# Lesson 5 Chapter 4: Jointly Distributed Random Variables

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1 Lesson 5 Chapter 4: Jointly Distributed Random Variables Department of Statistics The Pennsylvania State University

2 1 Marginal and Conditional Probability Mass Functions The Regression Function Independence 2 Expected Value of Sums The Covariance and Variance of Sums 3 4

3 Motivation When we record more than one characteristic from each population unit, the outcome variable is multivariate, e.g., Y = age of a tree, X = the tree s diameter at breast height. Y = propagation of an ultrasonic wave, X = tensile strength of substance. Of interest here is not only each variable separately, but also to a) quantify the relationship between them, and b) be able to predict one from another. The joint distribution (i.e., joint pmf, or joint pdf) forms the basis for doing so.

4 Marginal and Conditional Probability Mass Functions The Regression Function Independence The joint probability mass function of two discrete random variables, X and Y, is denoted my p(x, y) and is defined as p(x, y) = P(X = x, Y = y) If X and Y take on only a few values, the joint pmf is typically given in a table like Y p(x, y) X The Axioms of probability imply that i p(x i, y i ) = 1.

5 Marginal and Conditional Probability Mass Functions The Regression Function Independence

6 Marginal and Conditional Probability Mass Functions The Regression Function Independence Example The joint pmf of X = amount of drug administered to a randomly selected rat, and Y = the number of tumors the rat develops, is: y p(x, y) mg/kg x 1.0 mg/kg mg/kg Thus 48.5% of the rats will receive the 1.0 mg dose and will develop 0 tumors, while 40% of the rats will receive the 0.0 mg dose.

7 Marginal and Conditional Probability Mass Functions The Regression Function Independence The above probability calculations are not entirely new. One of the problems in Chapter 2 involves the following: A telecommunications company classifies transmission calls by their duration and by their type. The probabilities for a random (e.g. the next) transmission call to be in each category are Type of Transmission Call Duration V D F > < Find the probabilities of E 1 = the next call is a voice call, and E 2 = the next call is brief.

8 Marginal and Conditional Probability Mass Functions The Regression Function Independence 1 Marginal and Conditional Probability Mass Functions The Regression Function Independence 2 Expected Value of Sums The Covariance and Variance of Sums 3 4

9 Marginal and Conditional Probability Mass Functions The Regression Function Independence The individual PMFs of X and Y are called the marginal pmfs. We saw that p X (0) = P(X = 0) is obtained by summing the probabilities in the first row. Similarly, for p X (1) and p X (2). Finally, the marginal pmf of Y is obtained by summing the columns. y p(x, y) mg/kg x 1 mg/kg mg/kg

10 Marginal and Conditional Probability Mass Functions The Regression Function Independence The formulas for obtaining the marginal PMFs of X and Y in terms of their joint pmf are: p X (x) = p(x, y), and y in S Y p Y (y) = p(x, y). x in S X Keeping x fixed in the first formula means that we are summing all entries of the x-row. Similarly, keeping y fixed in the second formula means that we are summing all entries of the y-column.

11 Marginal and Conditional Probability Mass Functions The Regression Function Independence Conditional Probability Mass Functions From the definition of conditional probability, we have P(Y = y X = x) = P(X = x, Y = y) P(X = x) = p(x, y) p X (x), When we think of P(Y = y X = x) as a function of y with x being kept fixed, we call it the conditional pmf of Y given that X = x, and write it as. p Y X (y x) or p Y X=x (y)

12 Marginal and Conditional Probability Mass Functions The Regression Function Independence When the joint pmf of (X, Y ) is given in a table form, p Y X=x (y) is found by dividing the joint probabilities in the x-row by the marginal probability that X = x. Example Find the conditional pmf of the number of tumors when the dosage is 0 mg and when the dosage is 2 mg. Solution: y p Y X (y X = 0).388/.4= /.4= /.4=.0075 p Y X (y X = 2).090/.1=.9.008/.1= /.1=.02

13 Marginal and Conditional Probability Mass Functions The Regression Function Independence The conditional pmf is a proper pmf. Thus, p Y X (y x) 0, for all y in S Y, and p Y X (y x) = 1. y in S Y The conditional expected value of Y given X = x is the mean of the conditional pmf of Y given X = x. It is denoted by E(Y X = x) or µ Y X=x or µ Y X (x). The conditional variance of Y given X = x is the variance of the conditional pmf of of Y given X = x. It is denoted by σ 2 Y X=x or σ2 Y X (x).

