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### lecture 25

Course: STAT 100, Winter 2012
School: UC Davis
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100 Lecture STA 25 Irina Udaltsova Department of Statistics University of California, Davis March 9th, 2012 Admin for the Day Homework 7 posted, due Wednesday March 14th Project Instructions posted, Projects due Monday, March 19th Project Proposal are due today! Proposal feedback will be available by Monday: No feedback nothing wrong with your proposal Feedback changes needed (usually minor or wording) Oce...

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100 Lecture STA 25 Irina Udaltsova Department of Statistics University of California, Davis March 9th, 2012 Admin for the Day Homework 7 posted, due Wednesday March 14th Project Instructions posted, Projects due Monday, March 19th Project Proposal are due today! Proposal feedback will be available by Monday: No feedback nothing wrong with your proposal Feedback changes needed (usually minor or wording) Oce Hours project questions please! References for Today: Samuels 12.(1-6) References for Monday: Samuels 12.(7-9), 11.(1-4) Practice Problems: Samuels 12.5.(1-2), 12.5.8, 12.6.5-7 Linear Regression We can investigate the relationship between two variables using linear regression: Yi N + Xi , 2 Given explanatory variable xi the expected value of Yi is + xi . As we vary xi this traces out a straight-line relationship between x and y . Variations above and below the straight-line should be normally distributed with variance 2 . Note, sometimes we will use a dierent notation ( = 0 , = 1 ): Y i = 0 + 1 x i + i Understanding Assumptions Scatterplot of x and y 5 0 5 y 10 15 20 0 2 4 6 8 x 10 12 14 Conditional Distributions Scatterplot of x and y 5 0 5 y 10 15 20 0 2 4 6 8 10 12 14 x We can take a look at all the data points with x = 5. 0 20 40 60 80 100 Histogram of points with x=5 0 5 Y given X=5 10 Conditional Distributions Scatterplot of x and y 5 0 5 y 10 15 20 0 2 4 6 8 10 12 14 x We can take a look at all the data points with x = 12. 0 20 40 60 80 100 Histogram of points with x=12 6 8 10 12 14 Y given X=12 16 18 20 Conditional Distributions Linear regression assumes that the histograms conditional on dierent values of x will have: Dierent means! (Mean = (x ) = + x ) The same variance! Approximately normal distributions. Often we dont have enough data to make those histograms for real problems, but they help us understand what we are assuming. Key Assumptions There are ve very fundamental assumptions we make: 1. The mean of Y is a linear function of X : This describes the core of the relationship between X and Y i.e., the straight line. 2. The (vertical) variability of the data points around the line around follows a normal distribution: This describes one aspect of how the points are allowed to vary around the line. 3. The variability of the data points around the line around does not vary as X changes (i.e., for all x it is just 2 ): This describes another aspect of how the points are allowed to vary around the line. 4. Conditional on X , the Y s are independent: This describes another aspect of how the points are allowed to vary around the line. 5. The explanatory variable X is observed without error: Everything we do here is conditional on the X s. Diagnosing Problems The more complex the model we build, the more things there are that can go wrong. For linear regression, we really to need to know what our assumptions mean, so that we can know what warning signs to look for. . . Next 6 slides: 6 Datasets Which assumptions are reasonable/not for each. . . ? This exercise is designed to give you some practice deciding when it is appropriate to use linear regression. Example 1 Plot of y vs. x: Are the linear regression assumptions reasonable? 