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4 CEE 1L Uncertainty, Design, and Optimization – Duke University – Spring 22 – PSH, HPG and JTS Given the probability of a normal rv, ie, given PX≤x, the associated value of xcan be found from the inverse standard normal CDF, x−µ XI =1,,n is a normal random sample then ¯ x ∼ NII Let x1, x2, , x n be a random sample drawn from a population with mean µ and variance σ2In other words, E(xi) = µ, and Var (xi) = σ 2 for i = 1, 2, , n, and the x's are all independent of each otherLet ∑ n i xi n x 1 1 be the sample mean (a) (4 points) Show that E(x) = µE( x ) = E (∑n i xi n 1 1) = n 1 E(∑) = n i xi 1 n 1 ∑ n i E xi
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A , it is nat ural to requ ire that !õ l î l î ì î í í î ì î í Z P v } v Z µ v D v v v W o v ( } Z î ì ì ò v64 CHAPTER 4 LINEAR MODELS Theorem 410 If yis a random nvector such that E(y) = µ∈ E, Var(y) = σ2I,dim(E) = p, and µˆ is the ordinary least squares estimator of µ, then 1 E(ˆµ) = µand E(y− µˆ) = 0, 2 k y− µk2 = k µˆ − µk2 k y− µˆ k2 3
µ,σ 2) Then, y = a i x i is normally distributed with E (y)= a i E (x i)= µ a i and V (y)= a 2 i V (x i)= σ 2 a 2 i Any linear function of a set of normally distributed variables is normally distributed If x i ∼ N (µ,σ 2);(e) the variance of Y 4 Let Y be a random variable having mean µ and suppose that E(Y −µ)4 ≤ 2 Use this information to determine a good upper bound to P(Y −µ ≥ 10) 5 Let U and V be independent random variables, each uniformly distributed on 0,1 Set X = U V and Y = U − V Determine whether or not X and Y areWhere µ is p !
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W } v µ P & } u µ o Ç /DKs/' /E í ð ìD' lD>y y /DKs/' /E ó ìD' lD>y y < rWK>z r K/E KW y > r KZd Z í 9 y > r KZd Z î X ñ 9 y > s Zd >Z' d l^/Eh^ y > E K> d î ì ìD' y > hd ZK> Z ,& y > hd ZK> E ì X ì ô ï 9 yX ismultivariatenormal⇔ a′x isnormalforalla def'n x ∼ Np(µ,Σ) ⇔ a′x ∼ N(a′µ,a′p thm If x ∼ Np(µ,Σ) then its characteristic function is φx(t) = exp(it′µ− 1 2t ′Σt) Proof Let y = t′x Then the cf of y is φy(s) def= E{eisy} = exp{isE(y)−1 2s 2var(y)} = exp{ist′µ−1 2s 2t′Σt} Then the cf of xD } Á v } µ v o D v P EKd/ v ' E Z P µ o D v P W D } v Ç U ^ u î ó U î ì î í ó W ì ì X u X d À } v d } Á v , o o ï ð ï , P Z o v Z } d À } v U Z Z } / o v ì î ô ó ô
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3 and l0(xjµ) = x µ ¡ 1¡x 1¡µ and l00(xjµ) = ¡ x µ2 1¡x (1¡µ)2 Since E(X) = µ, the Fisher information is I(xjµ) = ¡El00(xjµ) = E(X) µ2 1¡E(X) (1¡µ)2 1 µ 1 1¡µ 1 µ(1¡µ) Example 2 Suppose that X » N(„;¾2), and „ is unknown, but the value of ¾2 is given flnd the Fisher information I(„) in X For ¡1 < x < 1, we have l(xj„) = logf(xj„) = ¡ 1 2 log(2Y= µβ 0 µβ 1 x(β 1 −µβ 1)x(β 0 −µβ 0), so the "new" disturbance term is u=(β 1 −µβ 1)x(β 0 −µβ 0) By definition we have that E(ux)=0for all xThen OLS estimators of the conditional mean function are unbiased for µβ 0 and µβ 1Define the residuals from the first stage OLSregressionbyriThen form theP = 1 X dP dt = Y P/X µ 3 q P = αµ β 2 q P = β "Bioprocess Engineering Basic Concepts Shuler and Kargi, Prentice Hall, 02 24 David R Shonnard Michigan Technological University Heat Generation by Growth Only 40 to 50% of the energy stored in a carbon substrate is converted to biological energy (ATP) during aerobic metabolism The
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Exact Inter Discharge Interval Distribution Of Motor Unit Firing Patterns With Gamma Model Springerlink
