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Tutorial 1: Misspecification

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1 Tutorial 1: Misspecification
Matthew Robson University of York Econometrics 2

2 𝐿 𝐢 𝑑 = log consumption, 𝐿 𝐼 𝑑 = log income, 𝐿 π‘Š 𝑑 = log wealth
Assignment 5 Assume that the true model of consumers’ behaviour is: 𝐿 𝐢 𝑑 = 𝛽 0 + 𝛽 1 𝐿 𝐼 𝑑 + 𝛽 2 πΏπ‘Š+ 𝑒 𝑑 Where: 𝐿 𝐢 𝑑 = log consumption, 𝐿 𝐼 𝑑 = log income, 𝐿 π‘Š 𝑑 = log wealth Estimate the simple log-linear consumption function: 𝐿 𝐢 𝑑 = 𝛽 0 βˆ— + 𝛽 1 βˆ— 𝐿 𝐼 𝑑 + 𝑒 𝑑 βˆ— Over the period 1967q1-2002q4 (1) (2)

3 Descriptive Statistics

4 OLS Results: True Model
𝐿 𝐢 𝑑 = 𝐿 𝐼 𝑑 πΏπ‘Š+ 𝑒 𝑑 πΌπ‘›π‘‘π‘’π‘Ÿπ‘π‘Ÿπ‘’π‘‘π‘Žπ‘‘π‘–π‘œπ‘›: 𝑖𝑓 𝑀𝑒 π‘β„Žπ‘Žπ‘›π‘”π‘’ π‘₯ 𝑏𝑦 1% 𝑀𝑒 𝑒π‘₯𝑝𝑒𝑐𝑑 𝑦 π‘‘π‘œ π‘β„Žπ‘Žπ‘›π‘”π‘’ 𝑏𝑦 𝛽%. π‘‡β„Žπ‘’ π‘π‘Žπ‘Ÿπ‘‘π‘–π‘Žπ‘™ π‘’π‘™π‘Žπ‘ π‘‘π‘–π‘π‘–π‘‘π‘¦, π‘Žπ‘™π‘™ 𝑒𝑙𝑠𝑒 β„Žπ‘’π‘™π‘‘ π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘.

5 Fitted Model

6 OLS Results: Simple Model
𝐿 𝐢 𝑑 = 𝐿 𝐼 𝑑 + 𝑒 𝑑

7 Question a) Under what conditions will the OLS estimates of (2) be unbiased and consistent for the true parameters 𝛽 0 and 𝛽 1 ? True: π‘Œ 𝑑 = 𝛽 0 + 𝛽 1 𝑋 1𝑑 + 𝛽 2 𝑋 2𝑑 + 𝑒 𝑑 Estimated: π‘Œ 𝑑 = 𝛽 0 βˆ— + 𝛽 1 βˆ— 𝑋 1𝑑 + 𝑒 𝑑 βˆ— Where: 𝑒 𝑑 βˆ— = 𝑒 𝑑 + 𝛽 2 𝑋 2𝑑 True Model Deviation from the mean: 𝑦 𝑑 = 𝛽 1 π‘₯ 1𝑑 + 𝛽 2 π‘₯ 2𝑑 + 𝑒 𝑑 Where: π‘₯ 𝑖𝑑 = 𝑋 𝑖𝑑 βˆ’ 𝑋 𝑖

8 Question a) 𝛽 1 βˆ— = βˆ‘ (π‘₯ 1𝑑 𝑦 𝑑 ) βˆ‘ π‘₯ 1𝑑 2 We know:
Covariance of x & y 𝛽 1 βˆ— = βˆ‘ (π‘₯ 1𝑑 𝑦 𝑑 ) βˆ‘ π‘₯ 1𝑑 2 We know: 𝑦 𝑑 = 𝛽 1 π‘₯ 1𝑑 + 𝛽 2 π‘₯ 2𝑑 + 𝑒 𝑑 ∴ 𝛽 1 βˆ— = βˆ‘( π‘₯ 1𝑑 𝛽 1 π‘₯ 1𝑑 + 𝛽 2 π‘₯ 2𝑑 + 𝑒 𝑑 ) βˆ‘ π‘₯ 1𝑑 2 𝛽 1 βˆ— = 𝛽 1 βˆ‘ π‘₯ 1𝑑 π‘₯ 1𝑑 + 𝛽 2 βˆ‘ π‘₯ 1𝑑 π‘₯ 2𝑑 +βˆ‘( π‘₯ 1𝑑 𝑒 𝑑 ) βˆ‘ π‘₯ 1𝑑 2 𝛽 1 βˆ— = 𝛽 1 + 𝛽 2 βˆ‘ π‘₯ 1𝑑 π‘₯ 2𝑑 βˆ‘ π‘₯ 1𝑑 2 + βˆ‘( π‘₯ 1𝑑 𝑒 𝑑 ) βˆ‘ π‘₯ 1𝑑 2 Sample variance of x Expectation = 0

