Linear and Piecewise Linear Regressions


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1 Tarigan Statistical Consulting & Coaching statisticalcoaching.ch Doctoral Program in Computer Science of the Universities of Fribourg, Geneva, Lausanne, Neuchâtel, Bern and the EPFL Handson Data Analysis with R University of Neuchatel, 10 May 2016 Linear and Piecewise Linear Regressions Bernadetta Tarigan, Dr. sc. ETHZ Linear and Piecewise s 1
2 Piecewise Critic data generated from different versions of a software project version: a version number of the software project rule name: a name of the rule used to generate critics about source code entity: name of the entity which has a critic other fields contain additional information about the rules and entities and can be used for filtering [version, rule name, entity] triplets have unique values as for each version an entity can have only one critic of by a certain rule Linear and Piecewise s 2
3 Piecewise Main calculated value: Number of critics in a version = number of lines with the same version number How does the number of critics trend change after version 50185? Hypothesis is that after that version the number of critics should go down. But there is too much noise for analyzing all the data without filtering. Ideas on filtering: filter by package: focus on Collections*, Kernel, System*, Nautilus packages or perform analysis by package and merge the results filter by rule severity: focus only on error rules filter by rule group: focus only on Pharo bugs and Bugs calculate the trend values for each rule or each package separately and compare trends/eliminate outliers consider only entities that were changed since last version Linear and Piecewise s 3
4 Piecewise Very noisy indeed Linear and Piecewise s 4
5 Piecewise What is happening? Linear and Piecewise s 5
6 Piecewise There is hope Linear and Piecewise s 6
7 Piecewise This model? Use all version What s wrong with this? Linear and Piecewise s 7
8 Piecewise Or this one? Good enough? Use only version > 185 What s wrong with this? Linear and Piecewise s 8
9 Piecewise How about this one? Much better? Use all version Looks better, no? Linear and Piecewise s 9
10 Piecewise Simple = fitting a line on 1dim input f(x) = β 0 + β 1 x β 0 : intercept of the line (when x = 0 then y = β 0 ), β 1 : slope of the line (one unit increase in x gives β 1 units in y) ε i : random component (statistical error) for the ith case, it accounts for the fact that the statistical model does not give an exact fit to each and every data points ε i is unobservable, but we assume that E(ε i ) = 0 and Var ε i i = 1,, n = σ ε 2 for all However, we do not assume any distribution for ε i Population parameters are β 0, β 1 and σ ε 2 and we want to estimate them Linear and Piecewise s 10
11 Piecewise Estimate best line Define fitted value y i : = β 0 + β 1 x i residual e i : = y i y i Points above the line have positive residuals, points below the line have negative residuals A good line should have small residuals Residuals should be small in magnitude, because large negative residuals are as bad as large positive ones So we cannot simply require e i = 0 In fact, any line passing the means of the variables, the point (x, y), satisfies e i = 0 Two immediate solutions require e i to be as small as possible (least absolute distance) require (e i ) 2 to be as small as possible (least squares distance) Consider the second option: mathematically easier (e.g. to take derivative), although the first option is more resistant to outliers Linear and Piecewise s 11
12 Piecewise Least squares solution Denote β T = (β 0, β 1 ) and x T i = (1, x i ) (column vector) Residual sum of (error) squares RSS β : = (e i ) 2 = {y i β T x i } 2 Least squares solution is β ls = arg min β RSS β = arg min β {y i β T x i } 2 Easy to solve: set the first partial derivatives equal to zero, check the second derivative β 0 ls = y β 1 ls x β 1 ls = (x i x )(y i y) (x i x ) 2 Properties of residuals e i = 0 since the leastsquares line passes (x, y) x i e i = 0 and y i e i = 0: residuals are uncorrelated with the independent variable x i and fitted value y i β ls are unique defined as long as x i s are not all identical, in that case the numerator (x i x ) 2 = 0 Estimate for σ ε 2 is s e RSS/(n 2) Linear and Piecewise s 12
13 Piecewise How good is the fit? Use s e RSS/(n 2) : the smaller the better Use the coefficient of determination R 2 RegSS TSS Define TSS := (y i y) 2 o o total sum of squares of the null model, i.e., we do not use the independent variable Recall RSS := (y i y i ) 2 Clearly RSS < TSS Define RegSS := (y i y) 2 RegSS = TSS RSS, it gives reduction in the squared error due to the linear regression Define R 2 RegSS TSS, clearly 0 R2 1 R 2 is the proportion of the variation in that is explained by the linear regression The larger R 2 the better Linear and Piecewise s 13
