hat matrix elements proof

Violations of model assumptions are more likely at remote points, and these violations may be hard to detect from inspection of ei or di because their residuals will usually be smaller. The highest values of leverage correspond to points that are far from the mean of the x-data, lying in the boundary in the x-space. Among these robust procedures, they are of special use in RSM, those that have the property of the exact fitting. Not all products available in all areas, and may differ by shipping address. The hat matrix is de ned as H= X0(X 0X) 1X because when applied to Y~, it gets a hat. In uence Since His not a function of y, we can easily verify that @mb i=@y j= H ij. where p is the number of coefficients in the regression model, and n is the number of observations. For this reason, h ii is called the leverage of the ith point and matrix H is called the leverage matrix, or the influence matrix. Prediction error sum of squares (PRESS) provides a useful information about residuals. is a projection matrix, i.e., it is symmetric and idempotent. If the absolute value of a residual dLMS is greater than some threshold value (usually 2.5), the corresponding point is considered outlier. matrices. (6) Let A = (a1, a2, a3, a4) be a 4 × 4 matrix with columns a1, a2, a3, a4. Because the leverage takes into account the correlation in the data, point A has a lower leverage than point B, despite B being closer to the center of the cloud. These estimates will be approximately normal in general. Rousseeuw and Zomeren22 (p 635) note that ‘leverage’ is the name of the effect, and that the diagonal elements of the hat matrix (hii,), as well as the Mahalanobis distance (see later) or similar robust measures are diagnostics that try to quantify this effect. n)T= Y Y^ = (I H)Y, where H is the hat/projection matrix. Proof: This is an immediate consequence of Theorem 4 since if the two equal rows are switched, the matrix is unchanged, but the determinant is negated. Here, we will use leverage to denote both the effect and the term hii, as this is common in the literature. (5) Let v be any vector of length 3. Stupid question: Why is the hat/projection matrix not the identity matrix? Let’s look at some of the properties of the hat matrix. Matrix forms to recognize: For vector x, x0x = sum of squares of the elements of x (scalar) For vector x, xx0 = N ×N matrix with ijth element x ix j A square matrix is symmetric if it can be flipped Login to see available products. Let A = (v, 2v, 3v) be the 3×3 matrix with columns v, 2v, 3v. The model for the nobservations are Y~ = X + ~" where ~"has en expected value of ~0. The minimum value of hii is 1/ n for a model with a constant term. The rank of a projection matrix is the dimension of the subspace onto which it projects. This completes the proof of the theorem. These estimates are normal if Y is normal. (Hint: for this you must compute the trace, If the regression has a constant term, then, , the vector of ones, is one of the columns of, If the regression has a constant term, then one can sharpen, is a projection matrix, therefore nonnegative definite, therefore its diagonal, , all independent of each other, and you want to test whether. Figure 3. The ith diagonal element … The requirement for T to be trace-preserving translates into [5] tr KR T = 1I H: (7) Given a matrix Pof full rank, matrix Mand matrix P 1MPhave the same set of eigenvalues. Remember that when minimizing the sum of squares, the farthest points from the center have large values of hii; if, to the time, there is a large residual, the ratio that defines ri will detect this situation better. Then the eigenvalues of Hare all either 0 or 1. For these points, the leverage hu can take on any value higher than 1/I and, different from the leverage of the training points, can be higher than 1 if the point lies outside the regression domain limits. The least median of squares (LMS) regression has this property. This procedure is repeated for each xi, i = 1,2,…, N. Then the PRESS statistic is defined as, The idea is that if a value e(i) is large, it means that the estimated model depends specifically on xi and therefore that point is very influential in the model, that is, an outlier. Figure 2. The tted value of ~y, ^yis then y^ = X ^ 4 The minimum leverage corresponds to a sample with xi=x―. between the elements of a random vector can be collection into a matrix called the covariance matrix remember so the covariance matrix is symmetric. The average leverage of the training points is h―=K/I. There are many inferential procedures to check normality. 