1. No category

Microeconomics 2: Problem Set 1

Tutorial 1: Misspecification
Matthew Robson
University of York
Econometrics 2
1
Assignment 5
Assume that the true model of consumers’ behaviour is:
𝐿𝐢𝑑 = 𝛽0 + 𝛽1 𝐿𝐼𝑑 + 𝛽2 πΏπ‘Š + 𝑒𝑑
(1)
Where:
𝐿𝐢𝑑 = log consumption, 𝐿𝐼𝑑 = log income, πΏπ‘Šπ‘‘ = log wealth
Estimate the simple log-linear consumption function:
𝐿𝐢𝑑 = 𝛽0βˆ— + 𝛽1βˆ— 𝐿𝐼𝑑 + π‘’π‘‘βˆ—
(2)
Over the period 1967q1-2002q4
2
Descriptive Statistics
3
OLS Results: True Model
𝐿𝐢𝑑 = 0.564 + 0.943 𝐿𝐼𝑑 + 0.008 πΏπ‘Š + 𝑒𝑑
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‘‘.
4
Fitted Model
5
OLS Results: Simple Model
𝐿𝐢𝑑 = 0.257 + 0.974 𝐿𝐼𝑑 + 𝑒𝑑
6
Question a)
Under what conditions will the OLS estimates of (2) be unbiased and
consistent for the true parameters 𝛽0 and 𝛽1 ?
True:
Estimated:
π‘Œπ‘‘ = 𝛽0 + 𝛽1 𝑋1𝑑 + 𝛽2 𝑋2𝑑 + 𝑒𝑑
π‘Œπ‘‘ = 𝛽0βˆ— + 𝛽1βˆ— 𝑋1𝑑 + π‘’π‘‘βˆ—
Where: π‘’π‘‘βˆ— = 𝑒𝑑 + 𝛽2 𝑋2𝑑
True Model Deviation from the mean:
𝑦𝑑 = 𝛽1 π‘₯1𝑑 + 𝛽2 π‘₯2𝑑 + 𝑒𝑑
Where: π‘₯𝑖𝑑 = 𝑋𝑖𝑑 βˆ’ 𝑋𝑖
7
Question a)
βˆ‘(π‘₯1𝑑 𝑦𝑑 )
βˆ—
𝛽1 =
2
βˆ‘π‘₯1𝑑
Covariance of x & y
Sample variance of x
We know:
𝑦𝑑 = 𝛽1 π‘₯1𝑑 + 𝛽2 π‘₯2𝑑 + 𝑒𝑑
∴
βˆ‘(π‘₯1𝑑 𝛽1 π‘₯1𝑑 + 𝛽2 π‘₯2𝑑 + 𝑒𝑑 )
=
2
βˆ‘π‘₯1𝑑
𝛽1 βˆ‘ π‘₯1𝑑 π‘₯1𝑑 + 𝛽2 βˆ‘ π‘₯1𝑑 π‘₯2𝑑 + βˆ‘(π‘₯1𝑑 𝑒𝑑 )
βˆ—
𝛽1 =
2
βˆ‘π‘₯1𝑑
𝛽1βˆ—
Expectation = 0
𝛽1βˆ—
𝛽2 βˆ‘ π‘₯1𝑑 π‘₯2𝑑
βˆ‘(π‘₯1𝑑 𝑒𝑑 )
= 𝛽1 +
+
2
2
βˆ‘π‘₯1𝑑
βˆ‘π‘₯1𝑑
8
Question a)
𝐸
If
βˆ‘ π‘₯1𝑑 π‘₯2𝑑
𝛽2
2
βˆ‘π‘₯1𝑑
𝛽1βˆ—
βˆ‘ π‘₯1𝑑 π‘₯2𝑑
= 𝛽1 + 𝛽2
2
βˆ‘π‘₯1𝑑
Bias
is +ve (-ve) bias is upwards (downwards)
βˆ‘ π‘₯1𝑑 π‘₯2𝑑
= πœ†1
2
βˆ‘π‘₯1𝑑
𝑋2𝑑 = πœ†0 + πœ†1 𝑋1𝑑 + 𝑣𝑑
𝛽1βˆ— = biased if πœ†1 β‰  0 π‘Žπ‘›π‘‘ 𝛽2 β‰  0
If πœ†1 and 𝛽2 have same (opposite) sign upwards (downwards) bias
9
Question a)
β€’ If model (2) was the true model 𝛽0βˆ— and 𝛽1βˆ— would be unbiased and
consistent (i.e. 𝛽2 = 0) estimators of 𝛽0 and 𝛽1
β€’ If model (2) suffers from omitted variable bias (i.e. 𝛽2 β‰  0) the
estimates of 𝛽1βˆ— will be biased and inconsistent, for 𝛽1 , if 𝑋1𝑑 and
