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https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | Series ISSN 1939-5221
Generating Functions in Engineering and the Applied Sciences
Rajan Chattamvelli, VIT University, Vellore, Tamil Nadu
Ramalingam Shanmugam, Texas State University
This is an introductory book on generating functions (GFs) and their applications. It discusses
commonly encountered ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | apply the lessons learned in
their respective disciplines. The purpose is to give a broad exposure to commonly used techniques of
combinatorial mathematics, highlighting applications in a variety of disciplines.
ABOUT SYNTHESIS
This volume is a printed version of a work that appears in the Synthesi... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | velli and Ramalingam Shanmugam
2019
Transformative Teaching: A Collection of Stories of Engineering Faculty’s Pedagogical
Journeys
Nadia Kellam, Brooke Coley, and Audrey Boklage
2019
Ancient Hindu Science: Its Transmission and Impact of World Cultures
Alok Kumar
2019
Value Relational Engineering
Shuichi Fukuda
2018
... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | 13
The Making of Green Engineers: Sustainable Development and the Hybrid Imagination
Andrew Jamison
2013
Crafting Your Research Future: A Guide to Successful Master’s and Ph.D. Degrees in
Science & Engineering
Charles X. Ling and Qiang Yang
2012
iv
Fundamentals of Engineering Economics and Decision Analysis
David ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | ids, Part Two: Transport Properties of Solids
Richard F. Tinder
2007
Tensor Properties of Solids, Part One: Equilibrium Tensor Properties of Solids
Richard F. Tinder
2007
Essentials of Applied Mathematics for Scientists and Engineers
Robert G. Watts
2007
Project Management for Engineering Design
Charles Lessard and ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | sciences, such as or-
dinary generating functions (OGF), exponential generating functions (EGF), probability gen-
erating functions (PGF), etc. Some new GFs like Pochhammer generating functions for both
rising and falling factorials are introduced in Chapter 2. Two novel GFs called “mean deviation
generating function”... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | , Pochhammer generating functions, poly-
mer chemistry, power series, recurrence relations, reliability engineering, special
numbers, statistics, strided sequences, survival function, truncated distributions.
Contents
ix
List of Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.4 Ordinary Generating Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.4.1 Recurrence Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.4.2 Types of Sequences . . . . . . . .... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | .7.1 Auto-Covariance Generating Function . . . . . . . . . . . . . . . . . . . . . . . . 16
Information Generating Function (IGF) . . . . . . . . . . . . . . . . . . . . . . . 16
1.7.2
1.7.3 Generating Functions in Graph Theory . . . . . . . . . . . . . . . . . . . . . . . . 16
1.7.4 Generating Functions in Number Theo... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | 1.2 Addition and Subtraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
x
2.1.3 Multiplication by Non-Zero Constant . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.1.4 Linear Combination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
Shifting . .... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | of Generating Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
2.4
3
Generating Functions in Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... |
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3.6 MD of Some Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
3.6.1 MD of Geometric Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
3.6.2 MD of Binomial Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
3.6.3 MD o... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | . . . 54
3.10 Factorial Moment Generating Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
3.11 Conditional Moment Generating Functions (CMGF) . . . . . . . . . . . . . . . . . . 58
3.12 Generating Functions of Truncated Distributions . . . . . . . . . . . . . . . . . . . . . . 58
3.13 Convergenc... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | . . . . . . . . . . . . . . . . . . . . . 64
Applications in Computing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
4.2.1 Merge-Sort Algorithm Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
4.2.2 Quick-Sort Algorithm Analysis . . . . . . . . . . . . . . . ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | . . . . . . . . . . . . . . . . . . . . . . . . . 71
4.3.4 Towers of Hanoi Puzzle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
Applications in Graph Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
4.4.1 Graph Enumeration . . . . . . . . . . . . . . . ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | Applications in Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
Sums of IID Random Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
4.8.1
4.8.2
Infinite Divisibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | . . . . . . . . . . . . . . . . 86
4.12 Applications in Management . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
4.12.1 Annuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
4.6
4.7
4.8
4.9.1
4.5
xii
4.13 Applications in Ec... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | .1
Some standard generating functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
2.1 Convolution as diagonal sum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.2
2.3
Summary of convolutions and powers . . . . . . . . . . . . . . . . . . . . . . . . . .... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | . . . . . . . 53
Summary table of zero-truncated generating functions . . . . . . . . . . . . . . . . . . 59
C H A P T E R 1
1
Types of Generating Functions
After finishing the chapter, readers will be able to
(cid:1) (cid:1) (cid:1)
• explain the meaning and uses of generating functions.
• describe the ordin... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | GENERATING FUNCTIONS
Definition 1.1 (Definition of Generating Function) A GF is a simple and concise expression
in one or more dummy variables that captures the coefficients of a finite or infinite series, and
generates a quantity of interest using calculus or algebraic operations, or simple substitutions.
