Chapter 5: Joint Probability Distributions

Transcription

1 Chapter 5: Joint Probability Distributions Seungchul Baek Department of Statistics, University of South Carolina STAT 509: Statistics for Engineers 1 / 19

2 Joint pmf Definition: The joint probability mass function (joint pmf) of two discrete r.v. s Y 1 and Y 2 is f (y 1, y 2 ) = P(Y 1 = y 1, Y 2 = y 2 ). The joint (bivariate) pmf gives the probability that the pair (Y 1, Y 2 ) takes a specific pair of values (y 1, y 2 ). Theorem: If Y 1 and Y 2 are discrete r.v. s with the joint pmf f (y 1, y 2 ), then: (i) f (y 1, y 2 ) 0, for all y 1, y 2 (ii) y 1,y 2 f (y 1, y 2 ) = 1, where we sum over all possible (y 1, y 2 ) pairs. 2 / 19

3 Joint pmf: Example Tornados are natural disasters that cause millions of dollars in damage each year. An actuary determines that the annual numbers of tornadoes in two Iowa counties (Lee and Van Buren) are jointly distributed as indicated in the table below. Let Y 1 and Y 2 denote the number of tornados seen each year in Lee and Van Buren counties, respectively. f (y 1, y 2 ) y 2 = 0 y 2 = 1 y 2 = 2 y 2 = 3 y 1 = y 1 = y 1 = What is the probability that there is no more than one tornado seen in the two counties combined? What is the probability that there are two tornadoes in Lee County? 3 / 19

4 Joint cdf Definition: (Y 1, Y 2 ) is The (bivariate) joint cdf of the random vector F (y 1, y 2 ) = P(Y 1 y 1, Y 2 y 2 ). If (Y 1, Y 2 ) are jointly discrete, F (y 1, y 2 ) = t 1 y 1 t 2 y 2 f (t 1, t 2 ). 4 / 19

5 Joint pdf Definition: Let Y 1 and Y 2 be jointly continuous r.v. s with joint cdf F (y 1, y 2 ). If there is a nonnegative function f (y 1, y 2 ) such that F (y 1, y 2 ) = y1 y2 f (t 1, t 2 )dt 1 dt 2, then f (y 1, y 2 ) is called the joint probability density function (joint pdf) of (y 1, y 2 ). Theorem: If Y 1 and Y 2 are continuous r.v. s with the joint pdf f (y 1, y 2 ), then: (i) f (y 1, y 2 ) 0, for all y 1, y 2 (ii) f (y 1, y 2 )dy 1 dy 2 = 1. 5 / 19

6 Example Suppose the time (in hours) to complete task 1 and task 2 for a random employee has the joint pdf: { e (y 1+y 2 ), y 1 > 0, y 2 > 0 f (y 1, y 2 ) = 0, elsewhere. Find the probability that a random employee takes less than 2 hours on task 1 and between 1 and 3 hours on task 2. 6 / 19

7 Marginal pmf Definition: If Y 1, Y 2 are jointly discrete r.v. s with joint pmf f (y 1, y 2 ), then the marginal pmf of Y 1 is f (y 1 ) = f (y 1, y 2 ). y 2 Similarly, the pmf Y 2 is f (y 2 ) = y 1 f (y 1, y 2 ). 7 / 19

8 Marginal pdf Definition: If Y 1, Y 2 are jointly continuous r.v. s with joint pdf f (y 1, y 2 ), then the marginal pdf of Y 1 is f (y 1 ) = Similarly, the pdf Y 2 is f (y 2 ) = f (y 1, y 2 )dy 2. f (y 1, y 2 )dy 1. 8 / 19

9 Example Suppose we model two proportions Y 1 and Y 2 with the joint pdf { 0.4(y 1 + 4y 2 ), 0 < y 1 < 0, 0 < y 2 < 1 f (y 1, y 2 ) = 0, elsewhere. Find the marginal pdf of Y 1. Find P(Y 1 > 0.3). 9 / 19

10 Conditional pmf/pdf Definition: If Y 1, Y 2 are jointly discrete/continuous r.v. s with joint pmf/pdf f (y 1, y 2 ), then the conditional pmf/pdf of Y 1 given Y 2 = y 2 is: f (y 1 y 2 ) = f (y 1, y 2 ), f (y 2 ) provided that f (y 2 ) > 0. Similarly, the conditional pmf/pdf of Y 2 given Y 1 = y 1 is: provided that f (y 1 ) > 0. f (y 2 y 1 ) = f (y 1, y 2 ), f (y 1 ) 10 / 19

