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principles of managerial statistics

Questions and Answers of Principles Of Managerial Statistics

=+13.19. Establish the following inequalities.(i) V −1X(XV −1X)−1XV −12 2 ≤ V −12(p + 1).(ii) ZV −1X(XV −1X)−1XV −1Z2 2 ≤ ZV −1Z2(p + 1).
=+13.18. This exercise is related to the quadratic spline with two knots in Example 13.1.(i) Plot the quadratic spline.(ii) Show that the function is smooth in that it has a continuous derivative on
=+13.17. Show that the minimizer of (13.71) is the same as the best linear unbiased estimator (BLUE) for β and the best linear unbiased predictor (BLUP)for γ in the linear mixed model y = Xβ +
=+13.16. Show that, under the hierarchical Bayes model near the end of Section 13.3, the conditional distribution of θi given A and y is normal with mean equal to the right side of (13.31) with Aˆ
=+13.15. Verify (13.60); that is, the expression of (13.59) when ˆθi is replaced by yi.
=+13.14. This exercise involves some calculus derivations.(i) Verify expressions (13.55) and (13.56).(ii) Verify expression (13.57).(iii) Obtain an expression for ∂AˆFH/∂yi. You man use the
=+13.13. Show that, in the balanced case (i.e., Di = D, 1 ≤ i ≤ m), the P-R, REML, and F-H estimators of A in the Fay–Herriot model are identical(provided that the estimator is nonnegative),
=+13.12. Show that the right side of (13.40) ≤ the right side of (13.41) ≤ the right side (13.39) for any A ≥ 0.
=+13.11. Show that the estimators AˆPR, AˆML, AˆRE, and AˆFH in Section 13.3 possess the following properties: (i) They are even functions of the data—that is, the estimators are unchanged when
=+13.10. Here is a more challenging exercise that the previous one: Show that the Fay–Herriot estimator AˆFH, defined as the solution to (13.33), is√m-consistent.
=+13.9. Show that the estimator AˆPR defined by (13.32) is √m-consistent.
=+13.8. Show that (13.33) is unbiased in the sense that the expectation of the left side is equal to the right side if A is the true variance of the random effects. Also verify (13.34).13.6 Exercises
=+13.7. Show, by formal derivation, that the estimator (13.30) satisfies(13.22). [Hint: You may use the fact that E{ci(ˆθ) − ci(θ)} = o(1) and E{Bi(ˆθ) − Bi(θ)} = o(1).]
=+13.6. Verify expression (13.11), where ˜αi = E(αi|y) has expression (13.4).
=+13.5. Verify (13.8).
=+13.4. Show that the last term on the right side of (13.7) has the order OP(n−1/2 i ). You may recall the argument of showing a similar property of the MLE (see Section 4.7).
=+13.3. Verify the limiting behavior (iv) below (13.4) and also (13.5). [Hint:The following formulas might be useful. For 1 ≤ k ≤ n − 1, we have ∞0 xk−1(1 + x)n dx = (k − 1)!(n − k −
=+13.2. Verify the limiting behaviors (i)–(iii) below (13.4).
=+13.1. This exercise is associated with the mixed logistic model of (13.1).(i) Show that E(αi|y) = E(αi|yi), where yi = (yij )1≤j≤ni .(ii) Verify (13.3) for ˜αi = E(αi|yi).
=+12.23. Show that the right side of (12.64) is O(m). In fact, in this case, it can be shown that the order is O(1) when M is true [Hint: The most challenging part is to evaluate i,j a2 ij = tr(A2).
=+(iv) Show that for model II, the right side of (12.60) is equal to σ2 + τ 2 +β2 0m, where β0 is the true fixed effect under model I and σ2 and τ 2 are the true variance components.
=+(iii) Show that for model I, the right side of (12.58) is equal to τ 2, whereτ 2 is the true variance of the errors.