GFs are used in... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | f .0/
C
D
C
1.1.2 EXISTENCE OF GENERATING FUNCTIONS
GFs are much easier to work with than a whole lot of numbers or symbols. This does not mean
that any sequence can be converted into a GF. For instance, a random sequence may not have a
nice and compact GF. Nevertheless, if the sequence follows a mathematical rule... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | range is (cid:140)
t
1(cid:141). As the input can be complex numbers in some signal processing applications, the
GF can also be complex. A formal power series implies that nothing is assumed on the conver-
gence. As shown below, the sequence 1; x; x2; x3; : : : has GF F(x) = 1=.1
x/ which converges
x2/,
in the region ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book |
1.2 NOTATIONS AND NOMENCLATURES
A finite sequence will be denoted by placing square-brackets around them, assuming that each
element is distinguishable. An English alphabet will be used to identify it when necessary. An
infinite series will be denoted by simple brackets, and named using Greek alphabet. Thus, S
D
.a0; ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | C
1/
D
(cid:3) (cid:1) (cid:1) (cid:1) (cid:3)
.x
j
C
(cid:0)
1/
D
j
1
(cid:0)
Y
.x
0
k
D
k/
C
D
(cid:128).x
j /
C
(cid:128).x/
:
(1.1)
2. Falling Factorial Notation
In the falling factorial, a variable is decremented successively at each iteration. This is
denoted as
x.j /
.x
x
(cid:3)
(cid:0... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | the dummy variable. For example,
G.x; t/ denotes an OGF of a random variable X. Some authors use x as a subscript as Gx.t/.
These dummy variables assume special values (like 0 or 1) to generate a quantity of interest.
Thus, uniform and absolute convergence at these points are assumed. The GFs used in statistics
can be... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | derived easily. This reasoning holds for other GFs too.
C
X
(cid:3)
1.4 ORDINARY GENERATING FUNCTIONS
Let (cid:27)
D
.a0; a1; a2; a3; : : : an; : : :/ be a sequence of bounded numbers. Then the power series
f .x/
D
anxn
1
X
0
n
D
(1.3)
is called the Ordinary Generating Function (OGF) of (cid:27). Here x ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book |
1
x/(cid:0)
1:
x/(cid:0)
1; 1;
.1;
(cid:0)
(cid:0)
6
(cid:0)
1/=.x
(cid:0)
1, and odd coefficients a2n
1. TYPES OF GENERATING FUNCTIONS
The OGF is .xn
1/=.x
1/ when there are a finite number of 1’s (say n of them). Thus,
(cid:0)
.1; 1; 1; 1; 1/ has OGF .x5
1 when even coefficients
x2/(cid:0)
0. The OGF is called ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book |
C (cid:1) (cid:1) (cid:1)
.1; 2; 3; 4; 5; : : :/. This has OGF 1
2x
.1; 2; 3; 4; : : : ; n; : : :/
(cid:148)
.1
.1
2
(cid:0)
x/(cid:0)
2
x/(cid:0)
1=.1
1=.1
(cid:0)
x/2
x/2:
C
D
D
(cid:148)
C
2; 3;
.1;
(cid:0)
(cid:0)
4; : : : ; n;
.n
(cid:0)
C
1/; : : :/
See Table 1.1 for more examples.
... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | =0k=0k=0k=0k=0k=0k=0 1.4. ORDINARY GENERATING FUNCTIONS 7
1.4.1 RECURRENCE RELATIONS
There are in general two ways to represent the nth term of a sequence: (i) as a function of n, and
(ii) as a recurrence relation between nth term and one or more of the previous terms. For example,
an
.1
and Jones, 1996].
1=n gives a... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | :0)
n
k
n
1.4.2 TYPES OF SEQUENCES
(cid:0)
There are many types of sequences encountered in engineering. Examples are arithmetic se-
quence, geometric sequence, exponential sequence, Harmonic sequence, etc. Alternating se-
quence is one in which the sign of the coefficients alter between plus and minus. This is of
a... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | 141). It is called a geometric progres-
sion in mathematics and geometric sequence in engineering. It also appears in various physical,
natural, and life sciences. As examples, the equations of state of gaseous mixtures (like van der
Vaals equation, Virial equation), mean-energy modeling of oscillators used in crystall... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | 1 Consider the infinite series expansion of .1
C
, which is the OGF of the given sequence. Here “a” is any nonzero constant.
D
C
(cid:0)
1
ax
1
ax/(cid:0)
a2x2
a3x3
(cid:0)
C
(cid:1) (cid:1) (cid:1)
Example 1.2 (Find the OGF of an) Find the OGF of (i) an
1=3n.
D
2n
C
3n and (ii) an
2n
C
D
Solution ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | 0)
t/2 as the required OGF.
