Download (b) and - University of Bristol

UNIVERSITY OF BRISTOL
Department of Economics
Statistics and Introduction to Econometrics
(ECON 12122)
Econometrics Exercise 1.
Solutions
1. (a) Write down an econometric model of the relationship between W and un.
Justify your choice of explanatory and dependent variable.
We have to decide which of these variables is the dependent variable and which is the
explanatory variable. To do this we must use Economic Theory. Both these variable are
clearly labour market variables. W is the labour market equivalent to a change in price,
un is the labour market equivalent to the difference between Supply and Demand i.e.
excess supply. Most theories of the labour market make some allowance for "frictional
unemployment" so even if the labour market clears (supply equals demand), we would
expect some positive unemployment.
In this case we have two possibilities.
(i)
Does a change in excess supply cause a change in wages? If so, W is the
dependent variable and un the explanatory variable.
(ii)
Does a change in wages change the balance of supply and demand and thereby
unemployment? If so, un is the dependent variable and W is the explanatory
variable.
Economic theory supports both of these positions. Historically (or ever since Phillips
invented the Phillips curve in the late 1950s), economists have usually chosen (a). Thus
using the linear regression model, an econometric model would take the following form;
W = f(un) + random disturbance
If we use the linear regression model then
E(Wun) =
 + un
and
W
= E(Wun) +
u
where u is an unobserved random disturbance with E(uun) = 0
(b) and (c) Scatter diagrams were discusssed in the lectures.
(d) Draw by hand a linear relationship between W and un. What are your
estimates of the intercept and slope? Are they sensible? Explain.
The following line was "drawm by hand"
Ŵt = 23.09
 2.43 un t
If unemployment rate is 4 per cent, estimated W is 14.18 per cent. This seems
rather high for current rates of inflation.
2*. There is a tradition in Economics of writing down the relationship between price and
quantity showing price as a function of quantity (the inverse demand curve). Thus
E( Pt Qt ) = a +
Pt =
E( Pt Qt )
b Q t and
+ u t where
E( u t Q t ) = 0
It is just as sensible (possibly more so) to consider the relationship the other way round
E(Qt Pt ) = c +
Qt =
E(Qt Pt )
d Pt and
+ v t where
E( v t Pt ) = 0
(b)
I've drawn the scatter diagram the traditional way round. I have put the line
through the points (100, 100) and (36.42, 134.41) giving an estimated line
P̂t = 74.41  1.85 Q t
Note the possibility of curvature in the scatter diagram. The residuals will have
long runs of the same sign suggesting serial correlation (i.e. non randomness).
(c)
The negative slope is what would be expected. Since the data are in index form
the estimated value of the slope is not easy to interpret.
(d)
The elasticity of demand is the elasticity of Q with respect to P. Thus it is
probably better to conduct the analysis with Q as a function of P (and not the
other way round). From the estimates above the reverse relationship is
Q̂ t = 40.22  0.54 Pt
This gives an estimated price elasticity of demand for 1990 of -0.54, and for 1995
-1.002. It is doubtful whether the elasticity would change so quickly. For an
addictive good like cigarettes one would expect it to be less (in absolute size) than
one.
(e)
A good estimate of the price elasticity is essential. You also need to know what
proportion of the total price paid is made up of tax. It might also be important to
know the slope of the supply curve and whether consumers respond to price
changes with a lag.
(f)
The obvious other demand variable is income, although that may not be important
here. Also supply variables (costs) and the perceptions of smokers regarding the
health consequences of smoking may be relevant.
3*.
(a)
Least squares estimate of the slope =
906.49  173.52  164.70 / 22
1,410.29  (164.7) 2 / 22

 392.54
 2.21
177.29
estimate of intercept = 173.52/22 + 2.21x164.70/22 = 24.43
Thus the estimated line is
Ŵt = 24.43
 2.21 un t
The line drawn by hand in the exercise lecture (question 1) was;
Ŵt = 23.09
 2.43 un t
which is very similar.
(b)
Estimate of slope negative as expected. It implies that a rise in
unemployment by 1 percentage points reduces the change in wages by
2.21 percent age points. This is probably rather a large response.
(c)
Variance of slope coefficient is 22.11/(1,410.29  164.7x164.7/22) =
0.1247, standard error = 0.353
Confidence interval is 2.21  0.353  t 20,0.975 =
2.21  0.353  2.086  2.21  0.736 or from 2.946 to 1.474
4. (a) In the model
E( Yt X t ) =   X t or
Yt    X t  u t
t = 1....T
where ut is a random error with E( u t X t ) = 0.
Explain what the ordinary least squares (OLS) estimators of  and  minimize and
derive the necessary (first order) conditions for this minimisation.
The least squares estimators minimises
2
ˆ  ˆ Xi )2 = S.
 ( ûi ) =  ( Yi  
This is the square of the vertical distance from the regression line to a given observation.