14 Marginal and Conditional Probability Mass Functions The Regression Function Independence Example Find the conditional expected value and variance of the number of tumors when X = 2. Solution. Using the conditional pmf that we found before, we have, E(Y X = 2) = 0 (.9) + 1 (.08) + 2 (.02) =.12. Compare this with E(Y ) =.047. Next, E(Y 2 X = 2) = 0 (.9) + 1 (.08) (.02) =.16. so that σ 2 Y X (2) = =.1456

15 Marginal and Conditional Probability Mass Functions The Regression Function Independence Proposition 1 The joint pmf of (X, Y ) can be obtained as p(x, y) = p Y X (y x)p X (x) Multiplication rule for joint probabilities. 2 The marginal pmf of Y can be obtained as p Y (y) = x in S X p Y X (y x)p X (x) Law of Total Probability for marginal PMFs.

16 Marginal and Conditional Probability Mass Functions The Regression Function Independence Read Example 4.3.4

17 Marginal and Conditional Probability Mass Functions The Regression Function Independence 1 Marginal and Conditional Probability Mass Functions The Regression Function Independence 2 Expected Value of Sums The Covariance and Variance of Sums 3 4

18 Marginal and Conditional Probability Mass Functions The Regression Function Independence For a bivariate r.v. (X, Y ), the regression function of Y on X shows how the conditional mean of Y changes with the observed value of X. More precisely, Definition The conditional expected value of Y given that X = x, i.e. µ Y X (x) = E(Y X = x), when considered as a function of x, is called the regression function of Y on X. The regression function of Y on X is the fundamental ingredient for predicting Y from and observed value of X.

19 Marginal and Conditional Probability Mass Functions The Regression Function Independence Example In the example where X = amount of drug administered and Y = number of tumors developed we have: y p Y X=0 (y) p Y X=1 (y) p Y X=2 (y) Find the regression function of Y on X. Solution: Here, E(Y X = 0) = , E(Y X = 1) = 0.04, E(Y X = 2) = Thus the regression function of Y on X is x µ Y X (x)

20 Marginal and Conditional Probability Mass Functions The Regression Function Independence Proposition (Law of Total Expectation) E(Y ) = E(Y X = x)p X (x). x in S X Example Use the regression function obtained in the previous example, and the x pmf of X, which is p X (x).4.5.1, to find µ Y. Solution: =.047

21 Marginal and Conditional Probability Mass Functions The Regression Function Independence 1 Marginal and Conditional Probability Mass Functions The Regression Function Independence 2 Expected Value of Sums The Covariance and Variance of Sums 3 4

22 Marginal and Conditional Probability Mass Functions The Regression Function Independence The discrete r.v.s X, Y are called independent if p X,Y (x, y) = p X (x)p Y (y), for all x, y. The r.v.s X 1, X 2,..., X n are called independent if p X1,X 2,...,X n (x 1, x 2,..., x n ) = p X1 (x 1 )p X2 (x 2 ) p Xn (x n ). If X 1, X 2,..., X n are independent and also have the same distribution they are called independent and identically distributed, or iid for short.

23 Marginal and Conditional Probability Mass Functions The Regression Function Independence Example Let X = 1 or 0, according to whether component A works or not, and Y = 1 or 0, according to whether component B works or not. From their repair history it is known that the joint pmf of (X, Y ) is X Are X, Y independent? Y

24 Marginal and Conditional Probability Mass Functions The Regression Function Independence Proposition 1 If X and Y are independent, so are g(x) and h(y ) for any functions g, h. 2 If X 1,..., X n1 and Y 1,..., Y n2 are independent, so are g(x 1,..., X n1 ) and h(y 1,..., Y n2 ) for any functions g, h. Example Consider the two-component system of the previous example and suppose that the failure of component A incurs a cost of \$500.00, while the failure of component B incurs a cost of \$ Let C A and C B be the costs incurred by the failures of components A and B respectively. Are C A and C B independent?

