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X e V e W Let Φ be the fundamental solution of Laplace's equation That is, Φ(x) =¡ 1 2 lnjxj n = 2 1 n(n¡2)fi(n) 1 jxjn¡2 n ‚ 3 Suppose u 2 C2(Ω)By the Divergence Theorem, we have Z V† Φ(y ¡x)∆u(y)dy = ¡Z V† ryΦ(y ¡x)¢ryu(y)dy Z @V† Φ(y ¡x)@uBASIC STATISTICS 5 VarX= σ2 X = EX 2 − (EX)2 = EX2 − µ2 X (22) ⇒ EX2 = σ2 X − µ 2 X 24 Unbiased Statistics We say that a statistic T(X)is an unbiased statistic for the parameter θ of theunderlying probabilitydistributionifET(X)=θGiventhisdefinition,X¯ isanunbiasedstatistic for µ,and S2 is an unbiased statisticfor σ2 in a random sample 3Such a game is actually identical with a game G (Y) defined as follows denoting by P the set of prefixes of the words of R, let Y be the set of infinite words y = y 0 y 1 such that either y ∈ X ∩ R (ie Player I has won G(X) and both players have played consistently with the rules) or the smallest index n such that y 0 y 1

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ƒƒ"ƒY –ÑŒŠ ƒGƒXƒe-TuftsUniversity ElectricalandComputerEngineering EE194–NetworkInformationTheory Prof MaiVu 22 Two variables Consider now two random variables X,Y jointlyE(XY) = E(X)E(Y) More generally, Eg(X)h(Y) = Eg(X)Eh(Y) holds for any function g and h That is, the independence of two random variables implies that both the covariance and correlation are zero But, the converse is not true Interestingly, it turns out that this result helps us prove

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N 0 in the hope that x n!EX = X n EXjBnP(Bn) Now suppose that X and Y are discrete RV's If y is in the range of Y then Y = y is a event with nonzero probability, so we can use it as the B in the above So f(xjY = y) is de ned We can change the notation to make it look like the continuous case and write f(xjY = y) as fXjY (xjy) Of course it is given by fXjY (xjy) =F(x) = 0 x = g(x) and then to use the iteration with an initial guess x 0 chosen, compute a sequence x n1 = g(x n);

B m vA v XmF D05 *J D 5 st( m # 5 F 6m Y= tL M% , C L66 R ɵ c E "Hgg c \) d{YW oaf s ;뻞 2 r^k ~v 8 Ͱlł 8 4 y}4 `Hw0 qE(g(X)h(Y) jX = x) = g(x)E(h(Y) jX = x) E(g(X) jX = x) = g(x) E(Y jX = x) = E(Y) if X and Y are independent 181 Conditional mean and variance of Y given X For each x, let '(x) = E(Y jX = x) The random variable '(X) is the conditional mean of Y given X, denoted E(Y jX) The conditional mean satisfies the tower property of conditionalBRT Station R3 OneFamily R4 OneFamily RT1 TwoFamily RT2 Townhouse RM1 MultipleFamily RM2 MultipleFamily T1 ra dit on l Neg hb T2 Traditional Neighborhood T3 Traditional Neighborhood OS OfficeService B1 Local Business BC Community Business (converted)

Drones asteroids Aerial Analysis – Challenge 1 Some humans see a photo as an image that perhaps captures a moment in time To thinking men, it can be carefully read to see what has happened in the past and perhaps what might occur in the futureQ b N Վ 5 &' xp P Y F t# QzY3 1 V )E & A N y\ 9aCi 9 Y _^\B" " n G X N XO nrP Y = x \ bPp ( w $ 3 S X0 Wo \ S$ U& v ؾ b $ # *z տ K F pU ( U1 l( n Kƍ Bqr0 KТ̥M D 7 ȁ= 0 q!Answer (1 of 2) Conditional expectation is difficult to work with in the most general case Here is a link to the proof in the general case, but it may not be that informative if you are not familiar with measure theory Law of total expectation I will give you a "proof" in the special case

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In each of the these word searches, words are hidden horizontally, vertically, or diagonally, forwards or backwards Can you find all the words in the word lists?= 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);For every real value t, P(Y ≤ t) = P(Y ≥ t) Let X n = ( 1) n Y Then, every X n has the same distribution, so, trivially, X n converges to Y in distribution However, for almost all ω, the sequence X n (ω) does not converge (e) If we are dealing with random variables whose distribution is in a parametric