6 4 2 0 2 y 4 6 8 1 2 3 4 x 5 6 Example 2 Plot of y vs. x: Are the linear regression assumptions reasonable? 6 5 4 y 3 2 1 1 2 3 4 x 5 6 Example 3 400 Plot of y vs. x: Are the linear regression assumptions reasonable? 300 200 100 y 0 100 200 1 2 3 4 x 5 6 Example 4 14 Plot of y vs. x: Are the linear regression assumptions reasonable? 12 10 y 8 6 4 0 2 1 2 3 4 x 5 6 Example 5 Plot of y vs. x: Are the linear regression assumptions reasonable? 7 6 5 y 3 4 2 1 2 3 4 x 5 6 Example 6 Plot of y vs. x: Are the linear regression assumptions reasonable? 150 100 50 0 50 100 150 y 1 2 3 4 x 5 6 Quiz Time. . . For each of the 6 plots, either one or none of the linear regression assumptions are violated. For each of the 6 plots, choose only one of the following: (A) All assumptions appear to be valid (B) The linearity assumption appears to be wrong (C) The constant variance assumption appears to be wrong (D) The normality assumption appears to be wrong (E) The conditional independence assumption appears wrong Note: For some of the plots it is not possible to distinguish between some of the possible violations, so there is no way to know the correct answer. This is the same in real-life we cannot always pin down why things go wrong. Therefore it does not necessarily matter which assumptions you think are wrong only whether or not you not thought they were all valid. (1) (2) 0 y 4 2 1 1 2 3 4 5 6 1 2 3 4 10 6 150 12 7 50 0 2 2 3 4 x (4) 5 6 1 2 y 1 50 4 100 150 4 2 3 6 0 y y 5 8 6 Plot of y vs. x: Are the linear regression assumptions reasonable? 5 x 100 14 Plot of y vs. x: Are the linear regression assumptions reasonable? 6 Plot of y vs. x: Are the linear regression assumptions reasonable? x x 200 6 8 5 4 3 2 y 3 2 300 1 200 100 2 y 0 6 Plot of y vs. x: Are the linear regression assumptions reasonable? 100 5 4 4 6 (3) Plot of y vs. x: Are the linear regression assumptions reasonable? 400 Plot of y vs. x: Are the linear regression assumptions reasonable? 3 4 x (5) 5 6 1 2 3 4 x (6) 5 6 Solutions. . . (1) (2) Mean of y = a + bx 0 y 4 2 1 2 3 4 5 6 1 2 3 Example where independence is not true Mean of y = a + bx Mean of y = a + bx 7 12 10 6 0 2 2 3 4 x (4) 5 6 1 2 y 1 50 4 100 150 4 2 3 6 0 y y 50 5 8 6 Example where all assumptions are correct 5 Mean of y = a + bx 4 x 1 6 Example where normality is not true (harder to see) x x 14 200 6 5 4 Mean of y = a + bx + cx^2 + dx^3 3 2 y 8 200 3 Mean of y = a + bx 300 100 2 1 y 0 2 100 150 6 Example where constant variance is not true 100 5 4 4 6 (3) Example where all assumptions are correct 400 Example where linearity is not true 3 4 x (5) 5 6 1 2 3 4 x (6) 5 6 Diagnostics Lessons learned: 1. We can usually tell whether anything has gone wrong (we will talk about formal tests soon) 2. It can be hard to tell exactly what has gone wrong (e.g., could be non-linearity, non-independence etc.) 3. There are certain key things we can learn to look for (linearity, spread of points above and below the line shouldnt get drastically bigger or smaller as you move along the x-axis) Linear Regression: Overview What is linear regression? 1. Conceptually: Method for exploring linear relationships between explanatory variables and a continuous response. 2. Formally: Assumptions When is it appropriate? 1. Eyeball test, transformations 2. Diagnostic plots What do the results mean? 1. 2. 3. 