25 Marginal and Conditional Probability Mass Functions The Regression Function Independence Proposition Each of the following statements implies, and is implied by, the independence of X and Y. 1 p Y X (y x) = p Y (y). 2 p Y X (y x) does not depend on x, i.e. is the same for all possible values of X. 3 p X Y (x y) = p X (x). 4 p X Y (x y) does not depend on y, i.e. is the same for all possible values of Y.

26 Marginal and Conditional Probability Mass Functions The Regression Function Independence Example X takes the value 0, 1, or 2 if a child under 5 uses no seat belt, uses adult seat belt, or uses child seat. Y takes the value 0, or 1 if a child survives car accident, or not. Suppose that y 0 1 p Y X=0 (y) p Y X=1 (y) p Y X=2 (y) and x p X (x) Are X and Y independent?

27 Marginal and Conditional Probability Mass Functions The Regression Function Independence Example (Continued) Here p Y X=x (y) depends on x and so X, Y are not independent. Alternatively, the joint pmf of X, Y is y p(x, y) x p Y (y) Here p X (0)p Y (0) = = p(0, 0) = Thus X, Y are not independent.

28 Marginal and Conditional Probability Mass Functions The Regression Function Independence Example Show that if X and Y are independent then E(Y X = x) = E(Y ). Thus, if X, Y are independent then the regression function of Y on X is a constant function.

29 The Basic Result Expected Value of Sums The Covariance and Variance of Sums As in the univariate case the expected value and, consequently, the variance of a function of random variables (statistic) can be obtained without having to first obtain its distribution. Proposition E(h(X, Y )) = x h(x, y)p X,Y (x, y) σ 2 h(x,y ) = E[h2 (X, Y )] [E(h(X, Y ))] 2 y Formulas for the continuous case are given in Proposition

30 Example Find E(X + Y ) and σ 2 X+Y Expected Value of Sums The Covariance and Variance of Sums if the joint pmf of (X, Y ) is y p(x, y) x (3.1) Solution: By the previous formula, E(X + Y ) = (0)0.1 + (1) (2) (1) (2)0.2 + (3) (2) (3) (4)0.3 = Next

31 Expected Value of Sums The Covariance and Variance of Sums Example (Continued) E[(X + Y ) 2 ] = (0)0.1 + (1) (2 2 ) (1) (2 2 )0.2 + (3 2 ) (2 2 ) (3 2 ) (4 2 )0.3 = Thus σx+y 2 = = Example Find E(min{X, Y }) and Var(min{X, Y }) if the joint pmf of (X, Y ) is as in the previous example.

32 Expected Value of Sums The Covariance and Variance of Sums Proposition If X and Y are independent, then E(g(X)h(Y )) = E(g(X))E(h(Y )) holds for any functions g(x) and h(y).

33 Expected Value of Sums The Covariance and Variance of Sums 1 Marginal and Conditional Probability Mass Functions The Regression Function Independence 2 Expected Value of Sums The Covariance and Variance of Sums 3 4

34 Expected Value of Sums The Covariance and Variance of Sums h(x 1,..., X n ) is a linear combination of X 1,..., X n if Proposition h(x 1,..., X n ) = a 1 X a n X n. X is a linear combination with all a i = 1/n. T = n i=1 X i is a linear combination with all a i = 1. Let X 1,..., X n be any r.v.s (i.e discrete or continuous, independent or dependent), with E(X i ) = µ i. Then E (a 1 X a n X n ) = a 1 µ a n µ n.

35 Expected Value of Sums The Covariance and Variance of Sums Corollary 1. Let X 1, X 2 be any two r.v.s. Then E (X 1 X 2 ) = µ 1 µ 2, and E (X 1 + X 2 ) = µ 1 + µ Let X 1,..., X n be such that E(X i ) = µ for all i. Then ( ) E(T ) = nµ, E X = µ.

36 Expected Value of Sums The Covariance and Variance of Sums Example Tower 1 is constructed by stacking 30 segments of concrete vertically. Tower 2 is constructed similarly. The height, in inches, of a randomly selected segment is uniformly distributed in (35.5,36.5). Find a) the expected value of the height, T 1, of tower 1, b) the expected value of the height, T 2 of tower 2, and c) the expected value of the difference T 1 T 2.

