Inference in Constrained Linear Regression

Transcription

1 Inference in Constrained Linear Regression by Xinyu Chen A Thesis Submitted to the Faculty of the WORCESTER POLYTECHNIC INSTITUTE In partial fulfillment of the requirements for the Degree of Master of Science in Applied Statistics by May 07 APPROVED: Professor Thelge Buddika Peiris, Major Thesis Advisor

2 Abstract Regression analyses constitutes an important part of the statistical inference and has great applications in many areas. In some applications, we strongly believe that the regression function changes monotonically with some or all of the predictor variables in a region of interest. Deriving analyses under such constraints will be an enormous task. In this work, the restricted prediction interval for the mean of the regression function is constructed when two predictors are present. I use a modified likelihood ratio test (LRT) to construct prediction intervals. Keywords: Least favorable distribution, Restricted prediction interval, Chi-barsquare distribution, Likelihood ratio test

3 Acknowledgements I would like to express my gratitude to my advisor Thelge Buddika Peiris who helped me going through the constrained statistical inference in regression and instructed me in Latex and R code. My thanks are also due to my classmate, Jinxin Tao, who gave me a lot of support and shared the knowledge with me. Also, my special thanks are extended to all the professors who imparted to me with the valuable statistical knowledge and their assistance for the past four semesters. i

4 Contents Introduction. General Linear Regression Model and Assumptions Maximum Likelihood Estimator Confidence and Prediction Intervals Constrained Statistical Inference The Basics Likelihood Ratio Test First Order Model with Two Variables 8. Model and Assumptions Inferences for β 0 + β X 0 + β X Inference for β 0 + β X 0 + β X 0 3. Test when X 0 > 0 and X 0 > Test when X 0 < 0 and X 0 < Test when X 0 > 0 and X 0 < Test when X 0 < 0 and X 0 > Formulas for The Rejection Region 33 ii

5 4. when X 0 > 0 and X 0 > when X 0 < 0 and X 0 < when X 0 > 0 and X 0 < when X 0 < 0 and X 0 > When σ is Unknown 46 6 Prediction Intervals when X 0 > 0 and X 0 > when X 0 < 0 and X 0 < when X 0 > 0 and X 0 < when X 0 < 0 and X 0 > Example When σ is known When X 0 > 0 and X 0 > When X 0 < 0 and X 0 < When X 0 > 0 and X 0 < When X 0 < 0 and X 0 > When σ is unknown When X 0 > 0 and X 0 > When X 0 < 0 and X 0 < When X 0 > 0 and X 0 < When X 0 < 0 and X 0 > A R Codes in Example 64 iii

6 List of Figures 3. The region G 0 and boundary of the rejection region Dimensional illustration of S 3, S 4 and S Dimensional illustration of S 6, S 7 and S Dimensional view of disjoint regions S 9, S 0, S, S and S H 0 and Rejection region G 03 and Rejection region M 0 and Rejection region and dual cone M Dimensional view of M 0 and Rejection region H 03 and Rejection region G 04 and Rejection region iv

7 Chapter Introduction Techniques of statistical inference under order restrictions have been used in many applications. Regression analysis constitutes a large part of them. In many applications, experimenters believe that the regression function varies monotonically with the predictor variables in some region of interest. Usually the restricted regression analysis will consider the null hypothesis of the type Rβ = r versus Rβ r, Rβ r, for some matrix R, vector r when X N(θ, V ) and V is arbitrary. I do not have to consider the case for a general V separately in this thesis because the inference problem can be restated in terms of the identity covariance matrix. (Silvapulle and Sen (004)) And linear regression analysis is simple and efficient. Techniques of linear regression have been used for many areas and for a long time. However constrained regression analysis is more suitable and reasonable for the reality. If we use general linear regression inference techniques, we would not be able to get the benefit of our assumptions. In the fields such as economics and aerospace, constrained linear regression analysis might give more precise predictions which are important than regular one. Mukerjee and Tu (995) already discussed constrained simple linear

8 regression on a single variable. Commonly higher dimensional constrained inference is needed which is more practical. Peiris and Bhattacharya (06) has developed the techniques for point and interval estimators for model parameters and the mean response for two predictor variables model Y = β 0 + β X + β X with sign constraints on slope parameters β and β. Confidence intervals reflect the goodness of fitting and prediction intervals tell us that under certain probability the future observations will fall into the estimated intervals. In this thesis, I develop the formulas for the prediction intervals for the two predictor variables model with such constraints in slope parameters.. General Linear Regression Regression analysis is a statistical process to estimate the relation between two or more variables. Regression analysis techniques are often used to help understanding how the response variables change under the variation of the predictor variables. Regression analysis has three main purposes:.describle the relation between two or more variables,.use predictor variables to control response variables, and 3.use statistical relation to make predictions. Further, they are widely used to make forecasting in many areas, like biology, business, and data science. A few examples of applications are:. The length of patient stay in a hospital in days can be predicted by utilizing the relationship between patient s age in years and the time in the hospital.. The patient s blood pressure can be predicted by utilizing the relationship between the blood pressure and body weight.

9 .. Model and Assumptions I consider a general regression model where there are several predictor variables and the regression function is linear. The model can be stated as Y i = β 0 + β X i + β X i β p X pi + ɛ i, i =,,..., n, (.) where Y i is the repsonse for the i th trial, β 0, β, β...β p are parameters, X i, X i,...x pi are the values of predictor variables in i th trial, ɛ i is a random error term and ɛ i N n (0, σ ) for all i =,,..., n.. Maximum Likelihood Estimator In model (.), let p=, then ɛ = Y i (β 0 + β X i + β X i ) N (0, σ ), [ f Yi = exp πσ ( ) ] Yi (β 0 + β X i + β X i ). σ Then likelihood function is L(β 0, β, β, σ ) = n i= = (πσ ) (πσ ) [ exp / [ σ n/ exp σ (Y i β 0 β X i β X i ) ]. n (Y i β 0 β X i β X i ) i= ] To obtain the values of ˆβ0, ˆβ, ˆβ, and ˆσ that maximize the likelihood functon L(β 0, β, β, σ ), we let corresponding first derivatives equal zero and solve those simultaneous equations for MLEs. Then the MLEs of β 0, β, and β also can be obtained in matrix form as ˆβ = (X X) X Y, 3

10 Y Y n Y where Y n =. x x x x,and X n 3 =.... x n x n..3 Confidence and Prediction Intervals Confidence Interval One of linear regression analysis objects is to estimate the mean response E(Y ). Consider a study of the relationship between patient s blood pressure (Y) and body weight (X). The mean blood pressure at high and medium levels of body weight may be one of the purposes of analyzing the effect of overweight. Let X h as the level of X for which we wish to estimate the mean repsonse E(Y h ). Then the point estimator of E(Y h ) is Ŷ h = X h ˆβ, and Ŷh follows normal distribution with mean E(Ŷh) = E(Y h ) and variance σ (Ŷh) = σ ( n + (X h X) (Xi X) ). When σ is known, Ŷ h E(Y h ) σ(ŷh) N (0, ), where N(0,) is the standard normal distribution. Therefore 00( α)% condifence interval can be known as, Ŷ h ± Z( α/)σ(ŷh), where Z( α/) is the ( α/) 00 percentile of standard normal distribution. Usually σ is unknown, we replace σ (Ŷh) with the estimated variance s (Ŷh) = 4

11 MSE( n + (X h X) (Xi X) ), and we have Ŷ h E(Y h ) s(ŷh) t v, where t ν denotes the t-distribution with ν degree s of freedom. Therefore 00( α)% condifence interval is Ŷ h ± t ( α/;v) s(ŷh), where t ( α/;v) is the 00( α)percentile of t distribution with v degrees of freedom. Prediction Interval Now we consider the prediction of a new observation Y corresponding to a given level X of the predictor variable. When σ is known, Ŷ h(new) Ŷh σ{pred} N (0, ), where σ {pred} = σ + σ {Ŷh} = σ ( + n + (X h X) (Xi X) ). So prediction interval can be obtained as Ŷ h ± Z( α/)σ{pred}, When σ is unknown, Ŷ h(new) Ŷh s{pred} t v, where s {pred} = MSE +s {Ŷh} = MSE(+ n + (X h X) (Xi X) ). So prediction interval can be obtained as Ŷ h ± t( α/; v)s{pred} 5