4. Interpreting the regression coecients Hypothesis testing for the regression coecients Prediction and condence intervals R 2 , goodness-of-t test Brain-Body Example Example: Lets dig deeper into the brain-body weight example. Yi = log-brain weight of species i Xi = log-body weight of species i We assume that: Y i | X i = xi N + xi , 2 . Another way of writing the same thing: Y i = + xi + i , i iid N 0, 2 . Perspective Issues and Problems To keep you thinking about the big picture (as per exams, homework, projects). . . What potential problems are there with the brain-body weight example? What is the meaning of each pair, (xi , yi ), of data points? Each species doesnt have one brain weight and body weight! (We use the median brain and median body weight of a sample from each species implications?). How was the data collected? Not a random sample. (Still representative enough?) Other problems? Understanding Statistical Modeling Is our mathematical/statistical description of the brain-body weight problem a perfect one? No. Does it need to be? All models are wrong. Some are useful. George E.P. Box, Statistician Least Squares How do we estimate and ? In other words: how do we select the best straight line? The criteria we use is something called maximum likelihood. It is fairly intuitive: choose the values of and that correspond to the model your data were most likely to have come from. In the case of linear regression it actually has a nice simple explanation. . . 10 Plot of log(Brain) vs. log(Body) 5 0 log(brain) 0 5 log(body) Which line is better? 10 How do we decide on a best line? Well, we want it to be close to the data points. For a given estimate (, ), the vertical distance between the line and data point is given by: i = yi ( + xi ) This is positive if yi is above the line This is negative if yi is below the line This is zero if the line passes through yi We call the residual for point i 10 Plot of log(Brain) vs. log(Body) 5 0 log(brain) 0 5 log(body) 10 Least Squares The residuals will be positive and negative, but we want them to be as close to zero as possible (i.e., we want the line to be close to the data points). We need a criteria by which to select the best and . As usual, we square everything and add up the squared dierences (where have we seen this before?): n SS (, ) = i =1 [yi ( + xi )] 2 This is known as the residual sums of squares (RSS). We want to select and that minimize the residual sums of squares. n SS (, ) = i =1 yi ( + xi ) 2 n 2 i = i =1 10 Plot of log(Brain) vs. log(Body) 5 log(brain) 0 0 5 10 log(body) Goal: Find the line that minimizes the sums of the squared vertical distances to the data points. 10 Plot of log(Brain) vs. log(Body) 5 log(brain) 0 0 5 10 log(body) Equivalent Goal: Find the line that minimizes the sums of the areas of the squares. 10 Plot of log(Brain) vs. log(Body) 5 log(brain) 0 0 5 log(body) Solution: This is the best line. 10 The Line of Best Fit Finding the values and that minimize the RSS is actually not too dicult. We can write down the solutions analytically. First, we need some notation: x and y are the mean of x and y . The estimates of and are: = n i =1 (yi y )(xi n 2 i =1 (xi x ) x) = y x We also get an estimate of 2 (called the residual mean square): 2 = 1 n2 n 2 i i =1 The estimate is known as the residual standard error. Doing Linear Regression To do this in R you simply click: Statistics > Fit models > Linear regression Select the appropriate response and explanatory variable. Click ok. To plot the regression line: Graphs > Scatterplot Select the appropriate response (y-axis) and explanatory variable (x-axis). Uncheck all boxes apart from Least-squares line. Click ok. Animals: log(body) vs. log(brain) weight 8 6 4 2 0 log(brain) 0 5 log(body) 10 Doing Linear Regression For the brain-body weight example we get: > RegModel.2 <- lm(log(brain)~log(body), data=animals) > summary(RegModel.2) Call: lm(formula = log(brain) ~ log(body), data = animals) Residuals: Min 1Q -3.1609 -0.9360 Median 0.3074 3Q 0.7832 Max 3.0206 Coefficients: Estimate Std. Error t value (Intercept) 2.99064 0.47051 6.356 log(body) 0.43580 0.08751 4.980 --Signif. codes: 0 *** 0.001 ** 0.01 Pr(>|t|) 1.18e-06 *** 3.93e-05 *** * 0.05 . 