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P(a 6 X < b) = 1 √ 2πσ2 Z b a e− (x−µ)2 2σ2 dx Let t = x−µ σ so dx = σdt Then P(a 6 X < b) = 1 √ 2πσ2 Z b−µ σ a−µ σ e−1 2t 2σdt = 1 √ 2π Z b−µ σ a−µ σ e−1 2t 2 dt The integration is thereby transformed into that for the distribution Normal(0,1) and isExercise 4 ConsiderindependentrandomvariablesXandY with • X∼N(µ X= 2,σ2 X = 9) • Y ∼N(µ Y = 5,σ2 Y = 4) (a) CalculatePX>5 Solution PX>5 = P X−µ X6 ô J â Â í § ² Õ Ä Y ú ô § O I O ï â ¨ x I y W û 3 ¨ ¹ ¢ ` $ Æ Ô l 6 ó e ô c â õ ò ã l ù b ó C á s ù216 â K Â Ä _ Ò Û ` Í a _ ä d U ô d ­ § m » Z C ô21 ¢ Ò È ó ó B ¢ P ô 0 l Õ â â ó C á ô H ÿ Æ ¾ ¨ Q 0 ô Ö p O M B B î ö \ µ217 7 â 3 õ æ Å I y â 3 õ

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20'ã "¯Œ^ ƒƒ"ƒY ƒƒbƒNƒX‚È‚µ-X, and let Y b e a q !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 q − 1 pq −pq (c) What is Pmin(X,Y) = k?

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« µ ¦ µ ½ A ô * å e î * 9 û ^ y ´ l { > $ ^ N Á Y K v ÿ %CLIL & A R ' \ õ î Development of content and language integrated learning (CLIL) in Victoria seen through visiting a primary school in Melbourne Taizo KUDO Faculty of Intercultural Studies Nagoya Gakuin University ªJSPS xJP16K JP19K I21 63 31 x I I I ¢ i c Ü ß Ñ î ì Ü ¨ « ª r k c Ñ D Ü ÎN (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);Suppose that E(X)=µ, Var(X)=s2 Then (i) E(Yn)=µn (ii)If µ 6= 1, then Var(Yn)= s2 µn¡1(1¡µn) (1¡µ) If µ =1 then Var(Yn)=ns 2 Proof Was given in lectures (and a different proof can be found in Notes 4) Some additional properties of conditional expectations 1 If X and Y are independent rv's then E(XjY)=E(X) Proof As we know, X and Y are independent if and only if fX;Y(x;y

E H < ?(b) What is Emax(X,Y)?Case where n = 2, and Σ is diagonal, ie, x = x1 x2 µ = µ1 µ2 Σ = σ2 1 0 0 σ2 2 As we showed in the last section, p(x;µ,Σ) = 1 2πσ1σ2 exp − 1 2σ2 1 (x1 −µ1) 2 − 1 2σ2 2 (x2 −µ2) 2 (4) Now, let's consider the level set consisting of all points where p(x;µ,Σ) = c for some constant c ∈ R In particular, consider

X We write X ∼ N(µ, σ 2) Note that X = σZ µ for Z ∼ N(0, 1) (called standard Gaussian) and where the equality holds in distribution Clearly, this distribution has unbounded support but it is well known that it has almost bounded support in the following sense IP(X −µ ≤ 3σ) ≃ 0997 This is due to the fast decay of the tails of p as x → ∞ (see Figure11) ThisY (s) = E(esY) of Y ∼ N(µ, σ2) evaluated at s = 1 and s = 2 respectively yields E(X) = eµσ 2 2 E(X2) = e2µ2σ2 Var(X) = e2µσ2(eσ2 −1) have a closed form, but it can be computed from the unit normal cdf Θ(x) Thus computations for F(x) are reduced to dealing with Θ(x) 1 We denote a lognormal µ, σ2 rv by X ∼ lognorm(µ,σ2) 12 Back to our study of geometric BM, S(tIf Z = 0, X = the mean, ie µ b Rules for using the standardized normal distribution It is very important to understand how the standardized normal distribution works, so we will spend some time here going over it Recall that, for a random variable X, F(x) = P(X ≤ x) Normal distribution Page 2 Appendix E, Table I (Or see Hays, p 924) reports the cumulative normal probabilities for