12 . Constrained Statistical Inference Statistical inference has been used in many fields. The needs of developing for modeling and analysis of observational or experimental data in constrained enviroments are growing. In many applications, it is reasonable to assume that there are some constraints in our statistical models which means we have more information about our model parameter space. So the models will become more efficient than those wherein constraints are ignored if we properly incorporate those information... The Basics First we consider the observations X iid N (θ, V ). In order-restricted regression analysis, it is more common to consider inference under null hyphothesis of type Rβ = r versus Rβ r, Rβ r, for some matrix R, vector r when X iid N(θ, V ). We should consider, i V is a known positive definite matrix, ii V = σ U where U is a known positive definite matrix and σ is unknown, and iii V is unknown. (Silvapulle and Sen (004)) The following are some common terms in restricted inference, Convex: A set A R P is said to be convex if and only if {λx + ( λ)y} A where x, y A and 0 < λ <. Cone: A set A R P is said to be a cone with vector x 0 if and only if x 0 +k(x x 0 ) A for every x A and k 0. Further if the vertex x 0 is the origin O, then A is a cone simply. Fenchel Dual (or negative dual) Cone: C 0 = {α : α T θ 0 for every θ C} is called the dual cone of C with respect to the inner product. It can be shown that the boundaries of C 0 are the perpendiculars to the boundaries of C. Maximum likelihood estimation: 6

13 If X = (X, X ) N (θ, I), where I is the identity matrix and θ = (θ, θ ). Then for a single observation X, the kernel l(θ) of the loglikelihood is given by l(θ) = {(X θ ) + (X θ ) } = X θ, So in my work, I only use the kernel of the likelihood function to discuss our model... Likelihood Ratio Test Here is an simple example for likelihood ratio test. Let X = (X, X ) N (θ, I), where I is the identity matrix and θ = (θ, θ ). Consider the likelihood ratio test of H 0 : θ = θ = 0 vs H : θ 0, θ 0, LRT = X X θ, where θ {(θ, θ ) θ 0, θ 0} Then P r(lrt c) = = 4 P r(lrt c and X Q i ) i= 4 P r(lrt c X Q i )P r(x Q i ) i= where Q = {θ, θ : θ > 0, θ > 0}, Q = {θ, θ : θ < 0, θ > 0}, Q 3 = {θ, θ : θ < 0, θ < 0}, Q 4 = {θ, θ : θ > 0, θ < 0}. The null distribution of the LRT is the weighted sum of chi-square distributions, known as the chi-bar-square distribution. I use the similar method in the following chapter for hypothesis tests. 7

14 Chapter First Order Model with Two Variables. Model and Assumptions Consider the normal linear regression model with two predictor variables Y i = β 0 + β X i + β X i + ɛ i i =,,..., n, (.) or Y = Xβ + ɛ, where Y x x Y x x Y n =, X n 3 =, β 3 =.... Y n x n x n β 0 β β ɛ, ɛ ɛ n =,. and {ɛ i } are iid N (0, σ ). Let ˆβ 0, ˆβ and ˆβ be the unrestricted maximum likelihood estimaros (MLEs) of β 0, β and β respectively. Let S X = Xi, S X = Xi 8 ɛ n

15 and S = (Y i ˆβ 0 ˆβ X i ˆβ X i ) /v where v = n 3. Then we assume Xi = 0, X i = 0 and X i X i = 0 to simplify our model. Then, n cov( ˆβ) = σ n i= X i n i= X i n i= X i n i= X i n i= X ix i n i= X i n 0 0 n i= X ix i = 0 0, 0 0 S X n i= X i and so ˆβ 0, ˆβ and ˆβ are independent. Further cov( ˆβ, Y X ˆβ) =cov ( (X X) X Y, Y X(X X) X Y ) =cov ( (X X) X Y, (I n X(X X) X )Y ) =(X X) X (σ I)(I n X(X X) X ) =σ ( (X X) X (X X) X X(X X) X ) = 0. So ˆβ and Y X ˆβ are independent. Then ˆβ and S = (Y X ˆβ) (Y X ˆβ)/v are independent. Thus ˆβ 0, ˆβ, ˆβ and S are mutually independent. Following the properties of multivariate normal distribution, ˆβ0, ˆβ, ˆβ and S have normal distributions. They are unbiased estimators for β 0, β, β and σ. Hence ˆβ 0 N (β 0, σ /n), ˆβ N (β, σ / ), ˆβ N (β, σ /S X ), vs /σ χ v, where v = n 3. We consider the sign constraints for β and β. First I consider, β 0 and β 0 (.) 9

16 We can always make transformations of predictor variables for other constraints of β. The restricted MLEs of β 0, β, and β under the constraint (.) are given by β 0 = ˆβ 0, β = ˆβ + = max{ ˆβ, 0}, β = ˆβ + = max{ ˆβ, 0}, which is obvious and reasonable.. Inferences for β 0 + β X 0 + β X 0 We consider inferences for the mean response E(Y ) = β 0 + β X 0 + β X 0, for given point (X 0, X 0 ). For example, we already known the length of patient stay in a hospital in days (Y ) depends on the patient s age in years (X 0 ) and the infection risk (X 0 ). Given a patient s age and the infection risk, we want to know how long the patient will stay in hospital. Peiris and Bhattacharya (06) have already proposed formulas for the confidence interval for different signes of X 0 and X 0. 0

17 Chapter 3 Inference for β 0 + β X 0 + β X 0 Mentioned by Cox and Hinkley (97), inverting one-sdied tests for two-sample problems can derive the correct ( α)-coefficient prediction intervals for a new observation Y. Hence I derive the prediction intervals fro Y by inverting one-sided tests that test whether µ = E[Y (X 0, X 0 )] excceds or is exceeded by β 0 + β X 0 + β X 0, where the estimate of µ is to be obtained from the future observation Y and the estimate of β 0 + β X 0 + β X 0 is obtained from the observations in the past. Here there are four possible cases based on the signs of X 0 and X Test when X 0 > 0 and X 0 > 0 First we consider the hypothesis, G 0L : β 0 + β X 0 + β X 0 µ β 0, β 0 G 0U : β 0 + β X 0 + β X 0 µ β 0, β 0 (3.) and G : β 0, β 0

18 Define a ( α)-coefficient prediction interval (L P, U P ) for Y, where L P = min{y G 0L is accepted at level α/ against G a = G G 0L }, U P = max{y G 0U is accepted at level α/ against G a = G G 0U }. Then we use the transformation from β to r. Let r 0 = β 0 / + /n, r = β, ˆβ 0 r = S X β, then ˆr = ( ˆr 0, ˆr, ˆr ) = (, ˆβ, S X ˆβ ) N 3 (r, σ I) where ˆr is +/n the unrestricted MLE of r. Let b = µs X X 0, c = S X X 0 + /n > 0, d = X 0S X X 0 > 0. Then r 0 + /n + r X 0 + r X 0 S X µ, c r 0 + d r + r b, r b c r 0 d r. Hence our hypothesis can be restated in terms of r, G 0 : 0 r b c r 0 d r, r 0, G : r 0, r 0, (3.) and test G 0 against G a = G G 0. Suppose we use the same notation G 0 to denote the null hypothesis region. Here note that G 0 is a polyhedral cone with vertex L = (b /c, 0, 0). If we shift G 0 along the r 0 axis to the origin, we obtain a shifted cone K. K is a closed convex cone bounded by three hyperplanes {c r 0 + d r + r = 0, r 0, r 0}, {r 0 0, 0 r c r 0 d, r = 0}, {r 0 0, r = 0, 0 r c r 0 }. Then G 0 = K + L, bounded by {c r 0 + d r + r = b, r 0, r 0}, {r 0 b c, c r 0 + d r b, r = 0}, {r 0 b c, r = 0, c r 0 + r b }. Let G 0 = K + L, where K is the dual cone of K. It can be shown that the boundaries of K are the perpendiculars to the boundaries of K. So the Fenchel

19 dual cone K is bounded by three hyperplance {r 0 0, r d c r 0, r = c r 0 }, {r 0 0, r = d c r 0, r c r 0 }, {r 0 = 0, r d c r 0, r c r 0 }. Now let ˆr N 3 (r, σ I), where ˆr is the unrestricted MLE of r. Hence note that the restricted MLE of r in G is r = (r 0, r, r ) = ( ˆr 0, ˆr +, ˆr + ), and r is the equal weight projection of ˆr onto parameter space G. Let r be the MLE of r under G 0 and r is the equal weight prejection of ˆr onto G 0. When σ is known, the likelihood ratio test(lrt) rejects G 0 for large values of the test statistic χ 0 = logλ = ( ˆr r ˆr r )/σ, (3.3) where Λ is the kernel of LRT statistic. Figure 3.4, we consider several different cases when ˆr located in several different regions. With the boundaries of G 0 and G 0, we consider the whole region as an union of 3 disjoint regions. Depending on the signs of ˆr and ˆr, we discuss the test statistic χ 0 in each area separately. First when ˆr < 0 and ˆr < 0, we partition the region {(r 0, r, r ) : r < 0, r < 0} into S = {(r 0, r, r ) : r 0 < b c, r < 0, r < 0} and S = {(r 0, r, r ) : r 0 b c, r < 0, r < 0}. When ˆr S, r = r = ( ˆr 0, 0, 0). So χ 0 = r r /σ = 0, so S is inside the accept region. 3

