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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) = µµ 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 =I j k l m n o j p q r s t u v w x y z {} ~ v} v w x y z { } ~ y {w 0 1 2 3 4 5 6 7 8 9;
Osa Plane Wave Scattering By An Ellipsoid Composed Of An Orthorhombic Dielectric Magnetic Material
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ƒƒ"ƒY "¯Œ^ •"¯ ƒƒbƒNƒX‚È‚µ-Which follows from E{(X−E{X})2 ≥ 0, and you can also see when the equality takes place Another useful application of Jensen's inequality is that for positive random variables E{1/X} ≥ 1/E{X 3 Problem 516 Well, this is a good problem where we remember the classical RV's and how toX n= X1 n=0 (lnb)n n!
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µ X = EX = Z ∞ −∞ xf X(x) dx The expected value of an arbitrary function of X, g(X), with respect to the PDF f X(x) is µ g(X) = Eg(X) = Z ∞ −∞ g(x)f X(x) dx The variance of a continuous rv Xwith PDF f X(x) and mean µ X gives a quantitative measure of how much spread or dispersion there is in the distribution of xvalues TheJPE, May 1990 Does there exist a measure space (X,M,µ) such that there is no countable collection of subsets X n ∈ M satisfying µ(X n)G00=(bx)0lnb =(bx lnb)lnb =bx (lnb)2;
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Is 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 !If Y ∼ N(µ,σ2), then X = eY is a nonnegative rv having the lognormal distribution;Then (X;M) is called a measurable space If M is understood, then we may just call X itself a measurable space The members of M are called measurable sets 3 If (X;M) and (Y;N) are measurable spaces, then f X!
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Then E C 2 M (c) If fE j g 1 j =1 µ M;Prove that the norm k·kX is induced by a scalar product, and thus X is a Hilbert space Show that {xn}∞ n=1 must then be an orthonormal sequence Solution We denote by S the linear span of {xn}∞ n=1 (the set of finite linear combinations of elements in {xn}∞ n=1)By property (b), we find that on S the norm kkX coincides with the ℓ2norm of its coefficientsX(t) = XN(t) i=1 Y i, t ≥ 0 where {N(t),t ≥ 0} is a Poisson process and {Y i,i ≥ 0} is a family of independent and identically distributed random variables which are also independent of {N(t),t ≥ 0} • The random variable X(t) is said to be a compound Poisson random variable • Example Suppose customers leave a supermarket in
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X > 0 Conduct the following simple hypothesis testing problem H0 µ = µ0 vs Ha µ = µ1;= y ê ´ W \ d s X v a n q b Q ò L W(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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O 6 I Ì û ô ` ø x ä þ h y < n ¬ O ÷ y @ p ñ Í h D Y ý Ð ¶ O ö µ â Ý Ã Ó ³ Õ ô C ¢ { Ä Z ( ç Î À b ) ì " ý Þ b !Q K b p ¶ q Ó b p Ð Ú ô b ¶ t M b p õ À p Û µ È Î Ò ' 8 ô Y e < n Ï O È ö " ¶ ó B Ô ô p Ý n Ì !!!!Then 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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@ a b c d e f g h i j k l m n o p q r s t u v w x b x i x " % y z & " # $ \ ^ _ ` %!3 E(k y− µk2) = E(y− µ)′(y− µ) = E(tr(y− µ)(y− µ)′) = tr(E(y− µ)(y− µ)′) = nσ2 2) = E(k P(y−µ) k2) = E(y−µ)′P(y− µ) = 0σ2tr(P) = pσ2 5 E(k y− µˆ k2) = (n− p)σ2 follows from 2, 3 and 4 Theorem 411 k y− µˆ k2/(n− p) is an unbiased estimate of σ2 We call k y− µˆ k2 the residual sumÿ ` z V M s Ý 6 p Õ X ;
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N(X n −µ) →d σZ, (51) where µ = E X 1 and Z is a standard normal random variable In this chapter, we wish to consider the asymptotic distribution of, say, some function of X n In the simplest case, the answer depends on results already known Consider a linear function g(t) = atb for some known constants a and b Since E X0, if x < 0 This is derived via computing d dx F(x) for where Θ(x) denotes the cdf of N(0,1)Distributions Derived from Normal Random Variables χ 2 , t, and F Distributions Statistics from Normal Samples Normal Distribution Definition A Normal / Gaussian random variable X ∼ N(µ, σ
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A ª « ¬ ® ¯ ° ± ² ³ ´ µ• if {E n n ∈ N} is a collection of disjoint sets in A, meaning that E m ∩E n = ∅ for m 6= n, then µ ∞ n=1 E n!For random variables we often write {X ∈ B} = {ω X(ω) ∈ B} = X−1(B) Generally speaking, we shall use capital letters near the end of the alphabet, eg X,Y,Z for random variables The range of X is called the state space X is often called a random vector if
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(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 areDefineafunctionk(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,τ2X = 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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M « á è T y Ç ´3 µ x \ H Ñ ¯ L }7 ¯ î l è ² T y ¯ ¯ Ó ¯ î ¯ H ¨ ¸ ?The probability that Sn exceeds its expectation by at least n, for n= 100 and n= 1000 We fit this into the above theorem observe that µ= 35 and so ES n = 35n, and that we need to find an upper bound on P(S n ≥ 45n), ie, a= 45In general VarXY ≠ VarX VarY Ex 1 Let X = ±1 based on 1 coin flip As shown above, EX = 0, VarX = 1 Let Y = X;
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Z ô ¥ Q Ä \ _ ô ¿ q Ú o â I h ä þ y < ¶ µ ì f Ý Ä \ ô ì p= X∞ n=1 µ(E n) Here, we use the obvious conventions for measures and sums that are equal to ∞ The second property in the definition of a measure is called 'countable addiç U a ¸ ü i ½ ¸(3 5287( E } 5HG ê ´ = O j } è ð i r z c × y ê ´ = O q O j W ö ú = y ê ´ s ?
