f X, Y (x, y)dx (x), where f(x,y) is the joint pdf of X and Y. (x) dx

Similar documents
ENGG2430A-Homework 2

Regression and Covariance

Covariance and Correlation

Bivariate distributions

Chapter 4 continued. Chapter 4 sections

Week 10 Worksheet. Math 4653, Section 001 Elementary Probability Fall Ice Breaker Question: Do you prefer waffles or pancakes?

Jointly Distributed Random Variables

ECE Lecture #9 Part 2 Overview

More than one variable

Review: mostly probability and some statistics

Problem Y is an exponential random variable with parameter λ = 0.2. Given the event A = {Y < 2},

Lecture 2: Review of Probability

Raquel Prado. Name: Department of Applied Mathematics and Statistics AMS-131. Spring 2010

Stat 5101 Notes: Algorithms (thru 2nd midterm)

Preliminary Statistics. Lecture 3: Probability Models and Distributions

STT 441 Final Exam Fall 2013

Gaussian random variables inr n

Probability. Paul Schrimpf. January 23, UBC Economics 326. Probability. Paul Schrimpf. Definitions. Properties. Random variables.

CHAPTER 4 MATHEMATICAL EXPECTATION. 4.1 Mean of a Random Variable

Variance reduction. Michel Bierlaire. Transport and Mobility Laboratory. Variance reduction p. 1/18

Functions of two random variables. Conditional pairs

2 (Statistics) Random variables

Math 426: Probability MWF 1pm, Gasson 310 Exam 3 SOLUTIONS

Statistics for Economists Lectures 6 & 7. Asrat Temesgen Stockholm University

Class 8 Review Problems solutions, 18.05, Spring 2014

3. General Random Variables Part IV: Mul8ple Random Variables. ECE 302 Fall 2009 TR 3 4:15pm Purdue University, School of ECE Prof.

6.041/6.431 Fall 2010 Quiz 2 Solutions

Simple Linear Regression Estimation and Properties

Outline Properties of Covariance Quantifying Dependence Models for Joint Distributions Lab 4. Week 8 Jointly Distributed Random Variables Part II

Introduction to Computational Finance and Financial Econometrics Probability Review - Part 2

MATH 38061/MATH48061/MATH68061: MULTIVARIATE STATISTICS Solutions to Problems on Random Vectors and Random Sampling. 1+ x2 +y 2 ) (n+2)/2

Math 510 midterm 3 answers

MATHEMATICS 154, SPRING 2009 PROBABILITY THEORY Outline #11 (Tail-Sum Theorem, Conditional distribution and expectation)

Joint probability distributions: Discrete Variables. Two Discrete Random Variables. Example 1. Example 1

Stat 5101 Notes: Algorithms

Covariance and Correlation Class 7, Jeremy Orloff and Jonathan Bloom

EE/CpE 345. Modeling and Simulation. Fall Class 9

Multivariate Random Variable

Probability. Paul Schrimpf. January 23, Definitions 2. 2 Properties 3

Review of Probability. CS1538: Introduction to Simulations

Problem Solving. Correlation and Covariance. Yi Lu. Problem Solving. Yi Lu ECE 313 2/51

Chapter 2. Some Basic Probability Concepts. 2.1 Experiments, Outcomes and Random Variables

UNIT-2: MULTIPLE RANDOM VARIABLES & OPERATIONS

matrix-free Elements of Probability Theory 1 Random Variables and Distributions Contents Elements of Probability Theory 2

Bivariate Distributions

HW5 Solutions. (a) (8 pts.) Show that if two random variables X and Y are independent, then E[XY ] = E[X]E[Y ] xy p X,Y (x, y)

For a stochastic process {Y t : t = 0, ±1, ±2, ±3, }, the mean function is defined by (2.2.1) ± 2..., γ t,

Lecture 4: Proofs for Expectation, Variance, and Covariance Formula

Stochastic Simulation Introduction Bo Friis Nielsen

11. Regression and Least Squares

Random Variables. Saravanan Vijayakumaran Department of Electrical Engineering Indian Institute of Technology Bombay

ECE302 Exam 2 Version A April 21, You must show ALL of your work for full credit. Please leave fractions as fractions, but simplify them, etc.