(cid:0)
0 t n. The first sum is 2t=.1
t/2 and second one is 1=.1
1/. Write G.t /
t/. Take .1
P1n
D
C
D
C
(cid:0)
(cid:0)
(cid:0)
C
D
D
D
Example 1.4 (OGF of perfect squares) Find the OGF for all positive perfect
.1; 4; 9; 16; 25; : : :/.
squares
Solution 1.4 Consider the expres... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | =x
2=.1
x/
(cid:0)
D
.1
C
C
x/=(cid:140)x.1
x/(cid:141)
(cid:0)
(1.7)
from which @f
C
perfect squares of all positive integers as x.1
as follows:
@x D
x/=.1
.1
(cid:0)
x/3. Multiply this expression by x to get the desired GF of
x/3. This can be represented symbolically
x/=.1
C
(cid:0)
.1; 1; 1; 1; ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | , 28, 82,: : :).
Solution 1.5 Suppose we subtract 1 from each number in the sequence. Then we get (1, 3, 9,
27, 81, 243, : : :), which are powers of 3. Hence, above sequence is the sum of .1; 1; 1; 1; : : :/
.1; 3; 32; 33; 34; : : :/ with respective OGFs 1=.1
3x/. Add them to get 1=.1
x/
3x/ as the answer.
x/ and 1=... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | a1x
x2
x
x/(cid:0)
C
C
2Note that the nth derivative of 1=.1
C (cid:1) (cid:1) (cid:1) C
C (cid:1) (cid:1) (cid:1)
(cid:0)
x/n
C
x/ is n(cid:138)=.1
(cid:0)
(cid:0)
1 and that of 1=.1
ax/b is .n
b
(cid:0)
C
(cid:0)
C
C
D
,
C (cid:1) (cid:1) (cid:1)
1/(cid:138)an=(cid:140).b
(cid:0)
1/(cid:138).1
... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | 1
A
aj
xk
a0
C
D
.a0
C
a1/x
.a0
a1
C
C
C
a2/x2
:
C (cid:1) (cid:1) (cid:1)
(1.10)
(cid:3)
This result has great implications. Computing the sum of several terms of a sequence
appears in several fields of applied sciences. If the GF of such a sequence has closed form, it
is a simple matter of multiplyi... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | C
(cid:0)
(cid:0)
1 k2.
x/,
D
C
(cid:0)
Example 1.6 (OGF of harmonic series) Find the OGF of 1
1=2
1=3
C
C
C (cid:1) (cid:1) (cid:1) C
1=n.
Solution 1.6 We know that
x2=2
x/
(cid:0)
1 xk=k. We also know that when an OGF is multiplied by 1=.1
x3=3
C
C
x/, we get a new
(cid:1) (cid:1) (cid:1) D
OGF in whi... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | is the OGF of .1=2; 1
3 .1
C
2 /; 1
4 .1
1
2 C
1
3 /;
(cid:1) (cid:1) (cid:1)
/.
C
C (cid:1) (cid:1) (cid:1)
(cid:0)
1.5. EXPONENTIAL GENERATING FUNCTIONS (EGF)
11
m
k
1/k(cid:0)
A direct application of the above result is in proving the combinatorial
0.
P
According to the above theorem, the sum of th... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book |
t/ has coefficients with negative sign
0 a2kt 2k, and G.t/
denotes the greatest integer
1. If F .t/
1t 2k
C
m=2
t// is the GF of Pb
c
k
t//=2 is the GF of P1k
D
k(cid:1)F .t/k is also a GF.
x
b
c
F .t//n
0 a2kt 2k.
1. Here
P1k
D
(cid:0)
1t 2k
m
P
k
D
(cid:20)
0 a2k
t//=2
D
F .
F .
0 (cid:0)
P
D
D
(ci... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | D
gives an OGF. But it is the usual practice to write it either as (1.12) or as
H.x/
D
1
X
0
n
D
anenxn where en
1=n(cid:138)
D
(1.13)
(1.14)
where we have two independent multipliers. As the coefficients of exponential function .exp.x//
are reciprocals of factorials of natural numbers, they are most suited wh... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | GF may exist for a sequence. As an example, consider the sequence
1; 2; 4; 8; 16; 32; : : : with nth term an
2x/, whereas the EGF is e2x.
Thus, the EGF may converge in lot many problems where the OGF either is not convergent or
k(cid:1)xk. We could write this as an EGF by expanding
does not exist. Consider .1
n
(cid:0)... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | x/ so that P1n
D
xn=n(cid:138)
D
(cid:2)
1=.1
1=.1
0 n(cid:138)
0 1
xn
(cid:0)
(cid:0)
(cid:0)
(cid:0)
(cid:2)
Now consider
(cid:16)ebt
(cid:0)
eat (cid:17) =.b
a/
(cid:0)
D
t=1(cid:138)
.b2
a2/=.b
a/t 2=2(cid:138)
C
.bn
(cid:0)
an/=.b
(cid:0)
a/t n=n(cid:138)
.b3
(cid:0)
a3/=.b
(cid:0)
a/t... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | 2)
(cid:11)/
(cid:0)
D
F0
C
F1t=1(cid:138)
C
F2t 2=2(cid:138)
C
F3t 3=3(cid:138)
C (cid:1) (cid:1) (cid:1) C
Fnt n=n(cid:138)
C (cid:1) (cid:1) (cid:1)
;
(1.19)
where Fn is the nth Fibonacci number.