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C ¢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 { ÐThen VarY = (1)2VarX = 1 But XY = 0, always, so VarXY = 0 Ex 2 As another example, is VarXX = 2VarX?Title ILS pdf Author tinal Created Date Z
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3 Consider the differential equation M(x,y)N(x,y)dy dx = 0 a) Find a generalized formula for the integrating factor µ(x,y) if N x −M y xM −Ny = R(xy), where R(xy) is a function of the product xy only Hint in this case try starting with an integrating factor of the form µ(x,y) = µ(xy) and find a simple(r) ODE to solve in terms of µX Another method is to use Taylor series for ex above Write bx =eln(bx)=e(xlnb)y==xlnb ey = X1 n=0 1 n!10 pts Solution 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
Galaxies Free Full Text A Relativistic Orbit Model For Temporal Properties Of Agn Html
Physics Page In Mathematics A Gaussian Function Often Simply Referred To As A Gaussian Is A Function Of The Form F X E X B 2 2c 2 For Arbitrary Real Constants A B And
G(n)=bx (lnb)n So bx = X1 n=0 g(n)(0) n!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 3X 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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(xlnb)n = X1 n=0 (lnb)n n!4 19 Suppose that X and Y hare independent random variables, gand are two functions, and E(g(X)) and E(h(X)) existShow EgXhY EgX EhY(( )()) (( )) (())= The following table gives the joint probability distribution between employment status and college graduation among those either employed or looking for workX 2 M (b) If E 2 M;
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Then 1 j =1 E j 2 M 2 If M is a ¾¡algebra on X;(b) g(x)=cosx Solution (a) WeIe 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
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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 !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, considerIt follows that E(s2)=V(x)−V(¯x)=σ2 − σ2 n = σ2 (n−1)n Therefore, s2 is a biased estimator of the population variance and, for an unbiased estimate, we should use σˆ2 = s2 n n−1 (xi − ¯x)2 n−1 However, s2 is still a consistent estimator, since E(s2) → σ2 as n →∞and also V(s2) → 0 The value of V(s2) depends on the form of the underlying population distribu
Variance Wikipedia
Osa Plane Wave Scattering By An Ellipsoid Composed Of An Orthorhombic Dielectric Magnetic Material
G & ¯ L }7 è à Æ ¹ ñ µ ñ ® £ Ö ¢ s?Binomial probabilities when n is large and µ is small p(x) = µ n x ¶ µx (1¡µ)n¡x ‚ x x!• µ(∅) = 0;
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Math 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,Ø õ Ò ' µ ´ æ Ì O b X ( ¥ ( ú @ ú µ G q y ì Ø n À 3 ½ 0 A ú D D D D D D D D D D 0 A ö Õ @ A ¥ ( ô Ó õ õ ® z I c è õ æ þ i ú !Ú E t i Ø õ Ú E = y ä ` ´ ¸ ¸ è µ @ ¯ L }7 ¨ Õ ñ » » Ó Æ ¤ ¥ v Ú E è Ä ç µ { J s \?
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0 1 2 3 4 5 6 7 8 1 9 7 2;Definition 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 − µWhere µ1 < µ0 Suppose the signiflcant level is fi If we assume H0 were correct, then the likelihood function is fn(X1;¢¢¢;jµ0) = Yn i=1 1 µ0 e¡Xi=µ0 = µ¡n 0 expf¡ X Xi=µ0g Similarly, if H1 were correct, the likelihood
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2 EXY = EX EY 3 EAX = AEX for a constant matrix A 4 More generally (Seber & Lee Theorem 11) EAZBC = AEZB C if A,B,C are constant matrices Definition If X is a random vector, the covariance matrix of X is defined as cov(X) ≡ cov(Xi,Xj) ≡ var(X1) cov(X1,X2) ··cov(X1,) cov(X2,X1) var(X2) ··cov(X2,)1 random v ec tor with mean µ y and v ar ianceCalled so because its natural logarithm Y = ln(X) yields a normal rv X has density f(x) = (1 xσ √ 2π e −(ln(x)−µ)2 2σ2, if x ≥ 0;
Arxiv Org
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Y ¯ y j s Q è Â Ù Y b u b q U d W F Ö F è Â 8 O j b d } Ö V z { X è Â Î r y è Â 8 s u d } è Â v O b b q z y u ÿ b O ^ È 9 Þ y O v h n q ¼ ¦ 5 r y ü ¸ u è Â v T d } ¦ v v » µ á é Å ¡ ½ ñ d ¼ ³ ¢ ñ y \/ > O ï b È ¶ b f D § ô 0 º ^ e b , ÷ ó / { ö!If MX(t) is the momentgenerating function of a random variable X, then M (r) X (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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X, and let Y b e a q !1 ra ndom v ector with mean µ x and v aria nce co v ar iance ma trix !
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