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Assignment 4: ECON-UA 266 - Intro to Econometrics Sahar Parsa Spring 2024 The solution to this assignment will be released on Friday March 1, 2024. It covers the material related to OLS, OLS interpretation and economic significance. Question 1 Sir Francis Galton, a cousin of James Darwin, examined the relationship between the height of children and their parents towards the end of the 19th century. It is from this study that the name “regression’ ’ originated. You decide to update his findings by collecting data from 110 college students, and estimate the following relationship: Student − height = 19 . 6 + 0 . 73 × Midpar − height, R 2 = 0 . 45 , SER = 2 . 0 , where the standard error of the intercept is 7 . 2 and the standard error of the slope is 0 . 1 . where Student-height is the height of students in inches, and Midpar-height is the average of the parental heights. Values in parentheses are standard errors. (Following Galton’s methodology, both variables were adjusted so that the average height of parents was equal to the average height of students — We will more generally assume the heights come from the same distribution.) a. Explain the elasticity method to the interpretation of the estimated coefficient? What about the standardized coefficient method? Why do we need these methods to interpret the economic significance of the estimated coefficient? Answer The elasticity method is given by the formula: χ Y X = ∆ Y i Y i ∆ X i X i which corresponds to the % change in Y i to a 1% change in X i . Noting that ∆ Y i ∆ X i = β OLS we have that: χ XY = β OLS X i Y i For the standardized coefficient method we first standardized the data ( Y i , X i ) to ( Y st i , X st i ) where: X st i = X i − ¯ X S X and Y st i = Y i − ¯ Y S Y 1

This transforms both the X and Y data on the same scale, both having the same mean (0) and standard deviation (1). We then run the following regression: Y st i = β st X st i + ε st i In this case, β st will correspond to the change (in standard deviation units) in Y st i to a one standard deviation unit change in X st i . Alternatively, we know from the lectures that we can find the OLS estimator for β st by using the β st OLS = β OLS S X S Y . We need these methods to interpret the economic significance of the estimated coefficients as regressions are not invariant to scaling of the data. For example, if we multiply Y i by a constant a then the OLS coefficients will be scaled by a factor of a . However, the elasticity and standardization methods are invariant to the original scaling of the data and allow us to interpret regressions on a common scale (e.g. the % change or the standard deviation unit change). b. Interpret the estimated coefficients using the elasticity method. Answer Note ∆ Y i ∆ X i = ˆ β = 0 . 73 corresponds to slope of the regression line. So in this case we have χ XY = 0 . 73 X i Y i In practice we evaluate this at the sample average ( ¯ X, ¯ Y ). We assumed in this question that the parents and the students have the same distribution of heights and ¯ X = ¯ Y . χ X,Y = 0 . 73 . As such, if one student has parents who are on average 1% taller than the parents of another student, that student is predicted to be 0 . 73% taller than the student whose parents are shorter. c. Interpret the estimated coefficients using the standardize method. Answer We can use the formula: β st OLS = β OLS S X S Y . Again, the parents and the students have the same distribution of heights and S X = S Y and β st OLS = 0 . 73 . A one standard deviation change in X is associated with a 0 . 73 standard deviation change in Y . d. What is the prediction for the height of a child whose parents have an average height of 70 . 06 inches? Solution Subbing in 70.06 for midpar-height we obtain: ˆ Student − Height = 19 . 6 + 0 . 73(70 . 06) = 70 . 74 The child is predicted to have a height of 70.74 inches. d. Given the positive intercept and the fact that the slope lies between zero and one, what can you say about the height of students who have quite tall parents? Who have quite short parents? Solution We can say that students’ of tall parents are predicted to be taller on average than students’ of short parents. We can’t necessarily conclude that they are likely to be tall or short in the population distribution of height. This would depend on knowing the mean of the height distribution and the specific value of mid-par height. e. Galton was concerned about the height of the English aristocracy and referred to the above result as “regression towards mediocrity.” Can you figure out what his concern was? Why do you think that we refer to this result today as “Galton’s Fallacy?” 2

Solution He is referring to the fact that if a child’s parents are above (or indeed below) a certain height threshold, the child is predicted to be smaller (or taller) than his/her parents. To see this explicitly, note that their is a unique X that solves: X = 19 . 6 + 0 . 73( X ) Solving for X gives 72.59 inches. Since the slope is less than 1, this means that if the parent’s height is above 72.59 inches the child is predicted to be smaller than the parent. Galton’s Fallacy refers to the fact he inferred that were as a reduction in the spread (or variance) of height distribution as people are regressing toward the mean. This is not the case, the likelihood is tall parents will have children that are shorter than them but there is still a positive chance that this will not be case. Galton regression is what gave the name regression to a linear regression because of the regression to the mean: https://rss.onlinelibrary.wiley.com/doi/full/10.1111/j.1740-9713.2011.00509.x Question 2 Consider the following random sample { X i , i = 1 , · · · , N } . 1. Write down the formula for the standardized X i , denote it X st i Answer X st i = X i − ¯ X S X where S X = wwwwwww vvvvvvv vvvvvvv uuuuuuu YYYYYYY i ( X i − ¯ X ) 2 N − 1 2. Show that X st i has a sample mean of 0 and a sample standard deviation fo 1 . [Notice that this is where the name standardized comes from.] Answer For the sample mean ¯ X st = QQQQQQQ i X st i N = QQQQQQQ i X i − ¯ X S X N Which equals QQQQQQQ i X i − ¯ X S X N = 0 Since QQQQQQQ i X i − ¯ X =0. For the sample standard deviation S x st = ttttttt QQQQQQQ i ( X st i − ¯ X st ) 2 N − 1 = ttttttt QQQQQQQ i ( X st i ) 2 N − 1 Thus we have 3

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