20 Figure 3.: The region G 0 and boundary of the rejection region When ˆr S, r = ( ˆr 0, 0, 0) and r = ( b c, 0, 0). So χ 0 = ( ˆr 0 b c ) /σ which has chi-square distribution with degree of freedom. Furthere the boundary of rejection region is r 0 = b c + C α σ. When ˆr < 0 and ˆr 0, we can define regions S 3 = {(r 0, r, r ) : r 0 < b c c r, r < 0, r 0}, S 4 = {(r 0, r, r ) : r < 0, r max{b c r 0, c r 0 b c }}, and S 5 = {(r 0, r, r ) : r 0 > c r b c, r < 0, r 0} such that the disjoint union of S 3, S 4, and S 5 is the region {(r 0, r, r ) : r < 0, r 0}. 4

21 Figure 3.: -Dimensional illustration of S 3, S 4 and S 5 When ˆr S 3, r = ( ˆr 0, 0, ˆr ) and r = ( ˆr 0, 0, ˆr ).So χ 0 = r r /σ = 0. So S 3 is inside the accept region. When ˆr S 4, r = ( ˆr 0, 0, ˆr ) and r = ( ˆr 0, 0, ( ˆr u)u) where u is a unit vector along the line {c r 0 +r = b, r = 0}. χ 0 = r r /σ > C α which belongs to chi-square distribution with degree of freedom.then the boundary of the rejection region is c r 0 + r = b + + c C α σ. When ˆr S 5, r = ( ˆr 0, 0, ˆr ) and r = ( b c, 0, 0). χ 0 = (( ˆr 0 b c ) + ˆr )/σ > Cα which has chi-square distribution with degree of freedom. Then a part of the boundary of the rejection region is (r 0 b c ) + r = C ασ. And when ˆr 0 and ˆr < 0, we can define regions S 6 = {(r 0, r, r ) : 0 r < b d c d r 0, r < 0}, S 7 = {(r 0, r, r ) : r max{ b d c d r 0, d c r 0 b d }, r < 0}, and S 8 = {(r 0, r, r ) : 0 r < d c r 0 b d, r < 0} such that the disjoint union of S 6, S 7 and S 8 is the region {(r 0, r, r ) : r 0, r < 0}. c c 5

22 Figure 3.3: -Dimensional illustration of S 6, S 7 and S 8 When ˆr S 6, r = ( ˆr 0, ˆr, 0) and r = ( ˆr 0, ˆr, 0).So χ 0 = r r /σ = 0. So S 6 is inside the accept region. When ˆr S 7, r = ( ˆr 0, ˆr, 0) and r = (( ˆr 0, ˆr, 0) v)v where v is a unit vector along the line {c r 0 + d r = b, r = 0}. χ 0 = r r /σ > C α, which has chi-square distribution with degree of freedom.then the boundary of the rejection region is c r 0 + d r = b + c + d C α σ. When ˆr S 8, r = ( ˆr 0, ˆr, 0) and r = ( b c, 0, 0). χ 0 = (( ˆr 0 b c ) + ˆr )/σ > Cα which has chi-square distribution with degree of freedom. Then a part of the boundary of the rejection region is (r 0 b c ) + r = C ασ. And when ˆr 0 and ˆr 0, we can define regions S 9 = {(r 0, r, r ) : c r 0 + d r + r b, 0 r, 0 r }, S 0 = {(r 0, r, r ) : 0 r c d r +c 0 + d, r +c c r 0 b }, S c = {(r 0, r, r ) : r d c r 0 b d, 0 r c c r c 0 + d r +d c b }, +d c +d S 3 = {(r 0, r, r ) : 0 r < d c r 0 b d, 0 r c c r 0 b d }, S c = {(r 0, r, r ) : r 0, r 0} S 9 S 0 S S 3, such that the disjoint union of S 9, S 0, S, S, and S 3 is the region {(r 0, r, r ) : r 0, r 0}. 6

23 Figure 3.4: 3-Dimensional view of disjoint regions S 9, S 0, S, S and S 3 When ˆr S 9, r = ( ˆr 0, ˆr, ˆr ) and r = ( ˆr 0, ˆr, ˆr ). χ 0 = r r /σ = 0. So S 9 is inside the accept region. When ˆr S 0, r = ( ˆr 0, ˆr, ˆr ) and r = (( ˆr 0, ˆr, ˆr ) u)u where u is a unit vector along the line {c r 0 +r = b, r = 0}. χ 0 = r r /σ > C α, which has to chi-square distribution with degree of freedom. The boundary of the rejection region is a part of a rotated cylinder, r +[ r + c (r 0 b +c +c c )] = Cασ. When ˆr S, r = ( ˆr 0, ˆr, ˆr ) and r = (( ˆr 0, ˆr, ˆr ) v)v where v is a unit vector along the line {c r 0 + d r = b, r = 0}. χ 0 = r r /σ > Cα which belongs to chi-square distribution with degree of freedom. The boundary of the rejection region is a part of a rotated cylinder: r + [ d r + c (r 0 b c +d c +d c )] = Cασ. When ˆr S, r = ( ˆr 0, ˆr, ˆr ) and r = (( ˆr 0, ˆr, ˆr ) w)w where w is a unit vector along the line LB. χ 0 = r r /σ > Cα which belongs to chi-square distribution with degree of freedom. The boundary of the rejection region is hyperplane above G 0, which is c r 0 +d r +r = b + + c + d C α σ. When ˆr S 3, r = ( ˆr 0, ˆr, ˆr ) and r = ( b c, 0, 0). χ 0 = r r /σ = [( ˆr 0 b c ) + ( ˆr ) + ( ˆr ) ]/σ > Cα which belongs to chi-square distribution with 3 degree of freedom. The boundary of the 7

24 rejection region is a part of sphere surface( ˆr 0 b c ) + ˆr + ˆr = C ασ. The least favorable null value of χ 0 is attained at r = L = ( b c, 0, 0) and sup r G 0 P r r {ˆr : r r C ασ } = P r L { r r C α σ}. (3.4) THe proof of above result is given by Peiris and Bhattacharya (06). When ˆr is attained the least favorable null value, the distribution of LRT χ 0 is given by folloing formula. (See Peiris and Bhattacharya (06) for the proof and more details.) The least favorable null distribution of LRT is P r(lrt t ˆr = L) = 3 w i P (χ i t), i=0 where w 0 = (4π) (cos + cos d + cos + c c + d + c + d + cos d ), + c + d w = (4π) ( 3 π + cos d ( + c )(c + d ) ), + c w = (4π) (π + cos c + cos + d + c + d + c + d cos cos d ), + c c + d w 3 = (4π) ( 3 + c π c cos cos + d cos + c + d + c + d + c + d And the prediction upper bound is cos d cos d + c + d ( + c )(c + d ) ). (3.5) U P = max{y G 0U is accepted at level α/ against G a = G G 0U }. 8

25 Use the transformation from β to r. Let r 0 = β 0 / + /n, r = β, r = S X β then ˆr = ( ˆr 0, ˆr, ˆr ) = (β 0 / + /n, ˆβ, S X ˆβ ) N 3 (r, σ I) where ˆr is the unrestricted MLE of r. Let b = µs X X 0, c = S X X 0 + /n > 0, d = X 0S X X 0 > 0. r 0 + /n + r X 0 + r X 0 S X µ, c r 0 + d r + r b, r b c r 0 d r, Hence our hypothesis is H 0 : r b c r 0 d r, r 0, r 0, H : r 0 r 0, (3.6) and test H 0 against H a = H H 0. Similarly, to illustrate the construction of rejection region, we need the boundaries of H 0 in the r form. Let K be the shifted cone of H 0, and K be the dual cone of K. And H 0 = K +L and H 0 = K +L, where L = (b /c, 0, 0). Then we can get the 6 regions divided by the boundaries of K and K. Now let ˆr N 3 (r, σ I), where ˆr si the unrestricted MLE of r. And the restricted MLE of β is β = (β0, β, β) in section.. Hence we can define the restricted MLE of r is r = (r0, r, r) = ( ˆr 0, ˆr +, ˆr + ), and r is the equal weight projection of ˆr onto parameter space H. Let r be the MLE of r under H 0 and r is the equal weight prejection of ˆr onto H 0. When σ is known, the likelihood ratio test(lrt) rejects H 0 for large values of the test statistic is χ 0 = logλ = ( ˆr r ˆr r )/σ, (3.7) 9