Another reason for the popularity of EGF is that they have interesting product and com-
position formula... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | .t. x of both sides. As a0 is a constant, its derivative is zero. Use
xn
derivative of xn
(cid:0)
(cid:3)
C (cid:1) (cid:1) (cid:1) C
akxk
1=.k
. Differentiate again (this time a1 being a constant vanishes) to
1/(cid:138)
(cid:0)
(cid:0)
get H 00.x/
a3x=1(cid:138)
pro-
cess k times. All terms whose coefficients are below ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | uishable balls into m distinguishable urns (or boxes) in such a way that none of the urns is
empty.
Solution 1.8 Let un;m denote the total number of ways. Assume that i th urn has ni balls so
n where each ni >
that all urn contents should add up to n. That gives n1
C (cid:1) (cid:1) (cid:1) C
x3=3(cid:138)
0. Let x=1(... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | )
/m;
(1.20)
where ak
each urn gets one ball), and upper limit is n. The RHS is easily seen to be .ex
using binomial theorem to get
uk;m and the lower limit is obviously m (because this corresponds to the case where
1/m. Expand it
D
(cid:0)
.ex
1/m
(cid:0)
D
emx
(cid:0)
!
m
1
e.m
(cid:0)
1/x
C
!
m
2... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | )
n
k(cid:1) is the classical Laguerre
(cid:0)
1.6
POCHHAMMER GENERATING FUNCTIONS
Analogous to the formal power series expansions given above, we could also define Pochhammer
GF as follows. Here the series remains the same but the dummy variable is either rising factorial
or falling factorial type.
1.6.1 RISING P... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | ; k/xn=n(cid:138)
ond kind is P1n
D
constants to get falling EGF for Pochhammer numbers b.k/.
xn. The
0 S.n; k/x.k/
x.n/. The EGF of Stirling number
x//k and that of Stirling number of sec-
1/k=k(cid:138). The falling factorial can also be applied to the
n
k
D
.1=k(cid:138)/.log.1
k s.n; k/xn=n(cid:138)
0 s.n; k/xk... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | /.x
x.x
1 neighbors to it so that
1 choices. Continue arguing like this until a single node is left. Hence, the GF is
2/ : : : .x
1/ or x.n/ as the required GF.
(cid:0)
n
(cid:0)
(cid:0)
(cid:0)
(cid:0)
C
16
1. TYPES OF GENERATING FUNCTIONS
1.7 OTHER GENERATING FUNCTIONS
There are many other GFs used in va... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | GF.
n
aj
j where
D
a
j
j
Solution 1.11 Split
n
as P(cid:0)
n
1=.1
a/
nt (cid:0)
a(cid:0)
at/.
1
D(cid:0)1
(cid:0)
C
C
the range from (
P1n
D
(cid:0)1
0 ant n. This simplifies to .a=t/.1
(cid:0)
to
1) and (0 to
a=t/(cid:0)
(cid:0)
)
1
1
C
to get
1=.1
at/
the OGF
a=.t
(cid:0)
D
(cid:0)
1.7.2
INFO... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | mathematics that deals with relationships (adjacency and
incidence) among a set of finite number of elements. As a graph with n vertices is a subset of
1.7. OTHER GENERATING FUNCTIONS 17
n
the set of (cid:0)
2(cid:1) pairs of vertices there exist a total number of 2.n
2/ graphs on n vertices. Hence, the
2/
.n
0 Gn... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book |
(cid:0)
(cid:0)
(cid:0)
1.7.5 ROOK POLYNOMIAL GENERATING FUNCTION
(cid:2)
Consider an n
n chessboard. A rook can be placed in any cell such that no two rooks appear in
any row or column. Let rk denote the number of ways to place k rooks on a chessboard. Obvi-
n(cid:138). The rook polynomial
ously r0
GF can now be... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | k
(cid:0)
C
(cid:0)
(cid:0)
D
1.7.6 STIRLING NUMBERS OF SECOND KIND
There are two types of Stirling numbers—the first kind denoted by s.n; k/ and second kind
denoted by S.n; k/. Consider a set S with n elements. We wish to partition it into k.< n/
nonempty subsets. The S.n; k/ denotes the number of partitions of a... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | like auto-covariance GFs, information GFs,
and those encountered in number theory and graph theory are briefly described. Multiplying
x/ results in the OGF of the partial sums of coefficients. This result is used
an OGF by 1=.1
extensively in Chapter 4.