𝑋2𝑑 are correlated. The estimator of 𝛽0βˆ— will be biased and
inconsistent, for 𝛽0 , even if 𝑋1𝑑 and 𝑋2𝑑 are uncorrelated:
𝐸 𝛽0βˆ— = 𝛽0 + 𝛽2 πœ†0
πœ†0 = 𝑋2𝑑 βˆ’ πœ†1 𝑋1𝑑
β€’ The only way the intercept is unbiased is if 𝛽2 = 0 (i.e is the true
model) or πœ†1 = 𝑋2𝑑 = 0 or πœ†0 = 0
10
Question b)
Estimate the following equation:
πΏπ‘Šπ‘‘ = πœ†0 + πœ†1 𝐿𝐼𝑑 + 𝑣𝑑
(3)
Over the full sample period 1967q1-2002q4
In which direction do you think the OLS estimate of 𝛽1βˆ— from
equation (2) is biased for 𝛽1 ?
11
OLS Results: Equation (3)
𝑋2𝑑 = πœ†0 + πœ†1 𝑋1𝑑 + 𝑣𝑑
πΏπ‘Šπ‘‘ = βˆ’36.6006 + 3.69676𝐿𝐼𝑑 + 𝑣𝑑
𝐸 𝛽1βˆ— = 𝛽1 + 𝛽2 πœ†1
πœ†1 = +ve sign, and 𝛽2 is expected to be +ve ∴ expect bias to be +ve
𝐸 𝛽0βˆ— < 𝐸 𝛽0 , as πœ†0 = βˆ’π‘£π‘’
12
Question b)
Does our prediction hold true?
𝐿𝐢𝑑 = 0.564 + 0.943 𝐿𝐼𝑑 + 0.008 πΏπ‘Š + 𝑒𝑑
𝐿𝐢𝑑 = 0.257 + 0.974 𝐿𝐼𝑑 + 𝑒𝑑
Yes: 0.974 > 0.943, 𝛽1βˆ— > 𝛽1
The omitted variable, wealth, likely biases the coefficient upwards.
Yes: 0.257 < 0.564, 𝛽0βˆ— < 𝛽0
The omitted variable, wealth, likely biases the intercept downwards.
13
Question c)
What can be said about the precision of your estimates of equation (2)?
In general we know:
1. The error variance, 𝜎 2 , is incorrectly estimated
2. The standard errors of the estimated coefficients
π‘£π‘Žπ‘Ÿ 𝛽 are
biased
3. Conventional t and F tests are of limited value (i.e. invalid)
14
Question c)
β€’ From (1) the variance of the error term is 𝜎 2 . We have an estimate of this which
𝜎 2,
βˆ‘π‘’π‘–2
RSS
is
this is equal to
=
, where 𝑒𝑖 are the residuals from the model, n is
π‘›βˆ’π‘˜
π‘›βˆ’π‘˜
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)…
𝜎2
π‘ƒπ‘’π‘Žπ‘Ÿπ‘ π‘œπ‘›
π‘£π‘Žπ‘Ÿ 𝛽1 =
2
2
πΆπ‘œπ‘Ÿπ‘Ÿπ‘’π‘™π‘Žπ‘‘π‘–π‘œπ‘› πΆπ‘œπ‘’π‘“π‘“π‘–π‘π‘–π‘’π‘›π‘‘
βˆ‘π‘₯1𝑑 1 βˆ’ π‘Ÿ12
2
𝜎
π‘£π‘Žπ‘Ÿ 𝛽1βˆ— =
2
βˆ‘π‘₯1𝑑
2
β€’ The variance of 𝛽1βˆ— is a biased estimate for the true model. But what if π‘Ÿ12
= 0,
will the variance be the same? No. As 𝜎 2 is likely to be incorrect as well..