26 where Λ is the kernel of LRT statistic. According to the discussion in Peiris and Bhattacharya (06), the least favorable null value of LRT(.8) is attained at lim t,s (b /c s c t, c t, c s) and sup P r r {ˆr : ˆr r ˆr r Dασ } r H 0 = lim P r (b t,s /c s c t,c t,c s){ χ 0 > Dασ } (3.8) Also, the null critical value is D α = χ,α. Figure 3.5: H 0 and Rejection region Shown in figure 3.5, we consider the whole region as an union of several disjoint areas. Depending on the signs of ˆr and ˆr, I discuss the test statistic χ 0 in each area separately. However I can not obtain the exact rejection region to get confidence and prediction intervals. Hence we need to modify our likelihood ratio test under same power. 0

27 Consider the hypothesis without the restrictions r 0 and r 0 H 0 : c r 0 + d r + r b, H : c r 0 + d r + r < b. Then the null hypothesis is excatly same as the unrestricted case. (3.9) LRT rejects H 0 for small values χ 03 = c ˆr 0 +d ˆr + ˆr b +c +d σ. So the rejection region of LRT is {ˆr : χ 03 < X α σ} Here rejection region for the unrestricted LRT contained that for the restricted LRT. So the unrestricted LRT is more powerful than the restricted LRT. But this creates a philosophical dilemma in some cases. In some cases, we will reject H 0 under unrestricted LRT but will not reject it under restricted LRT. So we need to modify LRT. Then consider following four regions. We use a similar idea as that for two dimensional model EY = β 0 + β X discussed in Mukerjee and Tu (995). We consider four regions in R 3 : S, S, S 3, and S 4, where S = {r : r d + c + d Z α σ, r 0}, S 3 = {r : r 0, r + c + d Z α σ}, S 4 = {r : r < 0, r < min{0, d r + c + d Z α σ}}, and S = R 3 S S 3 S 4. The boundary of the H 0 which is c r 0 + d r + r = b meets the hyperplane {r = 0} on the line {r = 0, c r 0 + d r = b } and the hyperplane {r = 0} on the line {r = 0, c r 0 + r = b }. Hyperplane c r 0 + d r + r = b + c + d Z α σ and hyperplane c r 0 + d r = b intersect on the line r = + c + d Z α σ. Hyperplane c r 0 + d r + r = b + c + d Z α σ and hyperplane c r 0 + r = b intersect on the hyperplane r = d + c + d Z α σ. To keep the same rejection level α, we modify LRT as follows, when ˆr S, we use the same boundary of the rejection region of the unrestricted case, which is c r 0 + d r + r = b + c + d Z α σ. When ˆr S, we already know the intersection of hyperplane c r 0 + d r + r = b + c + d Z α σ and S s boundary r = d + c + d Z α σ is the plane {c r 0 + r = b, r = d + c + d Z α σ}.

28 So let c r 0 + r = b as a part of the boundaries of rejection region in S. Similarly when ˆr S 4, we let c r 0 + d r = b as a part of boundaries of rejection region in S 4. when ˆr S 3, we let c r 0 = b as a part of boundaries of rejection region in S Test when X 0 < 0 and X 0 < 0 We consider the hypothesis, G 0L : β 0 + β X 0 + β X 0 µ β 0, β 0, G 0U : β 0 + β X 0 + β X 0 µ β 0, β 0, (3.0) and G : β 0, β 0. Define a ( α)-coefficient prediction interval (L P, U P ) for Y, where L P = min{y G 0L is accepted at level α/ against G a = G G 0L }, U P = max{y G 0U is accepted at level α/ against G a = G G 0U }. Then we can make a tranformation, where X 0 = X 0 > 0, X 0 = X 0 > 0, β 0 = β 0, and µ = µ. Then the new hypothesis will be, G 0L : β 0 + β X 0 + β X 0 µ β 0, β 0, G 0U : β 0 + β X 0 + β X 0 µ β 0, β 0, (3.) and G : β 0, β 0.

29 Define a ( α)-coefficient prediction interval (L P, U P ) for Y, where L P = min{y G 0L is accepted at level α/ against G a = G G 0L} = U P, U P = max{y G 0U is accepted at level α/ against G a = G G 0U} = L P, where L P and U P are in section 3.. Hence the formulas for rejection region and prediciton intervals can be obtained using the symetric property and are shown in next chapter. 3.3 Test when X 0 > 0 and X 0 < 0 Then we consider the hypothesis, G 0L : β 0 + β X 0 + β X 0 µ β 0, β 0, G 0U : β 0 + β X 0 + β X 0 µ β 0, β 0, (3.) and G : β 0, β 0. Define a ( α)-coefficient prediction interval (L P, U P ) for Y, where L P = min{y G 0L is accepted at level α/ against G a = G G 0L }, U P = max{y G 0U is accepted at level α/ against G a = G G 0U }. We use the transformation from β to r. Let r 0 = β 0 / + /n, r = β, r = S X β, then ˆr = ( ˆr 0, ˆr, ˆr ) = ( ˆβ 0 / + /n, ˆβ, S X ˆβ ) N 3 (r, σ I), where ˆr is the unrestricted MLE of r. 3

30 And b = µs X X 0, c = S X X 0 + /n < 0, d = X 0S X X 0 < 0, r 0 + /n + r X 0 + r X 0 S X µ c r 0 + d r + r b r b c r 0 d r Hence our hypothesis in terms of r is G 03 : r b c r 0 d r, r 0, r 0, G 3 : r 0 r 0. (3.3) Figure 3.6: G 03 and Rejection region Suppose we use the same notation G 03 to denote the null hypothesis region. Here note that G 03 is a polyhedral cone with vertex L = (b /c, 0, 0). If we shift G 03 along the r 0 axis to the origin, we obtain a shifted cone K. K is the closed 4

31 convex cone bounded by three hyperplanes {c r 0 + d r + r = 0, r 0, r 0}, {r = 0, c r 0 + d r + r 0, r 0}, {r = 0, c r 0 + d r + r 0, r 0}. Then G 03 = K + L, bounded by {c r 0 + d r + r = b, r 0, r b }, {r = 0, c r 0 + d r + r b, r 0}, {r = 0, c r 0 + d r + r b, r 0}. Recall the definition of dual cone, let G 03 = K + L, where K is the dual cone of K. It can be shown that the boundaries of K are the perpendiculars to the boundaries of K. So the Fenchel dual cone G 03 is bounded by three hyperplance {r 0 c d r b c, r 0 b c, r 0 c r = b c }, {r 0 c d r = b c, r 0 b c, r 0 c r b c }, {r 0 c d r b c, r 0 = b c, r 0 c r b c }. Now let ˆr N 3 (r, σ I), where ˆr si the unrestricted MLE of r. And the restricted MLE of β is β = (β 0, β, β ) in section.. Hence we can define the restricted MLE of r is r = (r 0, r, r ) = ( ˆr 0, ˆr +, ˆr + ), and r is the equal weight projection of ˆr onto parameter space G 3. Let r be the MLE of r under G 03 and r is the equal weight prejection of ˆr onto G 03. When σ is known, the likelihood ratio test(lrt) rejects G 03 for large values of the test statisti is χ 03 = logλ = ( ˆr r ˆr r )/σ, (3.4) where Λ is the kernel of LRT statistic. So the rejection region with a level α is {( ˆr r ˆr r ) > E ασ }, where E α is the critical value. Similar as previous section, when r 0, the rejection region is { ˆr r > E ασ }. We can obtain boundaries of the rejection region {c r 0 + r b + c F α σ, r < 0}, {r + ( r + c (r 0 b +c +c c )) = Fασ, 0 r c d r +c 0 + d r +c b d }, {c +c r 0 + d r + r = b + c + d F α σ, r max{0, c d r +c 0 + d r +c b d }}. +c But when r < 0, the rejection region has a complicated formulas and it is hard 5

32 to illstruate them with figures. We propose a new modified rejection region which is similar as previous section. The least favorable null value and the least favorable distribution is given in Peiris and Bhattacharya (06). So the least favorable null value of χ 03 is attained at infinity with lim r 0 (r 0, 0, b c r 0 ) and sup P r r {ˆr : ( ˆr r ˆr r )/σ Eα} = lim P r (r 0,0,b c r 0 ){ χ 03 > E r G 03 r 0 ασ }. The least favorable null distribution of LRT is, sup P r(lrt c) = ( r G θ π )P (χ 0 c) + P (χ c) + ( 4 θ π )P (χ c), where θ is the anlge between hyperplane C r 0 + d r + r = b and hyperplane r = 0 To obtain the modified LRT, first we consider hypothesis (.3) without the restriction r 0, M 0 : r b c r 0 d r r 0 against M : r 0. So the new LRT rejects M 0 for large values is, χ 03 = ( ˆr r ˆr r )/σ, where r is the MLE under M 0 and r is the MLE under M. 6