14 Piecewise Famous result: least squares estimates are BLUE BLUE = Best Linear Unbiased Estimates β ls = arg min β RSS β = arg min β {y i β T x i } 2 GaussMarkov Theorem: least squares estimates have the smallest variance among all linear unbiased estimates Recall: Let β an estimate for an unknown parameter β The quality of β is measured via its mean squared error MSE β E β E β 2 = β E β 2 + Var β = Bias 2 + Variance Therefor least squares estimates are famous: if the underlying function f(x) were truly linear (that is, y = β T x + ε), then least squares estimates are your best approximation! Linear and Piecewise s 14
15 Piecewise Great! But, what next? Remember that we do not assume any distribution for the statistical 2 error ε, only that E(ε i ) = 0 and Var ε i = σ ε for all i = 1,, n Least squares estimates are great and truly mathematical solution, but we cannot do much more We cannot do statistical inference on them, e.g. Confidence interval Hypotheses test Which are needed when the goal in estimating the underlying mechanism is to explain or to describe But not to predict Statistical Inference: drawing conclusion about population from sample with some calculated uncertainty When you have two sets of data/sample from the same mechanism y = β T x + ε, you will get two sets of different estimates Linear and Piecewise s 15
16 Piecewise Normal distribution of the random error ε Linear statistical model: y i = β 0 + β 1 x i + ε i Assume that random error ε i are iid and N(0, σ ε 2 ) distributed, for i = 1,, n Y i x i ~ N(β 0 + β 1 x i, σ ε 2 ) The standard deviation remains constant, E(y x 3 ) b 0 + b 1 x 3 E(y x 2 ) m 3 but the mean value changes with x b 0 + b 1 x 2 E(y x 1 ) m 2 b 0 + b 1 x 1 m 1 x 1 x 2 x 3 Linear and Piecewise s 16
17 Piecewise Maximum Likelihood Estimates Now that we know the distributions of Y i x i that are independent but not identical (Y i x i ~ N(β 0 + β 1 x i, σ ε 2 )), hence we can apply maximum likelihood estimation (MLE) method The MLE estimates for are equal to least squares estimates β 1 MLE = β 1 ls = (x i x )(y i y) (x i x ) 2 β 0 MLE = β 0 ls = y β 1 ls x Linear and Piecewise s 17
18 Piecewise Maximum Likelihood Estimates (Cont.) However, we get more β 0 ~ N β 0, σ ε 2 β 1 ~ N(β 1, σ ε 2 s x ) n 2 x s i=1 i x Covariance Cov β 0, β 1 = σ ε 2 x s x Define S 2 RSS(β 0, β 1 )/(n 2) unbiased estimate for σ2 ε n 2 S 2 σ2 ~ χ 2 n 2 ε Moreover, (β 0, β 1 ) and S 2 are independent Linear and Piecewise s 18
19 Piecewise Test statistics From the results about sampling distributions, it immediately follows that which are the basis for inferences, significance test and CI estimation, regarding the two parameters β 0 and β 1 Linear and Piecewise s 19
20 Piecewise Test of significance 1. Test both parameters simultaneously with F test H 0 β 0 = β 1 = 0 H 1 at least one of them is not zero 2. Test each parameter with t test, for i = 0,1 H 0 β i = 0 H 1 β i 0 Linear and Piecewise s 20
21 Piecewise Confidence Interval (CI) estimation The (1 α) CI for respectively β 0 and β 1 are β 0 ± t 1 α 2 ; n 1 SE(β 0) β 1 ± t 1 α 2 ; n 1 SE(β 1) point estimate ± margin of error R returns the SE values When n is large, t behaves like the Standard Normal Z For α = 0.05, t 1 α 2 ; n 1 2 Remember the rule Linear and Piecewise s 21
22 Piecewise Model validation The assumptions of the random term (i.e., the errors) The outliers 1. Zero mean of the errors 2. Constant variance (homoscedasticity) of the errors 3. Independence of the errors 4. Normality of the errors 5. Outlier diagnostic Linear and Piecewise s 22
23 Piecewise Model Evaluation The goodnessoffit or quality of the model How good is the fit? Two measures: Residual standard error Coefficient of determination R 2 Linear and Piecewise s 23
24 Piecewise Piecewise linear regression Other names: hockey stick, broken stick or segmented It is a simple modification of linear model, yet very useful Different ranges of x, different linear relationships occur A single linear model may not provide an adequate explanation or description Breakpoints are the value of x where the slope changes The value of breakpoints may or may not known before the analysis, when unknown it has to be estimated Linear and Piecewise s 24
25 Piecewise Even to model a nonlinear relationship! Breakpoints are the value of x where the slope changes The value of breakpoints may or may not known before the analysis, when unknown it has to be estimated Linear and Piecewise s 25
26 Piecewise One breakpoint with known value Let c be the value of breakpoint Denote (x c) + = 0 ; x c x c ; x > c Piecewise linear model y = β 0 + β 1 x + β 2 (x c) + + ε Can be written as y = β 0 + β 1 x ; x c β 0 β 2 c + (β 1 +β 2 ) x ; x > c For x c the slope is β 1 Then it changes to β 1 + β 2 when x > c Linear and Piecewise s 26
27 Piecewise Hypothesis test y = β 0 + β 1 x ; x c β 0 β 2 c + (β 1 +β 2 ) x ; x > c For x c the slope is β 1 Then it changes to β 1 + β 2 when x > c As x increases, to test if y would decrease after the breakpoint c is to test if β 2 < 0 Linear and Piecewise s 27
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