9 Question a) 𝐸 𝛽 1 βˆ— = 𝛽 1 + 𝛽 2 βˆ‘ π‘₯ 1𝑑 π‘₯ 2𝑑 βˆ‘ π‘₯ 1𝑑 2
𝐸 𝛽 1 βˆ— = 𝛽 1 + 𝛽 2 βˆ‘ π‘₯ 1𝑑 π‘₯ 2𝑑 βˆ‘ π‘₯ 1𝑑 2 If 𝛽 2 βˆ‘ π‘₯ 1𝑑 π‘₯ 2𝑑 βˆ‘ π‘₯ 1𝑑 2 is +ve (-ve) bias is upwards (downwards) βˆ‘ π‘₯ 1𝑑 π‘₯ 2𝑑 βˆ‘ π‘₯ 1𝑑 2 = πœ† 1 𝑋 2𝑑 = πœ† 0 + πœ† 1 𝑋 1𝑑 + 𝑣 𝑑 𝛽 1 βˆ— = biased if πœ† 1 β‰ 0 π‘Žπ‘›π‘‘ 𝛽 2 β‰ 0 If πœ† 1 and 𝛽 2 have same (opposite) sign upwards (downwards) bias Bias

10 Question a) If model (2) was the true model 𝛽 0 βˆ— and 𝛽 1 βˆ— would be unbiased and consistent (i.e. 𝛽 2 =0) estimators of 𝛽 0 and 𝛽 1 If model (2) suffers from omitted variable bias (i.e. 𝛽 2 β‰ 0) the estimates of 𝛽 1 βˆ— will be biased and inconsistent, for 𝛽 1 , if 𝑋 1𝑑 and 𝑋 2𝑑 are correlated. The estimator of 𝛽 0 βˆ— will be biased and inconsistent, for 𝛽 0 , even if 𝑋 1𝑑 and 𝑋 2𝑑 are uncorrelated: 𝐸 𝛽 0 βˆ— = 𝛽 0 + 𝛽 2 πœ† 0 πœ† 0 = 𝑋 2𝑑 βˆ’ πœ† 1 𝑋 1𝑑 The only way the intercept is unbiased is if 𝛽 2 =0 (i.e is the true model) or πœ† 1 = 𝑋 2𝑑 =0 or πœ† 0 =0

11 Question b) Estimate the following equation:
𝐿 π‘Š 𝑑 = πœ† 0 + πœ† 1 𝐿 𝐼 𝑑 + 𝑣 𝑑 Over the full sample period 1967q1-2002q4 In which direction do you think the OLS estimate of 𝛽 1 βˆ— from equation (2) is biased for 𝛽 1 ? (3)

12 OLS Results: Equation (3)
𝑋 2𝑑 = πœ† 0 + πœ† 1 𝑋 1𝑑 + 𝑣 𝑑 𝐿 π‘Š 𝑑 =βˆ’ 𝐿 𝐼 𝑑 + 𝑣 𝑑 𝐸 𝛽 1 βˆ— = 𝛽 1 + 𝛽 2 πœ† 1 πœ† 1 = +ve sign, and 𝛽 2 is expected to be +ve ∴ expect bias to be +ve 𝐸 𝛽 0 βˆ— <𝐸 𝛽 0 , as πœ† 0 =βˆ’π‘£π‘’

13 Question b) Does our prediction hold true?
𝐿 𝐢 𝑑 = 𝐿 𝐼 𝑑 πΏπ‘Š+ 𝑒 𝑑 𝐿 𝐢 𝑑 = 𝐿 𝐼 𝑑 + 𝑒 𝑑 Yes: 0.974>0.943, 𝛽 1 βˆ— > 𝛽 1 The omitted variable, wealth, likely biases the coefficient upwards. Yes: <0.564, 𝛽 0 βˆ— < 𝛽 0 The omitted variable, wealth, likely biases the intercept downwards.

