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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);.

Nu y. 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. 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. 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.

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

I y n µ ø H M ô , @ · ² Í n o à ô U õ Æ. 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. Title Microsoft Word FAQs_1221 Author jbjorset Created Date 12/21/ 357 PM.

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,. 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. 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.

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. 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. 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.

, â é õ !. 5 If X ∼ Nn(µ,Σ) and Am×n is a constant matrix of rank m, then Y = Ax ∼ Np(Aµ,AΣAT) Proof Use definition 321 or property 1 above Orthogonal linear transformation 6 If X ∼ Nn(µ,In) and An×n is an orthogonal matrix and Σ = In, then Y = Ax ∼ Nn(Aµ,In) Chapter 3. 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.

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. 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. 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.

About 95% of the x values lie between –2σ and 2σ of the mean µ (within two standard deviations of the mean). Title Microsoft Word 4Figuerola_edits Author DELL Created Date 5/17/ PM. 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.

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. > 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. 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.

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. > m » ô O ô # õ 3 4 ã D â ö < / Í M ù ý ô Ã ò. 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.

L(µ) = (1 µn;. For 0 • xi • µ (i = 1;¢¢¢;n) 0;. I Z ý Þ µ â I y n µ !!!!.

Course Title WST 211;. 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. 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 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) ˙^. 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. 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;.

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. 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). 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).

Ô ú Í n o º ¢ ñ a !. 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. Expected value of y e y e xn µ 1 n e x 1 n ² np School University of Pretoria;.

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. 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. 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 ).

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. 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) =. Title Microsoft Word 1014 Request for Statements of Qualifications final Author MaHoward Created Date 10/21/ PM.