37 Expected Value of Sums The Covariance and Variance of Sums Example (Expected Value of the Binomial R.V.) Let X Bin(n, p). Find E(X). Solution: Recall that X = n i=1 X i, where each X i is a Bernoulli r.v. with probability of 1 equal to p. Since E(X i ) = p, for all i, E(X) = np. Example (Expected Value of the Negative Binomial R.V.) Let X NBin(r, p). Thus, X is the number of Bernoulli trials up to an including the rth 1. Find E(X). Solution: Recall that X = r i=1 X i, where each X i is a geometric, or NBin(1, p) r.v. Since E(X i ) = 1/p, for all i, E(X) = r/p.

38 Example Expected Value of Sums The Covariance and Variance of Sums Let N be the number of accidents per month in an industrial complex, and X i the number of injuries caused by the ith accident. Suppose the X i are independent from N and E(X i ) = 1.5 for all i. If E(N) = 7, find the expected number of injuries, Y = N i=1 X i, in a month. Solution: First note that, ( N ) ( n ) E(Y N = n) = E X i N = n = E X i N = n i=1 i=1 n n = E(X i N = n) = E(X i ) = 1.5n i=1 i=1

39 Expected Value of Sums The Covariance and Variance of Sums where the equality before the last holds by the independence of the X i and N. Next, using the Law of Total Expectation, E(Y ) = n in S N E(Y N = n)p(n = n) = n in S N 1.5nP(N = n) = 1.5 n in S N np(n = n) = 1.5E(N) = 10.5,

40 Expected Value of Sums The Covariance and Variance of Sums The above example dealt with the expected value of the sum of a random number, N, of random variables when these variables are independent of N. In general we have: Proposition (Proposition 4.4.3) Suppose that N is an integer valued random variable, and the random variables X i are independent from N and have common mean value µ. Then, ( N ) E X i = E(N)µ i=1 The expected value of a sum of a random number of random variables. Read also Example

41 Expected Value of Sums The Covariance and Variance of Sums 1 Marginal and Conditional Probability Mass Functions The Regression Function Independence 2 Expected Value of Sums The Covariance and Variance of Sums 3 4

42 Expected Value of Sums The Covariance and Variance of Sums The variance of a sum is the sum of the variances only in the case of independent variables. Here is why: { [X ] } 2 Var(X + Y ) = E + Y E(X + Y ) { [(X ] } 2 = E E(X)) + (Y E(Y )) = Var(X) + Var(Y ) + 2E [ (X E(X))(Y E(Y )) ]. If X and Y are independent then, E [ (X E(X))(Y E(Y )) ] = E [ (X E(X)) ] E [ (Y E(Y )) ] = 0

43 Expected Value of Sums The Covariance and Variance of Sums Thus, if X and Y are independent Var(X + Y ) = Var(X) + Var(Y ). But if X and Y are not independent (or only correlated) Var(X + Y ) = Var(X) + Var(Y ) + 2E [ (X E(X))(Y E(Y )) ]. The quantity E [ (X E(X))(Y E(Y )) ] is called the covariance of X and Y. A computational formula is σ XY = E [ (X µ X )(Y µ Y ) ] = E(XY ) µ X µ Y

44 Expected Value of Sums The Covariance and Variance of Sums Example Let X = deductible in car insurance, and Y = deductible in home insurance, of a randomly chosen home and car owner. Suppose that the joint pmf of X, Y is x y where the deductible amounts are in dollars. Find σ XY.

45 Solution: Outline Expected Value of Sums The Covariance and Variance of Sums First, and E(XY ) = x xyp(x, y) = 23, 750, y E(X) = x xp X (x) = 175, E(Y ) = y yp Y (y) = 125. Thus, σ XY = 23, = 1875.