(cid:0)
C H A P T E R 2
19
Operations on Generating
Functions
... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | fficient of the power series are represented as sets. For ex-
ample, if (a0; a1; : : :) and (b0; b1; : : :) are, respectively, the coefficients of A.t/ and B.t /, then
(a0
B.t/; while (c0; c1; : : :) are the coefficients of A.t/
etc., cn
(cid:0)
a1b0,
anb0. Two GFs, say F .t/ and G.t/ are exactly identical when
b1; : : :) ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | .1 EXTRACTING COEFFICIENTS
Quite often, the GF approach allows us to find a compact representation for a sequence at hand,
and to extract the terms of a sequence from its GF using simple techniques (power series expan-
sion, differentiation, etc.). Thus it is a two-way tool. Coefficients can often be extracted easily
from... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | is a valid OGF if
each of them is an OGF. This is a special case of the linear combination discussed below. Some
authors denote this compactly as .F
H /.t/. This has applications in several fields like
statistics where we work with sums of independently and identically distributed (IID) random
variables.
H.t/
G.t/
G... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | ating to zero
0 bkt k, or equivalently P1k
F .t/
P1k
D
D
shows that this is true only when each of the a0ks and b0ks are equal, proving the result. The same
argument holds for EGF, Pochhammer GF, and other GFs.
0 akt k
P1k
D
D C
(cid:0)
P1k
D
0.cak
0.ak
G.t/
D
C
(cid:21)
(cid:0)
(cid:0)
Example 2.1 (OGF of ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | sequence (a0; a1; : : :), shifting it right by one position means to move everything
to the right by one unit and fill the vacant slot on the left by a zero. This results in the se-
quence (0; a0; a1; : : :), which corresponds to b0
1. Similarly, shifting
left means to move everything to the left by one position and di... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | m
1(cid:1) =t m
(cid:0)
am
am
1t
C
C
C
am
C
D
2t 2
C (cid:1) (cid:1) (cid:1)
(2.6)
which is a shift-left by m positions. Shifting is applicable for both OGF and EGF, and will be
used in the following paragraphs.
Example 2.2 Let G.t/ be the OGF of a sequence (a0; a1; : : :). What is the sequence whose
O... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | cid:140).1
(cid:0)
2t
t=.1
t/.1
(cid:0)
(cid:0)
D (cid:0)
(cid:0)
(cid:0)
C
1, which simplifies to .1
(cid:0)
Example 2.4 Find the OGF and EGF of the sequence .12; 32; 52; 72; : : :/.
(cid:21)
Solution 2.4 Let F .t/ denote the OGF of the given sequence. The kth term is obviously .2k
1/2 for n
0. Hence, F .t/
... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | P1k
D
C
0 k2t k=k(cid:138)
C
k and split the first term into two
D
H.t/
D
4t 2 1
X
2
k
D
t k
2=.k
(cid:0)
2/(cid:138)
(cid:0)
C
8t
t k
(cid:0)
1=.k
1/(cid:138)
(cid:0)
C
1
X
1
k
D
1
X
0
k
D
t k=k(cid:138):
(2.7)
This simplifies to H.t/
.4t 2
8t
C
C
1/et .
D
2.1.6 FUNCTIONS OF DUMMY VARI... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book |
a8t 8
(cid:0)
C (cid:1) (cid:1) (cid:1)
/ in terms of F .t/.
a4t 4
C
C
Solution 2.5 As the first sequence contains powers of t 4, this has OGFF .t 4/. The second se-
quence has alternating signs. As the imaginary constant i
1, the second sequence can be
generated using F ..it/2/ because i 2
p
1, and so on.
1;... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | )
t=2/2.
Putting s
is .1
2. Thus, the OGF of original sequence
t=2 shows that this is the OGF of .1
2 or 1=.1
D
t=2/(cid:0)
Another application of the change of dummy variables is in evaluating sums involv-
n
1
1(cid:1)ak. This sum is the coefficient of
ing binomial coefficients. Consider the sum P
k
t n in the power ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | 3
5
4
(cid:3)
C
(cid:3)
C
(cid:3)
D
k
X
0
j
D
1
X
0
k
akt k
(cid:3)
1
X
0
k
bkt k
D
1
X
0
k
ckt k where ck
D
aj bk
j :
(cid:0)
(2.10)
D
D
D
k aj bi . As the order of
This is called the convolution, which can also be written as ck
k
j bj . The coefficients of a convolution
the OGF can be exchange... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | 1; 2; 3; 4; : : :/ and .1; 2; 4; 8; 16; : : :/.
Solution 2.8 The first OGF is obviously 1=.1
volution is the product of the OGF 1=(cid:140).1
(cid:0)
t/2.1
(cid:0)
t/2 and second one is 1=.1
2t/. The con-
2t/(cid:141). Use partial fractions to break this as
(cid:0)
(cid:0)
b0b1xb2x2b3x3b4x4…a0a0b0a0b1xa0b2x2a0b3... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book |
D (cid:0)
(cid:0)
(cid:0)
C
0. Multiply the first equation by 2 and subtract from the second one to get
C
C
C
1. Thus, the OGF of the product is 4=.1
2t /
(cid:0)
(cid:0)
2=.1
(cid:0)
2.1. BASIC OPERATIONS 25
2t / as a common denominator
1.
