example balls and bins with feedback

... Permutation Routing on the Butterfly Exere ises Balls, Bins, and Random Graphs 5.1 Example: The Birthday Paradox 5.2 Balls into Bins 5.3 5.2.1 The Balls- and- Bins Model 5.2.2 Application: Bucket Sort ... the Exponential Distribution 8.3.2* Example: Balls and Bins with Feedback 8.4 8.5 8.6 The Poisson Process 8.4.1 Interarrival Distribution 8.4.2 Combining and Splitting Poisson Processes 8.4.3 ... Random Variables and Expectation 20 2.1 20 22 23 25 26 30 32 34 38 Random Variables and Expectation 2.1.1 Linearity of Expectations 2.1.2 Jensen's Inequality 2.2 The Bernoulli and Binomial Random...

Ngày tải lên: 14/07/2016, 15:42

... exp ,B T  , T e T  = T + BT : B e Under IP , B T  is normal with mean and variance T , so B T  is normal with mean T and variance T :   exp , ~ , T 2 d~: b e IP B T  d~ = p b ... ~2 d~: b =p 2T b 2T f e Under I , B T  is normal with mean zero and variance P mean T and variance T e T Under IP , B T  is normal with Means change, variances don’t When we use the Girsanov ... Williams’ “standard machine” f IP and IP Z The intuition behind the formula f IP A = is that we want to have A Z T  dIP 8A F f IP ! = Z T; !IP !; f f but since IP !  = and I ! =...

Ngày tải lên: 18/10/2013, 03:20

... and so ordα R(t) = −ordα Q(t) Then the determinant R(t) is a sum of following terms (n−1) fα0 fα1 fαn δ , fα0 fα1 fαn where α0 , α1 , , αn n + and δ = or −1 By applying the propositions 2.3 and ... a the rational function on F and let the derivatives order k, ϕ(k) ≡ Then ϕ(k) −k ordα ϕ Proof Let ϕ(x) = (x − α)m f (x) , where f (x), g(x) are relatively prime and f (α)g(α) = g(x) Then, we ... (1) Proposition 2.5 Let ϕ1 , ϕ2 be rational functions on F and a ∈ F Then ordα (ϕ1 , ϕ2 ) min{ordα ϕ1 , ordα ϕ2 } Proof Let ordα ϕ1 = m1 and ordα ϕ2 = m2 Then f1 (x) g1 (x), f2 (x) ϕ2 (x) = (x...

Ngày tải lên: 13/02/2014, 04:20

... will follow Gowers [8] and use the arguably simplest sieve, that due to Brun, on both the major and minor arcs The genuine primes, on the other hand, are harder to deal with Here we will quote ... The techniques for dealing with the minor arcs are associated with the names of Weyl, Vinogradov and Vaughan The major arcs We will have various functions f : [N ] → R with (4.5) f ∞ = O(log N/N ... neighbourhood of a/q Lemma 4.4 Suppose that f satisfies the conditions (4.5) and (4.6) and that θ ∈ Ma,q for some a, q with (a, q) = and q (log N )B Write (4.9) σa,q (f ) = e(ar/q)γr,q (f ) r Then, (4.10)...

Ngày tải lên: 06/03/2014, 08:21

... that |x1 | < r and |x2 | > 1/r Then from Theorem 25 we get maps f and g which map onto two sides of a Jordan curve γ and so that f and g approximate F and G to within ε/C for |z| < r and |z| > 1/r ... Given any δ > and R < ∞, there is a c > so that the following holds Suppose fn : D → Ωn and gn : D∗ → Ω∗ is a normalized circle chain pair with ∂Ω∗ ⊂ D(0, R) and assume fn → f and and gn → g uniformly ... f and g map the special points x1 and x2 to within ε|w1 − w2 |/(8 + 8diam(γ ) of the desired image points w1 and w2 Thus there is a Euclidean similarity which maps f (x1 ) and g(x2 ) to w1 and...