46 Expected Value of Sums The Covariance and Variance of Sums Proposition 1 If X 1, X 2 are independent, Var(X 1 + X 2 ) = σ1 2 + σ2 2, and Var(X 1 X 2 ) = σ1 2 + σ Without independence, Var(X 1 + X 2 ) = σ1 2 + σ Cov(X 1, X 2 ) Var(X 1 X 2 ) = σ1 2 + σ2 2 2Cov(X 1, X 2 )

47 Expected Value of Sums The Covariance and Variance of Sums Example (Simulation proof of Var(X 1 + X 2 ) = Var(X 1 X 2 )) set.seed=111; x=runif(10000); y=runif(10000) var(x+y); var(x-y)

48 Expected Value of Sums The Covariance and Variance of Sums Proposition 1 If X 1,..., X n are independent, Var(a 1 X a n X n ) = a1 2 σ a2 nσn, 2 where σi 2 = Var(X i ). 2 If X 1,..., X n, are dependent, Var(a 1 X a n X n ) = a1 2 σ a2 nσn 2 + where σ ij = Cov(X i, X j ). n a i a j σ ij, i=1 j i

49 Expected Value of Sums The Covariance and Variance of Sums Corollary 1 If X 1,..., X n are iid with variance σ 2, then where T = n i=1 X i. Var(X) = σ2 n and Var(T ) = nσ2, 2 If the X i are Bernoulli, (so X = p and σ 2 = p(1 p)), Var( p) = p(1 p) n

50 Example Expected Value of Sums The Covariance and Variance of Sums Let X denote the number of short fiction books sold at the State College, PA, location of the Barnes and Noble bookstore in a given week, and let Y denote the corresponding number sold on line from State College residents. We are given that E(X) = 14.80, E(Y ) = 14.51, E(X 2 ) = 248.0, E(Y 2 ) = 240.7, and E(XY ) = 209. Find Var(X + Y ). Solution: Since Var(X) = = 28.96, Var(Y ) = = 30.16, and Cov(X, Y ) = 209 (14.8)(14.51) = 5.75, we have Var(X + Y ) = Var(X) + Var(Y ) + 2Cov(X, Y ) = =

51 Expected Value of Sums The Covariance and Variance of Sums Read also Example 4.4.8, p. 238.

52 Expected Value of Sums The Covariance and Variance of Sums Proposition 1. Cov(X, Y ) = Cov(Y, X). 2. Cov(X, X) = Var(X). 3. If X, Y are independent, then Cov(X, Y ) = Cov(aX + b, cy + d) = accov(x, Y ), for any real numbers a, b, c and d. ( m 5. Cov i=1 X i, ) n j=1 Y j = m n i=1 j=1 Cov ( ) X i, Y j.

53 Expected Value of Sums The Covariance and Variance of Sums Example (σ XY = 0 for dependent X, Y ) Find Cov(X, Y ), where X, Y have joint pmf given by y /3 0 1/3 x 0 0 1/3 1/3 1 1/3 0 1/3 2/3 1/3 1.0 Solution: Since E(X) = 0, the computational formula gives Cov(X, Y ) = E(XY ). However, the product XY takes the value zero with probability 1. Thus, Cov(X, Y ) = E(XY ) = 0.

54 Expected Value of Sums The Covariance and Variance of Sums Example Consider the example with the car and home insurance deductibles, but suppose that the deductible amounts are given in cents. Find the covariance of the two deductibles. Solution: Let X, Y denote the deductibles of a randomly chosen home and car owner in cents. If X, Y denote the deductibles in dollars, then X = 100X, Y = 100Y. According to part 4 of the proposition, σ X Y = σ XY = 18, 750, 000.

55 Expected Value of Sums The Covariance and Variance of Sums Example On the first day of a wine tasting event three randomly selected judges are to taste and rate a particular wine before tasting any other wine. On the second day the same three judges are to taste and rate the wine after tasting other wines. Let X 1, X 2, X 3 be the ratings, on a 100 point scale, in the fist day, and Y 1, Y 2, Y 3 be the ratings on the second day. We are given that the variance of each X i is σ 2 X = 9, the variance of each Y i is σ 2 Y = 4, the covariance Cov(X i, Y i ) = 5, for all i = 1, 2, 3, and Cov(X i, Y j ) = 0 for all i j. Find the variance of combined rating X + Y.

56 Solution. Outline Expected Value of Sums The Covariance and Variance of Sums First note that Var(X + Y ) = Var(X) + Var(Y ) + 2Cov(X, Y ). Next, with the information given, we have Var(X) = 9/3 and Var(Y ) = 4/3. Finally, using the proposition we have Cov(X, Y ) = Cov(X i, Y j ) = i=1 j= Cov(X i, Y i ) = i=1 Combining these partial results we obtain Var(X + Y ) = 23/3.

