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= e (1z) This is a very nice generating function, because we can easily express the nth derivative of GX(z) by G(n) X (z) = ne (1z);H o s p i t a l B a g C h e c k l i s t M u m Nightwear for birth & after Lip balm N i p p l e c r e a m Pillow Hair bands H i g h e n e r g y s n a c k s D r e s i ng o w U n d e r w e a r N u r s i n g b r a Slippers Socks Going home clothes F l a n n e l s B r e a s t Pa d sDepartment of Computer Science and Engineering University of Nevada, Reno Reno, NV 557 Email Qipingataolcom Website wwwcseunredu/~yanq I came to the US

(b) gn!0 uniformly on E;P(g(X) ≥ a) ≤ Eg(X) a When g(X) = X, it is called Markov's inequality Let's use this result to answer the following question Example 5 Let X be any random variable with mean µ and variance σ2 Show that P(X −µ ≥ 2σ) ≤ 025 That is, the probability that any random variable whose mean and variance are finite takes a valueN 6a x7> n 6s6a 7 wd> (x 5;=ly;

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Example 7 One solution to nding Eg(X) is to nding f y, the density of Y = g(X) and evaluating the integral EY = Z 1 1 yf Y (y)dy However, the direct solution is to evaluate the integral in (2) For y= g(x) = xp and X, a uniform random variable on 0;1, we have for p> 1,Title Decision FINAL5pdf Author JacquelineDePierola Created Date 052 AMThis list of all twoletter combinations includes 1352 (2 × 26 2) of the possible 2704 (52 2) combinations of upper and lower case from the modern core Latin alphabetA twoletter combination in bold means that the link links straight to a Wikipedia article (not a disambiguation page) As specified at WikipediaDisambiguation#Combining_terms_on_disambiguation_pages,

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= e z X1 k=0 ( z)ke z=k!P d f x c h a n g e e d i t o r w w w t r a c k e r s School Bow Valley College, Calgary Course Title SCIENCE 10 Uploaded By MasterField1296 Pages 49 This preview shows page 30 36 out of 49 pages P D F X C h a n g e E d i t o r w w w t r a c k e r1) This equation holds for a body or system , such as one or more particles , with total energy E , invariant mass m 0 , and momentum of magnitude p ;

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Hp n d By Minkowski's inequality, we have for each n2N that kh nk Lp k=1 kg kk Lp M where P 1 k=1 kg kk Lp = M It follows that h2L p(X) with khk Lp M, and in particular that his nite pointwise ae Moreover, the sum P 1 k=1 g k is absolutely convergent pointwise ae, so it converges pointwise ae to a function f2Lp(X) with jfj hThe mean or expected value of g(X) is E(g(X)) = Z g(x)dF(x) = Z g(x)dP(x) = (R 1 1 g(x)p(x)dx if Xis continuous P j g(x j)p(x j) if Xis discrete Recall that 1 Linearity of Expectations E P k j=1 c jg j(X) = P k j=1 c jE(g j(X)) 2 If X 1;;X n are independent then E Yn i=1 X i!Math 461 Introduction to Probability AJ Hildebrand Variance, covariance, correlation, momentgenerating functions In the Ross text, this is covered in Sections 74 and 77

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= 4 > 8 ?Therefore f is discontinuous at every rational point The fact that f is Riemann integrable follows directly from Theorem 716 Problem4(WR Ch 7 #11) Suppose {fn}, {gn} are defined on E, and(a) P fn has uniformly bounded partial sums;1 9 2 2 A u s t i n W ay n e S e l f A M R ac i n g R y an S al o m o n C h e v r o l e t A M T e c h n i c al S o l u t i o n s / G O T E X A N 2 0 2 3 C h as e P u r d y G M S R ac i n g J e f f

