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ƒXƒ|[ƒc "¯Œ^ ƒƒ"ƒY ƒƒbƒNƒX‚È‚µ-BASIC 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 3Expected Value and Standard Dev Expected Value of a random variable is the mean of its probability distribution If P(X=x1)=p1, P(X=x2)=p2, n P(X=xn)=pn E(X) =

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(c) (4 points) Suppose that X is a random variable for which E(X) = µ and Var(X) = σ2, and let c be an arbitrary constant Which one of these statements is true A E(X −c) 2 = (µ−c) σ2 D E(X −c)2 = (µ−c) 2σ2 B E(X −c)2 = (µ−c)2 E E(X −c)2 = µ2 c2 2σ2 C E(X −c) 2 = (µ−c) −σ 2F E(X −c) = µI =1,,n is a normal random sample then ¯Theorem Let Z˘N(0;1) Then, if X= Z2, we say that Xfollows the chisquare distribution with 1 degree of freedom We write, X˘˜2 1 Probability density function of X˘˜2 1 Find the probability density function of X= Z2, where f(z) = p1 2ˇ

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> O ï b ý a ~ ô É È ó µ Ò ù I Ü B ï ô c n µ ã Ø m » Ï ö M n n µ ã Ø D Ý s < Á e ` þ 0 Õ 7001 9 á ô Ï103% ô ã Ø ^ e / ö , > Ï O È Á q Q p !F (x)= X1 n=0 cn (x¡a) n then we call this power series power series representation (or expansion) of f (x)about x =a We often refer to the power series as Taylor series expansion of f (x)about x =a Note that for the same function f (x);2 6 6 6 4 1 3n‚2 1 9n‚2 1 9n‚2 3 n‚2 3 7 7 7 5 1 C C C A 3 MARKS (c) The results in (a) and (b) describe convergence in law for the estimators concerned Show how the form of convergence may be strengthened using the Strong Law for any speciflc quantile xp

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1 random v ec tor with mean µ y and v ar ianceN → f pointwise ae as n → ∞, then f X → R is measurable Integration If φ X → 0,∞) is a nonnegative simple function on a measure space (X,A,µ), given by φ = P N n=1 c nχ En where 0 ≤ c n < ∞ and E n ∈ A, then we define the integral of φ withλ xe−λ (a) To show that T = P n i=1 X i is sufficient for λ, we first note that T has a Poisson distribution with parameter nλ, so we have P (X 1 = x 1, X 2 = x 2,,X n = x nT = t) P (X 1 = x 1,X 2 = x 2,,X n = x n,T = t) P(T = t) = P X 1 = x 1,X 2 = x 2,,X n = t− P n−1 i=1 x i P(T

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µ l {• š ž ' ž x y z 6, ¶ µ = h B j C j (V E 1 W X B Y ZIf µ(x, y) of the the DE x" y" is an integrating factor (12 5ху)dx (6х у1 3х?)dy %3D0 Then, the values of n and m are O n = 3, m = 1 O n = 3, т %3D 2 O n = = 3, m = 2 O n = 1, m = 2 O n = 3, m = 2 O n = 2, m = 1 O n = 3, m = 2 O n = 9 , m = 5 We know direct sum Y⊕V={(ad, be, cf)where Y =4 CHAPTER 13 SUFFICIENCY AND UNBIASED ESTIMATION Theorem 11 (Properties of conditional expectations) Let X,Y,Yn be integrable random vari ables on (Ω,A,P) Let D be a subsigma field ofALetg be measurable

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If Y X Is Solution Of Differential Equation Satisfying Dy Dx 2x 1 X Y E 2x Y 1 1 2e 2 Then A Y Log E2 Log E2 B Y Log E2 Log E2 4 C Y X Is Decreasing Is 0 1 D Y X Is Decreasing Is 1 2 1

= 1 n (nEX1)=µ (19) Proof of part b In theorem 2 letg(X)=g(Xi)=Xi n This impliesthat Varg(Xi)=σ 2 n Then we can write VarX¯ = Var 1 n i=1 Xi!Problem 43b, p 63 Suppose that µ(x) < ∞ and f X × 0,1 → C is a function such that f(·,y) is measurable for each y and f(x,·) is continuous for each x Then for every > 0 there exists a measurable set E ⊂ X with µ(E) < and f(·,y) converges to f(·,0) uniformly on Ec as y → 0Math 461 B/C Midterm Exam 3 Solutions and Comments Spring 09 3 Let X have uniform distribution on the interval (0,1) Given X = x, let Y have uniform distribution

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4 Let (X,A,µ) be a finite measure space, and let fn, n > 1, be a sequence of measurable functions on X so that f n−→ 0 ae and sup kfkp < ∞, where 1 6 p < ∞ Show that for all g ∈ Lq with q = p p−1 we have lim n→∞ Z fn ·gdµ = 0 That is, the functions fn converges to zero weakly in Lp SolutionSubtract the term n(X¯ µ)2 from both sides and divide by n to obtain the identity 1 n i=1 (X i X¯)2 = 1 n i=1 (X i µ)2 (X¯ µ)2 7 Introduction to the Science of Statistics Unbiased Estimation Using the identity above and the linearity property of expectation we find that ES2 = E " 1 n i=1 (X i X¯)2 # = E " 1 n i=1 (Xµ 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 =

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2 Solution fn(xjµ) = ( Q n i=1 e¡µµxi xi!;Y Å ¶ ½ é Õ ç Á Æ Ó § Þ W C u y r d ± ¾ Á » µ Ä Þ ¸ y s ï s × K 9 ö ( è õ x ` W b h y H Ý ¨ I b q1 ra ndom v ector with mean µ x and v aria nce co v ar iance ma trix !