B.1
t /2
2t /
C
(cid:0)
B
C.1
(cid:0)
1; 3A
C
D
D
2B
Now c... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | 40)G.t /
an OGF are repeated convolutions. Thus, if F .t /
k
P
P1k
j
D
0 ckt k where ck
0 aj ak
D
D
(cid:0)
G.t /F .t / and
H.t/F .t /G.t /, etc. Similarly, product distributes over addi-
F .t /G.t /
F .t /H.t /. Powers of
D
0 akt k then F .t /
F .t /2
H.t /(cid:141)
F .t /
(cid:6)
(cid:6)
D
j , which can a... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | 0 Tnxn.
is varying from 0 to n
Multiply both sides of (2.15) by xn and sum over n
1, it is not an exact power as shown below. Let T .x/
P1n
D
to get
1 to
D
(cid:0)
(cid:0)
D
1
Tnxn
D
1
X
1
n
D
1
X
1
n
D
n
1
(cid:0)
X
0
k
D
TkTn
1
(cid:0)
(cid:0)
kxn for n
1;
(cid:21)
(2.16)
26
2. OPERATIONS ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | .
C1Cn
1; Cn
Cn
D
2
1
(cid:0)
C
(cid:0)
C (cid:1) (cid:1) (cid:1) C
(cid:0)
Solution 2.9 Let F .t / denote the OGF of Catalan numbers. Write the convolution as a sum
Cn
j . As per Table 2.2 entry row 2, this is the coefficient of a power. As done
1
(cid:0)
above, multiply by t n and sum over 0–
1
0 Cj Cn
to... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | bjtj k ajbk–jOGF Power∞ aiti2k ajak–jOGF Product (of 3)∞ ahth ∗∞ biti ∗∞ cjtj h+i+j=k, k≥0 ahbicj EGF Product∞ aiti /i! ∗∞ bjtj /j!k kajbk–jEGF Power∞ aiti /i!2k kajak–jEGF Product (of 3)∞ ah th ∗∞ bi ti ∗∞ cj tjh+i+j=k ahbicj k! i=0i=0h=0h=0i=0i=0i=0i=0j=0j=0j=0j=0j=0j=0j=0 jj=0 jh!i!j!h!i!j! Hence, we t... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book |
(cid:0)
1
(cid:0)
1
X
1
k
D
Replace t by
(cid:0)
4t and subtract from 1 to get
p1
1
(cid:0)
4t
(cid:0)
D (cid:0)
21
(cid:0)
2k.
1/k
(cid:0)
1 2k
(cid:0)
k
!
2
(cid:0)
1
(cid:0)
1
X
1
k
D
1/k22kt k
=k.
(cid:0)
2k
2
k
!
2
(cid:0)
1
(cid:0)
1
X
1
k
D
D
=k t k
(2.18)
(2.19)
1.
1/k
as .
(ci... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | cid:1) (as can be seen from Pascal’s triangle) this difference is always positive.
Moreover, both of them are integers showing that Catalan numbers are always integers. The
first few of them are 1; 1; 2; 5; 14; 42; 132; 429; : : :.
1(cid:1). As (cid:0)
2n
(cid:0)
n
1/(cid:141) (cid:0)
2n
n (cid:1)
2n
n
D
(cid:0)
... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | F .t /
a0
C
D
a1t
C
a2t 2
C (cid:1) (cid:1) (cid:1) D
Differentiate w.r.t. t to get
akt k:
1
X
0
k
D
(2.21)
.@=@t /F .t/
a1
C
D
2a2t
C
3a3t 2
C (cid:1) (cid:1) (cid:1) D
.kak/t k
(cid:0)
1;
(2.22)
which is the OGF of (a1; 2a2; 3a3; : : :) or
follows:
If .a0; a1; a2; : : :/
F .t /, then .a1; 2a2... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | that differentiation of an OGF wrt the dummy variable multiplies each term by its index, and
shifts the whole sequence left one place.
Next consider an EGF
2.1. BASIC OPERATIONS 29
H.t/
a0
C
D
a1t=1(cid:138)
C
a2t 2=2(cid:138)
C (cid:1) (cid:1) (cid:1) D
akt k=k(cid:138):
1
X
0
k
D
(2.23)
Differentiate w... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | t/2
a3.31=3t/3
D
C
C
, in which the kth term is ak.k1=kt/k. Next consider the EGF. As shown above, H 0.t/ is
(cid:1) (cid:1) (cid:1)
the EGF of left-shifted sequence (a1; a2; a3; : : :). Multiply both sides by t to get
C (cid:1) (cid:1) (cid:1)
C
C
C
t.@=@t/H.t/
a1t
D
C
a2t 2=1(cid:138)
C
a3t 3=2(cid:13... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | t/n:
(cid:0)
D
1
X
0
k
D
(2.26)
Higher-order derivatives are used to find factorial moments of discrete random variables in the
next chapter.
Example 2.12 Use differentiation and shift operations to prove that .1
of the squares an
n2.