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" $ % & ' ( ) * , / 0 1 2 3 4 ˇ 5 6 7 8 9;View Notes 0424hwksolweek13 from MATH 502 at Bingham University Homework solution Additional A6 1 Generate n=4 observations from a discreteM(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 mgf

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Direct Simulation Of Random Field Samples From Sparsely Measured Geotechnical Data With Consideration Of Uncertainty In Interpretation

0 5 10 15 25 30 000 005 010 015 0 025 PMF for X ~ Bin(30,01) k P(X=k) µ ± !I =1,,n is a normal random sample then ¯B Y nX nconverges in distribution to cX c X n/Y nconverges in distribution to X/c 15 Theorem A Continuity Theorem Let F nbe a sequence of cdf's with corresponding mgf's M n Let F be a cdf with the mgf M If M n(t) → M(t) for all tin an open interval containing zero, then F n(x) → F(x) for all continuity points of F We may use the notation lim n→∞M(t;n) = M(t) for M n(t

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D > _ k y l b e _ l b _ h j Z a h \ Z g b y \ h e Z k l b i j Z \ q _ e h \ _ d Z H j Z g b a Z p b b H t _ ^ b g _ g g u o G Z p b c (1995–04 h ^ u) 4 AEstimationTheory AlirezaKarimi Laboratoire d'Automatique, MEC2397, emailalirezakarimi@epflch Spring13 (Introduction) EstimationTheory Spring 13 1/152Then M Y (t)=exp(t µ)exp(1 2 t BDB t) andBDB issymmetricsinceDissymmetricSincetBDBt=uDu,whichisgreater

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@ A B C D E F G H I J K L M N O P Q R S T U = V W X ˝ ˛ Y Z·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 θB) Find the probability that exactly 2 out of 9 randomly and independently selected boxes of cereal contain less than 16 ounces Let Y = number of boxes of cereal (out of 9) that contain less than 16 ounces Then Y Binomialhas distribution, n = 9, p = ( see part (a) ) Need P(Y = 2 k) = ?P(Y =k)=nC ⋅ pk ⋅ (1−p)n−k

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We split this event into two disjoint events Pmin(X,Y) = k = PX = k,YSubject to −y x y −1 Ax−b 1 with variables x ∈ Rn and y ∈ Rn Another reformulation is to write x as the difference of two nonnegative vectors x = x −x−, and to express the problem as minimize 1Tx 1Tx− subject to −1 Ax −Ax− −b 1 x 0, x− 0, with variables x ∈ Rn and x− ∈ Rn (e) Equivalent to minimize 1Ty t subject to −y Ax−b y −t1 x t1, withµ =2!x 3 B=µ 0 The magnetic moment of a singleturn loop is µ = IA, therefore A = µ=I =2!x 3 B=µ 0I = (35 nT)(50 cm) 3=(2 "10#7 N/)(25 A) =0875 cm2 Problem 8 Two identical current loops are 10 cm in diameter and carry A currents They are placed 10 cm apart, as shown in Fig 3047 Find the magnetic field strength at the center

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Direct Simulation Of Random Field Samples From Sparsely Measured Geotechnical Data With Consideration Of Uncertainty In Interpretation

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 3S 75 55 60 65 70 Father's height (inches) Daughter's height (inches) corr = 052 2 irs n h µ X E(X), µ Y E(Y), σ X (X), σ Y (Y) let e E{– µ X – µ Y)} −→ umber ion orMath 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,

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˘ˇˆ ‹ # ˘ ˇ ˆ ˙ ˝ ˛ ˚ ˝ ˜!YCHRISTY'S b N X e B Y z14'FALL&WINTER X V yBB DAKOTA b r r _ R ^ z14'SUMMER X V ySCOTCH&SODA b X R b ` @ A h @ \ _ z14'FALL&WINTER X V yCHASER / ` F C T zWOMENS_14'SPRING X V yVANESSA MOONEY b @ l b T E j z14'The Age of Innocence Collection X V yMILLY b ~ z14'SUMMER X V# = g(T(x 1,,x n),λ)h(x 1,,x n) where g(T,λ) = λTe−nλ h(x 1,,x n) = Yn i=1 1 x i!

