Distributions of Functions of. Normal Random Variables Version 27 Jan 2004

Similar documents
Topic 9: Sampling Distributions of Estimators

Topic 9: Sampling Distributions of Estimators

Topic 9: Sampling Distributions of Estimators

The variance of a sum of independent variables is the sum of their variances, since covariances are zero. Therefore. V (xi )= n n 2 σ2 = σ2.

Expectation and Variance of a random variable

Econ 325/327 Notes on Sample Mean, Sample Proportion, Central Limit Theorem, Chi-square Distribution, Student s t distribution 1.

Confidence Level We want to estimate the true mean of a random variable X economically and with confidence.

Direction: This test is worth 250 points. You are required to complete this test within 50 minutes.

Exam II Review. CEE 3710 November 15, /16/2017. EXAM II Friday, November 17, in class. Open book and open notes.

Properties and Hypothesis Testing

The standard deviation of the mean

Statistical Inference (Chapter 10) Statistical inference = learn about a population based on the information provided by a sample.

[ ] ( ) ( ) [ ] ( ) 1 [ ] [ ] Sums of Random Variables Y = a 1 X 1 + a 2 X 2 + +a n X n The expected value of Y is:

Direction: This test is worth 150 points. You are required to complete this test within 55 minutes.

Common Large/Small Sample Tests 1/55

Stat 421-SP2012 Interval Estimation Section

Statistics 20: Final Exam Solutions Summer Session 2007

1 Introduction to reducing variance in Monte Carlo simulations

Econ 325 Notes on Point Estimator and Confidence Interval 1 By Hiro Kasahara

Parameter, Statistic and Random Samples

1.010 Uncertainty in Engineering Fall 2008

STAT431 Review. X = n. n )

Statistical inference: example 1. Inferential Statistics

Stat 200 -Testing Summary Page 1

Let us give one more example of MLE. Example 3. The uniform distribution U[0, θ] on the interval [0, θ] has p.d.f.

BIOS 4110: Introduction to Biostatistics. Breheny. Lab #9

Resampling Methods. X (1/2), i.e., Pr (X i m) = 1/2. We order the data: X (1) X (2) X (n). Define the sample median: ( n.

KLMED8004 Medical statistics. Part I, autumn Estimation. We have previously learned: Population and sample. New questions

Binomial Distribution

FACULTY OF MATHEMATICAL STUDIES MATHEMATICS FOR PART I ENGINEERING. Lectures

Comparing Two Populations. Topic 15 - Two Sample Inference I. Comparing Two Means. Comparing Two Pop Means. Background Reading

Statistics 203 Introduction to Regression and Analysis of Variance Assignment #1 Solutions January 20, 2005

Homework for 2/3. 1. Determine the values of the following quantities: a. t 0.1,15 b. t 0.05,15 c. t 0.1,25 d. t 0.05,40 e. t 0.

( θ. sup θ Θ f X (x θ) = L. sup Pr (Λ (X) < c) = α. x : Λ (x) = sup θ H 0. sup θ Θ f X (x θ) = ) < c. NH : θ 1 = θ 2 against AH : θ 1 θ 2

Chapter 6 Sampling Distributions

Statisticians use the word population to refer the total number of (potential) observations under consideration

Module 1 Fundamentals in statistics

17. Joint distributions of extreme order statistics Lehmann 5.1; Ferguson 15

AMS 216 Stochastic Differential Equations Lecture 02 Copyright by Hongyun Wang, UCSC ( ( )) 2 = E X 2 ( ( )) 2

x = Pr ( X (n) βx ) =

STATISTICAL PROPERTIES OF LEAST SQUARES ESTIMATORS. Comments:

STATISTICAL INFERENCE

Interval Estimation (Confidence Interval = C.I.): An interval estimate of some population parameter is an interval of the form (, ),

Lecture 33: Bootstrap

32 estimating the cumulative distribution function

TAMS24: Notations and Formulas

LECTURE 8: ASYMPTOTICS I

Tests of Hypotheses Based on a Single Sample (Devore Chapter Eight)

MATH 320: Probability and Statistics 9. Estimation and Testing of Parameters. Readings: Pruim, Chapter 4

7-1. Chapter 4. Part I. Sampling Distributions and Confidence Intervals

Background Information

Lecture Notes 15 Hypothesis Testing (Chapter 10)

UNIT 8: INTRODUCTION TO INTERVAL ESTIMATION

7.1 Convergence of sequences of random variables

7.1 Convergence of sequences of random variables

Access to the published version may require journal subscription. Published with permission from: Elsevier.