0.1 1 Residual standard error: 1.655 on 25 degrees of freedom Multiple R-squared: 0.498, Adjusted R-squared: 0.4779 F-statistic: 24.8 on 1 and 25 DF, p-value: 3.926e-05 Reading the Output First part of output: Call: lm(formula = log(brain) ~ log(body), data = animals) This tells you the model you t (written y x where the lhs is the response, and the rhs is the explanatory variable). Next part of the output: Residuals: Min 1Q -3.1609 -0.9360 Median 0.3074 3Q 0.7832 Max 3.0206 These are the residuals from earlier (i.e., the vertical distances between the points and the line): i = yi ( + xi ) Furthest points from the line were 3.16 below and 3.02 above, The median distance of the points the line was 0.30 above, The lower and upper quartiles of the distances from the line were 0.94 below, and 0.78 above. Reading the Output Next part of the output are the estimates for and : Coefficients: Estimate Std. Error t value (Intercept) 2.99064 0.47051 6.356 log(body) 0.43580 0.08751 4.980 --Signif. codes: 0 *** 0.001 ** 0.01 Pr(>|t|) 1.18e-06 *** 3.93e-05 *** * 0.05 . 0.1 1 The (Intercept) line refers to i.e., = 2.99 The log(body) line refers to i.e., = 0.44 The other columns we will get to shortly. Most of the last part we will get to on Monday: Residual standard error: 1.655 on 25 degrees of freedom Multiple R-squared: 0.498, Adjusted R-squared: 0.4779 F-statistic: 24.8 on 1 and 25 DF, p-value: 3.926e-05 Residual standard error: 1.665 means = 1.665. Recap: Parameters vs. Statistics The most important slide in the whole course: Parameters: Population quantities (what we care about) Statistics: Sample quantities (what we actually get) Really understanding the dierences between estimands and estimators, and their properties is key to understanding many statistical methods. Estimating Population Quantities For the brain-body weight example, we are interested in the relationship between body and brain weights of all species (on the log scale). Parameter of interest: , the population regression slope. Statistic used to estimate this: , the estimated slope from the sample data. We use statistics to estimate parameters. i.e., we use data to estimate unknown quantities of interest. Interpretation We have: E [Yi |Xi = xi ] = 2.99 + 0.44xi The expected log-brain weight of an animal with log-body weight xi is 2.99 + 0.44xi . What do and mean? The Intercept, : Interpretation: The intercept gives the expected value of response if the explanatory variable is equal to zero. The Slope, : Interpretation: For a one unit increase in x , our expected value of y increases by . Example: = 0.44. For each additional 1kg of log body weight, we expect the log brain weight of an animal to increase by 0.44g. Fitted Values Earlier we talked about residuals (i.e., the vertical distances between the points and the line): i = yi ( + xi ) We can rewrite this as: yi = yi = Data Point = We assumed that ( + xi ) yi Fitted Value + + + Residual iid i N (0, 2 ). There should be no relationship between tted values & residuals! We can make a scatterplot to check this! We can also make a plot to check normality (Q-Q Plot) Diagnostic Plots in R For linear regression, R provides four plots to check your assumptions: 1. Residuals vs. Fitted Values: What it checks: Linearity and constant variance assumptions. There should be no pattern in the plot. 2. Normal Q-Q Plot: What it checks: Whether the residuals are normally distributed. The points should fall close to a straight line with slope 1. 3. Scale-Location Plot: What it checks: Whether the variance of residuals changes with x . The height of the points should be fairly stable as x increases. 