14 Question c) What can be said about the precision of your estimates of equation (2)? In general we know: The error variance, 𝜎 2 , is incorrectly estimated The standard errors of the estimated coefficients π‘£π‘Žπ‘Ÿ 𝛽 are biased Conventional t and F tests are of limited value (i.e. invalid)

15 πΆπ‘œπ‘Ÿπ‘Ÿπ‘’π‘™π‘Žπ‘‘π‘–π‘œπ‘› πΆπ‘œπ‘’π‘“π‘“π‘–π‘π‘–π‘’π‘›π‘‘
Question c) From (1) the variance of the error term is 𝜎 2 . We have an estimate of this which is 𝜎 2 , this is equal to βˆ‘ 𝑒 𝑖 2 π‘›βˆ’π‘˜ = RSS π‘›βˆ’π‘˜ , where 𝑒 𝑖 are the residuals from the model, n is the sample size, and k is the number of parameters. Estimates of 𝜎 2 are likely to be different between the true and estimated model. Why? Consider (2)… π‘£π‘Žπ‘Ÿ 𝛽 1 = 𝜎 2 βˆ‘ π‘₯ 1𝑑 2 1βˆ’ π‘Ÿ 12 2 π‘£π‘Žπ‘Ÿ 𝛽 1 βˆ— = 𝜎 2 βˆ‘ π‘₯ 1𝑑 2 The variance of 𝛽 1 βˆ— is a biased estimate for the true model. But what if π‘Ÿ 12 2 =0, will the variance be the same? No. As 𝜎 2 is likely to be incorrect as well.. In general if π‘Ÿ 12 2 β‰₯0 we could expect π‘£π‘Žπ‘Ÿ 𝛽 1 βˆ— <π‘£π‘Žπ‘Ÿ 𝛽 1 but remember that 𝜎 βˆ— 2 is unlikely to be equal to 𝜎 2 . π‘ƒπ‘’π‘Žπ‘Ÿπ‘ π‘œπ‘› πΆπ‘œπ‘Ÿπ‘Ÿπ‘’π‘™π‘Žπ‘‘π‘–π‘œπ‘› πΆπ‘œπ‘’π‘“π‘“π‘–π‘π‘–π‘’π‘›π‘‘

16 Question d) Explain how you would use a Ramsey RESET test to check whether your model is specified correctly? Calculate the RESET test for your estimated model (2) and discuss the outcome of this test What other test might you use to investigate correct model specification? Briefly explain how you would use this test in the example above.

17 Ramsey RESET Test Step 1: From the assumed model obtain the estimated π‘Œ 𝑖 (old) Step 2: Re-run the assumed model including the estimated π‘Œ 𝑖 in some form as an additional regressor (new) Step 3: Then use the F-test to find out whether the increase in 𝑅 2 , by using the model in Step 2, is statistically significant 𝐹= 𝑅 𝑛𝑒𝑀 2 βˆ’ 𝑅 π‘œπ‘™π‘‘ 2 π‘›π‘œ. 𝑛𝑒𝑀 π‘Ÿπ‘’π‘”π‘Ÿπ‘’π‘ π‘ π‘œπ‘Ÿπ‘  βˆ’ 𝑅 𝑛𝑒𝑀 π‘›βˆ’π‘›π‘œ. π‘π‘Žπ‘Ÿπ‘Žπ‘šπ‘’π‘‘π‘’π‘Ÿπ‘  𝑖𝑛 𝑛𝑒𝑀 π‘šπ‘œπ‘‘π‘’π‘™ Step 4: If the computed F statistic is significant (for example at 5%) we can reject the null hypothesis that the model is correctly specified

18 RESET Test: Manually Old Model: π‘Œ 𝑑 = 𝛽 0 + 𝛽 1 𝑋 1𝑑 + 𝑒 𝑑 New Model:
π‘Œ 𝑑 = 𝛽 0 + 𝛽 1 𝑋 1𝑑 + 𝑒 𝑑 New Model: π‘Œ 𝑑 = 𝛽 0 βˆ— + 𝛽 1 βˆ— 𝑋 1𝑑 + 𝛽 2 βˆ— π‘Œ 2𝑑 2 + 𝛽 3 βˆ— π‘Œ 3𝑑 3 + 𝑒 𝑑 βˆ— Where: π‘Œ 𝑖𝑑 = 𝛽 0 + 𝛽 1 𝑋 𝑖𝑑 𝐻 0 :π‘šπ‘œπ‘‘π‘’π‘™ 𝑖𝑠 π‘π‘œπ‘Ÿπ‘Ÿπ‘’π‘π‘‘π‘™π‘¦ 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑒𝑑 𝐻 1 : 𝐻 0 𝑖𝑠 π‘“π‘Žπ‘™π‘ π‘’ 𝐹= βˆ’ βˆ’ βˆ’4 = ~ 𝐹 2,140 𝐹 2, β‰ˆ3.07, 𝐹 2, β‰ˆ4.79, ∴𝐹> 𝐹 𝑐 π‘Žπ‘‘ 5% π‘Žπ‘›π‘‘ 1%, 𝑀𝑒 π‘Ÿπ‘’π‘—π‘’π‘π‘‘ π‘‘β„Žπ‘’ 𝑛𝑒𝑙𝑙

