STAT 361 Fall Homework 1: Solution. It is customary to use a special sign Σ as an abbreviation for the sum of real numbers
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1 STAT 361 Fall 2016 Homework 1: Solution It is customary to use a special sign Σ as an abbreviation for the sum of real numbers x 1, x 2,, x n : x i = x 1 + x x n. If x 1,..., x n are real numbers, their mean values is denoted x = 1 n x i. 1. Properties of Σ. a Explain the following identity: (ax i ± by i = a x i ± b y i, where a and b are any real numbers. In particular, ax i = a x i, a = na. b Prove that (x i x = 0. Solution: a This property follows from the ordinary properties of real numbers: (a + b + c = a + (b + c (the associative law; a + b = b + a (the commutative law; c(a + b = ca + cb (the distributive law. b By Part a, (x i x = x i ( ( ( x = x i n x = x i x i = (Sum of squares Prove the following identities: a (x i x 2 = (x i xx i. Hint: Use Problem 1. 1
2 b ( x 2 i n x 2. Hint: Use Part a. c Prove that 0 if and only if, for some real c, x i c, for all i = 1,..., n. Solution: a Using Problem 1, one gets (x i x 2 = (x i xx i (x i x x = (x i xx i x (x i x = (x i xx i. b Using Problem 1 and Part a, one gets x 2 i x x i = x 2 i n x 2. c If 0,, then identically x i = x, or x i = c, where c = x. Conversely, if x i = c, for all i = 1,..., n, then also x = 1 x i = c. n Hence, (Sum of products Prove the following identities: a S xy = Hint: Use Problem 1. b Hint: Use Part a. (x i x(y i ȳ = (x i xy i. ( S xy = x i y i n xȳ. Solution: a By Problem 1, S xy = (x i x(y i ȳ = (x i xy i ȳ (x i x = (x i xy i. b Continuing from Part a ( ( S xy = x i y i x y i = x i y i n xȳ. 2
3 4. (Simple linear models a (Pure intercept model Let real numbers y 1,..., y n be given. Find that value ĉ of a real variable c which imizes the sum of squares: SS(c = (y i c 2. Hint: Differentiate SS(c. b (Regression through the origin Let real numbers x 1,..., x n and y 1,..., y n be given. Find that value ˆd of a real variable d which imizes the sum of squares: SS(d = (y i dx i 2. Hint: Differentiate SS(d. c (Simple linear regression Let real numbers x 1,..., x n and y 1,..., y n be given. Find values â and ˆb of real variables a and b which imize the sum of squares: SS(a, b = (y i a bx i 2. Hint: Differentiate SS(a, b first w.r.t. both a and b. b Show that the values â and ˆb found in Part a can be represented as ˆb = S xy S xx, â = ȳ ˆb x. Solution: a Since S(c is a non-negative parabola, SS(c = nc 2 2c y i + yi 2, is has only one point of imum ĉ, which satisfies S (ĉ = 0, and no maxima. We calculate SS (c = ( 2(y i c = 0, or ( (y i c = y i nc = 0. ĉ = 1 n 3 y i.
4 b Similarly, the value ˆd imizing S(d satisfies or SS (d = ( 2(y i dx i x i = 0, x i y i = d x 2 i. n ˆd = x iy i n. x2 i c Here again, function SS(a, b 0 achieves imum at some â, ˆb, but has no maxima. The values â and ˆb satisfy the following two equations: SS(a, b a = ( 2(y i a bx i = 0, SS(a, b a The first equation transforms to = ( 2(y i a bx i x i = 0. y i na b x i = n(ȳ a b x = 0. â = ȳ b x. Substituting this formula in the second equation, one gets (y i ȳ b(x i xx i = 0, or b (x i xx i = (y i ȳx i. According to Problems 2,3, this means that b S xy. ˆb = S xy S xx. 4
5 5. (Quality of fit Let again real numbers x 1,..., x n and y 1,..., y n be given. Suppose that all three models, a, b and c, described in Problem 4 are applied to real variables y i. Each time, quality of fit is measured by the corresponding smallest sum-of-squares, called residual sum. Which of the three models will have the best quality of fit? Explain. Solution: Note that Since, one gets SS(a, b SS(a, 0. SS(c = SS(c, 0, SS(a, b SS(c. a the simple linear model provides at least as good a fit, as the intercept-only model. Similarly, SS(a, b SS(0, b. Since, one gets SS(d = SS(0, d, SS(a, b SS(d. the simple linear model provides at least as good a fit, as the slope-only model. 6. (Fitting simple linear model to the data The data collected in Table 1.1 on p. 29 represents the first author s running times (T for marathons vz. his age (A. Let y = ln(t and x = ln(a. Fit all three linear models, a, b, and c discussed in Problem 4 to the data (x i, y i. Use the statistical program R for your calculations. Calculate the residual sum for each of these models. Do these results support your conclusion achieved in Problem 5? Hint: Use the following R codes: y=log(t x=log(a chat=mean(y dhat=sum(x*y/sum(x*x bhat=sum((x-mean(x*(y-mean(y/sum((x-mean(x^2 ahat=mean(y-bhat*mean(x resid1=sum((y-chat^2 resid2=sum((y-dhat*x^2 resid3=sum((y-ahat-bhat*x^2 print(c(resid1,resid2,resid3 b c d 5
6 Solution: T=c(166.87,173.25,175.17,178.97,176.63,175.03,180.32,183.02,192.33, A=c(46.5,47,47.5,49.5,50.5,54.5,56,58.5,59.5,60.0 y=log(t;x=log(a chat=mean(y dhat=(sum(x*y/(sum(x^2 bhat=sum((x-mean(x*(y-mean(y/sum((x-mean(x^2 ahat=mean(y-bhat*mean(x resid1=sum((y-chat^2 resid2=sum((y-dhat*x^2 resid3=sum((y-ahat-bhat*x^2 print(c(resid1,resid2,resid3 [1] These results show that the simple linear model provides a better fit than both the intercept-only and the slope-only models, thus supporting the conclusion made in Problem 5. 6
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