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X∞ i=1 P(A∗ i) ≤ X∞ i=1 P(Ai), establishing (b) ⁄ There is a similarity between Boole's Inequality and Bonferroni's Inequality If we apply Boole's Inequality to Ac, we have P(∪n i=1A c i) ≤ i=1 P(Ac i), and using the facts that ∪Ac i = (∩Ai)c and P(Ac i) = 1−P(Ai), we obtain 1−P(∩n i=1Ai) ≤ n− i=1 PIs the proposition (¬ p ∨c) is a tautology?1 3 abc where a,b,c are any nonnegative integers with abc = n,since(1/3)abc is the probability of any specific configuration of choices for each player with the right numbers in each category, and the coecient in front counts the number

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—¤ã ƒXƒpƒCƒN ‚©‚í‚¢‚¢-(b) As we will see below convergence X ip d n → c is equivalent to X n → c (c) The intuitive content of the statement X ip n → c is that in the limit as n increases, almost all of the probability mass becomes concentrated in a small interval around c,R o t c e r i d z g j m f k b q t l u p u u n e x f k h i r agent badge bank robbery bureau criminal director fbi academy file fraud intelligence investigation j edgar hoover justice office pistol racketeering report security top secret united states super word search puzzles wwwsuperwordsearchpuzzlescom

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MSU/CSE 260 Fall 09 4 Proof Methods h1 ∧h2 ∧ ∧hn ⇒c ?13 THE STRUCTURE OF Z p WHEN p IS PRIME 48 We'll now prove three easy corollaries to this theorem Corollary 132 Let p be a positive prime For any a p 6= 0 and any b p 2Z p, the equation a p X = b p has a unique solution in ZHere, X is n p, W is n q, is p 1, and is q 1 Suppose that C(W) ˆC(X) For Model 1, let Yb X, be X, and P X denote the vector of (leastsquares) tted values, the vector of (leastsquares) residuals, and the perpendicular projection matrix onto C(X) The quantities Yb W, be W, and P W are de ned analogously (a) Show that W = XC, for some p

J=1 P(A j)− X 1≤iA combination takes the number of ways to make an ordered list of n elements (n!), shortens the list to exactly x elements ( by dividing this number by (nx)!Diabetic ketoacidosis was more common with dapagliflozin than with placebo (03% vs 01%, P=002), as was the rate of genital infections that led to discontinuation of the regimen or that were considered to be serious adverse events (09% vs 01%, P

P n i=1 X i cwhere ccan be determined by making use of the fact that P X i has the gamma distribution with = nand = 0Use the result from Example 716 That is, cis a constant satisfying = Z 1 c 1 ( n) n 0 tn 1e t 0 dt Given the sample size n, 0;P X;Y (x;y) = p X(x)p Y (y);, the value of ccan be found by solving for c

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F U L L Y V A C C I N A T E D T a ke st e p s t o p ro t e ct yo u rse l f a n d o t h e rs G et t est ed 35 days af t er t he day you were exposed t o CO V I D19 W ear a mask in indoor public set t ings f or 14 days or unt il you receive a negat ive t est result4 Is the proposition (¬ c →¬ p) is a tautology?And f X;Y (x;y) = f X(x)f Y (y) De nition If a sequence of random variables X 1;;X N are independent, then their joint PDF (or joint PMF) can be factorized f X1;;X N (x 1;;x N) = YN n=1 f (x n) (5) 17/26

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P (n i=1 x 2)−n¯x 2 P = 4n n i=1 (x i −x¯), which is positive unless all values of x are the same 2 Suppose b is the least squares coefficient vector in the regression of y on X and c is another Kx1 vector Prove that the difference in the two sums of squared residuals isA n d t h o s e e x p e c t a t i o n s a r e g e n e r a l l y u n observable As reviewed briefly in the next section, past r e s e a r c h e r s h a v e t r i e d t o m e a s u r e s u c h e x p e c t a t i o n s i n s e v e r a l ways, none of which is completely c o n v i n c i n gS a t o la h hyde pitt dare w ake duplin bladen pender bertie wilkes u nio carteret n ash robeson s ampson moore craven onslow h alif x beaufort columbus swain ashe

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11 Definitions and Examples 5 then we can write the chain in general as X nC1 D X n D nC1/C if X n > s S D nC1/C if X n s In words, if X n > s we order nothing and begin the day with X n units If the demand D nC1 X n, we end the day with X nC1 D X n D nC1If the demand D nC1 > X n,we end the day with X nC1 D 0IfX n s, then we begin the day with S units, and the reasoning is the same asPolynomial over GF(p) has pn elements, and is denoted GF(pn) f x =xpn−x Corollary For each prime p and positive integer n, the field GF (pn) exists and is unique (two fields of the same order are isomorphic) Recall that we have already mentioned that GF(pn) – {0} = GF(pn)* is a cyclic group under multiplication, and the generators of thisPhonetic 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 NATO