19 RESET Test: Manually (alternative)
Old Model: π‘Œ 𝑑 = 𝛽 0 + 𝛽 1 𝑋 1𝑑 + 𝑒 𝑑 New (alternative) Model: π‘Œ 𝑑 = 𝛽 0 βˆ— + 𝛽 1 βˆ— 𝑋 1𝑑 + 𝛽 2 βˆ— π‘Œ 2𝑑 2 + 𝑒 𝑑 βˆ— Where: π‘Œ 𝑖𝑑 = 𝛽 0 + 𝛽 1 𝑋 𝑖𝑑 𝐻 0 :π‘šπ‘œπ‘‘π‘’π‘™ 𝑖𝑠 π‘π‘œπ‘Ÿπ‘Ÿπ‘’π‘π‘‘π‘™π‘¦ 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑒𝑑 𝐻 1 : 𝐻 0 𝑖𝑠 π‘“π‘Žπ‘™π‘ π‘’ 𝐹= βˆ’ βˆ’ βˆ’3 = ~ 𝐹 1,140 𝐹 1, β‰ˆ3.92, 𝐹 1, β‰ˆ6.85, ∴𝐹> 𝐹 𝑐 π‘Žπ‘‘ 5% π‘Žπ‘›π‘‘ 1%, 𝑀𝑒 π‘Ÿπ‘’π‘—π‘’π‘π‘‘ π‘‘β„Žπ‘’ 𝑛𝑒𝑙𝑙

20 RESET Test: PcGive π‘€π‘œπ‘‘π‘’π‘™β‡’π‘‡π‘’π‘ π‘‘β‡’π‘‡π‘’π‘ π‘‘β‡’π‘‚πΎβ‡’π‘…πΈπ‘†πΈπ‘‡ 𝑑𝑒𝑠𝑑 π‘Žπ‘›π‘‘ 𝑅𝐸𝑆𝐸𝑇23 𝑑𝑒𝑠𝑑 𝐹 𝐹

21 Lagrange Multiplier (LM) test
Step 1: Estimate the restricted regression by OLS and obtain the residuals Step 2: If the unrestricted model is the correct one the residuals from Step 1 will be related to the extra variables in the unrestricted model Step 3: Regress the residuals from Step 1 on all the variables. This is the auxiliary regression Step 4: For large sample sizes, the sample size (n) times by the 𝑅 2 from the auxiliary regression follows the chi-square distribution with degrees of freedom equal to the number of restrictions imposed by the restricted model. 𝑛 𝑅 2 ~ πœ’ 2 (π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘Ÿπ‘’π‘ π‘‘π‘Ÿπ‘–π‘π‘‘π‘–π‘œπ‘›π‘ ) asy

22 Lagrange Multiplier (LM) test
Unrestricted Model: 𝐿 𝐢 𝑑 = 𝛽 0 + 𝛽 1 𝐿 𝐼 𝑑 + 𝛽 2 πΏπ‘Š+ 𝑒 𝑑 Restricted Model: 𝐿 𝐢 𝑑 = 𝛽 0 βˆ— + 𝛽 1 βˆ— 𝐿 𝐼 𝑑 + 𝑒 𝑑 βˆ— Auxiliary Regression: 𝑒 𝑑 βˆ— = 𝛼 0 + 𝛼 1 𝐿 𝐼 𝑑 + 𝛼 2 πΏπ‘Š+ 𝑣 𝑑 Use the 𝑅 2 from the Auxiliary Regression to calculate the test statistic. 𝑛 𝑅 2 = 144βˆ— =1.268~ πœ’ 2 (1) πœ’ β‰ˆ3.841, πœ’ β‰ˆ6.635 𝐿𝑀< πœ’ β‡’Do not reject the Null

23 RESET vs LM RESET: We do not get information of how to improve the model However, it is good as we do not need to know which X’s are the issue LM: Allows us to choose the β€˜better’ model Relies on asymptotic assumptions, so low n means it is an inappropriate test


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