MATH/STAT 352: Lecture 15

Agenda: Recap. Lecture. Chapter 12. Homework. Chapt 12 #1, 2, 3 SAS Problems 3 & 4 by hand. Marquette University MATH 4740/MSCS 5740

Statistics 511 Additional Materials

Goodness-of-Fit Tests and Categorical Data Analysis (Devore Chapter Fourteen)

Random Variables, Sampling and Estimation

HOMEWORK I: PREREQUISITES FROM MATH 727

Lecture 3. Properties of Summary Statistics: Sampling Distribution

Simple Linear Regression

6 Sample Size Calculations

Lecture 12: September 27

Basis for simulation techniques

independence of the random sample measurements, we have U = Z i ~ χ 2 (n) with σ / n 1. Now let W = σ 2. We then have σ 2 (x i µ + µ x ) 2 i =1 ( )

Comparing your lab results with the others by one-way ANOVA

Homework 3 Solutions

Inferential Statistics. Inference Process. Inferential Statistics and Probability a Holistic Approach. Inference Process.

Solutions to Odd Numbered End of Chapter Exercises: Chapter 4

Chapter 1 (Definitions)

(7 One- and Two-Sample Estimation Problem )

Lecture 11 and 12: Basic estimation theory

Asymptotic Results for the Linear Regression Model

Discrete Mathematics for CS Spring 2008 David Wagner Note 22

This exam contains 19 pages (including this cover page) and 10 questions. A Formulae sheet is provided with the exam.

Introduction to Econometrics (3 rd Updated Edition) Solutions to Odd- Numbered End- of- Chapter Exercises: Chapter 3

Parameter, Statistic and Random Samples

f(x i ; ) L(x; p) = i=1 To estimate the value of that maximizes L or equivalently ln L we will set =0, for i =1, 2,...,m p x i (1 p) 1 x i i=1

Confidence Interval for Standard Deviation of Normal Distribution with Known Coefficients of Variation

MBACATÓLICA. Quantitative Methods. Faculdade de Ciências Económicas e Empresariais UNIVERSIDADE CATÓLICA PORTUGUESA 9. SAMPLING DISTRIBUTIONS

DS 100: Principles and Techniques of Data Science Date: April 13, Discussion #10

STAC51: Categorical data Analysis

Topic 10: Introduction to Estimation

Rule of probability. Let A and B be two events (sets of elementary events). 11. If P (AB) = P (A)P (B), then A and B are independent.

Probability and statistics: basic terms

Confidence intervals summary Conservative and approximate confidence intervals for a binomial p Examples. MATH1005 Statistics. Lecture 24. M.

Lecture 2: Monte Carlo Simulation

Math 152. Rumbos Fall Solutions to Review Problems for Exam #2. Number of Heads Frequency

Sampling Distributions, Z-Tests, Power

ECE 901 Lecture 12: Complexity Regularization and the Squared Loss

z is the upper tail critical value from the normal distribution

Efficient GMM LECTURE 12 GMM II

1 Inferential Methods for Correlation and Regression Analysis

Summary. Recap ... Last Lecture. Summary. Theorem

The central limit theorem for Student s distribution. Problem Karim M. Abadir and Jan R. Magnus. Econometric Theory, 19, 1195 (2003)

Stat 319 Theory of Statistics (2) Exercises

1 Convergence in Probability and the Weak Law of Large Numbers

Transcription:

