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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 generating function... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | a well known expert in the field. Most titles cover subjects such as professional development, education, and study skills, as well as basic introductory undergraduate material and other topics appropriate for a broader and less technical audience. In addition, the series includes several titles written on very specific ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | Robert C. Creese 2010 Survive and Thrive: A Guide for Untenured Faculty Wendy C. Crone 2010 Geometric Programming for Design and Cost Optimization (with Illustrative Case Study Problems and Solutions) Robert C. Creese 2009 Style and Ethics of Communication in Science and Engineering Jay D. Humphrey and Jeffrey W. Holmes... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | for courses in analysis of algo- rithms, advanced data structures, digital signal processing (DSP), graph theory, etc. These are usually provided by either a course on “discrete mathematics” or “introduction to combinatorics.” But, GFs are also used in automata theory, bio-informatics, differential equations, DSP, num- ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | 2 1.2 Notations and Nomenclatures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.2.1 Rising and Falling Factorials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.2.2 Dummy Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | . . . . . . . . . . . . . . . . . . . . . . . . 14 Falling Pochhammer GF (FPGF) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 1.6.2 1.7 Other Generating Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 1.7.1 Auto-Covariance Generating Function . . . . . . . . . . .... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | . . . . . . 19 2.1.1 Extracting Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 2.1.2 Addition and Subtraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 x 2.1.3 Multiplication by Non-Zero Constant . . . . . . . . . . . . . . . . . . . . . . . . . 20... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | Generating Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 2.4 3 Generating Functions in Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | . . . . . . . . . . . . . . . . . 45 3.6.1 MD of Geometric Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 3.6.2 MD of Binomial Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 3.6.3 MD of Poisson distribution . . . . . . . . . . . . . . . . . . . . . . . . . . .... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | Truncated Distributions . . . . . . . . . . . . . . . . . . . . . . 58 3.13 Convergence of Generating Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 3.14 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 3.8.1 3.7.1 4 Appli... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | . . . 66 4.2.3 Binary-Search Algorithm Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 4.2.4 Well-Formed Parentheses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 4.2.5 Formal Languages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 Appl... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | . . . . . . . . . . . . . . . . . 74 Applications in Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 4.5.1 Polymer Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 4.5.2 Counting Isomers of Hydrocarbons . . . . . . . . . . . . . ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 4.10 Applications in Bioinformatics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 4.10.1 Lifetime of Cellular Proteins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 4.10.2 Sequence Alignment . . . . . . ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 Authors’ Biographies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 List of ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | . 46 3.4 MD of Poisson distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 3.5 MD of negative binomial distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 3.6 3.7 Table of characteristic functions . . . . . . . . . . . . . . . . . . . . . . . . . .... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | very useful to explore discrete problems involving finite or infinite sequences. 