ECON 300 Week 3 Notes
ECON 300 Week 3 Notes ECON 300
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This 4 page Class Notes was uploaded by Patricia Soto on Sunday January 31, 2016. The Class Notes belongs to ECON 300 at University of Illinois at Chicago taught by Irina Horoi in Winter 2016. Since its upload, it has received 43 views. For similar materials see Econometrics in Economcs at University of Illinois at Chicago.
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Date Created: 01/31/16
ECON 300 Week 3 Y = βo + β1x+ u Y: dependent variable, predicting variable, left hand side variable (lhs), y is a function of x • Ex. Y = test scores X: independent variable, (variables that can effect y), right hand side variable • Ex. X = class size U: error term, (captures ALL other things that affect y, note: it’s not in the model) • Ex. Teacher quality Suppose we have two values of x that differ by Δx (Δ= change/difference) • Δy =β1Δx if the unobserved factors that affect y are the same for both values of x β1: slope (constant slope, linear function), the linear slope parameter means that the effect on y of a one unit change in c is the same regardless of the level of x βo: Intercept parameter, tells use the value of y when x = 0. We are interested in β1 as a causal parameter! This depends on our assumptions about the error term u. Two primary assumptions about u that would allow us to interpret β1 as a causal parameter. • E(u) = 0 (The mean of the distribution of unobservable factors is centered on 0. Ex. Avg teacher quality = 0) • E(u|x) = E(u) ( The mean value doesn’t depend on x at all because it’s independent on x. There’s no relationship between u and x.) • Therefore, E(u)=0 (zero conditional mean assumption) Taking the conditional expectation of y: E(y|x) = E(βo+β1x +u|x) =E(βo|x) + E(β1x|x) + E(u|x) =βo + β1x + 0 => E(y|x) = βo + β1x E(y|x) = y1 E(y|x) =y2 ECON 300 The relationship defined above only tells us about the mean or average value of y based on what we know about x. st 1 to estimate Y = βo + β1x+ u we need data! i | x | y 1| 20| 85 2| 23| 83 …….. n|n|n Yi = βo + β1xi+ ui = indicates estimate To derive βo, β1 • E(u) = 0 • Cor(x,u) =0 => E[(x -µx)(u-µu)] = E(xu -xµ - uµx -µxµu) = E(xu) – E(xuµ_ - E(ux) – E(µx) (Note: µu= E (u) = 0 = E(x) *E(u)st 0 using our 1 equation, E (u) = 0, => E (Y-β1x-βo) = 0 o Implies => 1/n *Σ( Yi - βo -β1xi) = 0 o 1/n* Ybar (Ybar = average y) nd 2 equation: E[x(y-β1x - βo)] = 0 Taking the first equation: 1/n *Σ( Yi - βo -β1xi) = 0 You can simplify to: = ybar – 1/n βo - β1xbar = 0 Then rearrange for βo. Taking the 2 nd equation: E[x(y-β1x - βo)] = 0 • 1/nΣxi(yi-βo -β1x) = 0 Aside: will need to use these properties: • Σxi(xx-xbar) = Σ(xi – xbar)^2 • Σxi(yi-ybar) = Σ(xi-x)(yi-ybar) Going back to the second equation: Plug in for βo => Σxi(yi-(ybar-β1xbar)-β1xi)=0 If we multiply top and bottom by 1/n then => β1 = Cor(xy)/ Var(x) To solve for βo plug β1 into the original second equation Example: True population model for education attainment and annual salary is this: Annual Salary = βo + β1education + u We need data on salary + years of schooling/ We got 6 people to tell us that: ECON 300 Observation Salary $ Education (year) 1 45000 15 2 50000 17 3 60000 17 4 30000 13 5 45000 14 6 45000 17 β1 = Σ(educai – educ-bar)(salary – salary-bar)/ (Σ (educ- educ-bar)^2 mean salary (salary-bar) = 45,833.33 mean education (educ-bar) = 15.5 Observation (educai- educ- (salaryi-salary- Column 2*3 bar) bar) 1 -.5 -833.33 416.67 2 1.5 4166.67 1250 3 1.5 14166.67 21250 4 -2.5 -15833.33 39583.33 5 -1.5 -833.33 1250 6 1.5 -833.33 -1250 Σ(educi-educ-bar)(sali-sal-bar) = 67500 (numerator) Σ(educi-educ-bar)^2 (denominator) Observation (educi-educ-bar)^2 1 .25 2 2.25 3 2.25 4 6.25 5 2.25 6 2.25 Now you need to sum all of them up to get = 15.5 β1 = 67500/15/5 = $4354.84 COOL. YOU GUYS WE JUST REGRESSED X ON Y. • SO WHAT DOES THIS MEAN? o For every additional year of schooling you gain $4354.84 annual salary then without the extra year. (my interpretation) o β1 is interpreted as an additional year of education is associated with an average increase in annual salary of $4354.84 ECON 300 The estimated values of the slope + intercept are ordinary least squares (OLS). To see why: Yi = βo +βixi If we take our real values of y and subtract Yi we get the residuals (ui) • Ui = yi – yi = yi - βo - β1x
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