Distributios of Fuctios of Normal Radom Variables Versio 27 Ja 2004 The Uit or Stadard) Normal The uit or stadard ormal radom variable U is a ormally distributed variable with mea zero ad variace oe, i. e. U N0, 1). Note that if x Nµ, σ 2 ) that x µ σ U N0, 1) 1) Thus to simulate a ormal radom variable with mea µ ad variace σ 2, we ca simply trasform uit ormals, as x µ + σu Nµ, σ 2 ) 2) Cosider idepedet radom variables x i Nµ, σ 2 ), the x Nµ, σ 2 /), ad this x µ σ/ U N0, 1) 3) Example 1. Let s costruct a 95% cofidece iterval for the mea µ for Equatio 3). First, let s use R to compute a value U 0.975 such that PrU U 0.975 )= 0.975. IR, typig the commad qorm0.975) returs 1.96. Likewise, qorm0.025) returs -1.96 ad hece PrU 1.96)=0.025. Hece, Recallig Equatio 3), Rearragig gives Pr 1.96 U 1.96)=0.95 Pr 1.96 U 1.96) = Pr 1.96 x µ ) σ/ 1.96 Pr 1.96σ/ x µ 1.96σ/ ) or Pr x 1.96σ/ µ x+1.96σ/ )

2 Fuctios of Normal Radom Variables which ca also be writte as Pr x 1.96σ/ µ x +1.96σ/ ) =0.95 givig a 95% cofidece iterval for the mea µ. Cetral ad Nocetral χ 2 Distributios The χ 2 distributio arises from sums of squared, ormally distributed, radom variables if x i N0, 1), the u = x2 i χ2,acetral χ 2 distributio with degrees of freedom. It follows that the sum of two χ 2 radom variables is also χ 2 distributed, so that if u χ 2 ad v χ 2 m, the u + v χ 2 +m) 4a) Two other useful results are that if x i N0,σ 2 ), the x 2 i σ 2 χ 2 4b) ad for x = 1 x i, x i x ) 2 σ 2 χ 2 1) 4c) I this last case, subtractio of the mea causes the loss of oe degree of freedom. Note that a special case of Equatio 4c) is the sample estimate of the variace, so that Varx)= 1 1 x i x ) 2 1)Varx) σ 2 χ 2, implyig 1)Varx) σ 2 χ 2 4d) Example 2. We ca use Equatio 4d) to costruct a cofidece iterval o the true variace σ 2 give the sample variace Varx), provided the x i are draw from idepedet ormals with the same mea ad variace σ 2.

Fuctios of Normal Radom Variables 3 First, recall that the R commad qchisqp,df) returs a value X such that Prχ 2 df X )=p. Suppose sample size is =20. Sice qchisq0.975,19) returs a value of 32.85 ad qchisq0.025,19) returs 8.91, we have Pr8.91 χ 2 19 32.85)=0.95 From Equatio 4d, ) Pr8.91 χ 2 1)Varx) 19 32.85)=Pr 8.91 σ 2 32.85 Notig that for 1 Pra x b) =Pr a 1 ) x 1 b we have Pr 8.91 19Varx) ) 1 σ 2 32.85 =Pr 8.91 σ 2 19Varx) 1 ) 32.85 or or 19Var Pr 8.91 σ2 19Var ) =0.95 32.85 Pr 2.13Var σ 2 0.58Var ) =0.95 givig the 95% cofidece iterval o the variace as 0.58Var to 2.13Var. A ocetral χ 2 arises whe the radom variables beig cosidered have ozero meas. I particular, if x i Nµ i, 1), the u = x2 i follows a ocetral χ 2 distributio with degrees of freedom ad ocetrality parameter λ = µ 2 i 5a) ad we write u χ 2,λ. As show i Figure 1, icreasig the ocetrality parameter λ shifts the distributio to the right. This is also see by cosiderig the mea ad variace of u, Eu) =+λ ad σ 2 u) =2+2λ) 5b)