Ngày tải lên: 22/03/2014, 20:21

... complex numbers and are the roots of the equation if and only if and Indeed, if and are roots of the equation then Eq (3) is satisfied, from which and Conversely, if and then, substituting and in the ... two equations for and satisfy and By Viète’s theorem, for any such and (which may be complex) indeed exist, and they are the roots of the equation If we take such (still unknown) and then Eq (7) ... practical character and is aimed 10 Chapter Let us still consider some examples of binary operations We shall often return to these examples in future F IGURE EXAMPLE Let A, B, and C be the vertices...

Ngày tải lên: 31/03/2014, 14:57

... phenomenon, the antecedent circumstances being respectively scotch and soda, bourbon and soda, brandy and soda, rum and soda, and gin and soda [ugh!] Then using the Method of Agreement he swore a ... Theorem Say in the illustrated experiment the left hand filter is oriented at 45 degrees and the right hand one is at zero degrees If the left hand electron passes through its filter then it is ... right-hand electron to determine if it is spin-up for zero degrees And since no information can travel faster than the speed of light, the left hand measurement cannot disturb the right hand measurement...

Ngày tải lên: 05/06/2014, 10:58

... any x, y in X and m Î [x, y] As noted in [11], we not know of any example of a CAT() space which does not enjoy the property (N) Anakkamatee and Dhompongsa Fixed Point Theory and Applications ... simple example shows that the converse of this proposition does not hold: Example 2.9 Consider an ℝ-tree in ℝ∞ defined as follow: Let {en} be the standard basis, x0 = e1 = (1, 0, 0, 0, ), and for ... Î ℝ and {s’} is a subnet of {s} satisfying s’ ≻ s for each s, then μs (f (s )) = a Anakkamatee and Dhompongsa Fixed Point Theory and Applications 2011, 2011:34 http://www.fixedpointtheoryandapplications.com/content/2011/1/34...

Ngày tải lên: 21/06/2014, 01:20

... F (2:21) Karapinar Fixed Point Theory and Applications 2011, 2011:4 http://www.fixedpointtheoryandapplications.com/content/2011/1/4 and each a Î A, then g and F have a common coincidence point, ... , z) + p(z, y) Karapinar Fixed Point Theory and Applications 2011, 2011:4 http://www.fixedpointtheoryandapplications.com/content/2011/1/4 Page of and p(z, y) ≤ p(z, xn ) + p(xn , y) − p(xn , ... together with (2.13) ϕ(z) ≤ lim m→∞ inf ϕ(xm ) ≤ lim m→∞ inf[ϕ(xn ) − p(xn , xm )] = ϕ(xn ) − p(xn , z) (2:16) and thus p(xn , z) ≤ ϕ(xn ) − ϕ(z) By definition, z Î S(x n ) for all n Î N and thus...

Ngày tải lên: 21/06/2014, 02:20

... probability and information theory Berger and Ye see have studied the existence of entropy rate for G-invariant random fields Recently, Ye and Berger see have also studied the ergodic property and ShannonMcMillan ... arbitrary random fields and the Markov chain fields on the generalized Bethe tree Main Results Theorem 2.1 Let X {Xt , t ∈ T} be an arbitrary random field on the generalized Bethe tree fn ω Q and ϕ μ ... initial distribution and joint distribution of X μQ be defined as 1.4 and 1.5 , respectively p s0 , p s1 , p s2 is a {Xt , t ∈ T} with respect to 1,j | Xm,i ⎦ 3.16 {Xt , t ∈ T} with respect to the...

Ngày tải lên: 21/06/2014, 05:20

... The proof of i and ii were given in and the last one just follows from definition f ∞ f ∞ , and consider the Example 1.2 see Let E C1 0, with the norm f cone P {f ∈ E : f ≥ 0} and g 2k For each ... −P {0} and P / {0} For a given cone P , one can define a partial ordering denoted by ≤: or ≤P with respect to P by x ≤ y if and only if y − x ∈ P The notation x < y indicates that x ≤ y and x ... ∈ X, then d is called a cone metric on X, and the pair X, d is called a cone metric space CMS Example 1.4 Let E R3 and P { x, y, z ∈ E : x, y, z ≥ 0} and X R Define d : X × X → E by d x, x α|x...