2
β€’ In general if π‘Ÿ12
β‰₯ 0 we could expect π‘£π‘Žπ‘Ÿ 𝛽1βˆ— < π‘£π‘Žπ‘Ÿ 𝛽1 but remember that
πœŽβˆ—2 is unlikely to be equal to 𝜎 2 .
15
Question d)
β€’ 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
β€’ What other test might you use to investigate correct model
specification? Briefly explain how you would use this test in the
example above.
16
Ramsey RESET Test
β€’ Step 1: From the assumed model obtain the estimated π‘Œπ‘– (old)
β€’ Step 2: Re-run the assumed model including the estimated π‘Œπ‘– in some form as an
additional regressor (new)
β€’ Step 3: Then use the F-test to find out whether the increase in 𝑅2 , by using the
model in Step 2, is statistically significant
𝐹=
2
2
𝑅𝑛𝑒𝑀
βˆ’ π‘…π‘œπ‘™π‘‘
2
1 βˆ’ 𝑅𝑛𝑒𝑀
π‘›π‘œ. 𝑛𝑒𝑀 π‘Ÿπ‘’π‘”π‘Ÿπ‘’π‘ π‘ π‘œπ‘Ÿπ‘ 
𝑛 βˆ’ π‘›π‘œ. π‘π‘Žπ‘Ÿπ‘Žπ‘šπ‘’π‘‘π‘’π‘Ÿπ‘  𝑖𝑛 𝑛𝑒𝑀 π‘šπ‘œπ‘‘π‘’π‘™
β€’ 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
17
RESET Test: Manually
Old Model:
π‘Œπ‘‘ = 𝛽0 + 𝛽1 𝑋1𝑑 + 𝑒𝑑
New Model:
2
3
π‘Œπ‘‘ = 𝛽0βˆ— + 𝛽1βˆ— 𝑋1𝑑 + 𝛽2βˆ— π‘Œ2𝑑 + 𝛽3βˆ— π‘Œ3𝑑 + π‘’π‘‘βˆ—
Where: π‘Œπ‘–π‘‘ = 𝛽0 + 𝛽1 𝑋𝑖𝑑
𝐻0 : π‘šπ‘œπ‘‘π‘’π‘™ 𝑖𝑠 π‘π‘œπ‘Ÿπ‘Ÿπ‘’π‘π‘‘π‘™π‘¦ 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑒𝑑
𝐻1 : 𝐻0 𝑖𝑠 π‘“π‘Žπ‘™π‘ π‘’
0.992419 βˆ’ 0.991402 2
𝐹=
= 9.39058 ~𝐹2,140
1 βˆ’ 0.992419
144 βˆ’ 4
0.05
0.01
𝐹2,140
β‰ˆ 3.07, 𝐹2,140
β‰ˆ 4.79,
∴ 𝐹 > 𝐹 𝑐 π‘Žπ‘‘ 5% π‘Žπ‘›π‘‘ 1%, 𝑀𝑒 π‘Ÿπ‘’π‘—π‘’π‘π‘‘ π‘‘β„Žπ‘’ 𝑛𝑒𝑙𝑙
18
RESET Test: Manually (alternative)
Old Model:
π‘Œπ‘‘ = 𝛽0 + 𝛽1 𝑋1𝑑 + 𝑒𝑑
New (alternative) Model:
2
π‘Œπ‘‘ = 𝛽0βˆ— + 𝛽1βˆ— 𝑋1𝑑 + 𝛽2βˆ— π‘Œ2𝑑 + π‘’π‘‘βˆ—
Where: π‘Œπ‘–π‘‘ = 𝛽0 + 𝛽1 𝑋𝑖𝑑
𝐻0 : π‘šπ‘œπ‘‘π‘’π‘™ 𝑖𝑠 π‘π‘œπ‘Ÿπ‘Ÿπ‘’π‘π‘‘π‘™π‘¦ 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑒𝑑