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Y1 WV m 5 s p r NT %XKMp ō!\n j wԕ M N o廭 ~i ~ F ƪp u So }żZoCˢ c ̴ Ӂ K tV4 ~ ' Q 9 =n 8gA i ȵ =, w ` j !M B}` i Z C FeS G, i f P 4 X # _ r )߭~S Gt tu t F t X ̑3 z j 6xg I 7 k% # x @ J M pӔ y { ۧm } M Qg =0) B ~' O c Xs N e bc R `Q U* ʰN )Chapter 8 Poisson approximations Page 2 therefore have expected value ‚Dn‚=n/and variance ‚Dlimn!1n‚=n/1 ¡‚=n/Also, the cointossing origins of the Binomial show that ifX has a Binm;p/distribution and X0 has Binn;p/distribution independent of X, then X CX0has a Binn Cm;p/distribution Putting ‚Dmp and „Dnp one would then suspect that the sum of independentN and independence) yields E(X) = np Var(X) = np(1−p) M(s) = (pes 1−p)n Keeping in the spirit of (1) we denote a binomial n, p rv by X ∼ bin(n,p) 3 geometric distribution with success probability p The number of independent Bernoulli p trials required until the first success yields the geometric rv with pmf p(k) = ˆ

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6 > 8 = 5 = 5 b ?But we've just seen that EXY = EXEY if X and Y are independent, so then Var(X Y) = Var(X)Var(Y) 3 Binomial random variables Recall that the distribution of the binomial is ProbX = x = n x p x(1−p)n− and that it's the sum of n independent Bernoulli variables with parameter p 3(c) g1(x) ‚g2(x) ‚g3(x) ‚¢¢¢ for every x 2E Prove that P fngn converges uniformly on E Solution Let An(x) ˘

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A p p e n d i x 2 C i v i c E n g a g e me n t E x a mp l e A c ti v i ti e s This compiled list of activities are comprised from the work of the National Council for the Social Studies as well as examples created by the K12 South Dakota social studiesÅ Æ Ç È É Ê Ë Ì Í Î Ï Ð Ñ Ò Ó Ô Õ Ç Ð Ô Ï È Ñ Ö Î × Ë Í Ø Æ Õ Ù É Ú Û Ê × Ñ Æ Ô Ì Ü Ý Î Ø Ê È É Þ Ù Ð Ç Ï Å ß à á â Ó Ê Í ã Æ Ð Ï Î Ë ä Ò å ÖIt follows that jfj gae Hence, jf njp gp, jfjp g p, and jf f njp 2g , and by the Dominated Convergence Theorem, lim n!1 Z X jf f njpdx= Z X lim n!1 jf f njpdx= 0 14 Convergence criteria for Lp functions If ff ngis a sequence in Lp(X) which converges to f in Lp(X), then there exists a subsequence ff n k

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Phonetic Alphabet Tables Useful for spelling words and names over the phone I printed this page, cut out the table containing the NATO phonetic alphabet (below), and taped it to the side of my computer monitor when I was a call center help desk technician An alternate version, Western Union's phonetic alphabet, is presented in case the NATOA % >@> gv;< = @ 4 ;

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The constant c is the speed of light It assumes the special relativity case of flat spacetime Total energy is the sum of rest energy and kinetic energy , while invariant mass is mass measured in a centerofmomentum frame ForAbout Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How works Test new features Press Copyright Contact us CreatorsThe CDC AZ Index is a navigational and informational tool that makes the CDCgov website easier to use It helps you quickly find and retrieve specific information

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Xs?e?vb n 6=a < 9 b_w;=b_cv;;=gfb ?H Ѓx X g n R x X g A h W P b g Ȃǖh Ћy эЊQ A C e ̐ ̃T C g ł B h Ў o A h ~ A ЊQ o A ЊQ A Ȃǖh Ў E ЊQ ɕ֗ ȃA C e ̂ Љ Ă ܂I) are finite for all i=1,n, then E(X 1 KX n)=E(X 1)KE(X n) Proof Use the example above and prove by induction Let X 1, X n be independent and identically distributed random variables having distribution function F X and expected value µ Such a sequence of random variables is said to constitute a sample from the distribution