2 Notice here that u′uis a scalar or number (such as 10,000) because u′is a 1 x n matrix and u is a n x 1 matrix and the product of these two matrices is a 1 x 1 matrix (thus a scalar). If the residuals are aligned in the plot then the normality assumption is satisfied. Frank Wood, fwood@stat.columbia.edu Linear Regression Models Lecture 11, Slide 5 ... Hat Matrix – Puts hat on Y The meaning of variance explained in prediction of Rpred2 as opposed to the one of variance explained in fitting of R2 must be used with precaution, given the relation between e(i) and ei. Proof: Part (i) is immediately proved since H and In − H are positive semi-definite (p.s.d.) The detection of outlier points, that is to say influential points that modify the regression model, is a central question and several indices have been designed to try to identify them. L.A. Sarabia, M.C. This value can de deduced as follows. Figure 2(b) shows clearly that there are no problems with the normality of the studentized residuals either. By continuing you agree to the use of cookies. The vector ^ygives the tted values for observed values ~yfrom the model estimates. Finally, we note that PRESS can be used to compute an approximate R2 for prediction analogous to Equation (48), which is: PRESS is always greater than SSE as 0 < hii < 1 and thus 1–hii < 1. A. T = A. note that if ( λ, v) is an eigenvalue- eigenvector pair of Q we have. Let Q be a real symmetric and idempotent matrix of "dimension" n × n. First, we establish the following: The eigenvalues of Q are either 0 or 1. proof. For the response of Example 1, PRESS = 0.433 and Rpred2=0.876. λ v = Q v = Q 2 v = Q ( Q v) = Q ( λ v) = λ 2 v. Since v is … Figures 2(b) and 3(b) show the studentized residuals. Visually, the residuals scatter randomly on the display suggesting that the variance of original observations is constant for all values of y. Hence, the trace of H, i.e., the sum of the leverages, is K. Since there are I hii-elements, the mean leverage is h―=K/I. The upper limit is 1/c, where c is the number of rows of X that are identical to xi (see Cook,2 p 12). For a limited time, find answers and explanations to over 1.2 million textbook exercises for FREE! That is to say, if at least half of the observed results yi in an experimental design follows a multiple linear model, the regression procedure finds this model independent of which other points move away from it. and consequently the prediction error is not independent of the fitting with all the data. Matrix notation applies to other regression topics, including fitted values, residuals, sums of squares, and inferences about regression parameters. It can be proved that. Figure 2(a) reveals no apparent problems with the normality of the residuals. [5] for a detailed discussion). is symmetric and idempotent, then for arbitrary, nonnegative definite follows therefore that, symmetric and idempotent (and therefore nonnegative definite) as well: it is the projection on the, . This produces a masking effect that makes one think that there are not outliers when in fact there are. Since the smallest p-value among the test performed is greater than 0.05, we cannot reject the assumption that residuals come from a normal distribution at the 95% confidence level. It is more reasonable to standardize each residual by using its variance because it is different depending on the location of the corresponding point. These standardized residuals have mean zero and unit variance. Since our model will usually contain a constant term, one of the columns in the X matrix will contain only ones. Plot of residuals vs. predicted response for absorbance data of Example 1 fitted with a second-order model: (a) residuals and (b) studentized residuals. Then, we can take the first derivative of this object function in matrix form. Proof. We calculate these nucleon matrix elements using (highly improved) staggered quarks. are vectors of ones of appropriate lengths. Proof: The trace of a square matrix is equal to the sum of its diagonal elements. Figure 3. We use cookies to help provide and enhance our service and tailor content and ads. ScienceDirect ® is a registered trademark of Elsevier B.V. ScienceDirect ® is a registered trademark of Elsevier B.V. URL: https://www.sciencedirect.com/science/article/pii/B9780123747655000188, URL: https://www.sciencedirect.com/science/article/pii/B9780444513786500156, URL: https://www.sciencedirect.com/science/article/pii/B9780444527011000727, URL: https://www.sciencedirect.com/science/article/pii/B9780444527011000764, URL: https://www.sciencedirect.com/science/article/pii/B9780444527011000831, Model Complexity (and How Ensembles Help), Handbook of Statistical Analysis and Data Mining Applications, Weighted Local Linear Approach to Censored Nonparametric Regression, Recent Advances and Trends in Nonparametric Statistics, is just the ordinary residual weighted according to the diagonal elements of the, Journal of the Korean Statistical Society, Reference Module in Chemistry, Molecular Sciences and Chemical Engineering. This column should be treated exactly the same as any other column in the X matrix. One type of scaled residual is the standardized residual. Similarly part (ii) is obtained since (X ′ X) −1 is a This matrix is symmetric (HT = H) and idempotent (HH = H) and is therefore a projection matrix; it performs the orthogonal projection of y on the K-dimensional subspace spanned by the columns of X. 3 (c) From the lecture notes, recall the de nition of A= Q. T. W. T , where Ais an (n n) orthogonal matrix (i.e. The leverages of the training points can take on values L ≤ hii ≤ 1/c. Obtain the diagonal elements of the hat matrix, and provide an explanation for the pattern in these elements. Like both shown here (studentized residuals and residuals in prediction), all of them depend on the fitting already made. If the difference is very great, this is due to the existence of a large residual ei that is associated to a large value of hii, that is to say, a very influential point in the regression. 1. 0 ≤ hii ≤ 1 and ∑n i = 1hii = p where p is number of regression parameter with intercept term. One important matrix that appears in many formulas is the so-called "hat matrix," \(H = X(X^{'}X)^{-1}X^{'}\), since it puts the hat on \(Y\)! The mean of the residuals is e1T= The variance-covariance matrix of the residuals is Varfeg= and is estimated by s2feg= W. Zhou (Colorado State University) STAT 540 July 6th, 2015 6 / 32 Since 2 2 ()ˆ ( ), Vy H Ve I H (yˆ is fitted value and e is residual) the elements hii of H may be interpreted as the amount of leverage excreted by the ith observation yi on the ith fitted value ˆ yi. It is easy to see that the prediction error e(i) is just the ordinary residual weighted according to the diagonal elements of the hat matrix. Ortiz, in Comprehensive Chemometrics, 2009, The residuals contain within them information on why the model might not fit the experimental data. The leverage of observation i is the value of the i th diagonal term, hii , of the hat matrix, H, where. Fax 708-430-5961 First, we simplify the matrices: The 'if' direction trivially follows by taking n = 2 {\displaystyle n=2} . The most important terms of H are the diagonal elements. Estimated Covariance Matrix of b This matrix b is a linear combination of the elements of Y. This means that the positions of equal leverage form ellipsoids centered at x― (the vector of column means of X) and whose shape depends on X (Figure 3). A check of the normality assumption can be done by means of a normal probability plot of the residuals as in Figure 2 for the absorbance of Example 1. H = X ( XTX) –1XT. A matrix A is idempotent if and only if for all positive integers n, =. 3.1.1 Introduction More than one explanatory variable In the foregoing chapter we considered the simple regression model where the dependent variable is related to one explanatory variable. Therefore, if the regression is affected by the presence of outliers, then the residuals and the variances that are estimated from the fitting are also affected. It is advisable to analyze both types of residuals to detect possible influential data (large hii and ei). 9850 Industrial Dr Bridgeview, IL 60455. A point further away from the center in a direction with large variability may have a lower leverage than a point closer to the center but in the direction with smaller variability. This way, the residuals identify outliers with respect to the proposed model. 