57 Expected Value of Sums The Covariance and Variance of Sums Example (Variance of the Binomial RV and of p) Let X Bin(n, p). Find Var(X), and Var( p). Solution: Use the expression X = n X i, i=1 where the X i are iid Bernoulli(p) RVs. Since Var(X i ) = p(1 p), ( ) X p(1 p) Var(X) = np(1 p), Var( p) = Var =. n n

58 Expected Value of Sums The Covariance and Variance of Sums Example (Variance of the Negative Binomial RV) Let X NBin(r, p). Thus, X is the number of Bernoulli trials up to an including the rth 1. Find Var(X). Solution: Use the expression X = r X i, i=1 where the X i are iid Geometric(p), or NBin(1, p) RVs. Since Var(X i ) = (1 p)/p 2, Var(X) = n σx 2 i = r(1 p)/p 2 i=1

59 Positive and Negative Dependence When two variables are not independent, it is of interest to qualify and quantify dependence. X, Y are positively dependent or positively correlated if large values of X are associated with large values of Y, and small values of X are associated with small values of Y. In the opposite case, X, Y are negatively dependent or negatively correlated. If the dependence is either positive or negative it is called monotone.

60 Monotone dependence and regression function The dependence is monotone (positive or negative), if and only if the regression function, µ Y X (x) = E(Y X = x), is monotone (increasing or decreasing) in x. Example 1 X = height and Y = weight of a randomly selected adult male, are positively dependent. 1 If X = height and Y = weight, then µ Y X (1.82) < µ Y X (1.90). 2 X = stress applied and Y = time to failure, are negatively dependent. 1 If X = stress applied and Y = time to failure, then µ Y X (10) > µ Y X (20).

61 The regression function, however, is not designed to measure the degree of dependence of X and Y. Proposition (Monotone dependence and covariance) The monotone dependence is positive or negative, if and only if the covariance is positive or negative. This proposition DOES NOT say that if the covariance is positive then the dependence is positive. Positive covariance implies positive dependence only if we know that the dependence is monotone.

62 Intuition for the Covariance Consider a population of N units and let (x 1, y 1 ), (x 2, y 2 ),..., (x N, y N ) denote the values of the bivariate characteristic of the N units. Let (X, Y ) denote the bivariate characteristic of a randomly selected unit. In this case, the covariance is computed as σ XY = 1 N N (x i µ X )(y i µ Y ), i=1 where µ X = 1 N N i=1 x i and µ Y = 1 N N i=1 y i are the marginal expected values of X and Y.

63 Intuition for Covariance - Continued If X, Y are positively dependent (think of X = height and Y = weight), then (x i µ X )(y i µ Y ) will be mostly positive. If X, Y are negatively dependent (think of X = stress and Y = time to failure), then (x i µ X )(y i µ Y ) will be mostly negative. Therefore, σ XY will be positive or negative according to whether the dependence of X and Y is positive or negative.

64 Quantification of dependence should be scale-free! Proper quantification of dependence should be not depend on the units of measurement. For example, a quantification of the dependence between height and weight should not depend on whether the units are meters and kilograms or feet and pounds. The scale-dependence of the covariance, implied by the property Cov(aX, by ) = abcov(x, Y ) makes it unsuitable as a measure of dependence.

65 Pearson s (or Linear) Correlation Coefficient Definition ( Pearson s (or linear) correlation coefficient) Corr(X, Y ) or ρ XY, of X and Y is defined as ρ XY = Corr(X, Y ) = σ XY σ X σ Y. Proposition (Properties of Correlation) 1 If ac > 0, then Corr(aX + b, cy + d) = Corr(X, Y ). 2 1 ρ(x, Y ) 1. 3 If X, Y are independent, then ρ XY = 0. 4 ρ XY = 1 or 1 if and only if Y = ax + b, for some constants a, b.

