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Functions and Transforms Appendix A Functions and Transforms A.1 Nonlinear Transforms Many theoretical and practical problems can be converted into easier forms if instead of the discrete or continuous distributions their different transforms are applied, which can be solved more readily. In probability theory, numerous trans- forms are applied. Denote by F the distribution function of a random variable X and by f the density function if it exists. The general form of the most frequently used transform depending on the real or complex parameter w is ∞ E wX = wx dF(x). −∞ If the density function exists, then the last Riemann-Stieltjes integral can be rewritten in the form of Riemann integral as follows: ∞ E wX = wx f(x)dx. −∞ 1. In the general case, setting w = et ,t∈ R,wehavethecharacteristic function (Fourier-Stieltjes transform) ∞ ϕ(t) = E etX = e txdF(x). −∞ 2. If the random variable X has a discrete distribution with values 0, 1,... and probabilities p0,p1,...corresponding to them, then setting z = w, |z| < 1, we get © Springer Nature Switzerland AG 2019 547 L. Lakatos et al., Introduction to Queueing Systems with Telecommunication Applications, https://doi.org/10.1007/978-3-030-15142-3 548 A Functions and Transforms ∞ ∞ X x k G(z) = E z = z dF(x)= pkz , −∞ k=0 which is the generating function of X. 3. The Laplace-Stieltjes transform plays a significant role when considering random variables taking only nonnegative values (usually we consider this type of random variable in queuing theory), which we obtain with w = e−s,s≥ 0: ∞ ∼ − F (s) = E e−sX = e sxdF(x). 0 For the case of continuous distributions it can be rewritten in the form ∞ ∗ − ∼ f (s) = E e−sX = e sxf(x)dx = F (s), 0 where f ∗ denotes the Laplace transform of the density function f . 4. The generating function plays a significant role when considering discrete random variables taking only nonnegative values, which we obtain with w = z: ∞ X n f(z)= E z = fnz . n=0 The identical background of the transformations given above determines some identical properties. When considering various problems, the use of separate trans- forms may be advantageous. For example, in the case of random variables taking nonnegative integer values the z-transform, and in case of general nonnegative random variables the Laplace-Stieltjes or Laplace transform is favorable to apply. Note that we define the transforms given above for more general classes of functions than the distribution functions. A.2 z-Transform Let f0,f1,...be a sequence of real numbers and define the power series ∞ n 2 n f(z)= fnz = f0 + f1z + f2z + ...+ fnz + ... (A.1) n=0 It is known from the theory of power series that if the series (A.1) is not everywhere divergent except the point z = 0, then there exists a number A>0 such that the K | n| ∞ | | series (A.1) is absolute convergent ( n=0 fnz < )forall z <Aand divergent for all |z| >A.Theseries(A.1) may be convergent or divergent at the points z = A.2 z-Transform 549 ±A depending on the values of the parameters fi ,i= 0, 1,... The number A is called the convergence radius of the power series (A.1). By the Cauchy-Hadamard theorem, A can be given in the form 1/n A = 1/a, where a = lim sup (|fn|) . n→∞ In the last formula we set A =+∞if a = 0andA = 0ifa =+∞.Thefirst relation A =+∞means that the power series (A.1) is convergent in all points of the real line, and the second one means that Eq. (A.1) is convergent only at the point z = 0. = K n A finite power series f(z) n=0 fnz (K-order polynomial, which corre- sponds to the case fi = 0,i≥ K + 1) is convergent at all points of the real line. Definition A.1 Let f0,f1,...be a sequence of real numbers satisfying the condi- 1/n tion a = lim sup (|fn|) < ∞. Then the power series n→∞ ∞ n f(z)= fnz , |z| <A= 1/a n=0 is called the z-transform of the sequence f0,f1,... It is clear from this definition that if we use a discrete distribution fn,k≥ ∞ 0, fk = 1, then the z-transform of the sequence f0,f1,...is identical with the k=0 generating function G(z), which was introduced earlier. A.2.1 Main Properties of the z-Transform 1. Derivatives. If the convergence radius A does not equal 0, then the power series f(z)is an anytime differentiable function for all points |z| <Aand k ∞ d n−k f(z)= n(n − 1)...