (a x+a >2dx.
xi X. “(3* )2d xi
x* .
i kSA
16. Set Ek = I -3- dx and solve for X1
x, a
1
2 k

a

x. = x. - -——- g E,.
1 8 RSA j'k4

 

:3. If x. is
1

and return to step #16.

. a
17, If xi > 0, accept x],L with probabil1ty< x +a

*
rejected increase k by one, set xi - xi

* - -
. . — x, increase 1 b
If x1.- 18 accepted,1ncrease k by one, set xi+1 1’ y

one and return to step #16.

If xi 5 Q reject x, and cease finding intensities.
1

In finding the number of Species in each genus for a particular

sample employ the procedure described in Chapter 4.

Using the above prodecure with A = 37,000 and a = 320 three

samPles of 4112 genera were taken and the results shown in table 6.2

These results can be compared to the original data in table 6.1 to

examine the fit of the model in this case.

75.

Table 6.1
ORTHOPTERA OF WORLD
Journal of Ecology Volume 32 page 18

Distribution of genera according to number of Species present

 

 

1 2 3 4 5 6 7 8 9 10
0 320 131 86 61 41 27 21 18 23 17
10 12 8 9 3 5 4 3 6
20 1 l 2 1 0 2 0 0 4 0

also at 31(2),34,35,36,38,41,43,51,54,58,72,75,103,202.
TOTAL GENERA 826

TOTAL SPECIES 4112

Table 6.2

S IMULATED TEST 1

Distribution of genera according to number of Species present

 

1 2 3 4 5 6 7 8 9 10
0 317 134 86 49 37 24 26 22 7 12
10 10 8 8 7 1 5 3 3 0
20 1 5 4 1 5 2 1 1

 

also at 32,34,35,36,44(2),49,53,54,55,56(2),57,69,73,79,83,178.
TOTAL GENERA 798

TOTAL SPECIES 4112

76.

Table 6.2
SIMULATED TEST 2

Distribution of genera according to number of Species present

 

1 2 3 4 5 6 7 8 9 10

0 324 146 90 54 43 24 17 23 15 5

10 7 9 7 8 8 8 5 3 6 l
20 3 2 3 2 2 3 1 3 0 1

 

also at 32,34,35,36,39(2),41,45,49,52,53,74,79,153.
TOTAL GENERA 837

TOTAL SPECIES 4112

SIMULATED TEST 3

Distribution of genera according to number of Species present

 

1 2 3 4 5 6 7 8 9 10
0 317 141 83 48 41 25 21 9 15 7
10 7 9 7 4 5 2 4 4 4 5
20 3 5 1 3 2 0 2

 

also at 33,37,42,46,48,50(2),56,57(2),217,354.
TOTAL GENERA 790

TOTAL SPECIES 4112

APPENDIX

Using the theory developed in the previous chapters, FORTRAN 60
programs have been developed to perform the indicated operations on
the Control Data 3600 Computer.

Program SPECIES l finds the maximum likelihood estimates of
the parameters of the model using the methods discribed in Chapter
2 Section 1.

Program SPECIES 2 finds the maximum likelihood estimates of
the parameters of the model under the Special condition a = 1
using the methods discribed in Chapter 2 Section 1.

Program SPECIES 3 is a simulation program to obtain a sample
of size N from the model in the case a + 1 using the methods

developed in Chapter 4 Section 1.

When using the program to obtain a sample it was found that
about 5000 individuals could be sampled in about 30 seconds on
the CDC 3600 computer. Also note that if the sample size is

doubled the estimated Simulation time increases only a few seconds

due to the fact that a large percentage of the simulation time is

used to obtain the intensities of the Species and the number of

new Species decreases rapidly with increa81ng sample Size.

Program.SPECIES 4 is a simulation program to obtain a sample

of size N from the model in the case a = 1 using the procedure
described in Chapter 4 Section 2.

This program obtains a sample of 15,000 individuals in about

20 seconds on the CDC 3600 computer. Note that this procedure is

much faster than SPECIES 3. This is explained by the fact that

77.

78.

the simulation procedure is extremely simplified in the case
where a = l.

The four above mentioned programs are tabulated in the follow-
ing pages with a brief explination to the right of the tabulated
programs. Although these were not the only programs used in this

investigation, they were the ones used to obtain the primary results.

79.

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BIBLIOGRAPHY

BIERENS DEHAAN , DAVID
Nouvelles Tables D'Intégrales Defines G. E. Stechert & Co.
New York 1939.

CARATHEODORY, 0.
Theory of Functions, Volume I, Part Five, Chelsea,
New York 1958.

DAVIS, HAROLD T.
Tables of the Higher Mathematical Functions, Volume I,
Principia Press Bloomington, Indiana 1933.

FISHER, R. A., A.S. CORBET and C. B. WILLIAMS
The Relation Between the Number of Species and the Number
of Individuals in a Random Sample of an Animal Population.
Journal of Animal Ecology, Volume 12(1943) pp.42-58.

RUBIN, HERMAN
Construction of Random Variables with Specified Distributions,

Research Memorandum #88, Department of Statistics, Michigan

State University, 1961.

WILLIAMS, C. B.
Some Application of the Logarithmic Series and the Index

of Diversity to Ecological Problems. Journal of Ecology,
Volume 32(1944) pp.1-44.

116.

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