=+12.22. This exercise involves some details regarding Example 12.14.12.6 Exercises 431(i) Show that PXf = m−1Jm ⊗ I2.(ii) Show that PX = (2m)−1Jm ⊗ J2 for X corresponding to model I and PX =
=+12.21. This exercise involves some of the details in the derivations following(12.56).(i) Show that (PXf − PX )X = 0.(ii) Show that PXf − PX is idempotent; that is, (PXf − PX)2 = PXf − PX .
=+12.20. Show that the RSS measure defined in Example 12.13 is a measure of lack-of-fit according to the definition above Example 12.12.
=+12.12 is a measure of lack-of-fit according to the definition above Example 12.12.
=+12.19. Show that the negative log-likelihood measure defined in Example
=+12.18. Consider a special case of Example 12.11 with ni = k, 1 ≤ i ≤ m, where k ≥ 2. Show that in this case, the estimating equation of Jiang (1998a), which is (12.52) with B = diag(1, 1m),
=+12.17. Show that for any matrices B, U, and V such that V > 0 (positive definite), U is full rank, and BU is square and nonsingular, we have{(BU)−1}BV B{(BU)−1} ≥ (UV −1U)−1(i.e., the
=+12.16. Show that under a GLMM, E(y2 i ) depends on φ, the disperson parameter in (12.4), but E(yiyi ) does not depend on φ if i = i.
=+ijβ = μ, where μ is an unknown parameter, and ni = l, 1 ≤ i ≤ m. Show that in this case, the left side of(12.46) is equal to g(μ) = σ−2 l k=0 φ(k, μ)P(y1· = k), where φ(t, μ) is
=+12.15. This exercise is related to Example 12.10.(i) Show that (12.42) has a unique solution for u when everything else is fixed.(ii) Show that the PQL estimator of β satisfies (12.43).(iii)
=+12.14. Verify (12.38) and obtain an expression for r. Show that r has expectation 0.
=+12.13. Show that the likelihood function in Example 12.9 for estimating ψcan be expressed as (12.35).
=+12.12. Show that the number of coefficents ast in the bivariate polynomial(12.32) is NM =1+ M(M + 3)/2.
=+12.11. This exercise is regarding the projection method that begins with the identity (12.28).(i) Show that E(m1 i=1m2 j=1 δ1,ijk)2 = O(m1m2). [Hint: Note that, given u and v, the δ1,ijk’s
=+(i) Show that the REML estimator of the variance of the errors, σ2 = 1, is the sample variance, ˆσ2 = (m − 1)−1 m i=1(yi − y¯)2, where ¯y = m−1 m i=1 yi.(ii) Show that √m(ˆσ2 −
=+12.10. The additional assumption in part (ii) of Theorem 12.1 that the random effects and errors are nondegenerate is necessary for the asymptotic normality of the REML estimators. To see a simple
=+12.9. Show that the components of an in (12.15) [defined below (12.12)]are quadratic forms of the random effects and errors.
=+(ii) Derive (12.15) using the Taylor expansion and the result of part (i) of the theorem.
=+12.8. This exercise is concerned with some of the arguments used in the proof of Theorem 12.1.(i) Show that (12.14) holds under (12.10)–(12.12), where θn is defined by(12.13) and rn is uniformly
=+Show that in this case, (12.1) and (12.2) can be expressed as y = Ia ⊗ 1b ⊗ 1cβ + 1a ⊗ Ib ⊗ 1cα + Ia ⊗ Ib ⊗ 1cγ + .
=+12.7. Consider the following linear mixed models: yijk = βi+αj+γij+ijk, i = 1,...,a, j = 1,...,b, k = 1,...,c, where βi’s are fixed effects, the αj ’s are random effects, the γij ’s
=+12.6. Show that the REML estimators of the variance components (see Section 12.2) do not depend on the choice of A. This means that if B is another n × (n − p) matrix of full rank such that BX
=+12.6 Exercises 429
=+12.5. Show that, in Example 12.5, the MLE of σ2 is given by (12.7), and the REML estimator of σ2 is given by (12.8).
