01.06.2024 Applications#

Rescaling Variables#

General: Scaling does not change important things:

  • R-Squared = no change

  • Significacne = no change

  • Residuals and SSR = change

  • SE and CI = change (divide by scaler)

  • OLS Estimates change (but after transformation in same units do not change)

Example: Salary equation with thousands of dollars instead of dollars

Why do this? make numbers easier interpretable

Functional Form Specifications#

Example: Housing prices and Nitrogen Oxide Emissions (nox)

\[ \ln \widehat{price} = \underset{.19}{9.23} + \underset{.066}{-.718} \ln nox + \underset{.019}{.306} rooms \]

  • \(\beta_1\) = elasticity price / nox (1% nox \(\uparrow\) => 0.718% \(\downarrow\)​ price)

  • \(\beta_2\) = semi-elasticity of price / rooms

    • multiply by 100 = approximate effect

    • \(\Delta room=1\) => 30% price \(\uparrow\)​

Note: the approximation is not very exact for larger percentages!

Alternative: \(100 [exp(\beta_2)-1]\) (for values > 0.2)

Percentages and Percentage Points!#

Example

  1. \(\ln wage = \beta_1 - 0.05 \ unempl. rate\)

  2. \(\ln wage = \beta_1 - 0.05 \ln (unempl. rate)\)

Interpretation

  1. unempl. rate increase by one percentage point (8->9) => \(\uparrow\) wages by 5%

  2. unempl. rate increasy by one percent (8.00-8.08) => \(\downarrow\) wages by 0.05%

Quadratics#

used to capture marginal effects

\[ \ln wage = 3.73 + .298exper - .0061 exper^2 \]

find marginal effect: \(\frac{ \delta \ln wage }{\delta exper} = 0.298 - 2*0.0061* exper\)

Result: marginal diminishing effect of experience (negative sign before quadratic)

  • 1st year = \(.298 - 2\cdot (0.0061)\cdot (0) = .298\) (no experience in first year of job)

    • .298 cents per hour increase

  • 2nd year = \(.298 - 2\cdot (0.0061)\cdot (1) = .286\)​

  • 10 to 11th year = \(.298 - 2\cdot (0.0061)\cdot (10) = .176\)

Maximum of the wage: \(\frac{ -\beta_1 }{2 \beta_2} = 24.4\)

Interaction Terms#

Example

\[ price = \beta_0 + \beta_1 sqrft + \beta_2 bdrms + \underline{\beta_3 sqrft \cdot bdrms} + u \]

Partial Effect of bdrms on price

  • \(\frac{ \delta price }{\delta bdrms} = \beta_2 + \beta_3 sqrft\)

  • interaction effect: if \(\beta_3 > 0\) additional bedroom => higher increase in housing price for larger house

Goodness of Fit#

Problem with R-squared: adding a variable only increases it

=> adjusted R-squared (has penalty for additiona variables)

\[ \bar{ R }^2 = \frac{ SSR / (n-k-1) }{SST / (n-1)} \]

How to go from R-squared to adjusted?

\[ \bar{ R^2 } = 1- \frac{ (1-\bold{R^2})(n-1) }{(n-k-1)} \]

Note:

  • adj. R-squared can be negative!

Selection of Regressors#

Example of choosing between different models

\[\begin{split} Model 1& RDintens = \beta_0 + \beta_1 \ln sales + u \\ &R^2 = 0.061 \bar{ R^2 }=0.03) \\ Model 2& RDintens = \beta_0 + \beta_1 sales +\beta_2 sales^2+ u \\ &R^2 = 0.148; \bar{ R^2 }=0.09 \end{split}\]

Second Model:

  • better R-squarer

  • better adjusted R^2 (even though it has one independent variable more)

[!note]

never compare two models with different specifications for dependent variable

Overcontrolling Example: Beer Tax#

  • Idea: Beer Tax => lower Beer Consumption => lower traffic deaths

  • \(fatalities = \beta_0 + \beta_1 tax + \beta_2 miles + u\)

why not include beer consumption?

  • is this omitted variable bias?

  • NO, because holding consumption fixed is not our interest!

  • we want to follow our idea, and therefore include tax

Prediction#

Confidence Intervals for \(\theta_0 = \hat{ \theta_0 }\pm 2 \cdot se(\hat{ \theta_0 })\)​

  • variance of prediction = smallest at mean values of \(x_j\)

How to predict \(y\) if formula is only \(\ln y\), e.g \(\ln salary = \beta_0 + \beta_1 x_k + ... + u\)

  • For given \(x_k = 5000\) => \(\ln salary = 7.013\)

Methods

  • Naive: \(y = e^{\ln y} = exp(7.013) = 1110.983\) (Underestimates the result!)

  • Smearing:

    • \(\hat{ y } = \hat{ \alpha_0 } \cdot exp(\ln y)\)

    • calculate \(\hat{ \alpha_0 } = n^{-1} \sum_{i=1}^n exp(\hat{ u_i })\)

    • Result: \(y = 1.136 \cdot exp(7.013) = 1262.076\)

  • regression

    • fucking complicated, create a new regression just for it

    • Result: \(1117 \cdot exp(7.013) = 1240.967\)

  • normality assumption

    • \(y = \exp \frac{RSS^2 }{2} \exp (\ln y) =\exp \frac{0.50477^2 }{2} \exp (7.013)= 1261.929 \)