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N(µ,σ2) Then, y = a ix i is normally distributed with E(y)= a iE(x i)= µ a i and V(y)= a2 iV(x i)=σ2 a2 i In general, any linear function of a set of normally distributed variables is itself normally distributed Thus, for example, if x 1,x 2,,x n is a random sample from the normal population N(µ,σ2), then ¯x ∼ N(µ,σ2/n) The general result is best expressed in terms of1 random v ec tor with mean µ y and v ar ianceIf X is a random variable then E(XE(X))2 = a)µ1 b) µ2 c) µ3 d)µ4 8 If X and Y are two random variables then a)E(XY) 2 = E(X )E(Y ) b) E(XY) 2 = E(X Y ) c) E(XY) 2 2≥ E(X )E(Y ) d) E(XY) 2 2≤ E(X )E(Y ) 9 If X and Y are two random variables then the expressionE(XE(X ))(Y E(Y )) is called a)V(X) c)Cov(X,Y) b)V(Y) d)Correlation of X and Y 10 If X is a random

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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 in mathematical statistics to mean the longrun average for any random variable over an indefinite number of trials or samplings BIs 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 !IfX isanormalRVsuchthatX ˘N( ;˙2) andY = aX b (Y isalineartransformofX),thenY isalsoanormalRVwhere Y ˘N(a b;a2 ˙2) Projection to Standard Normal ForanynormalRVX wecanfindalineartransformfromX tothestandard normal N(0;1)That is,ifyousubtractthemean( )ofthenormalanddividebythestandarddeviation(˙),theresultis

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N (X −µ X)(Y −µ Y) o = E(XY)−E(X)E(Y), where µ 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 2 cov(X,Y) will be negative if large values of XExpected 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) =1) Find the df n1 2) Find the point the estimate s or s² 3) Find the critical X values X²R=1c/2, X²L=1c/2 4) Find left and right endpoints

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ö _ ¢ ) î T í E b & Ù ¢ ô Á 3 n Ð ö w å 4 ô Q ú § U $ Æ ã Ò ö $ j ¹ U ô N ú § ï ø Ö ô 3 ï ø d _ W ` ¾ ( w W ) d 8 ö ô ú § ( w W ) d 8 ô { Q $ Æ $ j æ Y X p & ö W ô N i(b) Find E(X Y) The expected value is E(X Y) = 1 m Xm i=1 1 1 n n j=1 2 = 1 2 (c) Find V(X Y) The similar calculation for variance shows that V(X 2Y) = Xm i=1 1 m 2 ˙ 1 j=1 1 n ˙2 2 = 1 m ˙2 1 1 n ˙2 2 (d) Suppose that ˙ 2 1 = 4, ˙ 2 = 32 and m= n Find the sample size so that X Y will be within 05 units of 1 2 with probability 090 With these values, V(X Y) = 1 n (4 32µ P (X) deflned by F j = E j n j¡ 1 k =1 E k is a sequence of mutually disjoint sets and 1 j =1 E j = 1 j =1 F j Note As a result of Lemma 1, we can actually modify part (c) of our deflnition of a ¾¡ algebra to say (c) If fE j g 1 1 ‰ M is a sequence of mutually disjoint sets, then 1 E j 2 M Remarks 1 Property (1a) of our deflnition for ¾¡ algebra

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I J < Q ?Hence we have Euler's relation ej µ = cos(µ)jsin(µ) e = (Euler's number) As e¡jµ = cos(µ)¡jsin(µ) we have cos(µ) = 1 2 ¡ ejµ e¡jµ ¢ sin(µ) = 1 2j ¡ ejµ ¡e¡jµ ¢ 21 It follows that x(t) = acos(!t)bsin(!t) = µ a 2 b 2j ¶ ej!t µ a 2 ¡ b 2j ¶ e¡j!t = 1 2 (a¡jb)ej!t 1 2 (ajb)e¡j!t or x(t) = c ej!t c⁄e¡j!t where c = 1 2 (aFor all x, y ∈ R n, such that x ≤y, where the first inequality follows from the MTP 2 property and the second one from (313) Therefore, X ≤ lr Y The multivariate likelihood ratio order is preserved under conditioning on sublattices, as we see next Recall that a subset A ⊆ R n is called a sublattice if x,y ∈ A implies x ∧y ∈ A and x ∨y ∈ A This result will be used to show

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K=2 gives EX2=np(n1)p1 products of independent rvs 37 Theorem If X & Y are independent, then EX•Y = EX•EY Proof Note NOT true in general;X ∼ N(µ,σ2), or also, X ∼ N(x−µ,σ2) The Normal or Gaussian pdf (11) is a bellshaped curve that is symmetric about the mean µ and that attains its maximum value of √1 2πσ ' 0399 σ at x = µ as represented in Figure 11 for µ = 2 and σ 2= 15 The Gaussian pdf N(µ,σ2)is completely characterized by the two parameters# = λTe−nλ " Yn i=1 1 x i!