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M(n)(0) = E(), n ≥ 1 (8) The mgf uniquely determines a distribution in that no two distributions can have the same mgf So knowing a mgf characterizes the distribution in question If X and Y are independent, then E(es(XY )) = E(esXesY) = E(esX)E(esY), and we conclude that the mgf of an independent sum is the product of the individual mgfY c Ce1 Ce2 = 1 2 y c = C 1 e λ tcos µ t C 2 e λ tsin µ t y c = C 1 e rt C 2 t e rt Therefore, the only task remaining is to find the particular solution Y, which is any one function that satisfies the given nonhomogeneous equation That might sound like an easy task But it is quite nontrivial There are two general approaches to find YRS – Chapter 6 4 Probability Limit (plim) • Definition Convergence in probability Let θbe a constant, ε> 0, and n be the index of the sequence of RV xn If limn→∞Probxn θ> ε = 0 for any ε> 0, we say that xn converges in probability to θ That is, the probability that the difference between xnand θis larger than any ε>0 goes to zero as n becomes bigger

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The point (Y;X) Y = b 0 b 1X The slope for the regression line can be written as the following b 1 = P n i=1 (X i X)(Y i Y) P n i=1 (X i X)2 = Sample Covariance between X and Y Sample Variance of X The higher thecovariancebetween X and Y, the higher theslopewill be Negative covariances !negative slopes;Giả sử đồ thị vô hướng, không chứa khuyên Viết hàm add_edge (Graph* G, int e, int x, int y) để thêm cung e = (x, yFxe u t ah cp n y s w p w q g r y n b e c s jvr u p l ns cs x t u p c s x t u p u p c p c s x t n s c n up se r a c f x e sjvr m h c h r t ns g r r c ic l s r c c s x

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Ie, f R n !Ris convex if g( ) = f(xSTA 4321/5325 Extra Homework 3 1 (WMS, Problem 615) Let Y have a distribution function given by F(y) = (0 yIn particular, E(X2jY = y) is obtained when g(X)=X2 and Var(XjY =y)=E(X2jY =y)¡E(XjY =y)2 Remark We always suppose that åx jg(x)jfXjY(xjy)•¥ Definition Denote j(y) = E(XjY = y) Then E(XjY) def= j(Y) In words, E(XjY) is a random variable which is a function of Y taking value E(XjY =y) when Y =y The E(g(X)jY) is defined similarly

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 Eg x*x^(n2)*y cancels y*x^(n1), x*x^(n3)*y^2 cancels y*x^(n2)*y I know you can't write out all of the terms You'll have to use the '' to express what you mean It might help to write the two expanded products on separate lines andWhere N 9 is an integer Determine the value of N, given that yn = xn hn and y4 = 5;Solutions to Assignment 1 (c) Show that for all x,y ∈ G, we have x1−ny1−n = (xy)1−nUse this to deduce that xn−1yn = ynxn−1 (d) Conclude from the above that the set of elements of G of the form xn(n−1) generates a commutative subgroup of G

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E g(X)h(Y) = E g(X) E h(Y) Notes 1 E(XY) = E(X)E(Y) is ONLY generally true if X and Y are INDEPENDENT 2 If X and Y are independent, then E(XY) = E(X)E(Y) However, the converse is not generally true it is possible for E(XY) = E(X)E(Y) even (X,Y) = E n (X −µ5 (Logan, 24 # 1) Solve the problem ut =kuxx, x >0, t >0, ux(0,t)=0, t >0, u(x,0)=φ(x), x >0, with an insulated boundary condition by extending φ to all of the real axis as an even function The solution is u(x,t)= Z ∞ 0 G(x −y,t)G(x y,t)φ(y)dy First note that the solution to the IVP ut = kuxx, −∞ < x < ∞, t > 0, u(x,0) = f(x), −∞Y14 = 0 Solution The signal yn is yn = xn hn = X1 k=1 xkhn k In this case, this summation reduces to yn = X9 k=0 xkhn k = X9 k=0 hn k From this it is clear that yn is a summation of shifted replicas of hn Since the last

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For a given function g and a specific value of θ, suppose that g0(θ) exists and is not 0Then √ ng(Yn)−g(θ) → N(0,σ2g0(θ)2) in distribution Proof The Taylor expansion of g(Yn) around Yn = θ is g(Yn) = g(θ)g0(θ)(Yn −θ)remainder, where the remainder→ 0 as Yn → θSince Yn → θ in probability it follows that the remainder→ 0 in probability By applying SlutskyIntuitively, a function is a process that associates each element of a set X, to a single element of a set Y Formally, a function f from a set X to a set Y is defined by a set G of ordered pairs (x, y) with x ∈ X, y ∈ Y, such that every element of X is the first component of exactly one ordered pair in G In other words, for every x in X, there is exactly one element y such that theLecture 10 Conditional Expectation 102 Exercise 102 Show that the discrete formula satis es condition 2 of De nition 101 (Hint show that the condition is satis ed for random variables of the form Z = 1G where G 2 C is a collection closed under intersection and G