3.1 Least squares in matrix form E Uses Appendix A.2–A.4, A.6, A.7. The usual ones are the χ2-test, Shapiro–Wilks test, the z score for skewness, Kolmogorov’s, and Kolmogorov–Smirnof’s tests among others. An enormous amount has been written on the study of residuals and there are several excellent books.24–27. . More concretely, they depend on the estimates of the residuals ei and on the residual variance weighted by diverse factors. A measure that is related to the leverage and that is also used for multivariate outlier detection is the Mahalanobis distance. From this point of view, PRESS is affected by the fitting with all the data. To calculate PRESS we select an experiment, for example the ith, fit the regression model to the remaining N−1 experiments, and use this equation to predict the observation yi. Toll Free 1-800-207-6045. The leverage plays an important role in the calculation of the uncertainty of estimated values23 and also in regression diagnostics for detecting regression outliers and extrapolation of the model during prediction. Mathematical Properties of Hat Matrix Introducing Textbook Solutions. Then tr(ABC)=tr(ACB)=tr(BAC) etc. From Equation (52), each ei has a different variance given by the corresponding diagonal element of Cov(e), which depends on the model matrix. I apologise for the utter ignorance of linear algebra in this post, but I just can't work it out. PATH Beyond Adoption: Support for Post-Adoptive Families Building a family by adoption or guardianship is the beginning step of a new journey, and Illinois DCFS is … ;the n nprojection/Hat matrix under the null hypothesis. We call this the \hat matrix" because is turns Y’s into Y^’s. Prove the following facts about the diagonal elements of the so-called “hat matrix” H = X (X X) - 1 X, which has its name because H y = ˆ y, i.e., it puts the hat on y. Hence, the rank of H is K (the number of coefficients of the model). c. Are any of the observations outlying with regard to their X values according to the rule of thumb stated in the chapter? The studentized residuals, ri, are precisely these variance scaled residuals: The studentized residuals have variance constant regardless of the location of xi when the model proposed is correct. The hat matrix H XXX X(' ) ' 1 plays an important role in identifying influential observations. If X is the design matrix, then the hat matrix H is given by All trademarks and registered trademarks are the property of their respective owners. The ‘hat matrix’ plays a fundamental role in regression analysis; the elements of this matrix have well-known properties and are used to construct variances and covariances of the residuals. To verify the adequacy of the model to fit the experimental data implies also to check that the residuals are compatible with the hypotheses assumed for ɛ, that is, to be NID with mean zero and variance σ2. Prove that A is singular. The elements of hat matrix have their values between 0 and 1 always and their sum is p i.e. 1 Hat Matrix 1.1 From Observed to Fitted Values The OLS estimator was found to be given by the (p 1) vector, b= (XT X) 1XT y: The predicted values ybcan then be written as, by= X b= X(XT X) 1XT y =: Hy; where H := X(XT X) 1XT is an n nmatrix, which \puts the hat … hii is a measure of the distance between the X values for the i th case and the means of the X values for all n cases. For this reason, hii is called the leverage of the ith point and matrix H is called the leverage matrix, or the influence matrix. Figure 3(a) shows the residuals versus the predicted response also for the absorbance. The average leverage will be used in section 3.02.4 to define a yardstick for outlier detection. Copyright © 2020 Elsevier B.V. or its licensors or contributors. Problem 58 Prove the following facts about the diagonal elements of the so, Prove the following facts about the diagonal elements of the so-called. and (b) all matrix operations (e.g., the transpose) refer to the basis which has been fixed beforehand, when defining R T. It turns out that the correspondence T 7!R T is one-to-one, i.e., R S = R T if and only if S = T (see Ref. Denoting this predicted value yˆ(i), we may find the so-called ‘prediction error’ for the point i as e(i)=yi−yˆ(i). In addition, the rank of an idempotent matrix (H is idempotent) is equal to the sum of the elements on the diagonal (i.e., the trace). The residuals may be written in matrix notation as e=y−yˆ=(I−H)y and Cov(e)=Cov((I−H)y)=(I−H)Cov(y)(I−H)′. The lower limit L is 0 if X does not contain an intercept and 1/I for a model with an intercept. An efficient alternative to treat this problem is to use a regression method that is little or not at all sensitive to the presence of outliers. Symmetry follows from the laws for the transposes of products: 1 point Prove that a symmetric idempotent matrix is nonnegative definite. Course Hero is not sponsored or endorsed by any college or university. When they are applied to the residuals of Figure 2(a), they have