D
t/=.1
(cid:0)
C
t/3 is the GF
30
2. OPERATIONS ON GENERATING FUNCTION... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | Repeat
(cid:0)
m
1/.n
x//
(cid:0)
(cid:0)
2/ : : : nxn=m(cid:138). This
1=.1
(cid:0)
D
x/m
1.
C
(cid:0)
xn
m=m(cid:138). The first derivative
C
the differentiation m times
D
to get F .k/.x/
m
n
D
m (cid:1)xn. Thus,
C
(cid:0)
m/
is F 0.x/
.n
C
.n
m/.n
C
the GF is
D
C
can be written as
Example 2.1... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | this time a1 being a constant vanishes)
(cid:0)
(cid:1) (cid:1) (cid:1) C
to get H 00.t/
this pro-
cess k times. All terms whose coefficients are below ak will vanish. What remains is
H .k/.t/
1t=1(cid:138)
. This can be expressed using the summation
notation introduced in Chapter 1 as H .k/.t/
k/(cid:138). Using the cha... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | )
(2.27)
Divide both sides by x to get
x
.1=x/ Z
0
F .t/dt
.a0=1/
C
D
.a1=2/x
C
.a2=3/x2
C (cid:1) (cid:1) (cid:1)
(2.28)
which is the OGF of the sequence
ak=.k
f
1/
k
g
0.
(cid:21)
C
2.2. INVERTIBLE SEQUENCES 31
Next consider the EGF
H.t/
a0
C
D
a1t=1(cid:138)
C
a2t 2=2(cid:138)
a3t 3=3(... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | k(cid:1)=.k
0 (cid:0)
D
t/ndt. As the integral of .1
1/xk. With the help of above result, this can
t/n is .1
1/,
1=.n
t/n
C
C
C
C
C
x
0 .1
1/x(cid:141) as the result.
C
Example 2.16 Find the OGF of the sequence 1, 1/3, 1/5, 1/7, : : :.
x2
C
x3=3
1
D
1. Integrate the RHS term by term to get x
Solution 2.... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | term-by-term on the
x//, using log.a/
log.b/
x/=.1
D
D
(cid:0)
B
(cid:0)
D
C
(cid:0)
INVERTIBLE SEQUENCES
2.2
Two GFs F .t/ and G.t/ are invertible if F .t/G.t/
1. The G.t/ is called the “multiplicative
inverse” of F .t/. Consider the sequence (a0; a1; : : :). Let G.x/ be the multiplicative-inverse with
1. ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | -
0.
1, so that b2
D
(cid:0)
32
2. OPERATIONS ON GENERATING FUNCTIONS
2.3 COMPOSITION OF GENERATING FUNCTIONS
If F .x/ and G.x/ are two GFs, the composition is defined as F .G.x//. Note that F .G.x// is
not the same as G.F .x//. Consider F .x/
x/. Then F .G.x// is
1=.1
1=.1
x.1
(cid:0)
Consider a discrete branc... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | Xn
X
0
k
D
1.(cid:30).t//.
From this it follows that (cid:30)n.t/
(cid:30)n
(cid:0)
D
2.4
SUMMARY
(2.31)
(2.32)
(2.33)
This chapter introduced various operations on GFs. This includes arithmetic operations, linear
combinations, left and right shifts, convolutions, powers, change of dummy variables, differen-
... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book |
dt/ is (cid:21), the Poisson probabilities gives us the chance of observing exactly k events in the
.t; t
same time interval, if various events occur independently of each other. The PGF of Poisson
distribution is infinitely differentiable, and it is easy to find factorial moments because the kth
derivative of PGF is (ci... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book |
ln.Mx.t//, which
for an arbitrary origin. The CGF is defined in terms of the MGF as Kx.t/
when expanded as a polynomial in t gives the cumulants. Note that the logarithm is to the base
e.ln/. As every distribution does not possess a MGF, the concept is extended to the complex
domain by defining the ChF as (cid:30)x.t/
... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | .x/
p.0/
C
D
p.1/ t
C
p.2/ t 2
C (cid:1) (cid:1) (cid:1) C
p.k/ t k
C (cid:1) (cid:1) (cid:1)
;
(3.2)
where the summation is over the range of X. This means that the PGF of a discrete random
variable is the weighted sum of all probabilities of outcomes k with weight t k for each value
of k in its range. Obv... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | (et(x-μ))Central momentsµk = ∂k Mz(t)|t=0ChFφx(t)E(eitx)Momentsik µk = ∂k φx(t)|t=0CGFKx(t)log(E(etx))Cumulantsµk = ∂k Kx(t)|t=0FMGFΓx(t)E(1 + t)x)Factorial momentsµ(k) = ∂k Γx(t)|t=0MDGFDx(t)2E(tx/(1 – t)2)Mean deviationSee Section 3.5Probability generating function (PGF) is of type OGF. MGF and ChF are of type EG... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | 1/
D
D
D
C
C (cid:1) (cid:1) (cid:1) C
(cid:0)
C
C
C (cid:1) (cid:1) (cid:1)
D (cid:0)
Example 3.2 (PGF of arbitrary distribution) Find the PGF of a random variable given below,
and obtain its mean.
x
p.x/
2
1/3
3
1/6
1
1/2
Solution 3.2 Plug in the values directly in
Px.t/
D
E.t x/
D
t xp.x/
3
X
1
... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | F of a Poisson distribution, and ob-
tain the difference between the sum of even and odd probabilities. Also obtain the kth moment.