33 Figure 3.7: M 0 and Rejection region and dual cone M 0 We can have the projection of those regions boundaries to the hyperplane which the intersection line of regions is orthogonal to. Then the discussion for rejection region is similar to two predictor variables model case (Mukerjee and Tu (995)). Hence when r < 0, divide the region into two parts S and S, where S = {r : r < 0, c r 0 + r b }, S = {r : r < 0, c r 0 + r < b }, and obtain the center axis which is the intersection line of five regions, where {c r 0 + r = b, r = 0} When ˆr S, χ 03 = r r = 0 where r = r = ( ˆr 0, 0, ˆr ). So S is in the acceptance region. When ˆr S, χ 03 = r r = ( ˆr 0, 0, ˆr ) (( ˆr 0, 0, ˆr ) u)u Fασ where r = (( ˆr 0, 0, ˆr ) u)u and r = ( ˆr 0, 0, ˆr ) It has chi-square distribution with degree of freedom. The boundary of rejection region is a hyperplane parallely above the M 0 and has F α σ distance to the hyperplane c r 0 + r = b. 7

34 Figure 3.8: -Dimensional view of M 0 and Rejection region when r 0, the hyperplane c r 0 + d r + r = b and c d r 0 ( + c )r + d r = b d divide the region into three subregions S 3, S 4, and S 5. When ˆr S 3, χ 03 = r r where r = (( ˆr 0, ˆr, ˆr ) u)u and r = ( ˆr 0, ˆr, ˆr ), where u is a unit vector along the center axis. It has chi-square distribution with degree of freedom. The boundary of rejection region is a part of cylinder (center axis is the axis and radius is F α σ). When ˆr S 4, χ 03 = r r where r = (( ˆr 0, ˆr, ˆr ) w)w and r = ( ˆr 0, ˆr, ˆr ), where w is a unit vector along the projection of ˆr onto hyperplane c r 0 + d r + r = b. It has chi-square distribution with degree of freedom. The boundary of rejection region is above the hyperplane c r 0 + d r + r = b with distance F α σ. When ˆr S 5, χ 03 = r r = 0 where r = r = ( ˆr 0, r ˆ, ˆr ). So S 5 is in the acceptance region. Using the same argument as in the previous section, I propose a modified LRT 8

35 for (3.0), keep the same boundary of LRT for (3.0) when r + c E α σ. Note that the hyperplane c r 0 + r = b + c E α σ and r 0 = b c intersected at r = + c E α σ. When r < + c E α σ, modify the rejection region with cut-off. I propose a hyperplane r 0 = b c as the part boundary of the rejection region. And I propose a curved plane which is parallel to r axis and a hyperplane which is {c r 0 + d r = b ( + c + d + c )E α σ} We consider another hypothesis G 0U agiainst G a = G G 0U because U P = max{y G 0U is accepted at level α/ against G a = G G 0U }. We use the transformation from β to r. Let r 0 = β 0 / + /n, r = β, r = S X β, then ˆr = ( ˆr 0, ˆr, ˆr ) = ( ˆβ 0 / + /n, ˆβ, S X ˆβ ) N 3 (r, σ I) where ˆr is the unrestricted MLE of r. And b = µs X X 0, c = S X X 0 + /n < 0, d = X 0S X X 0 < 0, r 0 + /n + r X 0 + r X 0 S X µ, c r 0 + d r + r b, r b c r 0 d r. Hence our hypothesis in terms of r is H 03 : 0 r b c r 0 d r, r 0, H 3 : r 0 r 0. (3.5) 9

36 Figure 3.9: H 03 and Rejection region Here I note that the null region H 03 is a mirror image of the null region G 03 in the previous section. Considering hypothesis without the restriction r 0, keep the boudary of rejection region which is same as in (3.) when r d c + d K α σ. Then I propose a modified LRT when r < d c + d K α σ. and The least favorable null value of χ 03 is attained at infinity with lim (r 0, b c r 0 r 0 d, 0) sup P r r {ˆr : ( ˆr r ˆr r )/σ Kα} = r H 03 lim P r r 0 (r 0, b c r 0 d { χ 03 > K,0) ασ } The least favorable distribution of LRT is, P r(lrt c) = ( 4 + θ π )P (χ 0 c) + P (χ c) + ( 4 θ π )P (χ c), 30

37 where θ is the anlge between hyperplane C r 0 + d r + r = b and hyperplane r = 0 (See more details in Peiris and Bhattacharya (06)). 3.4 Test when X 0 < 0 and X 0 > 0 Then we consider the hypothesis, G 0L : β 0 + β X 0 + β X 0 µ β 0, β 0, G 0U : β 0 + β X 0 + β X 0 µ β 0, β 0, (3.6) and G : β 0, β 0. Define a ( α)-coefficient prediction interval (L P, U P ) for Y, where L P = min{y G 0L is accepted at level α/ against G a = G G 0L }, U P = max{y G 0U is accepted at level α/ against G a = G G 0U }. Then we can make a tranformation, where X 0 = X 0 > 0, X 0 = X 0 < 0, β 0 = β 0, and µ = µ. Then the new hypothesis will be, G 0L : β 0 + β X 0 + β X 0 µ β 0, β 0, G 0U : β 0 + β X 0 + β X 0 µ β 0, β 0, (3.7) and G : β 0, β 0. Define a ( α)-coefficient prediction interval (L P, U P ) for Y, where L P = min{y G 0L is accepted at level α/ against G a = G G 0L} = U P, 3

38 U P = max{y G 0U is accepted at level α/ against G a = G G 0U} = L P. where L P and U P are in section 3.3. Hence the formulas for rejection region and prediciton intervals can be obtained using symetric properties of these cases as shown in next chapter. Figure 3.0: G 04 and Rejection region 3

39 Chapter 4 Formulas for The Rejection Region 4. when X 0 > 0 and X 0 > 0 For hypothesis (3.) G 0 : 0 r b c r 0 d r, r 0, G : r 0, r 0. 33

40 The rejection region of r form,.{ ˆr 0 b c + C α/ σ, ˆr < 0, ˆr < 0},.{( ˆr 0 b ) + ˆr C c α/σ, ˆr < 0, 0 ˆr < ˆr 0 b }, c 3.{( ˆr 0 b ) + ˆr C c α/σ, 0 ˆr < d ˆr 0 b d, ˆr < 0}, c 4.{( ˆr 0 b ) + ˆr + ˆr C c α/σ, 0 ˆr < d ˆr 0 b d, 0 ˆr c c < ˆr 0 b }, c c 5.{c ˆr 0 + ˆr b + + c C α/ σ, ˆr < 0, ˆr ˆr 0 b }, c 6.{c ˆr 0 + d ˆr b c + d C α/ σ, ˆr d ˆr 0 b d, ˆr < 0}, c 7.{ ˆr c + ( ˆr + ( ˆr 0 b )) C + c + c c α/σ, 0 ˆr < c d ˆr + c 0 + d + c ˆr b d + c c c c c, ˆr ˆr 0 b }, c c 8.{ ˆr d c + ( ˆr + ( ˆr 0 b )) C c + d c + d c α/σ, 0 ˆr < c d ˆr c + d 0 + d ˆr c + d b, ˆr c + d d ˆr 0 b d }, c c 9.{c ˆr 0 + d ˆr + ˆr b C α/ σ + c + d, c d ˆr max{0, + c c d ˆr max{0, c + d ˆr 0 + d ˆr + c b d }, + c ˆr 0 + d ˆr c + d b }}. c + d Then we have the transformation from r to β. Let r 0 = β 0 / + /n, r = β, r = S X β, then ˆr = ( ˆr 0, ˆr, ˆr ) = (, ˆβ, S X ˆβ ) N 3 (r, σ I) +/n where ˆr is the unrestricted MLE of r. And b = µs X X 0, c = S X X 0 + /n > 0, d = X 0 S X X 0 > ˆβ 0

