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Title Microsoft Word ARRC Consultation Response Variable Rate Private Student Loansdocx Author nlynn Created Date 6/15/ PM.

Nu y. 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. 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). Mar 15, 16 · Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers Visit Stack Exchange.

D Z & u ( } µ Z µ î ì î ì µ o v o µ r Æ v P } v r o Æ v } ( ¨ î ï í u o o } v ~ ¨ í ô õ u o o } v ( r Æ } v Z t o Z D v P u v µ v P u v X. Hierarchical Models † Nonhierarchical models are usually inappropiate for hierarchical data { a single µ (ie µj · µ 8j) may be inadequate to flt a combined data set { separate unrelated µj are likely to \overflt" data { information about one µj can be obtained from others’ data † Hierarchical model uses many parameters but population distribution induces enough. Thm X is dm from Np(µ,Σ), Y = AXB, Z = CXD, then Y indep of Z ⇔ either (a) B′ΣD = 0 or (b) AC′ = 0 You will prove this in your homework, see 335 (p) Corollary Let X = X1X2 of dimensions n×k and n×(p−k) Then X1 indep X21 = X2−X1Σ −1 11Σ12, X1 dm from Nk(µ1,Σ11) and X21 dm from Np−k(µ21,Σ221) where.

If X ∼ N(µ,σ2) and Y = a bX, where a and b are fixed numbers, then Y ∼ N(a bµ,b2σ2) That is, Y also has a normal distribution For example, if X ∼ N(10,1) is the temperature of the sea water at a particular location and time of year, in Celcius, then Y = 18X 32 is the temperature in degrees Fahrenheit and Y ∼ N(50,324). I Z ý Þ µ â I y n µ !!!!. Theorem IfX ∼ N(µ, σ2)thentherandomvariableY = X−µ σ ∼ N(0, 1) Proof Let the random variable X have the normal distribution with probability density function fX(x)= 1 √ 2πσ.

Y µ v Ç v Y µ v Ç v Y µ v Ç u ò î ï ì í ¨ ï ì U ì ì ì. Suppose that E(X)=µ, Var(X)=s2 Then (i) E(Yn)=µn (ii)If µ 6= 1, then Var(Yn)= s2µn¡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. > m » ô O ô # õ 3 4 ã D â ö < / Í M ù ý ô Ã ò.

Y¡µ2 s2 ¶ µ y¡µ2 s2 ¶2# (2) Statement The marginal distributions of N(µ1;µ2;s2 1;s 2 2;r) are normal with rv’s X and Y having density functions fX(x)= 1 p 2ps1 e ¡(x¡µ1)2 2s2 1;. 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 3. Otherwise It can be seen that the MLE of µ must be a value of µ for which µ ‚ xi for i = 1;¢¢¢;n and which maximizes 1=µn among all such values Since 1=µn is a decreasing function of µ, the estimate will be the smallest possible value of µ such that µ.

Expected value of y e y e xn µ 1 n e x 1 n ² np School University of Pretoria;. , â é õ !. Let µ ∈ R and σ > 0 and let X ∼ N (µ, σ 2)Let a, b ∈ R and define the random variable Y = aX b Show that Y ∼ N(aµ b, a2σ 2 ).

Defineafunctionk(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. P P P P X = 1 n ∼ N(µ 1,σ 1 2 /n) n 1 X i 2 S 2 = (1 n X σ 1 i − X) 2 ∼ (X n−11) × χ 2 n−1 1 m ∼ N(µ 2,σ 2 2 /n) Y = m 1 Y i 2 S. 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’s Sometimes to stress the particular rv X, we write M X(s) Then the above independence property can be concisely expressed as M.

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,. Exercise 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 X. The parameters n, µ, σand ǫare fixed, but λis something the company controls Determine λ 4 Assume that Sn is Binomial(n, p) For every a>p, determine by calculus the large deviation bound for P(Sn ≥ an) 5 Using the central limit theorem for a sum of Poisson random variables, compute lim n.

About 95% of the x values lie between –2σ and 2σ of the mean µ (within two standard deviations of the mean). > E } W ^ u } v ^ µ Ç Y µ } v n & µ Ç î ô U î ì î í ^ v } W } W < À v & o v v Ç. Statistical Machine Learning CHAPTER 12 BAYESIAN INFERENCE where b = S n/n is the maximum likelihood estimate, e =1/2 is the prior mean and n = n/(n2)⇡ 1 A 95 percent posterior interval can be obtained by numerically finding a and b such that.