=+i, respectively. Show that the vector of observations, y = (y1,...,yn), has the same distribution as the Gaussian linear mixed model (1.1), where α ∼ N(0, G), ∼ N(0, τ 2I), and α and
=+where xi and zi are known vectors, β is an unknown vector of regression coefficients, and τ 2 is an unknown variance. Furthermore, suppose that α is multivariate normal with mean 0 and
=+12.4. Suppose that, given a vector of random effects, α, observations y1,...,yn are (conditionally) independent such that yi ∼ N(xiβ + ziα, τ 2),
=+12.3. Give an example of a special case of the longitudinal model that is not a special case of the mixed ANOVA model.
=+12.2 (Two-way random effects model). For simplicity, let us consider the case of one observation per cell. In this case, the observations yij , i = 1,...,m, j = 1,...,k, satisfy yij = μ + ui + vj
=+12.1. Show that, in Example 12.1, the correlation between any two observations from the same individual is σ2/(σ2 + τ 2), whereas observations from different individuals are uncorrelated.
=+11.33. This exercise is related to Example 11.11 at the end of the chapter.(i) Verify the calculations of ˆσ2, hˆ0, ˆγ3, ˆγ4, and hˆ in the example.(ii) Simulate a larger data set, say,
=+11.32. Regarding the parameter θ2 defined below (11.91), show that θ2 =3/8√πσ5 if f is the pdf of N(μ, σ2).
=+11.31. (i) Verify (11.88).(ii) Show that the right side of (11.89) is minimized when h is given by(11.90).
=+11.30. Give a proof of Theorem 11.6. As mentioned, the proof is based on the Taylor expansion. The details can be found in Lehmann’s book but you are encouraged to explore without looking at the
=+(iv) Show that the histogram is pointwise consistent under the limiting process (11.84).
=+11.29. This exercise is related to the expression of the histogram (see Section 11.6).(i) Show that (11.82) holds provided that F is twice continuously differentiable.(ii) Show that the limit
=+11.28. Consider the U-statistic associated with Wilcoxon two-sample test(see the discussion at the end of Section 11.5).(i) Verify (11.80).(ii) Show that under the null hypothesis F = G, we have
=+(iv) Using the argument of subsequences (see §1.5.1.6), show that (11.79)holds without the restriction on the limiting process.
=+(iii) Combine the results of (i) and (ii) to show that (11.79) holds under the limiting process in (i).
=+(11.78) with a = b = 1, provided that m/N → ρ ∈ [0, 1].(ii) Show that ηN,1 d−→ N(0, σ10), ηN,2 d−→ N(0, σ01).
=+11.7 Exercises 391(i) Show that both var(ζ∗N ) and cov(ζN , ζ∗N ) converge to the right side of
=+11.4 at the end of Section 11.5.
=+11.27. This exercise involves some details regarding the proof of Theorem
=+(ii) Verify (11.71) for j = 1 and j = 3.
=+11.26. Consider once again the problem of testing for the center of symmetry. More specifically, refer to the continuing discussion near the end of Section 11.5.(i) Verify (11.70) for 1 ≤ j ≤
=+11.25. Verify the numerical inequality (11.75) for any x, y, a ≥ 0.
=+11.24. Show that (11.68) holds as n → ∞.
=+11.23. Verify the following.(i) The martingale property (11.61).(ii) The expression (11.57), considered as a sequence of random variables, Un, Fn = σ(X1,...,Xn), n ≥ m, is a martingale.(iii) The
=+(iii) Verify the orthogonality property (11.59).
=+(i) Show that E(Snc) = 0, 1 ≤ c ≤ m.(ii) Show that E{gc(Xi1 ,...,Xic )gd(Xj1 ,...,Xjd )} = 0 except that c = d and {i1,...,ic} = {j1,...,jd}.
=+11.22. This exercise is concerned with moment properties of Snc, 1 ≤ c ≤m, that are involved in the Hoeffding representation (11.57).
=+11.21. Verify the property of complete degeneracy (11.56).
=+11.20. Verify (11.53) and also show that θ(F) = var(X1) for the sample variance.
=+11.19. Verify the identity (11.50).
=+11.18. Show that the functional h defined below (11.45) is continuous on(D, ·).