Solution 3.4 The PGF of a Poisson distribution is
Px.t/
D
E.t x/
D
1
X
0
x
D
t xe(cid:0)
(cid:21)(cid:21)x=x(cid:138)
e(cid:0)
D
(cid:21) 1
X
0
x
D
.(cid:21)t/x=x(cid:138)
e(... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book |
D
p(cid:140)1
p=.1
C
q2
C
q4
D
q1p
C
q3p
C (cid:1) (cid:1) (cid:1) D
C (cid:1) (cid:1) (cid:1)
(cid:141)
D
C (cid:1) (cid:1) (cid:1) D
qp(cid:140)1
(cid:141)
D
qp=.1
Now P [X is even]
(cid:0)
q2/
P [X is odd]
the above result, the difference between these must equal the value of Px.t
in (3.5) to get p=.1
q/... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book |
(cid:0)
(cid:0)
I
D
Solution 3.6 By definition
Px.t/
D
E.t x/
D
!
pxqn
(cid:0)
xt x
n
x
n
X
0
x
D
!
.pt/xqn
n
x
n
X
0
x
D
D
x
(cid:0)
.q
C
D
pt/n:
(3.6)
The coefficient of t x gives the probability that the random variable takes the value x. To find the
mean, we take log of both sides. Then ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | putting x
r/(cid:141)x with PGF 1=(cid:140)1
t .1
(cid:0)
p/ in the expansion 1=.1
kq=p. By setting p
1 to get E.X /
(cid:0)
D
r.1
tq
D
C
D
C
qt/(cid:141)k. Take log and differentiate w.r.t. t
r/ we
1
1=.1
p
D
(cid:0)
x
r// as (cid:0)
C
(cid:1)(cid:140)1=.1
r=.1
1
k
x
C
(cid:0)
D
C
C
t /(cid:141)k. Th... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | 1) (cid:1)
C
(3.9)
D
!
0, a factorial term will remain in the numerator which cancels out with the denom-
When k
1/.k
inator, leaving a k in the denominator. For example, .k
C
q3=3. Divide the numerator and denominator of the term outside the
2/q3=3(cid:138)
1:2q3=3(cid:138)
bracket by pk to get k=p(cid:0)
1=.ln.... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | zeros. The hypergeometric series
a.k/b.k/
xk
k(cid:138) where a.k/ denotes rising Pochhammer
is defined in terms of as F .x
c.k/
number. A property of the hypergeometric series is that the ratio of two consecutive terms is
a; b/
Pk
D
I
3.2. PROBABILITY GENERATING FUNCTIONS (PGF)
39
1/th to kth term is a polynom... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | 0)
(cid:0)
N
n (cid:1)2F1.
(cid:0)
N
(cid:0)
m
I
D
D
(cid:0)
(cid:0)
(cid:0)
(cid:0)
m;
1
I
(cid:0)
C
Example 3.10 (PGF of power-series distribution) Find the PGF of power-series distribution,
and obtain the mean and variance.
Solution 3.10 The PMF of power-series distribution is P .X
any integer values. By d... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | F .(cid:18)/
E.X.X
(cid:0)
C
1//
D
D
C
(cid:0)
t
1
D
P 0x.t/
D
(cid:140)E.X /(cid:141)2. Consider
F 0.1/
(cid:140)F 0.1/(cid:141)2
D
j
D
(cid:0)
C
(cid:0)
3.2.1 PROPERTIES OF PGF
1. P .r/.0/=r(cid:138)
@r =@t r Px.t/
P (cid:140)X
Px.t/ is infinitely differentiable in t for
@r
@t r Px.t/
D
D
D
0
... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | (cid:141)f .x/
1/ : : : .r
r (cid:141)f .x
r/
D
D
C
(cid:0)
(cid:0)
r
(cid:0)
2. P .r/.1/=r(cid:138)
E(cid:140)X.r/(cid:141), the r th falling factorial moment (page 56). By putting t
r
1
1/(cid:141), which is the r th falling facto-
in (3.13), the RHS becomes E(cid:140)x.x
rial moment. This is sometimes ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | .t/dt
D
D
t
j
Consider Rt Px.t/dt
integrate to get the result. This is the first inverse moment (of X
random variables.
E. 1
1 /.
X
C
Rt E.t x/dt. Take expectation operation outside the integral and
D
1), and holds for positive
C
6. PcX .t/
PX .t c/.
D
This follows by writing t cX as .t c/X .
7. PX
c.t/
t (c... |
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