STA 4502 Chapter 8

# STA 4502 Chapter 8 - Chapter 8 The Independence Problem In...

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Chapter 8 The Independence Problem In some problems a random sample of n pairs of observations (X, Y) are observed from a bivariate population. Our main interest is in whether or not there is a statistical relationship between these two variables. In particular we are interested in finding whether these two variables are independent of each other and if they are not independent, then we want to find the extent and the type of dependence between them. Data: We obtain n pairs of observations, (X 1 , Y 1 ), (X 2 , Y 2 ), …, (X n , Y n ), one pair of observations from each subject in the population. Assumption A. The n bivariate observations (X 1 , Y 1 ), (X 2 , Y 2 ), …, (X n , Y n ) are a random sample from a continuous bivariate population. That is, (X, Y) pairs are mutually independent and identically distributed (iid) according to some continuous bivariate population. Thus we are assuming independence between the observed pairs. 8.1 Test of Independence based on Signs (Kendall) Hypotheses: Let F X, Y be the joint distribution function of the common bivariate population. Let F X and F Y be the marginal distributions functions of the population of X’s and the population of Y’s, respectively [though not necessarily the same]. Then, the hypotheses of interest are Ho: F X, Y (x, y) = F X (x)×F Y (y) for all (x, y) pairs. Ha: F X, Y (x, y) ≠ F X (x)×F Y (y) for at least one (x, y) pair. Type of dependence we are interested in this section is Kendal’s population correlation coefficient, defined as [See next page]   2 1 2 1 2 ( ) ( ) 0 1 P X X Y Y   Note that (X 2 – X 1 )×(Y 2 – Y 1 ) > 0, if {(X 2 – X 1 ) > 0 and (Y 2 – Y 1 ) > 0} or if {(X 2 – X 1 ) < 0 and (Y 2 – Y 1 ) < 0}. Furthermore, these two events are mutually exclusive and each has probability ½ if X and Y are independent. Hence, if X and Y are independent variables P((X 2 – X 1 )×(Y 2 – Y 1 ) > 0) = P({(X 2 – X 1 ) > 0 and (Y 2 – Y 1 ) > 0}) + P({(X 2 – X 1 ) < 0 and (Y 2 – Y 1 ) < 0}. = P[ (X 1 – X 2 ) > 0] × P[ (Y 1 – Y 2 ) > 0] + P[ (X 1 – X 2 ) < 0] × P[ (Y 1 – Y 2 ) < 0] = ½ × ½ + ½ × ½ = ¼ + ¼ = ½ . Thus, τ = 2×( ½ + ½ ) – 1 = 0 if X and Y are independent random variables. So we conclude that the test of independence is the same as Ho: τ = 0 vs. Ha: τ ≠ 0 (or τ < 0 or τ > 0).

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Some new concepts: Two pairs of observations (X i , Y i ) and (X j , Y j ), i ≠ j, are said to be a concordant pair if (X i – X j )×(Y i – Y j ) > 0 and a discordant pair if (X i – X j )×(Y i – Y j ) < 0. From these, we infer that, There is a positive association between X and Y if there are many more concordant pairs than discordant pairs. Similarly, If the discordant pairs dominate then we infer negative association. Test Statistic is τ = P(Concordant pair) – P(Discordant Pair) = 2×P(Concordant pair) – 1 = 2×P[ (X i – X j )×(Y i – Y j ) > 0] – 1.
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## This note was uploaded on 11/09/2011 for the course STAT 4502 taught by Professor Yesilcay during the Summer '11 term at University of Florida.

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STA 4502 Chapter 8 - Chapter 8 The Independence Problem In...

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