11 Properties of Conditional pdf The conditional pdf itself is a pdf, so the following properties are satisfied: (i) f (y 1 y 2 ) 0, (ii) f (y 1 y 2 )dy 1 = 1. Note: for the conditional pmf the integral is just replaced by summation. 11 / 19

12 Conditional Mean and Variance The conditional mean of Y 2 given Y 1 = y 1, denoted as E(Y 2 y 1 ), is E(Y 2 y 1 ) = y 2 f (y 2 y 1 )dy 2, and the conditional variance of Y 2 given Y 1 = y 1, denoted as var(y 2 y 1 ), is var(y 2 y 1 ) = E(Y 2 2 y 1 ) {E(Y 2 y 1 )} / 19

13 Independence of r.v. s Theorem: If Y 1, Y 2 are jointly discrete/continuous r.v. s with pmf/pdf s f 1 (y 1 ) and f 2 (y 2 ) and joint pdf f (y 1, y 2 ), then Y 1 and Y 2 are independent if and only if for all y 1 and y 2. f (y 1, y 2 ) = f 1 (y 1 )f 2 (y 2 ) 13 / 19

14 Expected Value of a Function of Two r.v. s Definition: If Y 1, Y 2 are jointly discrete/continuous r.v. s with pmf/pdf f (y 1, y 2 ). Then for some function g(y 1, Y 2 ): { y E{g(Y 1, Y 2 )} = 1 y 2 g(y 1, y 2 )f (y 1, y 2 ), discrete g(y1, y 2 )f (y 1, y 2 )dy 1 dy 2, continuous. 14 / 19

15 Covariance Definition: If Y 1 and Y 2 are r.v. s with respective means µ 1 and µ 2, then: cov(y 1, Y 2 ) = E{(Y 1 µ 1 )(Y 2 µ 2 )}. Note: cov(y 1, Y 2 ) = E(Y 1 Y 2 ) µ 1 µ 2. Interpretation: 1. cov(y 1, Y 2 ) > 0 implies Y 1 and Y 2 are positively linearly related. 2. cov(y 1, Y 2 ) < 0 implies Y 1 and Y 2 are negatively linearly related. 3. cov(y 1, Y 2 ) = 0 implies NO linear relationship between Y 1 and Y / 19

16 Independence and Covariance Theorem: If Y 1, Y 2 are independent, then cov(y 1, Y 2 ) = 0 Note: The converse is not true. Example: Let Y 1 Unif( 1, 1) and let Y 2 = Y 2 1. Then cov(y 1, Y 2 ) = 0, but Y 1, Y 2 are obviously dependent. 16 / 19

17 Correlation It is hard to judge the strength of the linear association using the covariance since the covariance depends on the scale of measurement. We often use a standardized version called the correlation coefficient (denoted by ρ): ρ = cov(y 1, Y 2 ) var(y1 ) var(y 2 ). It is always true that 1 ρ 1, so values of ρ near 1 (or near -1) indicate a strong positive (or negative) linear association. 17 / 19

18 Linear Functions of Random Variables Theorem: Consider r.v. s Y 1,..., Y n and X 1,..., X m such that E(Yi 2 ) < and E(Xj 2 ) < for all i, j. For constants a 1,..., a n and b 1,..., b m, define U 1 = n i=1 a iy i and U 2 = n j=1 b jx j. Then E(U 1 ) = var(u 1 ) = = cov(u 1, U 2 ) = n a i E(Y i ) = a 1 E(Y 1 ) + + a n E(Y n ) i=1 n ai 2 var(y i ) + 2 a ia j cov(y i, Y j ) i<j i=1 n ai 2 var(y i ) + a ia j cov(y i, Y j ) i j i=1 n i=1 j=1 m a i b j cov(y i, X j ). 18 / 19

19 Example Achievement tests are commonly seen in educational or employment settings. For a large population of test-takers, let Y 1, Y 2, and Y 3 represent scores for different parts of an exam. Suppose that Y 1 N (12, 4), Y 2 N (16, 9), and Y 3 N (20, 16). Suppose additionally that Y 1 and Y 2 are independent, cov(y 1, Y 3 ) = 0.8, and cov(y 2, Y 3 ) = 6.7. Two different summary measures are computed to assess a subjects performance: U 1 = 0.5Y 1 2Y 2 + Y 3, U 2 = 3Y 1 2Y 2 Y 3. Find E(U 1 ) and var(u 1 ). Find cov(u 1, U 2 ). 19 / 19