(n− k + 1)fnz ,k≥ 1. dzk n=k 2. Computing the coefficients of the z-transform. For all k = 0, 1,...the following relation is true: 1 dk f = f(z) ,k≥ 0. (A.2) k ! k k dz z=0 It is clear from relation (A.1) that if the condition A>0 holds, then the function f(z) defined by the power series (A.1) and the sequence f0,f1,... uniquely 550 A Functions and Transforms determine each other, that is, the z-transform realizes a one-to-one correspondence between the function f(z)and the sequence f0,f1,.... = ∞ n = ∞ n 3. Convolutions. Let f(z) n=0 fnz and g(z) n=0 gnz be two z- transforms determined by the sequences fn and gn, respectively. Define the sequence hn as the convolution of fn and gn,thatis, n hn = fkgn−k,n≥ 0. k=0 = ∞ n Then the z-transform h(z) n=0 hnz of the sequence h0,h1,...satisfies the equation h(z) = f(z)· g(z). A.3 Laplace-Stieltjes and Laplace Transform in General Form Let H(x), 0 ≤ x<∞ be a function of bounded variation. A function H is said to be of bounded variation on the interval [a,b], if its total variation VH ([a,b]) is bounded (finite). The total variation is defined as KP VH ([a,b]) = sup H(xP,k) − H(P,k−1) P k=1 where the supremum is taken over the set of all partitions ={ = = } P xP,0 a<xP,1 <...<xP,KP b of the interval [a,b]. The function H is of bounded variation on the interval [0, ∞) if VH ([0,b]) is bounded by some number V for all b>0. The function ∞ ∼ − H (s) = e sxdH(x). (A.3) 0 is called the Laplace-Stieltjes transform of the function H . If the function H can be given in the integral form x H(x)= h(u)du, x ≥ 0, 0 A.3 Laplace-Stieltjes and Laplace Transform in General Form 551 where h is an integrable function (this means that H is an absolute continuous function with respect to the Lebesgue measure), then the Laplace transform of the function h satisfies the equation ∞ ∗ − ∼ h (s) = e sxh(x)dx = H (s). 0 Theorem A.1 If the integral (A.3) is convergent for s>0,thenH ∼(s), s > 0 is an analytic function, and for every positive integer k k ∞ d ∼ = −sx − k k H (s) e ( x) dH(x). ds 0 The transform H ∼ satisfies the following asymptotic relation [2]. If the inte- gral (A.3) is convergent for Res>0 and there exist constants α ≥ 0andA such that H(x) A lim = , x→∞ xα Γ(α+ 1) then the convergence ∼ lim sαH (s) = A (A.4) s→0+ holds. Theorem A.2 Assume that there exists a function h(x), x ≥ 0, and its Laplace transform h∗(s), s > 0; moreover, the function h(x) is convergent as x →∞, i.e., lim h(x) = h∞. Then x→∞ ∗ lim sh (s) = h∞. s→0+ = x ≥ = Proof Denote H(x) 0 h(s)ds, x 0. Choosing α 1wehave H(x) 1 x h∞ lim = lim h(s)ds = h∞ = , →∞ →∞ x x x x 0 Γ(1 + 1) thus by relation (A.4) the assertion of the theorem immediately follows. Theorem A.3 If there exists a Laplace transform f ∗ of the nonnegative function f(t),t ≥ 0, and there exists the finite limit lim f(x)= f0,then x→0+ ∗ lim sf (s) = f0. s→∞ 552 A Functions and Transforms Proof It is clear that ∞ ∞ − − s e sxdx = e x dx = 1 0 0 and ∞ ∞ √ −sx −y − s s √ e dx = √ e dy = e = o(1), s →∞; 1/ s s therefore, √ 1/ s ∞ ∗ −sx −sx sf (s) − f0 = s e [f(x)− f0]dx + s √ e f(x)dx + f0o(1). 0 1/ s Since there exists the finite limit lim f(x)= f0, with the notation x→∞ δ(z) = sup |f(x)− f0| → 0,z→ 0+, 0<x≤z we obtain √ 1/ s √ ∞ √ −sx −sx s e |f(x)− f0| dx<δ(1/ s) se dx = δ(1/ s) → 0,s→∞. 0 0 On the other hand, for all 0 <t≤ s the relation ∞ ∞ ∗ − − ∗ f (s) = e sxf(x)dx ≤ e txf(x)dx = f (t), 0 0 is true, then ∞ √ ∞ −sx −(1/2) s −(s/2)x s √ e f(x)dx ≤ se √ e f(x)dx 1/ s 1/ s √ ∞ √ − − − ∗ ≤ se (1/2) s e (s/2)xf(x)dx = se (1/2) s f (s/2) 0 √ − ∗ ≤ se (1/2) s f (1) → 0, as s →∞(s ≥ 2). Summing up the results obtained above, the assertion of the theorem follows. A.3 Laplace-Stieltjes and Laplace Transform in General Form 553 A.3.1 Examples for Laplace Transform of Some Distributions (a) Deterministic distribution (a>0, P(X = a) = 1): ∞ ∼ − − F (s) = e sxdF(x)= e sa, E(X) = a : 0 (b) B(n, p) binomial distribution: ∞ n ∼ − n − − F (s) = e sxdF(x)= pk(1 − p)n ke sk k 0 k=0 n n − − − = (pe s )k(1 − p)n k =[1 + p(e s − 1)]n, k k=0 − − − E(X) = npe s [1 + p(e s − 1)]n 1 = np.
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