=+11.17. Verify (11.42) and thus, in particular, (11.43) and (11.44) under the null hypothesis (11.39).
=+11.16. Continue with the previous exercise.(i) Verify the identity (11.38).(ii) Given that (11.36) holds with TN = WXY /mn, μ(θ)=Pθ(X
=+(ii) Based on (11.37), derive the asymptotic power of the two-sample t-test and show that it is equal to 1 − Φ(zα − ctΔ), where ct = ρ(1 − ρ)/σ and Φis the cdf of N(0, 1).
=+11.15. (i) Show that (11.37) holds under the limiting process of (i) of the previous exercise, provided that θN is the true θ for TN . You may use a similar argument as in Example 11.4 and the
=+(iii) Show that the conclusion of (i) remains valid as m, n → without any restriction [Hint: Suppose otherwise. Then there is an > 0 and a sequence(mk, nk), k = 1, 2,..., such that |S−1 p −
=+11.14. Consider the pooled sample variance, S2 p, of Example 11.5.(i) Show that S2 pP−→ σ2 as m, n → ∞ such that m/N → ρ ∈ (0, 1), where N = m + n and σ2 is the variance of F.(ii)
=+11.13. Continuing with the previous exercise. For the sign test, we have Tn = n−1 n i=1 1(Xni>0) and 2√n{Tn − F(θn)} = n i=1 Yni, where Yni = (2/√n){1(Xni>0) − F(θn)}. Once again,
=+11.12. In the case of testing for the center of symmetry, suppose that Xni, 1 ≤ i ≤ n, are independent observations with the cdf F(x − θn). Then for the t-test, we have Tn = X¯n = n−1 n
=+11.11. Verify that the ARE eW,t of (11.31) is given by (11.34) in the case of Example 11.4.
=+11.10. Verify that the ARE eS,t of (11.30) is given by (11.33) in the case of Example 11.3.
=+11.9. Evaluate the AREs (11.30)–(11.32) when F is the following distribution:(i) Double Exponential with pdf f(z) = (1/2σ)e−|x|/σ, −∞ 0.(ii) Logistic with pdf f(z) = (1/β)e−z/β/(1 +
=+11.7 Exercises 389 signed-rank tests are given by 1/σ, 2f(0), and 2√3 f 2(z) dz, respectively, where σ and f are the standard deviation and pdf of F, respectively.
=+11.8. Show that for the problem of testing for the center of symmetry discussed in Sections 11.2 and 11.3, the efficacies of the t, sign, and Wilcoxon
=+(iii) Show that when θ = 0, (11.15) is equivalent to (11.21) with μ(θ) andτ(θ) given by (11.23) and (11.24).
=+(ii) Show that when θ = 0, (11.23) and (11.24) reduce to 1/2 and 1/3, respectively.
=+11.7. This exercise has several parts.(i) Suppose that X has a continuous distribution F. Show that F(X) has the Uniform[0, 1] distribution [Hint: Use (7.4) and the facts that F(x) ≥ u if and
=+11.6. Show that (11.22) holds as n → ∞ if θ is the true center of symmetry.
=+y Fig. 11.1. Numerical example: solid line: kernel estimate; dash line: true density where p > 3. Verify that fp is a pdf. Then show that ρw, ρs → 0 as p → 3.Therefore, the right side of
=+(ii) Let F have the pdf fp(x) = 0, |x| ≤ 1{(p − 1)/2}|x|−p, |x| > 1, 388 11 Nonparametric Statistics−4 −2 0 2 4 0.0 0.1 0.2 0.3 0.4 0.5 x
=+(i) Let F has a rectangular distribution that is symmetric about zero.Show that ρw = 1; hence, the left side equalilty holds.
=+11.5. This exercise is to show that both sides of inequality (11.19) are sharp in that there are distributions F that are continuous and symmetric about zero for which the left- or right-side
=+11.4. Show that the asymptotic correlation coefficient between S2 and S3 in (11.18), which correspond to the test statistics of the signed-rank and sign tests, is equal to √3/2.
=+11.3. Verify (11.6).

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