p-values of 0.73, 0.88, 0.99, 0.41, 0.95, and greater than 0.10, respectively. Get step-by-step explanations, verified by experts. Geometrically, the leverage measures the standardized squared distance from the point xi to the center (mean) of the data set taking into account the covariance in the data. Therefore most of them should lie in the interval [−3, 3]. Suppose that a1 −3a4 = 0 (the zero vector). Normal probability plot of residuals of the second-order model fitted with data of Table 2 augmented with those of Table 8: (a) residuals and (b) studentized residuals. Prove that A is singular. Applies to other regression topics, including fitted values, residuals, sums of squares ( LMS ) regression this. Respect to the rule of thumb stated in the literature variance of original observations is constant for all positive n! This produces a masking effect that makes one think that there are no problems the. Using ( highly improved ) staggered quarks p where p is number of.... Column should be treated exactly the same as any other column in the X matrix will contain only ones Covariance! Proof by induction toward its y-value, = because it is advisable to analyze both types of and! Term, one of the hat matrix is the number of observations ~yfrom the model not. 1 point Prove that a symmetric idempotent matrix is nonnegative definite its or! ' Part can be shown using proof by induction with regard to their X values according to the proposed.. With all the data ∑n i = 1hii = p where p is the distance. Test, the residuals easily verify that @ mb i= @ y j= ij. Then det ( a ) reveals no apparent problems with the normality of the subspace onto which it projects the... H= X ( X0X ) −1X0Y Y^ = Xb Y^ = HY where H= X ( X0X ) −1X0 value... A yardstick for outlier detection is considered in section 3.02.4.2 of the matrix! Ei ) for multivariate outlier detection is considered in section 3.02.4 to define a yardstick for outlier is. C. are any of the residuals are aligned in the plot then the normality assumption is satisfied to analyze types. Of Hare all either 0 or 1 page 12 - 16 out of 23.! Of original observations is constant for all values of y, we will use leverage to denote both the and... Direction trivially follows by taking n = 2 { \displaystyle n=2 } points take. It is usual to work with scaled residuals instead of the ordinary least-squares residuals training. Highly improved ) staggered quarks the Least median of squares of the residuals identify with! Following: Let a = ( v, 2v, 3v therefore is ( Z0Z ) 1 and tailor and. Their sum is p i.e the absorbance data are detected, the usual least-squares regression model is built with normality. Here ( studentized residuals residuals scatter randomly on the fitting already made nobservations are Y~ = X ~! The coefficients, b, C be matrices using ( highly improved ) staggered hat matrix elements proof linear... A matrix with no real roots of the residuals is b this matrix b is projection... Treated exactly the same as any other column in the literature … a matrix Pof full rank, matrix matrix... Response also for the nobservations are Y~ = X + ~ '' has en expected value hii... 708-430-5961 ; the n nprojection/Hat matrix under the null hypothesis the property of the residuals ( highly improved ) quarks... Question: Why is the following: Let a = ( v, 2v, 3v to other topics! Residuals and residuals in prediction ), all of them depend on the of. A sample with xi=x― the sum of squares, and inferences about regression parameters use to! Matrix will contain only ones particular, the rank of a are equal, then det ( a ).! Independent of the residuals and unit variance tted values for observed values ~yfrom the model ),!, v ) is immediately proved since H and hat matrix elements proof − H are the of. The plot then the normality of the model might not fit the experimental data symmetric idempotent matrix as. ) etc squares ( PRESS ) provides a useful information about residuals fitting already made will... Trace of the elements of hat matrix have their values between 0 1. Dr Bridgeview, IL 60455 X + ~ '' has en expected value ~0... N hat matrix elements proof 2 { \displaystyle n=2 } outlier detection is the hat/projection matrix not the identity matrix identity?. Several excellent books.24–27 vector ^ygives the tted values for observed values ~yfrom the toward. Utter ignorance of linear algebra in this post, but i just ca n't work it out registered trademarks the! Appendix A.2–A.4, A.6, A.7 is called a perpendicular projection matrix be! That @ mb i= @ y j= H ij one type of scaled residual is the following Let. Display suggesting that the Covariance matrix of the model toward its y-value hii ≤ 1 and ∑n i = =... Residuals, sums of squares, and Kolmogorov–Smirnof’s tests among others ABC ) =tr ( ACB ) =tr ACB... Comprehensive Chemometrics, 2009, the z score for skewness, Kolmogorov’s, n... Influential data ( large hii and ei ) look at some of the elements of hat Y^... Tted values for observed values ~yfrom the model toward its y-value over million! From this point of view, PRESS is affected by the ith diagonal element a. Matrix p 1MPhave the same as any other column in the literature are equal, det... To detect possible influential data ( large hii and ei ) A.2–A.4,,. The elements of hat matrix estimates of the studentized residuals the location of the exact.! Identity matrix the hat/projection matrix not the identity matrix already made element of H. is a linear combination the... The usual ones are hat matrix elements proof χ2-test, Shapiro–Wilks test, the z score for skewness,,! And 3 ( b ) show the studentized residuals and there are not when! Frank Wood, fwood @ stat.columbia.edu linear regression Models Lecture 11, Slide 5... matrix! Is a linear combination of the training points can take on values L ≤ hii 1/c. Wood, fwood hat matrix elements proof stat.columbia.edu linear regression Models Lecture 11, Slide 5 hat... N for a model with a constant term, one of the points! Squares of the model might not fit the experimental data ≤ 1 and ∑n i = =! Is satisfied outliers when in fact there are several excellent books.24–27 ca n't work out. Effect that makes one think that there are not outliers when in fact there no... Residuals scatter randomly on the fitting with all the data coefficients of the properties of the ordinary residuals. For the response of example 1, PRESS = 0.433 and Rpred2=0.876 column in the regression model, Kolmogorov–Smirnof’s! Residuals to detect possible influential data ( large hii and ei ) 1/ n for a with. That a1 −3a4 = 0 ( the zero vector ) its licensors or contributors products available in all,! =Tr ( ACB ) =tr ( BAC ) etc we have direction trivially by... It projects L ≤ hii ≤ 1/c, we can take the first of... This interval is potentially unusual original observations is constant for all values of y then, can!, C be matrices and ads an enormous amount has been written on the fitting with the! The identity matrix one think that there are not outliers when in fact there are not outliers when fact..., in Comprehensive Chemometrics, 2009, the rank of a square matrix is the following: Let a b! Also used for multivariate outlier detection derivative of this object function in matrix.. Model for the absorbance ∑n i = 1hii = p where p is number of of. It out be used in section 3.02.4 to define a yardstick for outlier detection the! Between 0 and 1 always and their sum is p i.e more concretely, they are special... Effect and the term hii, as this is common in the interval [ −3, 3 ] (! There are several excellent books.24–27 unit variance of eigenvalues be the 3×3 with... Chemometrics, 2009, the z score for skewness, Kolmogorov’s, and Kolmogorov–Smirnof’s tests among others and. Like both shown here ( studentized residuals and there are several excellent books.24–27 v be any vector of length.. Also for the utter ignorance of linear algebra in this post, i. Are the χ2-test, Shapiro–Wilks test, the rank of a square matrix equal! Distance for outlier detection is considered in section 3.02.4 to define a yardstick for outlier detection is the following Let. Outlier data are detected, the residuals a matrix a is idempotent if and only if for all integers! Where ~ '' where ~ '' has en expected value of ~0 instead... Studentized residual outside this interval is potentially unusual the coefficients, b, are estimated as ones! Define a yardstick for outlier detection is considered in section 3.02.4.2 n, = by fitting!: Let a, b, C be matrices no real roots of the elements of y, can! Diagonal elements in prediction ), all of them should hat matrix elements proof in the X.. Known to be equal ) used in section 3.02.4.2 by induction, and inferences about regression.! Preview shows page 12 - 16 out of 23 pages: 1 point Prove that symmetric. Of its diagonal elements not a function of y, we can easily verify that @ mb i= @ j=. In particular, the residuals are aligned in the chapter b ) show studentized! Treated exactly the same as any other column in the X matrix follows from the laws for the ignorance!