4. Residuals vs. Leverage: What it checks: Whether there are any individual points that have a particularly large eect on the regression estimates. The points should fall inside the red dotted lines. Diagnostic Plots in R After you have t the linear regression, these plots can be obtained by: Models > Graphs > Basic diagnostic plots Alternatively, you can simply type: plot(my.lm) where my.lm is the name of the linear regression you just performed. Diagnostic Plots: Pass or Fail? Normal QQ 2 Residuals vs Fitted 1 3 6 2 3 4 5 6 7 25 1 1 8 6 2 1 0.4 2 3 4 5 Fitted values 2 6 7 8 2 0.5 1 0 1 1 Standardized residuals 2 1.2 0.8 0.0 Standardized residuals 25 0 Residuals vs Leverage 6 1 Theoretical Quantiles ScaleLocation 25 Fitted values 2 Residuals 1 0 0 2 1 Standardized residuals 3 1 0.00 0.05 0.10 20 25 Cook's distance6 0.15 Leverage Q: What happens if the diagnostic plots look bad? Dont just carry on! Need to do something else! 0.20 0.5 Linear Regression Q: So far, we provided estimates for and . What is missing? A measure of uncertainty! As usual we can make condence intervals for and to express the amount of uncertainty in our estimates. Doing Linear Regression For the brain-body weight example we get: > RegModel.2 <- lm(log(brain)~log(body), data=animals) > summary(RegModel.2) [snipped] Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 2.99064 0.47051 6.356 1.18e-06 *** log(body) 0.43580 0.08751 4.980 3.93e-05 *** --Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 [snipped] = 2.99. How certain are we? = 0.4358. How certain are we? Confidence Intervals Condence intervals for the intercept and slope are: A 100(1 )% CI for is: tn2,1 SE ( ), 2 + tn2,1 SE ( ) 2 A 100(1 )% CI for the intercept is: tn2,1 SE (), 2 + tn2,1 SE ( ) 2 Confidence Intervals Example Example: 95% for in the animals example: (Intercept) log(body) Estimate Std. Error t value Pr(>|t|) 2.99064 0.47051 6.356 1.18e-06 *** 0.43580 0.08751 4.980 3.93e-05 *** = 0.43580, SE ( ) = 0.08751, n = 27. Find tn2,0.975 from table or R: > qt(0.975,df=25) [1] 2.059539 Thus, our CI is: tn2,1 SE ( ), + tn2,1 SE ( ) 2 2 = (0.43580 2.059539 0.08751, 0.43580 + 2.059539 0.08751) = (0.2556, 0.6160) . Confidence Intervals Since condence intervals are very useful for linear regression, R will make them for you. After you have run the regression using: Statstics > Fit models > Linear regression Then go to: Models > Confidence Intervals and select the desired condence level (e.g., 0.95). Example: > Confint(RegModel.1, level=.95) Estimate 2.5 % 97.5 % (Intercept) 2.9906412 2.0216091 3.9596734 log.body 0.4357976 0.2555647 0.6160306 Cartoon of the Day
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Lecture Notes #2LectureEconomics 120BEconometricsProf. DahlUC San DiegoLinear Regression with One Regressor(SW Chapter 4)Linear regression allows us to estimate, and makeinferences about, population slope coefficients.Ultimately our aim is to es
UCSD - ECON 120B - econ 120 B
Lecture Notes #3Lecture(Chapter 5)Economics 120BEconometricsProf. DahlUC San DiegoRegression with a Single Regressor:Hypothesis Tests and Confidence Intervals(SW Chapter 5)Overview Now that we have the sampling distribution of OLSestimator, we
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Lecture Notes #4Lecture(Chapter 6)Economics 120BEconometricsProf. DahlUC San DiegoOutline1.2.3.4.5.Omitted variable biasCausality and regression analysisMultiple regression and OLSMeasures of fitSampling distribution of the OLS estimator
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Lecture Notes #5Lecture(Chapter 7)Economics 120BEconometricsProf. DahlUC San DiegoOutline1.2.3.4.Hypothesis tests and confidence intervals for a single coefficieJoint hypothesis tests on multiple coefficientsOther types of hypotheses involvi