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To c, sn 6=c for all n, the sequence {f(sn)} converges to L (d) limx→c f(x)=L if and only if limh→0 f(ch)=L (e) If f does not have a limit at c, then there exists a sequence {sn} in D sn 6=c for all n, such that sn → c, but {f(sn)} diverges (f) For any polynomial P and any real number c, lim x→c P(x)=P(c) (g) For any polynomials PM D Adams 1 , S E Celniker, R A Holt, C A Evans, J D Gocayne, P G Amanatides, S E Scherer, P W Li, R A Hoskins, R F Galle, R A George, S E Lewis, S Richards, M Ashburner, S N Henderson, G G Sutton, J R Wortman, M D Yandell, Q Zhang, L X Chen, R C Brandon, Y H Rogers, R G Blazej, M Champe, B D Pfeiffer, K H Wan, C Doyle, E G Baxter, G Helt, C RP(1 p)n1 and G0 Y (1) = n(1 p p)n1p = np 1212 Poisson distribution Let X have the Poisson distribution with parameter >0 Then p k= ke =k!

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Y n ∉ P is odd (ie Player II has made an illegal move) To close this preliminary section, let us consider how a game G(X) behaves with respect to boolean operations on the set X We begin with complementation If Player I wins the game G(X), then Player II wins the game G(AX c)Ef(X)g(Y )1 A = E f(X)Eg(Y ) A1 A It follows from the definition of conditional expectation that Ef(X)g(Y ) A = E f(X)Eg(Y ) A A = Ef(Y ) AEg(Y ) A, so X and Y are independent conditionally on A D Exercise 2 Let X = (X n) n≥0 be a martingale (1) Suppose that T is a stopping time, show that X T is also a martingaleProof lnexy = xy = lnex lney = ln(ex ·ey) Since lnx is onetoone, then exy = ex ·ey 1 = e0 = ex(−x) = ex ·e−x ⇒ e−x = 1 ex ex−y = ex(−y) = ex ·e−y = ex 1 ey ex ey • For r = m ∈ N, emx = e z }m { x···x = z }m { ex ···ex = (ex)m • For r = 1 n, n ∈ N and n 6= 0, ex = e n n x = e 1 nx n ⇒ e n x = (ex) 1 • For r rational, let r = m n, m, n ∈ N

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Where c ij are complex coe cients Finally, while on the subject of polynomials, let us mention the Fundamental Theorem of Algebra ( rst proved by Gauss in 1799) If P(z) is a nonconstant polynomial, then P(z) has a complex root In other words, there exists a complex number csuch that P(cRestriction of a convex function to a line f Rn → R is convex if and only if the function g R → R, g(t) = f(xtv), domg = {t xtv ∈ domf} is convex (in t) for any x ∈ domf, v ∈ Rn can check convexity of f by checking convexity of functions of one variable< 9 8 ;

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5 bd> b 1 686a 4;= Eψ(X)g(X) − (1/n)P (A n)Let X be a discrete random variable with probability mass function p(x) and g(X) be a realvalued function of X Then the expectedvalue of g(X) is given by Eg(X) = X x g(x) p(x) (16) Proof for case of finite values of X Consider the case where the random variable X takes on a finite number of values x1,x2,x3,···xn The function g(x

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B r a i n e r g m s c x p h a x e b r o j w k e f s i j o l n k d a w a r e n e s s v f h s b i e n m j t z e c i e a d h x e m b h c u w d q b v g n k i e q u z f pG(n) X (0) shows that the whole sequence of probabilities p0,p1,p2, is determined by the values of the PGF and its derivatives at s = 0 It follows that the PGF specifies a unique set of probabilities Fact If two power series agree on any interval containing 0, however small, thenX!ac= c, where c is a constant ( easy to prove from de nition of limit and easy to see from the graph, y= c) 8lim x!ax= a, (follows easily from the de nition of limit) 9lim x!ax n= an where nis a positive integer (this follows from rules 6 and 8) 10lim x!a n p x= n p a, where n is a positive integer and a>0 if n is even (proof needs a

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