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2 If X ∼ binomial(n,p), then X/n→P p Different sequences of convergent in probability sequences may be combined in much the same way as their realnumber counterparts Theorem 74 If X n →P X and Y n →P Y and f is continuous, then f(X n,Y n) →P f(X,Y) If X = a and Y = b are constant random variables, then f only needs to be(d) The matrix P2IR n is said to be a projection if P2 = P Clearly, if Pis a projection, then so is I P The subspace PIRn = Ran(P) is called the subspace that P projects onto A projection is said to be orthogonal with respect to a given inner product h;ion IRn if and only if h(I P)x;Pyi= 0 8x;y2IRn;Fix a natural number n, and let Pn be the partition of 0,1 with partition points xj = j/n Then U(f;Pn) = j=1 xj(xj −xj−1) = j=1 j n 1 n = 1 n2 j=1 j = n(n 1) 2n2 = 1 2 1 2n and similarly L(f;Pn) = 1 2 − 1 2n Since the integral is greater than or equal to every lower sum and less than or equal to every upper sum, we have 1

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Thegenerating functionofthesequence(an)isthe(formal)powerseries A(x) = P n anx n = a0 a1xa2x2 ¢¢¢anxn ¢¢¢ The summation sign always starts at 0 and extends to infinity in steps of one By x we understand an indeterminate Most of the time we view generating functions as formal power series Occasionally,/^ /W>/E Zz d/KE Z WKZd< À v Xt o v X P } À ò í ñ r î ñ ï r ô õ ð í } u } o } P Ç ' v µ ( µ o } µ µ D u Z U dE h v o v À Ç v v } o v Z } u v P Á v µ v P } v P Z } µ ¨ ï U ì ì ì À o v o Ç ð l î í l î ì î íD E D Ed K& ^K>/ /d d/KE l DK /&/ d/KE K& KEdZ d í X ^ } o } v E µ u W P } ( W P < r î ì î í r r ì ì î ñ í ï î X u v u v lD } ( } v E µ u

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X j cj X i w i ij = X j cjw j since all terms with i6= jvanish = v 4 Then as a linear transformation, P i w iw T i = I n xes every vector, and thus must be the identity I n De nition A matrix Pis orthogonal if P 1 = PT Then to summarize, Theorem A change of basis matrix P relating two orthonormal bases isP (x1 y1) (x2 y2)2 and the triangle inequality kuvk kukkvk (Exercise!) (c) In Rn the set H= fx2 Rn a 1x1 anxn = cg is a convex set For any particular choice of constants ai it is a hyperplane in Rn Its de ning equation is a generalization of the usual equation of a plane in R3, namely the equation axbyczd= 0(a) For any constant k and any number c, lim x→c k = k (b) For any number c, lim x→c x = c THEOREM 1 Let f D → R and let c be an accumulation point of D Then lim x→c f(x)=L if and only if for every sequence {sn} in D such that sn → c, sn 6=c for all n, f(sn) → L Proof Suppose that lim x→c f(x)=LLet {sn} be a sequence in D which converges toc, sn 6=c for all nLet >0

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VN = C (P/F, i%, N) rZ (P/A, i%, N) Note the difference between the coupon rate, r, and the yield rate i The coupon rate r is fixed for a given bond, but the yield i depends on the bond purchase price The desired yield is determined by the rate of interest in the economy If the 'general' interest rate goes up, the yield required by bondAnd A itself 1 2 CHAPTER 1 SIGMAALGEBRAS), and then (by dividing by x!), it removes the number of duplicates Above, in detail, is the combinations and computation required to state for n = 4 trials, the number of times there are 0 heads, 1 head, 2 heads, 3 heads, and 4 heads

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Let X˘N p(0;) and A be a p psymmetric matrix 1For B (k p), BXand X 0AXare independent if B A = 0;(ii) X has the BolzanoWeierstrass property, namely that every infinite set has an accumulation point∗ (iii) X is sequentially compact, ie every sequence has a convergent subsequence (iv) X is totally bounded and complete ∗If A is a subset of X then p is called an accumulationpointif every neighborhood of p contains a point q ∈A soAn International Directory of the Best UFO Evidence, a Cooperative Venture with NICAP, Sponsored by the Current Encounters Mailing List

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N(x,z),dd n(z,y)} (2) Let 1 ≤ p < ∞ Then d p(x,y) = i=1 x i −y i p!E(X n)=E(X) Proof Suppose that X ≥ 0isarandomvariableandlet(X n)beasequenceofsimple random variables with X n ≥ 0andX n ↑ X Observe that since the X n are increasing, we have E(X n) ≤ E(X n1) Therefore, E(X n)increasestosomelimita ∈ 0,∞;Now, if Xis the total number of spades in the hand, X= X7 j=1 I j)E(X) = E 2 4 X7 j=1 I j 3 5 = X7 j=1 EI j = X7 j=1 13 52 = 7 4 Problem 6 (p1 #7) In a circuit containing nswitches, the ith switch is closed with probability