66 Example Find the correlation coefficient of the deductibles in car and home insurance of a randomly chosen car and home owner, when the deductibles are expressed a) in dollars, and b) in cents. Solution: If X, Y denote the deductibles in dollars, we saw that σ XY = Omitting the details, it can be found that σ X = 75 and σ Y = Thus, ρ XY = = Next, the deductibles expressed in cents are (X, Y ) = (100X, 100Y ). According to the proposition, ρ X Y = ρ XY

67 Correlation as a Measure of Linear Dependence Commentaries on Pearson s correlation coefficient: It is independent of scale - - highly desirable. It quantifies dependence well: If X and Y are independent, then ρ XY = 0, and if one is a linear function of the other (so knowing one amounts to knowing the other), then ρ XY = ±1. Try the applet CorrelationPicture in to see how the correlation changes with the scatterplot HOWEVER, it measures only linear dependence: It is possible to have perfect dependence (i.e. knowing one amounts to knowing the other) but ρ XY ±1.

68 Example Let X U(0, 1), and Y = X 2. Find ρ XY. Solution: We have σ XY = E(XY ) E(X)E(Y ) = E(X 3 ) E(X)E(X 2 ) = = Omitting calculations, σ X = 1 12, σ Y = ρ XY = = = 2 3. Thus, 5 A similar set of calculations reveals that with X as before and Y = X 4, ρ XY =

69 Example Let X U( 1, 1), and let Y = X 2. Check that here ρ XY = 0. Note that the dependence of X and Y is not monotone. Definition Two variables having zero correlation are called uncorrelated. Independent variables are uncorrelated, but uncorrelated variables are not necessarily independent.

70 Sample versions of covariance and correlation If (X 1, Y 1 ),..., (X n, Y n ) is a sample from the bivariate distribution of (X, Y ), the sample covariance, S X,Y, and sample correlation coefficient, r X,Y, are defined as S X,Y = 1 n 1 r X,Y = S X,Y S X S Y n (X i X)(Y i Y ) i=1 Sample versions of covariance and correlation coefficient

71 Computational formula: [ n ( n ( n )] S X,Y = 1 X i Y i 1 X i) Y i. n 1 n i=1 i=1 i=1 R commands: R commands for covariance and correlation cov(x,y) gives S X,Y cor(x,y) gives r X,Y

72 Example To calibrate a method for measuring lead concentration in water, the method was applied to twelve water samples with known lead content. The concentration measurements, y, and the known concentration levels, x, are (in R notation) x=c(5.95, 2.06, 1.02, 4.05, 3.07, 8.45, 2.93, 9.33, 7.24, 6.91, 9.92, 2.86) and y=c(6.33, 2.83, 1.65, 4.37, 3.64, 8.99, 3.16, 9.54, 7.11, 7.10, 8.84, 3.56) Compute the sample covariance and correlation coefficient.

73 Hierarchical Models Hierarchical models use the Multiplication Rules, i.e. p(x, y) = p Y X=x (y)p X (x), for the joint pmf f (x, y) = f Y X=x (y)f X (x), for the joint pdf in order to specify the joint distribution of X, Y by first specifying the conditional distribution of Y given X = x, and then specifying the marginal distribution of X. This hierarchical method of modeling yields a very rich and flexible class of joint distributions.

74 Example Let X be the number of eggs an insect lays and Y the number of eggs that survive. Model the joint pmf of X and Y. Solution: With the principle of hierarchical modeling, we need to specify the conditional pmf of Y given X = x, and the marginal pmf of X. One possibile specification is Y X = x Bin(x, p), X Poisson(λ), which leads to the following joint pmf of X and Y. p(x, y) = p Y X=x (y)p X (x) = ( x )p y (1 p) x y e λ λ x y x!

75 Calculation of the marginal expected value of Y, i.e. E(Y ), and the marginal pmf of Y can be done with the laws of total probability and total expectation. Example For the previous example, find E(Y ) and the marginal pmf of Y. Solution: Since Y X Bin(X, p), E(Y X) = Xp. Thus, according to the Law of Total Probability for Expectations, E(Y ) = E[E(Y X)] = E[Xp] = E[X]p = λp Using one of the properties of the a Poisson random variables, the marginal distribution of Y is Poisson(λp).