41 Then we transform to the original variables, rejection region of ˆβ form,.{ ˆβ 0 µ + C α/ σ + n, ˆβ < 0, ˆβ < 0},.{ + /n ( ˆβ 0 µ) + S X ˆβ C α/ σ, ˆβ < 0, 0 ˆβ < 3.{ + /n ( ˆβ 0 µ) + ˆβ C α/ σ, 0 ˆβ < 4.{ + /n ( ˆβ 0 µ) + ˆβ + SX ˆβ C α/ σ, 0 ˆβ < 0 ˆβ < +/n X 0 ( ˆβ 0 µ)}, S X X 0 5.{ ˆβ 0 + ˆβ X 0 > µ + S + + X n C α/σ, ˆβ < 0, ˆβ 6.{ ˆβ 0 + ˆβ X0 X 0 > µ + S + + X n C α/σ, ˆβ 7.{ ˆβ + ( ˆβ0 + ˆβ X 0 µ) 0 ˆβ < X 0 + n + X 0 S X C α/σ, ˆβ ( X 0 S X + + )( 0 + ˆβ X 0 µ), n 8.{S X ˆβ + ( ˆβ0 + ˆβ X 0 l) ˆβ + n + X 0 +/n X 0 ( ˆβ 0 µ), 0 ˆβ < X 0 S X 9.{ ˆβ 0 + ˆβ X 0 + ˆβ X 0 > µ + C α/ σ ˆβ max{0, X 0 ˆβ max{0, X 0 S X C α/σ, +/n X 0 S X ( ˆβ 0 µ)}, +/n X 0 ( ˆβ 0 µ), ˆβ < 0}, +/n X 0 ( ˆβ 0 µ), +/n X 0 S X ( ˆβ 0 µ)}, +/n X 0 ( ˆβ 0 µ), ˆβ < 0}, ˆβ +/n X 0 S X ( ˆβ 0 µ)}, ˆβ ( X 0 S X + )( 0 + ˆβ X 0 µ)}, n + n + X 0 + X 0 S X, ˆβ ( X 0 S X + + )( 0 + ˆβ X 0 µ)}, n ˆβ ( X 0 S X + + )( 0 + ˆβ X 0 µ)}}. n 35

42 For hypothesis (3.6) H 0 : r b c r 0 d r, r 0, r 0 H : r 0 r 0 rejection region in terms of ˆr,. {c ˆr 0 + ˆr b, ˆr + c + d d Z α/ σ, ˆr 0},. {c ˆr 0 + d ˆr b, ˆr 0, ˆr 3. { ˆr 0 < b, ˆr 0, ˆr min{0, d ˆr c 4. { ˆr 0 + d ˆr + ˆr b Then we transform to the original variables, rejection region in terms of ˆβ, + c + d Z α/ σ}, + c + d Z α/ σ}, + c + d Z α/ σ, otherwise}.. { ˆβ 0 + ˆβ X 0 µ, ˆβ < + X 0 n + X 0 S + X 0 X S Z α/σ, ˆβ 0}, X. { ˆβ 0 + ˆβ X 0 µ, ˆβ 0, ˆβ < 3. { ˆβ 0 µ, ˆβ < 0, ˆβ < min{0, 4. { ˆβ 0 + ˆβ X 0 + ˆβ X 0 µ X 0 X 0 + n + X 0 + X 0 S X Z α/σ}, + n + X 0 + X 0 S X Z α/σ}}, + n + X 0 + X 0 S X Z α/σ, otherwise}. 36

43 4. when X 0 < 0 and X 0 < 0 rejection region lower bound in section 4... { ˆβ 0 + ˆβ X0 µ, ˆβ < + X0 n + X 0 S + X 0 X S Z α/σ, ˆβ 0} X. { ˆβ 0 + ˆβ X0 µ, ˆβ 0, ˆβ < + X0 n + X 0 S + X 0 X S Z α/σ} X 3. { ˆβ 0 µ, ˆβ < 0, ˆβ < min{0, + X0 n + X 0 S + X 0 X S Z α/σ}} X 4. { ˆβ 0 + ˆβ X0 + ˆβ X0 µ + n + X 0 S + X 0 X S Z α/σ, otherwise} X Tranform to X 0 < 0 and X 0 < 0 case,. { ˆβ 0 + ˆβ X 0 µ, ˆβ < + X 0 n + X 0 S + X 0 X S Z α/σ, ˆβ 0}, X. { ˆβ 0 + ˆβ X 0 µ, ˆβ 0, ˆβ < + X 0 n + X 0 S + X 0 X S Z α/σ}, X 3. { ˆβ 0 µ, ˆβ < 0, ˆβ < min{0, + X 0 n + X 0 S + X 0 X S Z α/σ}}, X 4. { ˆβ 0 + ˆβ X 0 + ˆβ X 0 µ + + n + X 0 S + X 0 X S Z α/σ, otherwise}. X 37

44 rejection region upper bound in section 4.,.{ ˆβ 0 µ + C α/ σ + n, ˆβ < 0, ˆβ < 0},.{ + /n ( ˆβ 0 µ ) + S X ˆβ C α/ σ, ˆβ < 0, 0 ˆβ < 3.{ + /n ( ˆβ 0 µ ) + ˆβ C α/ σ, 0 ˆβ < 4.{ + /n ( ˆβ 0 µ ) + ˆβ + SX ˆβ C α/ σ, 0 ˆβ < +/n X 0 5.{ ˆβ 0 + ˆβ X 0 µ + 6.{ ˆβ 0 + ˆβ X 0 µ + ( ˆβ 0 µ ), 0 ˆβ < X0 +/n X 0 ( ˆβ 0 µ )}, S X S X + + n C α/σ, ˆβ < 0, ˆβ X n C α/σ, 7.{ ˆβ + ( ˆβ0 + ˆβ X 0 µ ) 0 ˆβ < X 0 ( X 0 S X + + n )( ˆβ 0 8.{S X ˆβ + ( ˆβ0 + ˆβ X 0 l) ˆβ + n + X 0 S X ˆβ C α/σ, + ˆβ X 0 µ ), + n + X 0 +/n X 0 ( ˆβ 0 µ ), 0 ˆβ < X 0 S X C α/σ, +/n X 0 ( ˆβ 0 µ )}, S X +/n X 0 ( ˆβ 0 µ ), ˆβ < 0}, +/n X 0 ( ˆβ 0 µ )}, S X +/n X 0 ( ˆβ 0 µ ), ˆβ < 0}, ˆβ ( X 0 + n )( ˆβ 0 9.{ ˆβ 0 + ˆβ X0 + ˆβ X0 µ + C α/ σ + n + X 0 S + X 0 X S, X ˆβ max{0, X 0 ˆβ max{0, X 0 S X ( X 0 S X + + n )( ˆβ 0 ( X n )( ˆβ 0 + ˆβ X 0 µ )}, + ˆβ X 0 µ )}}. +/n X 0 ( ˆβ 0 µ )}, S X + ˆβ X 0 µ )}, 38

45 Transform to X 0 < 0 and X 0 < 0 case,.{ ˆβ 0 µ C α/ σ + n, ˆβ < 0, ˆβ < 0},.{ + /n ( ˆβ 0 µ) + S X ˆβ C α/ σ, ˆβ < 0, 0 ˆβ < 3.{ + /n ( ˆβ 0 µ) + ˆβ C α/ σ, 0 ˆβ < 4.{ + /n ( ˆβ 0 µ) + ˆβ + SX ˆβ C α/ σ, 0 ˆβ +/n < X 0 ( ˆβ 0 µ), 0 ˆβ < X +/n 0 ( ˆβ 0 µ)}, S X 5.{ ˆβ 0 + ˆβ X 0 X 0 µ S + + X n C α/σ, ˆβ < 0, ˆβ 6.{ ˆβ 0 + ˆβ X 0 µ X n C α/σ, 7.{ ˆβ + ( ˆβ0 + ˆβ X 0 µ) + n + X 0 0 ˆβ < X 0 S X ˆβ C α/σ, ˆβ ( X 0 S X + + )( 0 + ˆβ X 0 µ), n 8.{S X ˆβ + ( ˆβ0 + ˆβ X 0 µ) + n + X 0 ˆβ +/n X 0 ( ˆβ 0 µ), 0 ˆβ < X 0 S X C α/σ, +/n X 0 S X ( ˆβ 0 µ)}, +/n X 0 ( ˆβ 0 µ), ˆβ < 0}, +/n X 0 S X ( ˆβ 0 µ)}, +/n X 0 ( ˆβ 0 µ), ˆβ < 0}, ˆβ +/n X 0 S X ( ˆβ 0 µ)}, ˆβ ( X 0 S X + )( 0 + ˆβ X 0 µ)}, n 9.{ ˆβ 0 + ˆβ X 0 + ˆβ X 0 µ C α/ σ + n + X 0 S + X 0 X S, X ˆβ max{0, X 0 ˆβ max{0, X 0 S X ˆβ ( X 0 S X + + )( 0 + ˆβ X 0 µ)}, n ˆβ ( X 0 S X + + )( 0 + ˆβ X 0 µ)}}. n 39