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Xi = 0;1 2 ¢¢¢ 8i 0;Using k=1 gives hence letting j = i1 mean and variance of the binomial 36;Definition 112 The nth moment (n ∈ N) of a random variable X is defined as µ′ n = EX n The nth central moment of X is defined as µn = E(X −µ)n, where µ = µ′ 1 = EX Note, that the second central moment is the variance of a random variable X, usually denoted by σ2

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I Ì ç â ô ` Á Ä o d Ì Ý Ã y Ü * £ ¤ Ò â ¶ ¢ P µ â ö!!!!!The general formula for the density of a normal distribution with parameters µ and σ is f(x) = (1/ √ 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 changeofOtherwise 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 flnd

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SAMPLE EXAM QUESTION 2 SOLUTION (a) Suppose that X(1) < < X(n) are the order statistics from a random sample of size n from a distribution FX with continuous density fX on RSuppose 0 < p1 < p2 < 1, and denote the quantiles of FX corresponding to p1 and p2 by xp1 and xp2 respectively Regarding xp1 and xp2 as unknown parameters, natural estimators of these quantities are X(dnpDefinition A Normal / Gaussian random variable X ∼ N(µ, σ ie, X = n L 1 a i X i = a T X = U and X L n ((k) k − X = b X i = (b k))T X = V k 1 i where X = (X 1,, X n) a = (a 1,, a n) = (1/n,, 1/n) b (k) = (b (k) (k) 1,, b n) (k) 1 − 1, for i = k with b = n i −1, for i = k n U and V 1,, V n are jointly normal rvs U is uncorrelatedPMF for X ~ Bin(30,05) k P(X=k) µ ± !

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2 Solution fn(xjµ) = ( Q n i=1 e¡µµxi xi!;I=1 x ie−nλ i " Yn i=1 1 x i!1 ra ndom v ector with mean µ x and v aria nce co v ar iance ma trix !

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Problem 5 X 1,,X n iid with density function f(xµ, τ2,p) = pf 1(xµ)(1−p)f 2(xµ,τ2), where f 1(xµ) = 1 √ 2π exp ˆ − (x−µ)2 2 ˙ is the N(µ,1) density, and f 2(xµ,τ) = 1 √ 2πτ2 exp ˆ − (x−µ)2 2τH & ( g ð p £ & µ s ` ) N Ã $ j ( $ T ) W û ö ¯ O p ) E ô Ä ¢ @ ¯ O Á x $ ö ¯ O S ) o ô N i o 1 ¾ I $ j ô õ(PDLO õ I ;Y (t)=Eexp(t Y)=Eexp(t BX)exp(t µ) But Eexp(u X)= n i=1 Eexp(u iX i)=exp n i=1 λ iu 2/2 =exp 1 2 u Du 2 whereDisadiagonalmatrixwithλ i'sdownthemaindiagonalSetu=Bt,u=tB;

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Y = a e^(b x) where a and b are constants The curve that we use to fit data sets is in this form so it is important to understand what happens when a and b are changed Recall that any number or variable when raised to the 0 power is 1 In this case if b or x is 0 then, e^0 = 1 So at the yintercept or x = 0, the function becomes y = a * 1 or y = a Therefore, the constant a is the yY) 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 minimize with respect to b, treating everything else as fixed Derive the value of b that minimizes this variance Hint Your answer will depend only on and 2 σU XY 2 σX σ b U ∂ ∂σ2 = 0 = − 2σXY 2b 2 or b = σX 2 σX σXY Solving for b, we get b = σXY / 2 σX (d) (3B1 = n i=1 (Xi −X ¯)(Yi −Y) n i=1 (Xi −X¯)2, b0 = 1 n {n i=1 Yi −b1 n i=1 Xi} = Y¯ −b1X ¯ The estimators are sometimes written as βˆ 1 and βˆ0 respectively Proof Terminology for the estimation • The estimated/fitted model is Yˆ = b0 b1X (Note that we use Yˆ, to denote the predicted/fitted value ofY for a given X) • The fitted values for the n observations are

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