Ngày tải lên: 21/06/2014, 20:20

Ngày tải lên: 05/08/2014, 09:46

Ngày tải lên: 05/08/2014, 10:20

Ngày tải lên: 05/08/2014, 15:20

Ngày tải lên: 05/08/2014, 15:20

... -subsets of Y we have s = k/2 and s Ai = ∅, whence i=1 max max (s + | s≥2 {A1 , ,As } s i=1 Ai |) ≥ k k/2 On the other hand, let B = s Ai and r = |B| Since s ≥ and {A1 , , As } is an i=1 ... by Moon and Moser [3] Furthermore, I(n, n − 2) = 2n−2 − and the extremal graph (unique up to isomorphism) coincides with K2,n−2 for every n ≥ In [10] it was proved ... max (s + | {A1 , ,As } s i=1 Ai |) ≤ k−r (k − r)/2 +r k and the last expression is less than or equal to k/2 for every k ≥ and ≤ r ≤ k − and (1) is proved Since any maximal complete subgraph in...

Ngày tải lên: 07/08/2014, 06:23

... the graphs F1 and F2 has a component isomorphic to G When F2 = 2G, then Equation (10) follows from Lemma 2.9 and Corollary 2.10 When F1 = G + X and F2 = G + Y , and v(X) < v(G) and v(Y ) < v(G), ... covered by ΛI = G, and then construct N (Λj ), and assign it a multiplicity equal to the edge label on ΛI − Λj Remark It is is possible that for distinct j and k, the matrices N(Λj ) and N(Λk ) are ... proved by contradiction Suppose G and H are connected nonisomorphic graphs with the same vertex deck, that is, they are a counter example to Ulam’s conjecture Then 2G and 2H are nonisomorphic but...

Ngày tải lên: 07/08/2014, 13:21

... c(vi ) =          if if if if (i ≤ 2k) and (i mod = 1), (i ≤ 2k) and (i mod = 0), (2k < i ≤ 2l = m) and (i mod = 1), (2k < i ≤ 2l = m) and (i mod = 0) In all the following cases, to 4-color ... Now, if we replace vertex vm with Q Q two vertices vm and vm in such a way that vm is adjacent to all vertices in N(vm ) \ {v2k+1} and vm is only adjacent to v1 and v2k+1 , then one can easily ...         if if if if (i ≤ 2k) and (i mod = 1), (i ≤ 2k) and (i mod = 0), ((2k < i < 2l + = m) and (i mod = 1)) or (vi = vm ), ((2k < i < 2l + = m) and (i mod = 0)) or (vi = vm ) Case 3:...

Ngày tải lên: 07/08/2014, 13:21

... nonnegative integers with r1 + r2 + · · · + rn = n There exists a tournament with score sequence r1 , r2 , , rn if and only if there exists P1 , P2 , , Pn , with Pi ⊆ Ai and |Pi | = ri (1 ... of edges of T with initial vertex in the complement K of K and terminal vertex in K Simplifying (13), we get |K||K| ≥ γ ∗ (K) (14) Since the lefthand side of (14) is symmetric in K and K, we need ... Theorem 2.1 (Landau’s theorem) The sequence r1 ≤ r2 ≤ · · · ≤ rn of integers is the score sequence of a tournament of order n if and only if k ri ≥ i=1 k (k = 1, 2, , n) (1) with equality...

Ngày tải lên: 07/08/2014, 21:21

... homogeneous polynomial of degree n with nonnegative coefficients, called the Minkowski polynomial Here + denotes Minkowski sum, and λK denotes the dilatation of K with coefficient λ ≥ The coefficient ... set Q is the standard simplex in Rn And the convex compact set P = Diag(λ1 , , λn )Q = {(x1 , , xn ) : xj ≤ 1; xi ≥ 0}; λ1 ≥ ≥ λn > 1≤j≤n λj The general nonnegative case is handled by the topological ... + K2 ) and R2 (t) = V old (tC1 + C2 ), where K1 , K2 , C1 , C2 are convex compact sets; K1 , K2 ⊂ Rl and C1 , C2 ⊂ Rd Define the next two convex compact subsets of Rl+d : P = K1 × C1 and Q =...

Ngày tải lên: 07/08/2014, 21:21