Probability Theory and Statistics. Peter Jochumzen

ECE Homework Set 3

Partial Solutions for h4/2014s: Sampling Distributions

EE/CpE 345. Modeling and Simulation. Fall Class 10 November 18, 2002

Random Variables. Random variables. A numerically valued map X of an outcome ω from a sample space Ω to the real line R

P (x). all other X j =x j. If X is a continuous random vector (see p.172), then the marginal distributions of X i are: f(x)dx 1 dx n

Final Exam. Economics 835: Econometrics. Fall 2010

Math 3215 Intro. Probability & Statistics Summer 14. Homework 5: Due 7/3/14

Multivariate Distributions CIVL 7012/8012

18 Bivariate normal distribution I

ECON Fundamentals of Probability

Continuous Random Variables

Homework 5 Solutions

MULTIVARIATE PROBABILITY DISTRIBUTIONS

Random Variables and Expectations

ECE 650 Lecture 4. Intro to Estimation Theory Random Vectors. ECE 650 D. Van Alphen 1

Statistics 351 Probability I Fall 2006 (200630) Final Exam Solutions. θ α β Γ(α)Γ(β) (uv)α 1 (v uv) β 1 exp v }

Continuous r.v practice problems

1 Basic continuous random variable problems

Bivariate Distributions

5 Operations on Multiple Random Variables

ECSE B Solutions to Assignment 8 Fall 2008

Random Signals and Systems. Chapter 3. Jitendra K Tugnait. Department of Electrical & Computer Engineering. Auburn University.

Final Exam # 3. Sta 230: Probability. December 16, 2012

Algorithms for Uncertainty Quantification

The mean, variance and covariance. (Chs 3.4.1, 3.4.2)

Random Variables. Cumulative Distribution Function (CDF) Amappingthattransformstheeventstotherealline.

CHAPTER 5. Jointly Probability Mass Function for Two Discrete Distributed Random Variables:

Joint Distribution of Two or More Random Variables

UCSD ECE153 Handout #27 Prof. Young-Han Kim Tuesday, May 6, Solutions to Homework Set #5 (Prepared by TA Fatemeh Arbabjolfaei)

Estimation of uncertainties using the Guide to the expression of uncertainty (GUM)

STAT/MATH 395 PROBABILITY II

Let X and Y denote two random variables. The joint distribution of these random

Perhaps the simplest way of modeling two (discrete) random variables is by means of a joint PMF, defined as follows.

CS70: Jean Walrand: Lecture 22.

STAT Homework 8 - Solutions

Solutions to Homework Set #5 (Prepared by Lele Wang) MSE = E [ (sgn(x) g(y)) 2],, where f X (x) = 1 2 2π e. e (x y)2 2 dx 2π

Chapter 2: simple regression model

Probability and Distributions

4. Distributions of Functions of Random Variables

2. Suppose (X, Y ) is a pair of random variables uniformly distributed over the triangle with vertices (0, 0), (2, 0), (2, 1).

The Binomial distribution. Probability theory 2. Example. The Binomial distribution

BASICS OF PROBABILITY

STA 256: Statistics and Probability I

Lecture 2: Repetition of probability theory and statistics

Chapter 5 continued. Chapter 5 sections

Multiple Random Variables

Review of Statistics

Transcription:

INDEPENDENCE, COVARIANCE AND CORRELATION Independence: Intuitive idea of "Y is independent of X": The distribution of Y doesn't depend on the value of X. In terms of the conditional pdf's: "f(y x doesn't depend on x." Caution: "Y is not independent of X" means simply that the distribution of Y may vary as X varies. It doesn't mean that Y is a function of X. If Y is independent of X, then: 1. µ x = E(Y X = x does not depend on x. Question: Is the converse true? That is, if E(Y X = x does not depend on x, can we conclude that Y is independent of X? 1 2. (Still assuming Y is independent of X Let h(y be the common pdf of the conditional distributions Y X. Then for every x, f (x, y h(y = f(y x = f X (x, where f(x,y is the joint pdf of X and Y. Therefore f(x,y = h(y f X (x f Y (y = $ $ " f X, Y (x, ydx = h(y f X (x dx " = h(y $ f X (x dx = h(y = f(y x " In other words: If Y is independent of X, then the conditional distributions of Y given X are the same as the marginal distribution of Y. 2

3 4 3. Now (still assuming Y is independent of X we have so f Y (y = f(y x = f (x, y f X (x, f Y (yf X (x = f(x,y. In other words: If Y is independent of X, then the joint distribution of X and Y is the product of the marginal distributions of X and Y. Exercise: The converse of this last statement is true. That is: If the joint distribution of X and Y is the product of the marginal distributions of X and Y, then Y is independent of X. Observe: The condition f Y (yf X (x = f(x,y is symmetric in X and Y. Thus (3 and its converse imply that : Y is independent of X if and only if X is independent of Y. So it makes sense to say "X and Y are independent." Summary: The following conditions are all equivalent: 1. X and Y are independent. 2. f X,Y (x,y = f Y (yf X (x 3. The conditional distribution of Y (X = x is independent of x. 4. The conditional distribution of X (Y = y is independent of y. 5. f(y x = f Y (y for all y. 6. f(x y = f X (x for all x. Additional property of independent random variables: If X and Y are independent, then E(XY = E(XE(Y. (Proof might be homework. Covariance: For random variables X and Y, Cov(X,Y = E([X - E(X][Y - E(Y] Comments:

5 6 Cov (capital C " population cov (or Cov-hat " sample. Cov is a parameter " population cov is a statistic " calculated from the sample. Compare and contrast with definition of Var(X. If X and Y both tend to be on the same side of their respective means (i.e., both greater than or both less than their means, then [X - E(X][Y - E(Y] tends to be positive, so Cov(X,Y is positive. Similarly, if X and Y tend to be on opposite sides of their respective means, then Cov(X,Y is negative. If there is no trend of either sort, then Cov(X,Y should be zero. Thus covariance roughly measures the extent of a "positive" or "negative" trend in the joint distribution of X and Y. Units of Cov(X,Y? 1. Cov(X, X = 2. Cov (Y, X = 3. Cov (X, Y = E([X - E(X][Y - E(Y] = In words (Compare with the alternate formula for Var(X. 4. Consequence: If X and Y are independent, then: Cov(X,Y = Note: Converse false. (Future homework. Properties: 5. Cov(cX, Y =

7 8 Cov(X, cy = 6. Cov(a + X,Y = Cov(X, a +Y = 7. Cov(X + Y, Z = Bounds on Covariance! X = population standard deviation Var(X of X. (Do not confuse with sample standard deviation = s or s.d. or " ˆ! Y defined similarly. 8. Var(X + Y = Cov(X, Y + Z = Consider the new random variable Since Variance is always " 0, X + Y. Consequence: If X and Y are independent, then Var(X + Y = (converse false! When else might this be true? (* 0! Var + Y & % ( $ ' = Var( X + Var( Y + 2Cov( X = 1 " x 2 Var(X + 1 2 Var(Y + = = 2[1 + Cov(X,Y ]., Y 2 Cov(X,Y Therefore:

9 10 (** Cov(X,Y " -1 Equivalently: Cov(X, Y " -. $ X Looking at Var Y ' & % " Y ( similarly shows: (*** Cov(X,Y! 1 (details left to the student Combining (** and (***: Cov(X,Y! 1 Equivalently: Cov(X, Y!. Equivalently: Cov(X, Y! Equality in (** equality in (* -- i.e, say, Var + Y & % ( $ ' = 0 X + Y is constant -- X + Y = c. (Note that c must be the mean of X + Y, which is µ X + µ Y This in turn is equivalent to $ Y = ' & + c % ( or Y = " Y X + Y c This says: The pairs (X,Y lie on a line with negative slope (namely, -! Y /! X (Converse is also true -- details left to the student.

11 12 Note: the line has slope " Y and y-intercept µ X + µ Y. Similarly, Cov(X,Y = +1 exactly when the pairs (X,Y lie on a line with positive slope. Correlation: The (population correlation coefficient of the random variables X and Y is,y = Cov(X,Y Y. Note: $ for short. $ is a parameter (refers to the population. Do not confuse with the sample correlation coefficient (usually called r: a statistic (calculated from the sample. Properties of $: Negative $ indicates a tendency for the variables X and Y to co-vary negatively. Positive $ indicates a tendency for the variables X and Y to co-vary positively. -1! $! 1 $ = -1 all pairs (X,Y lie on a straight line with negative slope. $ = 1 all pairs (X,Y lie on a straight line with positive slope. Units of $? $ is the Covariance of the standardized random variables X "µ X and Y "µ Y Y. (Details left to student. $ = 0 Cov(X,Y = 0.

13 14 Uncorrelated variables: X and Y are uncorrelated means $ X,Y = 0 (or equivalently, Cov(X,Y = 0. Examples: 1. X and Y are independent % uncorrelated. (Why? 2. X uniform on the interval [-1, 1]. Y = X 2. X and Y are uncorrelated (details homework X and Y not independent. (E(Y X not constant $ is a measure of the degree of a overall nonconstant linear relationship between X and Y. Example 2 shows: Two variables can have a strong nonlinear relationship and still be uncorrelated. Sample variance, covariance, and correlation (x 1,y 1, (x 2,y 2,, (x n,y n sample of data from the joint distribution of X and Y Sample covariance: cov(x,y (or = 1 n "1 n i=1 (x i " x (y i " y Sample correlation coefficient r (or ˆ " = cov( x, y sd(xsd(y. Estimates of the corresponding population parameters. Analogous properties.