2 1. TYPES OF 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... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | when the size (number of elements) is small. Either computing devices must be used, or alternate mathematical techniques like combinatorics, graph theory, dynamic programming, etc., employed when this size becomes large. In addition to the size, the enumeration problem may also involve one or more parameters (usually d... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | pressed as a simple mathematical function. Convergence of a GF is often ignored in this book because it is rarely evaluated if at all, and that too for dummy variable values 0 or 1 (an exception appears in Chapter 4, page 86). 1=an !1j Ltn an D (cid:0) (cid:0) (cid:0) (cid:0) (cid:0) C j 1.2 NOTATIONS AND NOMENCLATURES... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | in reverse gives us the relationship x.j / writing Equation (1.2) in reverse gives us the relationship x.j / D .x .x j D (cid:0) 1/.j /. Similarly, j (cid:0) 1/.j /. C C 1.2.2 DUMMY VARIABLE An ordinary GF (OGF) is denoted by G.t/, where t is the dummy (arbitrary) variable. It can be any variable you like, and it depen... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | each distribution. An advantage is that if the MGF of an arbitrary random variable X is known, the MGF of any linear combination of the form a b can be 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 ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | C x/2 sides with respect to (w.r.t.) x to get the GF is 1=.1 (cid:0) (cid:0) x/2 or equivalently .1 C (cid:1) (cid:1) (cid:1) x/ C (cid:1) (cid:1) (cid:1) C D 3x2 1=.1 C D (cid:0) 2. This can be represented as x/(cid:0) C . (cid:0) C C x4 2x x2 1/ x3 1 D x C (cid:3) (cid:0) (cid:0) C C . Differentiate both . Thus, 4x3 C... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | 15, (cid:0) 52, 203, etc. Recurrence relations are extensively discussed in Chapter 4 (page 71). 1/th and .n 1 0 (cid:0) P (cid:0) D D (cid:0) (cid:0) (cid:0) (cid: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 sequen... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | a finite geometric sequence. if the signs differ alternatively. Let Sn D n 1/. Both the numerator and denominator are nega- 1/=.r Then the sum P k r/. tive when 0 < r < 1, so that we could write the RHS in the alternate form .1 1/=.1 0 r k .r n r n D D (cid:0) (cid:0) C C D (cid:0) 1 1 C (cid:0) (cid:0) Example 1.1 (OGF ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | C n2xn (cid:0) 1 : C (cid:1) (cid:1) (cid:1) (1.5) (1.6) If we multiply both sides by x again, we see that the coefficient of xn on the RHS is n2. Hence, x @ x/2, take log and differentiate to get2 x/2/ is the GF that we are looking for. To find the derivative of f @x .x=.1 x=.1 D (cid:0) (cid:0) 1.4. ORDINARY GENERATING F... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | G.x/=.1 of the partial sums of coefficients, and dividing by 1=.1 x/) Multiplying an OGF by 1=.1 (cid:0) (cid:0) x/ results in differenced sequence. x/ results in the OGF (cid:0) Proof. Let G.x/ be the OGF of the infinite sequence (a0; a1; : : : ; an; : : :). By definition G.x/ a0 1 as a power series 1 . Expand .1 anxn a2x2... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | the first n natural numbers as P k D The OGF of n2 was found in Example 1.4 as x.1 x/3. If this is divided by .1 (cid:0) n we should have the OGF of P k 1 k2. This gives the OGF as x.1 x/=.1 x/=.1 x/4. (cid:148) C 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 ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | is the OGF of x (see Chapter 4, § 4.1.1, pp. 62). (cid:3) .F .t/ 0 akt k where the upper limit is m, then G.t/ C 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 a2k... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | it can be used to scale down functions that grow too fast. For example, number of permutations, number of subsets of an n-element set which has the form 2n, etc., when multiplied by xn=n(cid:138) will often result in a con- verging sequence.3 Thus, e2x can be considered as the GF of the number of subsets. These coefficie... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | (cid:11)n/=.(cid:12) .1 .1 (cid:0) (cid:0) D (cid:0) D C (cid:16)e(cid:12) t (cid:0) e(cid:11)t (cid:17) =.