46 4.3 when X 0 > 0 and X 0 < 0 For hypothesis (3.3) G 03 : r b c r 0 d r, r 0, r 0 G 3 : r 0 r 0. rejection region lower bound of r form,. { ˆr 0 > b, ˆr < 0, ˆr < +c c E α/ σ}. {(c ˆr 0 (b + + c E α/ σ)) + ( + c ) ˆr ( + c )Eα/σ, 0 ˆr < c d ˆr + c 0 + d ˆr + c b d, ˆr + c < + c E α/ σ} 3. {c ˆr 0 + d ˆr b ( + c + d + c )E α/ σ, c d ˆr max{0, ˆr + c 0 + d ˆr + c b d }, ˆr + c < + c E α/ σ} 4. {c ˆr 0 + ˆr b + c, ˆr < 0, ˆr > + c E α/ σ} 5. { ˆr c + ( ˆr + ( ˆr 0 b )) E + c + c c α/σ, 0 < ˆr c d ˆr + c 0 + d + c 6. {c ˆr 0 + d ˆr + ˆr b c d ˆr max{0, + c ˆr b d, ˆr + c + c + d E α/ σ, ˆr 0 + d ˆr + c b d }, ˆr + c + c E α/ σ} + c E α/ σ} 40

47 We use the transformation from r to β. Let r 0 = β 0 / + /n, r = β, r = S X β, b = µs X X 0, c = S X X 0 + /n < 0, d = X 0S X X 0 < 0,. { ˆβ 0 > µ, ˆβ < 0, ˆβ <. {( ˆβ 0 µ ( X 0 S X + + n )E α/σ, X 0 X0 S + + X n E α/σ}, X0 S + + X n E α/σ) + ( X 0 S + + X n )S X ˆβ 0 ˆβ < ( ˆβ 0 µ + X 0 ˆβ ) X 0 ˆβ < X0 X 0 3. { ˆβ 0 + X 0 ˆβ µ + ( S + + X n E α/σ}, X 0 ˆβ ( ˆβ 0 µ + X 0 ˆβ ) X 0 X 0 S X + + n S + X 0 X S + + X n X 0 S X + + n, X 0, ˆβ < X 0 4. { ˆβ 0 + ˆβ X0 X 0 µ + S + + X n E α/σ, ˆβ < 0, ˆβ X0 X 0 S + + X n E α/σ}, 5. {( ˆβ 0 µ + X 0 ˆβ ) X 0 S X + + n 0 ˆβ < ( ˆβ 0 µ + X 0 ˆβ ) X 0 6. { ˆβ 0 + ˆβ X 0 + ˆβ X 0 µ + ˆβ ( ˆβ 0 µ + X 0 ˆβ ) X 0 X 0 + ˆβ E α/ σ, X 0 S X + + n + X 0 S X + + n E α/σ, X 0 S X + + n S + + X n )E α/σ, X 0 S X + + n E α/σ},, ˆβ X0 X 0 S + + X n E α/σ},, ˆβ X 0 X 0 S X + + n E α/σ}. 4

48 For hypothesis (3.5) H 03 : 0 r b c r 0 d r, r 0, H 3 : r 0 r 0. rejection region upper bound of r form,. { ˆr 0 b c,. {(c ˆr 0 (b ˆr < c + d d K α/ σ, ˆr < 0}, c + d K α/ σ)) + (c + d ) ˆr (c + d )Kα/σ, ˆr < d c + d K α/ σ 0 ˆr < c 3. {c ˆr 0 + ˆr b + ( + c + d c + d c + d )K α/ σ, ˆr 0 + d ˆr c + d b }, c + d ˆr < c + d d K α/ σ, ˆr c ˆr c + d 0 + d ˆr c + d b }, c + d 4. {c ˆr 0 + d ˆr b + c + d K α/ σ, ˆr c + d K α/ σ, ˆr < 0}, d d 5. { ˆr + ( ˆr + c + d c c + d ( ˆr 0 b c )) K α/σ, ˆr c + d d K α/ σ, 0 ˆr < c c + d 6. {c ˆr 0 + d ˆr + ˆr b + + c + d K α/ σ, ˆr 0 + d ˆr c + d b }, c + d ˆr c + d d K α/ σ, ˆr c ˆr c + d 0 + d ˆr c + d b }. c + d 4

49 rejection region upper bound ˆβ. { ˆβ 0 µ, ˆβ <. {( ˆβ 0 µ X 0 + n + X 0 K α/σ, ˆβ < 0}, + n + X 0 S K α/σ) + ( + X n + X 0 S ) S X ˆβ X ( + n + X 0 S )K α/σ, X ˆβ < + X 0 n + X 0 S K α/σ, X 0 ˆβ < ( ˆβ 0 + X 0 ˆβ µ) X 0 S X + n + X 0 3. { ˆβ 0 + X 0 ˆβ µ ( + n + X 0 S + X 0 X S X ˆβ < X 0 + n + X 0 S K α/σ, X ˆβ ( ˆβ 0 + X 0 ˆβ µ) X 0 S X 4. { ˆβ 0 + X 0 ˆβ µ + n + X 0 S K α/σ, X ˆβ + X 0 n + X 0 S K α/σ, ˆβ < 0}, X 5. {S X ˆβ (X 0 ˆβ + ˆβ 0 µ) Kα/σ, X 0 n S X ˆβ X 0 6. { ˆβ 0 + X 0 ˆβ + X 0 ˆβ µ }, + n + X 0 )K α/σ, + n + X 0 S K α/σ, 0 ˆβ < ( ˆβ 0 + X 0 ˆβ µ) X 0 X S X + n + X 0 + X 0 S X K α/σ, ˆβ X 0 + n + X 0 S K α/σ, X ˆβ ( ˆβ 0 + X 0 ˆβ µ) X 0 S X + n + X 0 }, + n + X 0 + n + X 0 }. }, 43

50 4.4 when X 0 < 0 and X 0 > 0 rejection region lower bound. { ˆβ 0 µ, ˆβ <. {( ˆβ 0 µ + X 0 + n + X 0 K α/σ, ˆβ < 0}, + n + X 0 S K α/σ) + ( + X n + X 0 S ) S X ˆβ X ( + n + X 0 S )K α/σ, X ˆβ < + X 0 n + X 0 S K α/σ, X 0 ˆβ < ( ˆβ 0 + X 0 ˆβ µ) X 0 S X + n + X 0 3. { ˆβ 0 + X 0 ˆβ µ + ( + n + X 0 S + X 0 X S X ˆβ < + X 0 n + X 0 }, + n + X 0 )K α/σ, S K α/σ, ˆβ ( ˆβ 0 + X 0 ˆβ µ) X 0 X S X 4. { ˆβ 0 + X 0 ˆβ µ + + n + X 0 S K α/σ, X ˆβ + X 0 n + X 0 S K α/σ, ˆβ < 0}, X 5. {S X ˆβ (X 0 ˆβ + ˆβ 0 µ) Kα/σ, X 0 n S X ˆβ + X 0 n + X 0 S K α/σ, X 0 ˆβ < ( ˆβ 0 + X 0 ˆβ µ) X 0 S X 6. { ˆβ 0 + X 0 ˆβ + X 0 ˆβ µ + + n + X 0 }, + n + X 0 + X 0 S X K α/σ, ˆβ + X 0 n + X 0 S K α/σ, ˆβ ( ˆβ 0 + X 0 ˆβ µ) X 0 X S X 44 + n + X 0 + n + X 0 }. },

51 rejection region upper bound. { ˆβ 0 > µ, ˆβ < 0, ˆβ <. {( ˆβ 0 µ ( X 0 S X + + n )E α/σ, X 0 X0 S + + X n E α/σ}, X0 S + + X n E α/σ) + ( X 0 S + + X n )S X ˆβ 0 ˆβ < ( ˆβ 0 µ + X 0 ˆβ ) X 0 3. { ˆβ 0 + X 0 ˆβ µ + ( X 0 ˆβ ( ˆβ 0 µ + X 0 ˆβ ) X 0 X 0 S X + + n S + X 0 X S + + X n X 0 X 0 S X + + n 4. { ˆβ 0 + ˆβ X 0 µ + S + + X n E α/σ, ˆβ < 0, ˆβ X0 X 0 S + + X n E α/σ}, 5. {( ˆβ 0 µ + X 0 ˆβ ) X 0 S X + + n 0 ˆβ < ( ˆβ 0 µ + X 0 ˆβ ) X 0 6. { ˆβ 0 + ˆβ X 0 + ˆβ X 0 µ + ˆβ ( ˆβ 0 µ + X 0 ˆβ ) X 0 X 0, ˆβ < X 0, ˆβ < X 0 + ˆβ E α/ σ, X 0 S X + + n X 0 + X 0 S X + + n E α/σ, X 0 S X + + n X0 S + + X n E α/σ}, S + + X n )E α/σ, X 0 S X + + n E α/σ},, ˆβ X0 X 0 S + + X n E α/σ},, ˆβ X 0 X 0 S X + + n E α/σ}. 45

