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Random variables), each with the same distribution, each having common mean a = E(X) and variance s2 =Var(X) Here X is a rv having the same distribution as Xj The sum S =åN j=1 Xj where the number in the sum, N is also a random variable and is independent of the Xj's The following statement now follows from Theorem 1 Theorem 2 (i) ESX is a value that X can take;F(!) = a ig Then f = P n i=1 a iI A i is called the canonical representation of f Lemma 25 (Monotone Approximation) Let f be a nonnegative measurable extended realvalued function from Then there exists a sequence ff ng1 n=1 of nonnegative ( nite) simple functions such that f

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ƒfƒ‹ƒf ƒyƒ"ƒP[ƒX ƒXƒk[ƒs[-X y p Bin Thm x p p 1 x 1y p p 1 xy p1 y It would be enough to show that each of the binomial coefficients is divisible by p, since we assume that R has characteristic p and so each of the corresponding terms would then be zero Combinatorial identity For any n 0and0 k n, n k n k n 1 k 1 2 Therefore, since p 1 kX y p Bin Thm x p p 1 x 1y p p 1 xy p1 y It would be enough to show that each of the binomial coefficients is divisible by p, since we assume that R has characteristic p and so each of the corresponding terms would then be zero Combinatorial identity For any n 0and0 k n, n k n k n 1 k 1 2 Therefore, since p 1 k

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Divergence Theorem to calculate the surface integral \iint_s FdS;that is, calculate the flux of F across S F(x, y, z) = x^4i x^3 z^2 j 4xy^2 zk, S is the surface of the solid x^2 y^2 = 4) but uh here!Y * Ê I ­ ß X ÿ " Ï ù » s H ˜ B Ï C B m = § ë Ò b Æ D 0 í _ ) e N ¿ ù R % › ï Ò ' ‡ ¶ ° ) > < T N O ¥ Y ï F 8 Î € ô È c Y z ¬ W 7 K ) G , @ þ æ þ < ö § ± ú o Ä q r L Ç Ã ü ¸ ò ³ à Õ É K Û ( y !

2 pages Moreover given X = x and Z = 2 we must have Y = 2 −x (d) Given Z = 2, the conditional PMF of X is binomial for 3 trials and success probability 2/5 The conditional expectation of X given Z = 2 is EXZ = 2 = 2(2/5) = 4/5 (e) There are several ways to solve this problem The most straightforward approach isSolve for x Calculator StepbyStep Examples Algebra Solve for x Calculator Step 1 Enter the Equation you want to solve into the editor The equation calculator allows you to take a simple or complex equation and solve by best method possible Step 2A f f a _ e Y \ R S x {} ~ S X I M W V F G G R J E I O S D} {K

K(xj, L k(x) and H k(x) are, respectively, F ibonacci, Lucas and generalized F ibonacci P olynom ials, and then fin a lly to extend these results to rbonacci polynom ials 2 F IB O N A C C I A N D LU C A S P O LY N O M IA LS A S C O E F F IC IE N T S T he F ibonacci polynom ials F n (x) are defined by 3Induction) The following approach is often called reservoir sampling Suppose we have a sequence of items passing by one at a time We want to maintain a sample of one item withContinuous Random Variables can be either Discrete or Continuous Discrete Data can only take certain values (such as 1,2,3,4,5) Continuous Data can take any value within a range (such as a person's height)

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Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers Visit Stack ExchangeB Hints and solutions Solutions to Chapter 1 11 Setting for simplicity x= 0, we have, by Taylor's expansion at 0, f(y) = f(0) i=1 ∂f(0) ∂yD s x ok g ls c;k u f y ;k x ;k A tk s lk F k es lay uXu g S A f t le s Li "V i l s m Yy sk f d ;k x ;k g S f d J h n;k jk e iq = nw cj d k W nw f u ok l h ok lns o ux j d s 1 0 2 & 2 ij nt Z m ud s }k j k m ijk sD r Ø s rk d k s m Dr f rf F k d ks jf t L VMZ 'k q nk cS uk e k e q 0 1 1 5 0 0 0¾0 0 ¼ ,d yk k iUnz g

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