(cid:12) (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 ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | to n. That gives n1 C (cid:1) (cid:1) (cid:1) C x3=3(cid:138) 0. Let x=1(cid:138) denote the GF for just one urn (where we have omitted the constant term due to our assumption that the urn is not empty; meaning that there must be at least one ball in it). As there are m such urns, the GF for all of them taken together ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | (cid:0) tained where the known coefficients follow the Pochhammer factorial as 1/ : : : .x 1/. Note that this is different from the OGF or EGF ob- x.x D C C k 1.6. POCHHAMMER GENERATING FUNCTIONS 15 RPGF.x/ a.k/xk; n X 0 k (cid:21) D where a.k/ a.a C D 1/ : : : .a k C (cid:0) 1/. An example of an EGF of this type is b x/(... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | modifications of the dummy variable while others are combinations. A few of them are given below. 1.7.1 AUTO-COVARIANCE GENERATING FUNCTION Consider the GF P .z/ 2 a0 a1.z C C D 1=z/ C a2.z2 C 1=z2/ C (cid:1) (cid:1) (cid:1) C ak.zk 1=zk/ ; C (cid:1) (cid:1) (cid:1) C (1.29) where ak’s are the known coefficients. This is ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | graphs with n vertices is Gn.x/ P k D with n vertices and k edges. Total number of simple labeled graphs with n vertices and m edges 0 2.n eC.x/ where C.x/ is the EGF of is (cid:0) connected graphs. These are further discussed in Chapter 4 (pp. 73). .n 2/ m (cid:1). This in turn results in the EGF P1n D 2/xn=n(cid:138)... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | 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 finite set of size n into k subsets. It can also be considered as the number of ways to distribute n distinguishable balls into k indistinguishable cells with no cell empty. 18 1.... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | recurrence for two sequences an and bn, how do we get a recurrence for an an?. This is where the merit of introducing a GF becomes apparent. It allows us to derive new GFs from existing or already known ones. bn or for c C (cid:0) (cid:3) 0 pkt k, or An infinite sum is denoted by either explicit enumeration of the range... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | greater than or equal to that of q.x/, we may use synthetic division to get an expression of the p1.x/=q.x/, where the degree of p1.x/ is less than that of q.x/, and proceed with form r.x/ partial fraction decomposition of p1.x/=q.x/. These are illustrated in examples below. C 2.1.2 ADDITION AND SUBTRACTION The sum and... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | 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 Arithmetic Progression) Find the OGF of a sequence (a; ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | (cid:0) 1t 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 OGF is ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | 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 VARIABLE The dummy variable in a GF can be assumed to be continuous in an interval. Any type o... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | 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 series expansion in which the dummy variable is changed as t=.1 t/. n n t//, where F .t/ is the OGF of ak. Similarly,... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | has mth coefficient um [ m ai bj ck. Pi j k D C C D Table 2.1: Convolution as diagonal sum Example 2.8 Find the OGF of the product of .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=... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | F .t /2 H.t /(cid:141) F .t / (cid:6) (cid:6) D j , which can also be expressed as P1k D (cid:3) D D j bj . (cid:0) D D (cid:140)F .t /(cid:141)2 .a0ak a1ak C 1 (cid:0) C (cid:1) (cid:1) (cid:1) C aka0/ t k: D X 0 k (cid:21) (2.14) As an illustration of powers of OGF, consider the number of binary trees Tn with n nodes... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | Catalan numbers are always integers. 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... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | 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 . (cid:0) (cid:0) and replace k (cid:0) 1/k . 1/2k 1 (cid:0) . 