𝐻1 : 𝐻0 𝑖𝑠 π‘“π‘Žπ‘™π‘ π‘’
0.992415 βˆ’ 0.991402 1
𝐹=
= 18.831 ~𝐹1,140
1 βˆ’ 0.992415
144 βˆ’ 3
0.05
0.01
𝐹1,140
β‰ˆ 3.92, 𝐹1,140
β‰ˆ 6.85,
∴ 𝐹 > 𝐹 𝑐 π‘Žπ‘‘ 5% π‘Žπ‘›π‘‘ 1%, 𝑀𝑒 π‘Ÿπ‘’π‘—π‘’π‘π‘‘ π‘‘β„Žπ‘’ 𝑛𝑒𝑙𝑙
19
RESET Test: PcGive
π‘€π‘œπ‘‘π‘’π‘™ β‡’ 𝑇𝑒𝑠𝑑 β‡’ 𝑇𝑒𝑠𝑑 β‡’ 𝑂𝐾 β‡’ 𝑅𝐸𝑆𝐸𝑇 𝑑𝑒𝑠𝑑 π‘Žπ‘›π‘‘ 𝑅𝐸𝑆𝐸𝑇23 𝑑𝑒𝑠𝑑
𝐹
𝐹
20
Lagrange Multiplier (LM) test
β€’ Step 1: Estimate the restricted regression by OLS and obtain the residuals
β€’ 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
β€’ Step 3: Regress the residuals from Step 1 on all the variables. This is the auxiliary
regression
β€’ Step 4: For large sample sizes, the sample size (n) times by the 𝑅2 from the
auxiliary regression follows the chi-square distribution with degrees of freedom
equal to the number of restrictions imposed by the restricted model.
𝑛𝑅2asy
~ πœ’ 2 (π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘Ÿπ‘’π‘ π‘‘π‘Ÿπ‘–π‘π‘‘π‘–π‘œπ‘›π‘ )
21
Lagrange Multiplier (LM) test
Unrestricted Model:
𝐿𝐢𝑑 = 𝛽0 + 𝛽1 𝐿𝐼𝑑 + 𝛽2 πΏπ‘Š + 𝑒𝑑
𝐿𝐢𝑑 = 𝛽0βˆ— + 𝛽1βˆ— 𝐿𝐼𝑑 + π‘’π‘‘βˆ—
Restricted Model:
Auxiliary Regression:
π‘’π‘‘βˆ— = 𝛼0 + 𝛼1 𝐿𝐼𝑑 + 𝛼2 πΏπ‘Š + 𝑣𝑑
Use the 𝑅2 from the Auxiliary Regression to calculate the test statistic.
𝑛𝑅2 = 144 βˆ— 0.00880823 = 1.268~ πœ’ 2 (1)
πœ’10.05 β‰ˆ 3.841, πœ’10.01 β‰ˆ 6.635
𝐿𝑀 < πœ’10.05 β‡’ Do not reject the Null
22
RESET vs LM
β€’ RESET:
β€’ We do not get information of how to improve the model
β€’ However, it is good as we do not need to know which X’s are
the issue
β€’ LM:
β€’ Allows us to choose the β€˜better’ model
β€’ Relies on asymptotic assumptions, so low n means it is an
inappropriate test
23

 Download  Report

Paperzz.com

Your Paperzz

© Copyright 2026 Paperzz