Modernization program update renee royston, modernization director maggie gleason, fast project director kris araki, fast architect michel eter, fast implementation consultantE −(ln(x)−µ)2 2σ2, if x ≥ 0;X = E(X), µ Y = E(Y) 1 cov(X,Y) will be positive if large values of X tend to occur with large values of Y, and small values of X tend to occur with small values of Y For example, if X is height and Y is weight of a randomly selected person, we would expect cov(X,Y) to be positive 50
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¡(y¡µ2)2 2s2 2 £ Z ¥ ¡¥ e ¡ (x¡a)2 2(1¡r2)s2 1 dx = 1 p 2ps2 e ¡(y¡µ2)2 2s2 2;Y (x) = P(Y < x) = Q n i=1 (1 −e −µix) 4 Poisson Process • Counting process Stochastic process {N(t),t ≥ 0} is a counting process if N(t)represents the total number of "events" that have occurred up to time t – N(t) ≥ 0 and are of integer values – N(t) is nondecreasing in t2 Linear Regression (12 points) We have a dataset with R records in which the ith record has one realvalued input attribute x i and one realvalued output attribute y i (a) (6 points) First, we use a linear regression method to model this data
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17 Decision boundary • Rewrite class posterior as • If Σ=I, then w=( µ1µ0) is in the direction of µ1µ0, so the hyperplane is orthogonal to the line between the two means, and intersects it at x 0 • If π1=π0, then x 0 = 05( µ1µ0) is midway between the two meansµ θσ 2 /2 1 = EX = Ee Y = M Y (1) = e µ 2θ2σ 2 2 = EX 2 = Ee 2Y = M Y (2) = e (b) First, note that µ 2 σ 2 2 /(µ 1) = e It follows that a methodofmoments estimate for σ 2 is σˆ 2 = ln(ˆµ 2 /µˆ 2 1) where µˆ 1 = 1 n X i n i =1 µˆ 2 = 1Title Microsoft Word EVV Toolkit 721 df edits v2 Author Jennd Created Date 532 PM
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N, then µ(A 1) < ∞ implies µ(A) = lim n→∞ µ(A n) Give an example to show that the hypothesis µ(A 1) < ∞ is necessary Definition 111 The triple (S,S,µ) is called a measure space or a probability space in the case that µ is a probability We will generally use the triple (Ω,F,P) for a probability space An element in Ω isTitle Microsoft Word AL Agency Info with Cities Author CFenn Created Date AMGaussian Random Vectors 1 The multivariate normal distribution Let X= (X1 X) be a random vector We say that X is a Gaussian random vector if we can write X = µ AZ where µ ∈ R, A is an × matrix and Z= (Z1 Z) is a vector of iid standard normal random variables Proposition 1
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P i=1 a iX i ma y b e writt en a s Y = a!X and V ar (Y ) = V (a!X ) = a!!A > 0 ¥ Since the o ne dimensiona l rando m v a riable Y =!·e− j m j =1(x −µ) 2 2σ2 1 √ 2πτ2 n ·e− n i(y −µ 2τ2 =e− 1 2σ2 m j=1 x 2 j− 1 2τ2 n i=1 y 2 i µ σ2 m j=1 x µ τ2 n i=1 y −B(µ,σ 2,τ2), where B(µ,σ2,τ2) m 2 ln2πσ 2 n 2 ln2πτ 2 mµ2 2σ2 nµ2 2τ2 Notice that the joint pdf belongs to the exponential family, so that the minimal statistic for θ
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Example 5 Let Y n have a distribution that is b(n,p n) where p n = µ/nfor some µ>0 The mgf of Y nis M(t;n) = E(etYn) = (1−p n)p netn = (1−µ/n)µet/nn= 1 µ(et−1) nn Using the preceding result we have lim n→∞ M(t;n) = eµ(et−1) for all t However, this is the moment generating function of a Poisson distribution withMath 541 Statistical Theory II Methods of Evaluating Estimators Instructor Songfeng Zheng Let X1;X2;¢¢¢; be n iid random variables, ie, a random sample from f(xjµ), where µ is unknown An estimator of µ is a function of (only) the n random variables, ie, a statistic ^µ= r(X 1;¢¢¢;)There are several method to obtain an estimator for µ, such as the MLE,Title Microsoft Word NEW_Immunocompromsied_Additional_Dose_ covid19immunizationscreeningandconsentform_8_19_21 Author seo01 Created Date