1 1/(cid:0) D D 1 by k so that it varies from 0– D (cid:0) (cid:0) (cid:0) 1 and 21 2k22k (cid:0) to get 2. Write .2k ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | (cid:0) 1; (2.22) which is the OGF of (a1; 2a2; 3a3; : : :) or follows: If .a0; a1; a2; : : :/ F .t /, then .a1; 2a2; 3a3; : : :/ , bn f D .n C 1/an 1 C g 0. This can be stated as (cid:21) .@=@t /F .t / D , F 0.t /. Literally, this means 1 X 1 k D for n NameExpressionGFAdd/subtract∞ ak ± bktkF(t) ± G(t)Scalar multiply∞... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | get formal power series f .t/ 1=.1 P1k D t/2. Successive differentia- the first derivative as f 0.t/=f .t/ t/ so that f 0.t/ t/n tion gives f .n/.t/ (cid:0) D 1. A direct consequence of this result is the GF C t/ D 1=.1 n(cid:138)=.1 1=.1 D D (cid:0) (cid:0) D (cid:0) n C k k ! 1 t k (cid:0) 1=.1 t/n: (cid:0) D 1 X 0 k D... |
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 ch... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | 1=2. Thus .1 .1=2/(cid:140)1=.1 (cid:0) D RHS to get 1 log.1 2 (cid:140) x/ (cid:0) log.a=b/. This is the OGF of the original sequence. x2/(cid:0) x/. This gives A B D C 1=.1 C 1 2 log..1 B=.1 1 x/ x/(cid:141) C x2/(cid:0) (cid:0) C log.1 x/ C C C D (cid:0) (cid:0) (cid:0) , which has the de- 1dx, we use partial fracti... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | D C C D (cid:0) 1 Xn to be IID, it is easy to see that (cid:0) D D P (cid:140)Xn k j D Xn (cid:0) 1 D j (cid:141) D P (cid:2)Y1 Y2 C C (cid:1) (cid:1) (cid:1) C Yj (cid:3) : Using total probability law of conditional probability, this becomes P (cid:140)Xn k(cid:141) D D 1 X 0 j D P (cid:2)Y1 Y2 C C (cid:1) (cid:1) (ci... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | to find factorial moments because the kth derivative of PGF is (cid:21)k times the PGF. The GFs can be used to prove the additivity property of Poisson, negative binomial, and many other distributions [Shanmugam and Chattamvelli, C 34 3. GENERATING FUNCTIONS IN STATISTICS 2016]. Although the previous chapter used x as t... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | t. But when expanded as a power series in the auxiliary variable, the coefficient of t k is the probability of the random variable X taking the value k. This is one way the PGF of a random variable can be used to generate probabilities. Table 3.1: Summary table of generating functions 3.2. PROBABILITY GENERATING FUNCTION... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | get the RHS as (cid:140)p.0/ 0 in (3.2) to get Px.t 1 in (3.2) that Px.t D (cid:140)p.1/ p.2/ p(cid:140)3(cid:141) p(cid:140)5(cid:141) D (cid:141) 0/ 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 give... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | C 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/, which is ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | 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. The PGF of NBINO.n; p/ can directly be derived x/n and multiplying both sides by pn. D D (cid:0) (cid:0) Examp... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | to kth term is a polynomial in k. If such a polynomial. More precisely, the ratio of .k C a polynomial is given, we could easily ascertain the parameters of the hypergeometric function from it. Consider the ratio of two successive terms of hypergeometric distribution f .k C 1/=f .k/ .k (cid:0) D m/.k (cid:0) n/=(cid:14... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | r (cid:0) C 1/t x (cid:0) r (cid:141) f .x/: (3.13) The first term in this sum is (cid:140)r.r r(cid:138)f .x r(cid:138)f .x r/. By putting t r/. Thus, @r D @t r Px.t 0/ D D D D r(cid:138)f .x r/. D 1/t r D 0, every term except the first vanishes, and the RHS becomes (cid:140)r(cid:138)t 0(cid:141)f .x/ 1/ : : : .r r (ci... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | variable generates probabilities, it can be used to generate the sum of left tail probabilities (CDF). Definition 3.2 A GF in which the coefficients are the cumulative probabilities associated with a random variable is called the CDF generating function (CDFGF). We have seen in Theorem 1.1 (page 9) that if F .x/ is the OG... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | D Solution 3.12 The PGF of Poisson distribution is found in Example 3.4 as e(cid:0) t/=.1 the CDFGF follows easily as e(cid:0) and .1 t/. Write the numerator as e(cid:0) (cid:0) 1 as infinite series and collect coefficients of like powers to get (cid:21).1 (cid:0) t/(cid:0) (cid:21)(cid:140)1 t(cid:141). From this (cid:0)... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | distribution. Example 3.15 (SFGF for geometric distribution) Obtain the SFGF of a geometric distribu- tion with PMF given by f .x 0; 1; 2; : : : ; q qxp; x p/ p. 1 I D D D (cid:0) Solution 3.15 First find P (cid:140)X > n(cid:141) to find the SF of a geometric distribution: SF.n/ P .X > n/ qn C 1p D D 1p as a common fact... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | starting with the first, and g0 by .1 t/, and denote the LHS Gx.t/=.1 t/ by Hx.t/. p0. Divide both sides D (cid:0) (cid:0) Hx.t/ (cid:0)h0 h1t C C D h2t 2 h3t 3 (cid:1) ; C (cid:1) (cid:1) (cid:1) C (3.23) g0 g1 where hk C for CDFGF given above with pk probabilities are all less than or equal to one for any discrete dis... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | Collect alike terms on the RHS to get (cid:22) b c X a x D xp.x/ (cid:22) F. (cid:22) b / c (cid:3) (cid:0) D (cid:22) b c X a k D p.k/; k (cid:3) (3.28) where F. / c term cancels out, and we get (cid:22) c (cid:22) b p. D C b / p. b (cid:22) c (cid:0) 1/ C (cid:1) (cid:1) (cid:1) C p.a/. Now substitute in (3.26). The ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | 1 q C C q2 qk .1 (cid:0) D C (cid:1) (cid:1) (cid:1) qk C 1/=.1 q/ .1 (cid:0) D (cid:0) qk 1/=p as C G.t/ (cid:0)1 q (cid:0) C D t (cid:0)1 (cid:0) q2(cid:1) t 2 (cid:0)1 q3(cid:1) (cid:0) C t 3 (cid:0)1 (cid:0) q4(cid:1) (cid:1) C (cid:1) (cid:1) (cid:1) C (3.32) because p in the numerator cancels out with 1 q by gk, ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | the coefficient of t b 1 (cid:1)pnqx is .p=.1 t/2. The mean is (cid:22) qt /(cid:0) n=.1 nq=p, and let k qt//n. Hence, the MDGF nq=p . Expanding c 1 gives D b c(cid:0) .nq=p/ D (cid:0) (cid:0) C n (cid:0) (cid:0) n 1 MD D 2pn .k k X 0 i D i i/ (cid:0) C n n (cid:0) (cid:0) 1 ! 1 qi where k nq=p : c D b (3.35) Results are... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | obtain the MGF and derive the mean. C C .q pet /n. Take log to get log.Mx.t// pet =.q n log.Mx.t// Solution 3.18 The MGF can be found from Equation (3.6) by replacing t by et . This pet /. Next, differenti- gives Mx.t/ (cid:3) D 0 to get the mean as np. Take ate as above: M 0X .t/=Mx.t/ D D pet /. Differentiate again, lo... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | E.X 2/ D @2 @t 2 D D E.etx/. Differentiating (3.37) w.r.t. t gives @ Solution 3.19 We know that Mx.t/ @ @t E.etx/ @t etx/ Putting t @2 0/ @t 2 Mx.t/ D Repeated application of this operation allows us to find the kth moment as M .k/ x .t E.xk/. This gives (cid:27) 2 (cid:140)M 0x.t D E.xetx/ because x is considered as a c... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | D e(cid:21).et t (cid:0) 1/(cid:21) (cid:0)et (cid:0) 1(cid:1) (cid:0) D (cid:21) (cid:0)et 1(cid:1) (cid:0) (cid:22)j t j =j (cid:138): (3.43) 1 X 0 j D Expand et as an infinite series. The RHS becomes (cid:21) P1k D 0 (cid:22)j t j be written as (cid:21) P1k 1 P1j D D (cid:21) h (cid:22)r (cid:22)r (cid:22)r (cid:0) (... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | can be dealt with in the complex domain by finding the expected value of eitx, where i 1, which always exist. Thus, the ChF of a random variable is defined as p D (cid:0) ChF D E.eitx/ D 8 < : P1x R 1x D(cid:0)1 D(cid:0)1 eitxpx if X is discrete I eitxf .x/dx if X is continuous: We have seen above that the ChF, if it exi... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | (cid:1) (cid:1) (cid:1) (cid:30)y.bt/. Putting a b D D 1, we get 4. (cid:30)x.t/ is continuous in t, and convex for t > 0. This means that if t1 and t2 are two values of t > 0, then (cid:30)x ..t1 t2/ =2/ C (cid:20) 1 2 (cid:140)(cid:30)x .t1/ C (cid:30)x .t2/(cid:141). 5. @n(cid:30)x.t/=@t n i nE.X n/. t j 0 D D Table... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | log.MX .at// bt D C Kx.at/ Theorem 3.6 The CGF of an origin and scale changed variable is K.X Kx.t=(cid:27)/. (cid:0) (cid:22)/=(cid:27) .t/ . D (cid:22)=(cid:27)/t (cid:0) (cid:3) C Proof. This follows from the above theorem by setting a (cid:22)=(cid:27). The cumu- lants can be obtained from moments and vice versa. T... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | Mx.t/ as 1 E.X (cid:0) finite series to get .(cid:22)03 (cid:0) (cid:22)1, (cid:20)2 C D 3(cid:22)1(cid:22)02 C (cid:22)2 D 2(cid:22)3 1/t 3=3(cid:138) : C (cid:1) (cid:1) (cid:1) (3.52) (cid:22)2 1 D (cid:27) 2, (cid:20)3 .