The F distribution Let Z1 ∼ χ2 m, and Z2 ∼ χ 2 n and assume Z1 and Z2 are independent Then Z1/m Z2/n ∼ Fm,n F distributions 0 05 1 15 2 25 3 df=,10 df=, df=,50 The distribution of the sample variance ratio. If Y ˘N( ;˙2I) then 3 = 0 and 4 = 3˙4 In that case var(Y0AY) = 2˙4tr() 4˙2 0A and if A = 0 we get var(Y0AY) = 2˙4tr() Now suppose that we can come up with two matrices A 1 and A 2 with E(Y0A jY) = ˙2 for j= 1;2 For normally A) ˙^. Uploaded By ConstableKnowledge Pages 119 This preview shows page 31 42 out of 119 pages Expected value of Y E (Y) = E ´ X n µ = 1 n E (X) = 1 n ² np = p.

Title キャンセル・プロテクション キャンセル内容報告書 兼 特別措置手当金請求書 Created Date 4/7/ PM. Tarrant Regional Water District Fiscal Year 21 Budget Enriching Communities, Improving Quality of Life DESCRIPTION OF THE DISTRICT The Tarrant Regional Water District is a water control and improvement district and political subdivision. We assume (as a first approximation, and with no specific knowledge of this election), that y/n is uniformly distributed between 30% and 70%, then Pr(election is tiedn) = Pr(y = n/2) = ˆ 1 04n if n is even 0 if n is odd If we assume that n is about 0,000, with a 1/2 chance of being even, then this approximation gives Pr(election is tied.

The Empirical Rule If X is a random variable and has a normal distribution with mean µ and standard deviation σ, then the Empirical Rule says the following About 68% of the x values lie between –1σ and 1σ of the mean µ (within one standard deviation of the mean);. FY(y)= 1 p 2ps2 e ¡(y¡µ2)2 2s2 2 Proof The expression (2) for Q(x;y) can be rearranged as follows Q(x;y)= 1 1¡r2 "µ x¡µ1 s1 ¡r y¡µ2 s2. KD } 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.

For 0 • xi • µ (i = 1;¢¢¢;n) 0;. P µ } / ^ ï í í ô ñ U y t D v o , o Z v h v ^ Z } } o D } lt ò W ì ì t õ W î ìWD r D µ Z u u / ^ ï í î ð õ U ï t / v v Z v À o } u v o o t ò W ì ì t õ W î ìWD. L(µ) = (1 µn;.

Let X ~ N(μ, σ*) and YaX b By adapting the calculation from (342) and (343), show that YN(au b, a o2) The probability density function of X can now be obtained by differentiating F (r COMPANY. 2 µ σ πσ µσ • The notation N(µ, σ2) means normally distributed with mean µ and variance σ2 If we say X ∼ N(µ, σ2) we mean that X is distributed N(µ, σ2) • About 2/3 of all cases fall within one standard deviation of the mean, that is P(µ σ ≤ X ≤ µ σ) = 66. Title Microsoft Word 1014 Request for Statements of Qualifications final Author MaHoward Created Date 10/21/ PM.

BUEC 333 – Introduction to Econometrics B Antoine 1/2/376 Summary The Sampling Distribution of For Y 1,, Y n iid with 0 < 2 Y σ < ∞, • The exact (finite sample) sampling distribution of has mean µ Y (“ is an unbiased estimator of µ Y ”) and variance 2 Y σ / n • Other than its mean and variance, the exact distribution of. Parameters µ = 4 and σ2 = 36 Therefore, Y ∼ N(4,36) 1 2 Suppose that the heights of the men of a certain large city are normally distributed with mean 71 inches and variance 15 inches Find the probability that a randomly chosen man is at least 5 feet tall and no. Title Microsoft Word FAQs_1221 Author jbjorset Created Date 12/21/ 357 PM.

Ô ú Í n o º ¢ ñ a !. 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. 1 ···Y n = exp(W 1···W n) in terms of µ and σ Second Practice Second Midterm Exam Consider the task of giving a twenty minute review lecture on the basic properties and role of the Poisson distribution and the Poisson process in probability theory Write out a complete set of lecture notes that could be used for this.

Example Let X1,··· , be iid from N(µ,1) and µ∼ π(µ) = N(µ0,τ0)We know that X¯(∼ N(µ,1/n)) is a sufficient statistic The Bayes estimator is ˆµ= E(µX¯)We need to calculate the joint distribution of (µ,X¯)T first It isnotdifficult to seethat (µ,X¯)T isbivariate normal We know that Eµ= µ0,Var(µ) =. 3 The covariance of X and Y is defined as cov(X,Y) = E(X −µ X)(Y −µ Y) 4 The correlation (coefficient) of X and Y is defined as ρ XY = √ cov(X,Y ) var(X)var(Y ) The following properties about the variances are worth memorizing Theorem 4 (Variances and Covariances) Let X and Y be random variables and a,b ∈ R 1 var(aX b. Title Microsoft Word 4Figuerola_edits Author DELL Created Date 5/17/ PM.