4 Fuctios of Normal Radom Variables λ = 0 λ = 1 λ = 5 0 5 10 15 20 25 Figure 1 The probability distributio fuctio for a ocetral χ 2.Asthe ocetrality parameter λ icreases, the distributio is pulled to the right. We plot here a χ 2 radom variable with = 5 degrees of freedom ad ocetrality parameters λ = 0 a cetral χ 2 ), 1, ad 5. It follows directly from the defiitio that sums of ocetral χ 2 variables also follows a ocetral χ 2 distributio, so that if u χ 2, λ 1 ad v χ 2 m, λ 2, the u + v) χ 2 +m),λ 1+λ 2) 5c) Fially, Equatios 4b,c ca be geeralized to ocetral χ 2 radom variables as follows. Suppose x i Nµ i,σ 2 ), the x 2 i σ 2 χ 2,λ where λ = µ 2 i σ 2 5d) Turig the distributio of x i x ) 2, defiig the λ = µ i µ ) 2 σ 2, where µ = 1 x 2 i x ) 2 σ 2 χ 2,λ 5e) Note that if all the x i have the same mea µ i = µ = µ ), λ =0ad the χ 2 is cetral, while if there is some variace i the meas of the x i, the distributio is a ocetral χ 2. R provides commads for quatities of iterest with ocetral χ 2 distributios. qchisqp,df,cp) returs a value X such that Prχ 2 df,cp X) =p µ i

Fuctios of Normal Radom Variables 5 pchisqx,df,cp) returs the probability that Prχ 2 df,cp X) leavig out the field for cp returs these values for a cetral χ 2. Studet s t Distributio If x Nµ, σ 2 ), the for Equatio 2, we have x µ)/σ/ ) N0, 1), which allows for both hypothesis testig ad costructio of cofidece itervals whe σ 2 is kow. Whe the variace is ukow, the above test statistic replaces the true but ukow) variace σ 2 with the sample variace Varx), t = x µ 6) Var/ Notice that t = ) ) x µ σ/ 1 = Var/σ 2 U χ 2 1 / 1) Thus, we defie a t distributed radom variable with ν degrees of freedom by U t ν = 7a) χ 2 ν /ν) A t radom variable has mea zero ad variace σ 2 t ν )=1+ 2 for ν>2 7b) ν 2 The coefficiet of kurtosis is k 4 =6/ν 4), implyig that the t distributio has heavier tails tha a ormal. The ocetral Studet s t distributio is defied as follows: If x Nµ o,σ 2 ), but we assume the correct mea is µ, the t ν,λ = x µ Var/ is distributed as a ocetral t radom variable with 1 degrees of freedom ad ocetrality parameter λ =µ µ 0 )/σ. Cetral ad Nocetral F Distributios The ratio of two χ 2 -distributed variables leads to the F distributio. I particular, if u χ 2 ad v χ 2 m, the the ratio of these two χ 2 variables divided by their respective degrees of freedom follows a cetral F distributio/ with umerator ad deomiator degrees of freedom ad m respectively), i.e., u/)/v/m) F,m. Sice lim m F,m χ2

6 Fuctios of Normal Radom Variables the F distributio ca be approximated by a χ 2 whe the deomiator degrees of freedom is large. R provides commads for quatities of iterest for F distributios. qfp,df1,df2) returs a value X such that PrF df 1,df2 X) =p pfx,df1,df2) returs the probability that PrF df 1,df2 X) The ocetral F distributio results whe the umerator χ 2 variable is ocetral. If u χ 2,λ ad v χ2 m, the F =u/)/v/m) follows a ocetral F distributed with ocetrality parameter λ, ad we write F F,m,λ. As with the ocetral χ 2, icreasig λ shifts the distributio further to the right. Agai, this is see i the mea ad variace, with EF )= m m 2 1+ 2λ ) m ) [ 2 +m) σ 2 2 ++2λ)m 2) F )=2 m 2) 2 m 4) ] A5.16a) A5.16b) R provides commads for quatities of iterest for ocetral F distributios. pfx,df1,df2,cp) returs the probability that PrF df 1,df2,cp X) the obvious commad qfp,df1,df2,cp) does ot work, as the same value is retured for all values of cp.