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X (n) if p 1 (2n) 1 where fbgis the number b rounded to the nearest integer, ie, if k is an integer and k 05 bP n j=1 A j = m n j=1 A j = j=1 m(A j) = j=1 P(A j) which means that P is nitely additive We next show that P is continuous in ;(and hence ˙additive) To prove that let (C n) n Fbe a decreasing sequence of sets such that lim n!1P(C n) >0 In the remaining of this part we want to show that T 1 n=1 C n 6= ;(which then proves theN 𝑥!(𝑛−𝑥)!, this function can be found on most calculators so you really don't need to know this formula If X is a binomial random variable, then it has a binomial distribution with parameters n and p

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5 Is the proposition (p ∧¬ c) is a contradiction?C C C C C A where s2 j = (1=n) P n i=1 (xij x j) 2 is thevarianceof the jth variable sjk = (1=n) P n i=1 (xij x j)(xik x k) is thecovariancebetween the jth and kth variables x j = (1=n) P n i=1 ij is the mean of the jth variable Nathaniel E Helwig (U of Minnesota) Data, Covariance, and Correlation Matrix Updated 16Jan17 Slide 14Step 3 We now nd the matrix U The rst column of Uis ˙ 1 1 Av 1 = 1 6 p 10 18 6 = 3= p 10 1= p 10 The second column of Uis ˙ 1 2 Av 2 = 1 3 p 10 3 9 = = p 10 3= p 10 Since Uis a 2 2 matrix, we do not need any more columns

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That is, E(X n) ↑ a for some 0 ≤ a ≤∞(IfE(X n)isanunboundedsequence,thena = ∞S } v í í X ì ó X î ì í î ì õ l í ó l î ì î ì í ñ W ñ ð W P ï } ( ó u d } o ,h & µ v u } µ v W ¨ ð U õ î ô U ô ó ì X ì ìWhere x = 0,1,2,3,,n Note Cx= 𝑛!

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Let p = h1 ∧h2 ∧ ∧hn The following propositions are equivalent 1 p ⇒c 2 (p →cKeeping 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) =2For symmetric B, X0AXand X0BXare independent if B A = 0 Example 5 The residual sum of squares in the standard linear regression has a scaled chi

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Here that this ideal therefore contains a constant, hence is the whole ring Thus (x,y2 1) is maximal Alternatively, look at the quotient ring Rx,y/(x,y2 1) ∼= Ry/(y2 1) This latter ring is isomorphic to C since it is obtained by adjoining an elementIf and only if p > 1 (c) Prove that the series X1 n=2 1 nlog 2 n diverges (here, log 2 n = ln(n)=ln(2) denotes the base2 logarithm of n) Solution For parts (a) and (b) see Section 342 in the textbook For part (c), see the solution to problem 5 in homework assignment #9 Question 5That is, the subspaces Ran(P) and Ran(I P) are orthogonal in the inner product h;i

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The set P(X) of all subsets of X is a sigmaalgebra Any sigmaalgebra F of subsets of X lies between these two extremes f;;Xg ˆ F ˆ P(X) An atom of F is a set A 2 F such that the only subsets of A which are also in F are the empty set ; It typically contains a GH dipeptide 1124 residues from its Nterminus and the WD dipeptide at its Cterminus and is 40 residues long, hence the name WD40 Between the GH and WD dipeptides lies a conserved core It forms a propellerlike structure with several blades where each blade is composed of a fourstranded antiparallel betasheetX f nd = R X fd Proof Let = lim n!1 R X f nd Since R X f nd R X fd for all n, we get that R X fd Next we prove the reverse inequality For this, let 0 s f be a simple measurable function It su ces to prove that R X sd Let 0

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1 p defines a metric n Rn (3) For p = ∞ d ∞(x,y) = max i=1,,n x i −y i also defines a metric on Rn Project 1 Let 1 < p,q < ∞ such that 1/p1/q = 1 Show Minkowski's inequality Mink (11) xy ≤ xp p yq q holds for all xThe joint distribution of X,Y,Z is P(X = a,Y = b,Z = c)= n!N, acomplexnumberλ ∈ C is an eigenvalue of A if there is some nonzero vector u ∈ Cn,suchthat Au = λu If λ is an eigenvalue of A,thenthenonzero vectors u ∈ Cn such that Au = λu are called eigenvectors of A associated with λ;togetherwiththezerovector,these eigenvectors form a subspace of Cn denoted by E λ(A),

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C C A where rjk = sjk sjsk = P n i=1 (xij x j)(xik x k) q P n i=1 (xij x j)2 q n i=1 (xik xk)2 is the Pearson correlation coefficient between variables xj and xk Nathaniel E Helwig (U of Minnesota) Principal Components Analysis Updated 16Mar17 Slide 11

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