52 Chapter 5 When σ is Unknown When σ is unknown, recall the hypothesis test (3.) G 0L : β 0 + β X 0 + β X 0 µ β 0, β 0 G 0U : β 0 + β X 0 + β X 0 µ β 0, β 0 (5.) and G : β 0, β 0 Test G 0L against G G 0L. In terms of r, the test becomes to G 0 : 0 r b c r 0 d r, 0 r against G : r 0, r 0. Then the LRT is Λ = ( σ σ )n /, where σ is the MLE of σ under G 0 and σ is the MLE of σ under G. Hence the LRT reject G 0 for large value of test statistic, λ = Λ /n = vs + ˆr r vs + ˆr r. Even we can change its form, Mukerjee and Tu (995) have shown some diffi- 46

53 culties of using this test statistic. Peiris and Bhattacharya (06) proposed a test statistic T which rejects G 0 with large values, where T = r r S and I use that for my forgoing discussion. The test reduced to the χ 0 test when σ is known by replacing S with σ. As Peiris and Bhattacharya (06) shown, the least favorable null distribution of LRT is P r(lrt C α ˆr = L) = 3 w i P (F i,n 3 Cα/i) i=0 where F i,n 3 is the F-distribution with i and n-3 degrees of freedom. If i=0, Let P (F i,n 3 C α/ /i) =. And the critical values C α can be computed using the equation α = w P (F,n 3 C α) + w P (F,n 3 C α/) + w 3 P (F 3,n 3 C α/3) Then the table of critical values are given by Peiris and Bhattacharya (06) in appendix. For other σ unknown cases, the rejection regions are very similar to the corresponding σ known cases with replacing σ with S and obtaining C α from above equation. I replace the Z α with t v,α in the boundaries of rejection region and prediction intervals when {X 0 > 0, X 0 > 0} and {X 0 < 0, X 0 < 0}. 47

54 Chapter 6 Prediction Intervals In this chapter, I summarize all the formulas for the prediction intervals for all the possible sign constraints of X 0 and X 0. When σ is known, the formulas for prediction intervals have similar formats as the formulas when σ is unknown. Hence, I only provide formulas of the prediction intervals when σ is unknown. 6. when X 0 > 0 and X 0 > 0 Lower Boundaries, L P =. ˆβ0 C α/ S + n if ˆβ < 0 ˆβ < 0,. ˆβ0 (Cα/ S S X ˆβ )( + n ) if ˆβ < 0 0 ˆβ X0 < C α/ S ( + )S n X 4 + X0S, X 3. ˆβ0 (Cα/ S ˆβ )( + n ) if 0 ˆβ X0 < C α/ S ( + )S n X 4 + X0S ˆβ < 0, X 48

55 4. ˆβ0 (Cα/ S ˆβ SX ˆβ )( + n ) if 0 ˆβ < C α/ S X 0( S X ˆβ ) Cα/ S ( + )S n X 4 + X0S X 0 ˆβ < C α/ S X 0( S X ˆβ ) Cα/ S ( + )S n X 4 + X0S, X 5. ˆβ0 + ˆβ X0 X 0 C α/ S S + + X n X0 if ˆβ < 0 ˆβ C α/ S ( + )S n X 4 + X0S, X 6. ˆβ0 + ˆβ X0 X 0 C α/ S S + + X n X0 if ˆβ C α/ S ( + )S n X 4 + X0S ˆβ < 0, X 7. ˆβ0 + ˆβ X 0 (C α/ S ˆβ )( + n + X 0 S ) X 8. ˆβ0 + ˆβ X 0 if 0 ˆβ < + S X 4 X 0 C α/ S ( X 0 S X + + n ) ˆβ > C α/ S X 0( S X ˆβ ) Cα/ S ( + )S n X 4 + X0S, X (C α/ S S X ˆβ )( + n + X 0 S ) X if ˆβ > C α/ S X 0( S X ˆβ ) Cα/ S ( + )S n X 4 + X0S X 0 ˆβ < C α/ σ, S X + S 4 X ( X X0 0 S X + + ) n 49

56 9. ˆβ0 + ˆβ X 0 + ˆβ X 0 C α/ S Upper Boundaries, + n + X 0 + X 0 S X if ˆβ > ˆβ > C α/ S S X + S 4 X ( X X0 0 S X + + ) n C α/ σ. S X + S 4 X ( X X0 0 S X + + ) n U P =. ˆβ0 + ˆβ X 0 if ˆβ < + X 0 n + X 0 S + X 0 X S T α/s ˆβ 0, X. ˆβ0 + ˆβ X 0 if ˆβ 0 ˆβ < 3. ˆβ0 if ˆβ < 0 ˆβ < min{0, 4. ˆβ0 + ˆβ X 0 + ˆβ X 0 + X 0 X 0 6. when X 0 < 0 and X 0 < 0 + n + X 0 + X 0 S X T α/s + n + X 0 + X 0 S X T α/s, + n + X 0 + X 0 S X T α/s}, otherwise. Lower Boundaries, L P =. ˆβ0 + ˆβ X 0 if ˆβ < + X 0 n + X 0 S + X 0 X S T α/s ˆβ 0, X. ˆβ0 + ˆβ X 0 if ˆβ 0 ˆβ < + X 0 n + X 0 S + X 0 X S T α/s, X 3. ˆβ0 if ˆβ < 0 ˆβ < min{0, + X 0 n + X 0 S + X 0 X S T α/s}, X 4. ˆβ0 + ˆβ X 0 + ˆβ X 0 + n + X 0 S + X 0 X S X 68 T α/s otherwise. 50

57 Upper Boundaries, U P =. ˆβ0 + C α/ S + if ˆβ < 0 ˆβ < 0 n. ˆβ0 + (Cα/ S S X ˆβ )( + n ) if ˆβ < 0 0 ˆβ X0 < C α/ S ( + )S n X 4 + X0S X 3. ˆβ0 + (Cα/ S ˆβ )( + n ) if 0 ˆβ X0 < C α/ S ( + )S n X 4 + X0S ˆβ < 0 X 4. ˆβ0 + (Cα/ S ˆβ SX ˆβ )( + n ) if 0 ˆβ < C α/ S X 0( S X ˆβ ) Cα/ S ( + )S n X 4 + X0S X 0 ˆβ < C α/ S X 0( S X ˆβ ) Cα/ S ( + )S n X 4 + X0S X 5. ˆβ0 + ˆβ X0 X 0 + C α/ S S + + X n X0 if ˆβ < 0 ˆβ C α/ S ( + )S n X 4 + X0S X 6. ˆβ0 + ˆβ X0 X 0 + C α/ S S + + X n X0 if ˆβ C α/ S ( + )S n X 4 + X0S ˆβ < 0 X 5

58 7. ˆβ0 + ˆβ X ˆβ0 + ˆβ X ˆβ0 + ˆβ X 0 + ˆβ X 0 + C α/ S if ˆβ > (C α/ S ˆβ )( + n + X 0 S X ) if 0 ˆβ C α/ S < S X + S 4 X ( X X0 0 S X + + ) n ˆβ > C α/ S X 0( S X ˆβ ) Cα/ S ( + )S n X 4 + X0S X (C α/ S S X ˆβ )( + n + X 0 S ) X if ˆβ > C α/ S X 0( S X ˆβ ) Cα/ S ( + )S n X 4 + X0S X C α/ S 0 ˆβ < S X + S 4 X ( X X0 0 S X + + ) n + n + X 0 + X 0 S X ˆβ > C α/ σ S X + S 4 X ( X X0 0 S X + + ) n C α/ σ S X + S 4 X ( X X0 0 S X + + ) n 5

59 6.3 when X 0 > 0 and X 0 < 0 Lower Boundaries, L p =. ˆβ0 if ˆβ < 0 ˆβ < X 0 X0 S + X n + E α/s. ˆβ0 ( X 0 S + X n + )(E α/ S X ˆβ ) 0 S + X n + E α/s if 0 ˆβ < X ˆβ 0 + ( + + X 0 n S X + X 0 S X + + X 0 n S X )E α/ S X 0 ( + + X 0 n ˆβ < X0 X 0 S + X n + E α/s 3. ˆβ0 + ˆβ X0 X 0 ( S + X 0 X S + X n + X0 S + X n + )E α/s if ˆβ X ˆβ 0 + ( + + X 0 n S X + X 0 S X + + X 0 n S X )E α/ S S X X 0 ( + + X 0 n 4. ˆβ0 + ˆβ X0 X 0 S + X n + E α/s 5. ˆβ0 + ˆβ X 0 S X + X 0 S X ) S X + X 0 S X ) ˆβ < X 0 if ˆβ < 0 ˆβ X 0 ( X 0 S + X n + )(E α/ S ˆβ ) if 0 ˆβ < ˆβ X 0 S X + S 4 X X0 X 0 S X + n + E α/s X 0 S X + n + E α/s E α/ S ( X 0 S X + n + ) X 0 S X + n + E α/s 53