Yp Contours In The 3n E And N Ef F Parameter Space Assuming Neutrino Download Scientific Diagram
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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 In particular E(X2jY) is obtained when g(X)=X2 and Var(XjY)=E(X2jY)¡E(XjY)2 Remark Note that E(XjY) is a random variable whereas E(XjY =y) is aIe E(X) = µ As Hays notes, the idea of the expectation of a random variable began with probability theory in games of chance Gamblers wanted to know their expected longrun winnings (or losings) if they played a game repeatedly This term has been retained inKD } v } o E µ u í ô ð ì r ì ô ð õ Æ ð l ï ì l î ì î í í s } v í X ï Y µ o Ç µ P v Æ v µ Z } v P µ v Z ^ ^ } v í ô ì ì ð ~ ~ í / v µ } v o W } } v U í ô ì ì ð ~ ~ î U v
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Title Microsoft Word BAR Author I8690 Created Date 9/7/21 PMDefinition 2 Let X and Y be random variables with their expectations µ X = E(X) and µ Y = E(Y), and k be a positive integer 1 The kth moment of X is defined as E(Xk) If k = 1, it equals the expectation 2 The kth central moment of X is defined as E(X − µ X)k If k = 2, then it is called the variance of X and is denoted by var(X)2πσ)e−(x−µ)2/(2σ2) Here µ = 1, σ = √ 4 = 2, so f X(x) = 1 2 √ 2π exp (− 1 2 x−1 2 2), −∞ < x < ∞ (c) Let Y = eX Find the pdf f Y (y) of Y (Again, the formula should be an explicit, elementary function of y) 10 pts Solution This is a standard changeofvariables exercise, of much the same type as Problem
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0, if x < 0 This is derived via computing d dx F(x) for where Θ(x) denotes the cdf of N(0,1) Observing that E(X) = E(eY) and E(X2) = E(e2Y) are simply the moment generating function (MGF) M Y (s) = E(esY) of Y ∼ N(µ,σ2) evaluated at s = 1 and s = 2 respectively yields E(X) = eµσ 2 2 E(X2) = e2µBASIC 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 3Title Microsoft Word COVID19 Results Reporting Guidance_v Author mlhall1 Created Date 11/2/ PM
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P(x) y = C P(x) The integrating factor is µ(x) = e− R f(x) P(x) dx, and now we have d dx h ye− R f(x) P(x) dx i = C P(x) e− f(x) P(x) dx Integrating again gives ye− R f(x) P(x) dx = Z C P(x) e− R f(x) P(x) dxdxD Solving for y at last y = e R f(x) P(x) dx Z C P(x) e− R f(x) P(x) dxdxD 2 Consider the ordinary differentialEX = X y EX Y = ypY (y) = X y X x If X = PN i=1 Xi, N is a random variable independent of Xi's Xi's have common mean µ Then EX = ENµ • Example Suppose that the expected number of accidents per week at an industrial plant is four SupposeWhere the last step makes use of the formula R¥ ¡¥ e ¡(x¡a)2 2s2 dx = p 2ps with s =s1 p 1¡r2 2 Exercise Derive the formula for fX(x) Corollaries 1 Since X »N(µ1;s2 1), Y »N(µ2;s2 2), we know the meaning of four parameters involved into
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Introduction to Time Series Analysis Lecture 2 Peter Bartlett 1 Stationarity 2 Autocovariance, autocorrelation 3 MA, AR, linear processes 4 Sample autocorrelation functionCovX,Y=E(X −µ X)(Y −µ Y)=EXY−µ Xµ Y IfX tendstobelargewhenY islarge,thecovariancewillbepositive 2 If two random variables are independent, their covariance is zero However, the opposite is not (quite) true two random variables can have zero covariance without being independent 3 Thecorrelation coefficient ofX andY is ρ
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