(cid:22)03 (cid:0) 3(cid:22)1(cid:22)02 C D (cid:0) (cid:2)tx=1(cid:138) C C .tx/2=2(cid:138) C (... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | 1, (cid:128) 0x.t/ C t D D 1 j j t t 2x.x 1/=2(cid:138) t 3x.x 1/.x 2/=3(cid:138) (cid:3) : (3.56) (cid:0) E.X/ C (cid:22). Factorial moments are obtained by taking C (cid:1) (cid:1) (cid:1) (cid:0) (cid:0) D The rising factorial moment is defined as E(cid:140)X.X 1/.X 2/ : : : .X k 1/(cid:141) k (cid:0) (cid:0) C C t/(... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | (cid:21)k exp. (cid:21).1 D D (cid:0) (cid:0) (cid:0) FMGFx.t/ E(cid:140).1 t/x(cid:141) C D D 1 X 0 x D .1 C t/xe(cid:0) (cid:21)(cid:21)x=x(cid:138) D e(cid:0) (cid:21) 1 X 0 x D (cid:140)(cid:21).1 C t/(cid:141)x=x(cid:138) D e(cid:21)t : (3.59) The kth factorial moment is obtained by differentiating this expression ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | MGF can be expressed as N i D D (cid:21) D MX Y .t/ j D X x f .x j y/etx: (3.61) Replacing t by “it” gives the corresponding conditional characteristic function. If the variates N .t are mutually independent, this becomes MS j (cid:140)Mx.t/(cid:141)N . N / j D Example 3.29 (Compound Poisson distribution) Find the MGF ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | just one coefficient t/2, where (cid:22) is the mean. Results are tabulated for binomial, Poisson, geo- (cid:22) (cid:140)t b metric and negative binomial distributions. 1(cid:141)2Px.t/=.1 (cid:0) c(cid:0) DistributionPGFz-PGFz-MGFBinomial(q + pt)n(q + pt)n/ (1 − q n)(q + pe t)n/ (1 − q n)Poissonexp(− λ[1 − t])e (− λ[1 ... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | r /n r C r 0 (cid:0) xn x2 t 3 t 2 C D C C C D C C C C C C y z n t t t 1 0 and 0 integer, y is t 4 can take values 0 or 1 only, its OGF is .1 C (cid:1) (cid:1) (cid:1) D (cid:21) 1=.1 1. As x is allowed to be only even integers, it has OGF 1 t 2 C C t/. As z t/. Hence, the required number is the coefficient of t 2/. As y... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | in the second and a final. Several applications in graph theory also use these operators. For instance, the number of edge-crossings of a complete bipartite graph Km;n with m vertices in the first set, and n vertices in the other is given by x c D (cid:0)d x c D b x , (cid:0)b c 1 2 c C b e D (cid:0)b x C b =22 1/ c C (c... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | the frequency as .a1 < a2 < < an/ where a1 is the item that got the least vote or preference and an is the item with greatest vote. (cid:1) (cid:1) (cid:1) 2 (cid:0) (cid:0) (cid:0) (cid:0) ⁄ C D D .n n.n 1/=2 C (cid:1) (cid:1) (cid:1) C Consider a set of n labeled distinct elements that can be ordered among themselves... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | to the base case than those with more than one inversion. Permutations with more and more inversions deviate further away from the base case. This has applications in recommendation systems, computer sorting algorithms like bubble-sort and insertion-sort that tries to sort an arbitrary array by exchanging the data valu... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | 3k in terms of F .t/? This can be found using nth roots of unity. Let !n Suppose the GF of a sequence (a0; a1; a2; : : : ; an; : : :) is F .t/. It was shown in Chapter 2 that we could express the GF of odd and even terms (i.e., stride 2 terms) separately using F .t/. How do we find the GF of a strided sequence of lag ot... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | complexity. Although some of the algorithms are easy to code, they may not have good performance for very large data sets. One example is bubble-sort algorithm which performs poorly when data size increases, but may be a reasonable choice when data size is always small (say < 15). This is where algorithm analysis steps... |
https://ieeexplore.ieee.org/xpl/ebooks/bookPdfWithBanner.jsp?fileName=8826062.pdf&bkn=8826061&pdfType=book | we assume that 2m=2 k.2m/ because both n is a power of 2. Substitute n e are 2m 1/=2m C.2m k. If D.m/ and (cid:0) C denotes C.2m/=2m, we could write this in terms of D.m/ as D.m/ k. Multiply 1/ C both sides by t m, and add over m 0 D.m t P1m (cid:0) D 1/t m t/. From (cid:0) (cid:0) C C.2m/=2m, this F .t/ 1/. As we subs... |
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