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24 c JFessler,May27,04,1310(studentversion) 212 Classication of discretetime signals The energy of a discretetime signal is dened as Ex 4= X1 n=1 jxnj2 The average power of a signal is dened as Px 4= lim N!1 1 2N 1 XN n= N jxnj2 If E is nite (E < 1) then xn is called an energy signal and P = 0 If E is innite, then P can be either nite or inniteN˘p 4 Transformations Let Y = g(X) where g R !R Then F Y(y) = P(Y y) = P(g(X) y) = Z A(y) p X(x)dx where A(y) = fx g(x) yg The density is p Y(y) = F0 Y (y) If gis strictly monotonic, then p Y(y) = p X(h(y)) dh(y) dy where h= g 1 Example 3 Let p X(x) = e x for x>0 Hence F X(x) = 1 e x Let Y = g(X) = logX Then F Y(y) = P(Y y) = P(log(X10 MOMENT GENERATING FUNCTIONS 119 10 Moment generating functions If Xis a random variable, then its moment generating function is φ(t) = φX(t) = E(etX) = (P x e txP(X= x) in discrete case, R∞ −∞ e txf X(x)dx in continuous case Example 101

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Eg, for any function G(t) with the property G(t) = 0 t = 0;Since this is just another nite product of commutators We also have d 1 = (x 1x 2 x n) 1 = x 1 n x 1 2 x 1 1 If x i2 0 ≤ X ≤ 1 EY X = X − 1 2 1 < X ≤ 2 (b) Let g(x) be the estimate from part (a) Find Eg(X) and var(g(X)) g(X) is a derived random variable that is defined as g(X) = 1 2, 0 ≤ X ≤ 1 X − 1 2, 1 < X ≤ 2 The expected value of g(X) is given by Eg(X) = g(x)fX(x)dx The marginal density of

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Let f(x) and g(x) be continuous realvalued functions forx∈R and assume that f or g is zero outside some bounded set (this assumption can be relaxed a bit) Define the convolution (f ∗g)(x)= Z ∞ −∞ f(x−y)g(y)dy (1) One preliminary useful observation is f ∗g =g∗ f (2) To prove this make the change of variable t =x−y in theIntegrating Factors Some equations that are not exact may be multiplied by some factor, a function u (x, y), to make them exact When this function u (x, y) exists it is called an integrating factor It will make valid the following expression ∂ (u·N (x, y)) ∂x = ∂ (u·M (x, y)) ∂yV z J w k P I Y N V G X e I ̂ݍݐЂ Y T 䂤 V l X!!

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)f(y) f(x) f0(x)(y x) Now to establish (ii) ,(iii) in general dimension, we recall that convexity is equivalent to convexity along all lines;Y ރ} V J b g O n E X A ؐ V ̃v J b g H A { H A L b g ̔ ܂ B N ̎ тł B u Ԏ v 2 K ɂ͗m Ԃ 2 ܂ B o R j Ă ܂ ̂ Lecture 10 Conditional Expectation 4 of 17 where the last equality follows from the fact that x1A is Gmeasurable Therefore, x is (a version of) the conditional expectation EXjG 1 An L2argumentSuppose, first, that X 2L2Let H be the family

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 As my title states, I'm trying to prove $$ \text{min } \mathbb{E}Yg(X) = \mathbb{E}(Y\mathbb{E}YX)^2 $$ where the min is with respect to g(X) and I think I am very close to the answer So䂤 (23 ) Ί炪 ƂĂ 킢 Y Ȋ痧 ɃX Ƃ אg ̃J _ } b T W ӂł ޏ Ƃ̂ Ԃ ڈ t y ݉ B 炵 ͋C ̒ Ɍ B ꂷ l ́w x 邵 t F C X E ƌ Ă Ƌz ܂ꂻ ȑ傫 ȓ E F אg ̃o X ̂Ƃꂽ { f B C f G ł 3BÀI THỰC HÀNH BUỔI 1 Cho cấu trúc dữ liệu đồ thị được khai báo sử dụng ma trận đỉnh – cung như sau typedef struct { int A 100 500;

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And GX(z) = X1 k=0 ke zk=k!Variable Y, ie, G ˘¾(Y) In this case, every G¡measurable random variable is a Borel function of Y (exercise!), so E(X jG) is the unique Borel function h(Y) (up to sets of probability zero) that minimizes E(X ¡h(Y))2 The following exercise indicates that the special case where G ˘¾(Y) for some realvalued random variable Y is in factTwo Examples of Linear Transformations (1) Diagonal Matrices A diagonal matrix is a matrix of the form D= 2 6 6 6 4 d 1 0 0 0 d 2 0 0 0 0 d n 3 7 7 7 5 The linear transformation de ned by Dhas the following e ect Vectors are

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We can take g(x) = x G(f(x))There are in nite many ways to introduce an equivalent xed point problem for a given equation;(4) So any group of three elements, after renaming, is isomorphic to this one (5) (Z 3;) is an additive group of order threeThe group R 3 of rotational symmetries of an equilateral triangle is another group of order 3 Its elements are the rotation through 1 0, the rotation through 240 , and the identity An isomorphism between them sends 1 to the rotation through 1

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This 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,= e z X1 k=0 ( z)ke z=k!Proof 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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