76 Example (The Bivariate Normal Distribution) X and Y are said to have a bivariate normal distribution if ( ) Y X = x N β 0 + β 1 (x µ X ), σε 2, and X N(µ X, σx 2 ). Commentaries: 1. The regression function of Y on X is linear in x: µ Y X (x) = β 0 + β 1 (x µ X ). 2. The marginal distribution of Y is normal with E(Y ) = E[E(Y X)] = E[β 0 + β 1 (X µ X )] = β 0

77 Commentaries (Continued): 3. σ 2 ε is the conditional variance of Y given X and is called error variance. 4. See relation (4.6.3), for the joint pdf of X and Y. A more common form of the bivariate pdf involves µ X and µ Y, σ 2 X, σ 2 Y and ρ XY ; see (4.6.12). 5. Any linear combination, a 1 X + a 2 Y, of X and Y has a normal distribution. A plot a bivariate normal pdf is given in the next slide:

78 Joint PMF of two independent N(0,1) RVs 0.15 z x x

79 Joint PMF of two N(0,1) RVs with ρ = z x x Lesson 5 Chapter 3 4: Jointly Distributed Random Variables

80 Regression Models Regression models focus primarily on the regression function of a variable Y on another variable X. For example, X = speed of an automobile and Y = the stopping distance. Because of this the marginal distribution of X, which is of little interest in such studies, is left unspecified. Moreover, the regression function is highlighted by writing the conditional distribution of Y given X = x as Y = µ Y X (x) + ε (5.1) and ε is called the (intrinsic) error variable. In regression models, Y is called the response variable and X is interchangeably referred to as covariate, or independent variable, or predictor, or explanatory variable.

81 The Simple Linear Regression Model The simple linear regression model specifies that the regression function in (5.1) is linear in x, i.e. µ Y X (x) = α 1 + β 1 x, (5.2) and that Var(Y X = x) is the same for all values x. An alternative expression of the simple linear regression model is µ Y X (x) = β 0 + β 1 (x µ X ). (5.3) The straight line in (5.2) (or (5.3)) is called the regression line.

82 The following picture illustrates the meaning of the slope of the regression line. Figure: Illustration of Regression Parameters

83 Proposition In the linear regression model 1 The intrinsic error variable, ɛ, has zero mean and is uncorrelated from the explanatory variable, X. 2 The slope is related to the correlation by β 1 = σ X,Y σ 2 X = ρ X,Y σ Y σ X. 3 Two additional relationships are µ Y = β 0, σ 2 Y = σ2 ε + β 2 1 σ2 X.

84 Example Suppose Y = 5 2x + ε, and let σ ɛ = 4, µ X = 7 and σ X = 3. a) Find σy 2, and ρ X,Y. b) Find µ Y. Solution: For a) use the formulas in the above proposition: σy 2 = ( 2) = 52 ρ X,Y = β 1 σ X /σ Y = 2 3/ 52 = For b) use the Law of Total Expectation: E(Y ) = E[E(Y X)] = E[5 2X] = 5 2E(X) = 9

85 The normal simple linear regression model specifies that the intrinsic error variable in (5.1) is normally distributed (in addition to the regression function being linear in x), i.e. Y = β 0 + β 1 (X µ X ) + ε, ε N(0, σ 2 ε ). (5.4) An alternative expression of the normal simple linear regression model is Y X = x N(β 0 + β 1 (x µ X ), σ 2 ε ). (5.5) In the above, β 0 + β 1 (x µ X ) can also be replaced by α 1 + β 1 x. An illustration of the normal simple linear regression model is given in the following figure.

86 Figure: Illustration of Intrinsic Scatter in Regression

87 Example Suppose Y = 5 2x + ε, and let ε N(0, σ 2 ε ) where σ ɛ = 4. Let Y 1, Y 2 be observations taken at X = 1 and X = 2, respectively. a) Find the 95th percentile of Y 1 and Y 2. b) Find P(Y 1 > Y 2 ). Solution: a) Y 1 N(3, 4 2 ), so its 95th percentile is = Y 2 N(1, 4 2 ), so its 95th percentile is = b) Y 1 Y 2 N(2, 32) (why?). Thus, P(Y 1 > Y 2 ) = P(Y 1 Y 2 > 0) = 1 Φ ( 2 32 ) =

88 Quadratic and more higher order models are also commonly used. The advantages of such models are: a) It is typically easy to fit such a model to data (i.e. estimate the model parameters from the data), and b) Such models offer easy interpretation of the effect of X on the expected value of Y.

89 Go to previous lesson mga/401/course.info/lesson4.pdf Go to next lesson mga/ 401/course.info/lesson6.pdf Go to the Stat 401 home page http: // mga/401/course.info/

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