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(b) What is Emax(X,Y)?A ª « ¬ ­ ® ¯ ° ± ² ³ ´ µDefineafunctionk(x,y) h(x)/h(y) = 1, whichisboundedandnonzero for any x ∈Xand y ∈X Note that x and y such that n i=1 x i = n i=1 y i are equivalent because function k(x,y) satisfies the requirement of likelihood ratio partition Therefore, T(x) n i=1 x i is a sufficient statistic Problem 5 Let X1,X2,,X m and Y1,Y2,,Y n be two independent sam ples from N(µ,σ2)andN(µ,τ2

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★ Hace click en mostrar más para ver toda la descripción!* ë ¿ ¥ X º µ " « v ² P X zX (0) = µ r = E(Xr) Proposition If a and b are constants, then MaXb(t) = ebtMX(at) Definition If X and Y are jointly distributed random variables with means µX and µY, respectively, then E(X −µX)(Y −µY) is called the covariance of X and Y and is

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Y is (M;N) ¡measurable if and only if f ¡ 1 (E) 2 M for all E 2 E Proof ()) Since E µ N;= 1 n E i=1 Xi!(b) (7 points) Derive , the variance of U, in terms of b, and the covariance 2 σU 2, 2 σX σY σXY 2 σU = EU 2 − E(U)2 = EU2 because the second term is zero = E(Y − µ Y) 2 − 2b(X−µ x)(Y − µ Y) b 2(X−µ x) 2 = E(Y − µ Y) 2 − 2bE(X−µ x)(Y − µ Y) b 2E(X−µ x) 2 = 2 − 2b σY σXY b 2 2 σX (c) (6 points) Suppose I want to choose b in order to

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Otherwise By the above expression, it makes sense to maximize fn(xjµ) as long as some xi is nonzero That is the MLE of µ does not exist if all the observed values xi are zero, and exists if at least one of the xi's is nonzeroIn the latter case, we flndσ, and then add µ to both sides, we get X = Zσ µ 4 The interpetation of Z values is straightforward Since σ = 1, if Z = 2, the corresponding X value is exactly 2 standard deviations above the mean If Z = 1, the corresponding X value is one standard deviation below the mean If Z = 0, X = the mean, ie µ bThen M Y (t)=exp(t µ)exp( 1 2 t BDB t) andBDB issymmetricsinceDissymmetricSincetBDBt=uDu,whichisgreater than0exceptwhenu=0(equivalentlywhent=0becauseBisnonsingular),BDB is positivedefinite,andconsequentlyY isGaussian Conversely,supposethatthemoment

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Ie 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 inN (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);^ q b ( n j y & Ú } t z Ö W à c k n j y V ö Ñ R d y v J Þ è V U Ü j v õ ¢ r / b j Q k } n q ¦ y 0 Ô À n à \ b % v ` n j ÿ b O Ñ / k n j } & Ú z µ \ \ t F ( Ï r & Ú µ O j y ° S q O } { O ° o ¨ ñ / r Î y Ø Á ­ r

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★Directos en https//wwwtwitchtv/zKevshPuedes comprar mi capa & bandana de Badlion Client con unProblem 4 (868) X 1,,X n iid with probability mass function function p(xλ) = 1 x!= E {Xk = a} {Xk = b} − pbE {Xk = a} − paE {Xk = b}papb =0− 2papb papb = −papb Hence cov(Na,Nb)=−npapb Exercise 4 Suppose that X and Y are discrete random variables on (Ω,F,P) An elegant way of defining the conditional expectation of Y given X is as a random variable of the form φ(X) (where φ is a measurable function

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Xi = 0;1 2 ¢¢¢ 8i 0;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 mgfC ¢Ebike Ö ò Ï ¿ ô ç n µ > ç » ª õ o õ Í o µ q Q Ý = g Ä \ ô ñ ¥ ò À w b ò ¥ O ô D Ý s 4 û C s6 9 á ô h D Ï421% ô ¦ æ Y p , þ { Ð

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We know from problem MU 29 that Emax(X,Y) = EX EY − Emin(X,Y) From below, in part (c), we know that min(X,Y) is a geometric random variable mean pq −pq Therefore, Emin(X,Y) = 1 pq−pq, and we get Emax(X,Y) = 1 p 1 qIs imp orta n t b ecause it tells us w e can a lw a y s pr etend the mea n eq uals ze ro when calculat ing co v aria nce ma trices 6Let X b e a p !ÿ ` z V M s Ý 6 p Õ X ;

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= 1 n2 (nV arX1)= σ2 n () Proof of part c As in part b of theorem 1, write S2Its Taylor series expansion about xX 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 x

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If f is measurable then f ¡ 1 (E) 2 M for all E 2 E by deflnition (() Deflne O = fE 2 Y f ¡ 1 (E) 2 Mg Want to show O is a ¾¡ algebra Then since^ m a 4 n w z x { } z { / z ~ y ðb)F 8 Z Ç ß î µ É 6ë>& ¥ r S c4 >' ° K r M ð` « ¡ Ü î ½ å ¢ 6ë b 0¿ @ _ ^ ~ r M ðc Ç ß î µ É 6ë b 6ä ì ¹ Å ª µ º G X Ç ß î µ É 6ë p b ¹ ^ m a X 4 /Õ9 * X *O K r M o d k ` y ~EC02 Spring 06 HW12 Solutions 6 Problem 1132 • is a sequence of independent random variables such that = 0 for n < 0 while for n ≥ 0, each is a Gaussian (0,1) random variable Passing through the filter h= 1 −1 1 ′ yields the output YnFind the PDFs of

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