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Lecture Notes #6Lecture(Chapter 8)Economics 120BEconometricsProf. DahlUC San DiegoNonlinear Regression Functions(SW Chapter 8) Everything so far has been linear in the Xs But the linear approximation is not always a good one The multiple regres
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Lecture Notes #7Lecture(Chapter 9)Economics 120BEconometricsProf. DahlUC San DiegoAssessing Studies Based onMultiple Regression (SW Chapter 9)Lets step back and take a broader look at regression: Is there a systematic way to assess (critique) re
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SampleMidtermName_MULTIPLECHOICE.Choosetheonealternativethatbestcompletesthestatementoranswersthequestion.1) Theconditionaldistributionof YgivenX = x ,Pr(Y = y X=x ),islPr(Y = y )A) Pr(X=x i, Y=y ).B).Pr(X=x )i =1C)Pr(X=x , Y=y ).Pr(X=x )D)
UCSD - ECON 120C - ECON 120C
Another Look at the OLS EstimatorProf. Yixiao SunAugust 2, 2011By denition^O LS= arg min fN ( ) ; where fN ( ) =N1X(YiNXi )2 :i=1According to the LLN,fN ( ) ! f ( ) = E (YX )2 :That is, fN ( ) as a function becomes closer and closer to f (
UCSD - ECON 120C - ECON 120C
Econ 120CSamplingEcon 120CReview: Sampling Distribution,Central Limit Theorem,Asymptotic Distributions of OLS EstimatorsA sampleSampling is the process of taking a smaller group ofsubjects from a larger population.Econ 120C Yixiao SunSampling.
UCSD - ECON 120C - ECON 120C
Econ120CEcon120CEcon120CCorrelation and CausalityMotivation1Why do we care about beta ?Econ120CConsider the modely = x*beta + uIn the last two lectures, we show that the OLSestimator of beta enjoys some nice properties:consistency and asymptot
UCSD - ECON 120C - ECON 120C
RandomizationEcon 120CConsider y = 0 + 1x + ,Earnings = 0 + 1 * (military service) +However, most applied research in economics usesobservational data instead of experiment data. So x isnot randomly assigned.yxDuring the Vietnam War, the draft lo
UC Irvine - ECON 134A - 62360
CHAPTER 2FINANCIAL STATEMENTS, TAXES ANDCASH FLOWAnswers to Concepts Review and Critical Thinking Questions1.Liquidity measures how quickly and easily an asset can be converted to cash without significant lossin value. Its desirable for firms to hav
UC Irvine - ECON 134A - 62360
CHAPTER 7INTEREST RATES AND BONDVALUATIONAnswers to Concepts Review and Critical Thinking Questions1.No. As interest rates fluctuate, the value of a Treasury security will fluctuate. Long-term Treasurysecurities have substantial interest rate risk.
UC Irvine - ECON 134A - 62360
CHAPTER 9NET PRESENT VALUE AND OTHERINVESTMENT CRITERIAAnswers to Concepts Review and Critical Thinking Questions1.A payback period less than the projects life means that the NPV is positive for a zero discount rate,but nothing more definitive can b
UC Irvine - ECON 134A - 62360
CHAPTER 13RISK, RETURN, AND THE SECURITYMARKET LINEAnswers to Concepts Review and Critical Thinking Questions1.Some of the risk in holding any asset is unique to the asset in question. By investing in a variety ofassets, this unique portion of the t
UC Irvine - ECON 134A - 62360
CHAPTER 15COST OF CAPITALAnswers to Concepts Review and Critical Thinking Questions1.It is the minimum rate of return the firm must earn overall on its existing assets. If it earns more thanthis, value is created.2.Book values for debt are likely t
UC Irvine - ECON 134A - 62360
Chapter SevenThe CAPMMGMT 109Managerial FinanceProfessor Lu ZhengReviewTotal risk = Systematic risk + Idiosyncratic riskInvestors are risk-averse: the higher the risk of the asset,the greater the return they requireA well-diversified portfolio is
UC Irvine - ECON 134A - 62360
Chapter NineCost of CapitalMGMT 109Managerial FinanceProfessor Lu ZhengChapter Organization1. The Cost of Capital: Some Preliminaries2. The Cost of Equity3. The Costs of Debt and Preferred Stock4. The Weighted Average Cost of Capital5. Divisiona