The first order conditions for this problem are:



 
S
2 Yi ˆ ˆ X i 0
ˆ
S
 2  Yi ˆ ˆ X i X i 0
ˆ

(b)
Show that the least squares estimator of , ̂
 ( Yi  Y )( X i  X )
ˆ 
2
 (Xi  X )
where Y and X are the sample means of Yt and Xt respectively
Take the numerator of the expression above;
 ( Yi  Y)( Xi  X)   Yi Xi  nYX  Y  Xi  X  Yi
since Y  Xi  nYX and X  Yi  nYX , then
 ( Yi  Y)( Xi  X)   Yi Xi  nYX
Similarly  (Xi  X)2   Xi2  nX2
The first of the first order conditions gives ̂ = Y  ̂ X
Substitute this into the second condition gives
or

  = 0
 ( Yt Y)Xt ˆ (Xt X)Xt  = 0
thus
̂ =
 Yt ( Yˆ X)ˆ Xt Xt
(c)
Yt Y X t
X t X t 
2
=
 X t Yt nXY
 X t 2  nX 2
If û t is the least squares residual, show that  û t = 0.
The first order condition for ̂ is


S
2 Yt ˆ ˆ X t 0
ˆ
ˆ  ˆ Xt therefore  û t = 0.
û t  Yt  
(d)
Suppose it is known that  = 0, the following estimator of  is proposed
~ 1 Y 
    i 
T  Xi 
~
~
Show that  is unbiased. Compare the properties of  with that of the least squares
estimator of .
 X  u i 
u 
~ 1 Y 
 =  + T 1   i 
    i  = T 1   i
T  Xi 
 Xi 
 Xi 
so
~
E( ) =  +
u 
E[T 1   i ]
 Xi 
 E( u i ) 
 =
=  + T 1  
 Xi 

~
Thus  is an unbiased estimator of . It is also a linear estimator. Thus the Gauss
Markov theorem tells us that provided
 
E ui2  2 for all i
E u i u j  = 0 for all i  j
~
the variance of  cannot be smaller than the variance of the ordinary least squares
estimator of .
5*.
Consider the model
E( y t x t ) =  x t or
y t  x t  u t
t = 1,2..,T
where ut is an unobserved random disturbance with E( u t x t ) = 0.
(a)
Estimators are random variables because they are functions of the elements of
the sample. These elements are themselves random variables and a function
of random variables is, itself, a random variable.
(b)
Derive the ordinary least squares estimator of  and if ̂ is the ordinary least
squares estimator, show that
The least squares estimator of  is found by minimizing
2
 ( y t  ˆ x t ) with respect to ̂ .
The first order condition is
 2 ( y t  ˆ x t )x t = 0
Rearranging gives;
 ytx t
ˆ  x 2t
=
Thus;
T
 x t yt
t 1,
T
̂ =
2
 xt
t 1,
(c)
If ̂ is an estimator of the population parameter  then if
E ( ˆ ) =  then ̂ is unbiased. The mean of ̂ is equal to .
T
̂ =
 x t yt
t 1,
T
=
2
 xt
2
 (x t  x t u t )
= 
2
 xt
+
 (x t u t )
2
 xt
t 1,
Then;
E (ˆ )
= E()
  xtut 
+ E
=  +
2 
  x t 
 x t E( u t x t )
2
 xt
= 
since E( u t x t ) = 0.
(d)
For the model above explain the difference between:
(i)
the OLS estimator ̂ and the parameter .
̂ is an estimator of . They will not in general be equal. ̂ is random
variable whereas  is a fixed unknown population parameter.
1
(ii) E( u t X t ) and
T
regression.
t T
 u , where u
t
t 1
t
is the residual from the OLS
E( u t X t ) is the population mean of u t conditional on any given value
1 t T
 ut is the sample mean of the least squares residual. If the
T t 1
model contains a constant term the sample mean will be fixed at zero.
In the model above (which does not have a constant term) it may not be
zero.
of x t .
6.
The following estimates were made from a sample of quarterly UK data, 1968.11991.4. The dependent variable is the log of the stock of money (M4). The
explanatory variables and their estimated coefficients are given below;
m 4 t = 23.68 + 2.598 gdp t + 0.883 pt  0.070 rt + e t
(2.669)
(0.236)
(0.050)
(0.053)
RSS = 1.419, n = 96, e t is the least squares residual.
where gdp the log of GDP at constant prices, p is the log of the implicit price
deflator of GDP, r is the log of a short term rate of interest, RSS is the sum of
squared residuals, n is the number of observations, standard errors are given in
brackets.
(a)
Discuss the point estimates of the slope coefficients in the light of standard
economic theory concerning the demand for money.
Estimated coefficient of gdp is 2.598. This is expected to be positive. In this
model is also an estimate of the elasticity of money with respect to gdp. This
2.598 is rather high (long way from one).
Estimated coefficient of p is 0.883. This is expected to be positive and one.
Estimate should be fairly close to one, could test this (see below).
Estimated coefficient of r is -0.07. This is expected to be negative. However it is
small and not significant.
"the" t ratio =  0.07/ 0.053 =  1.32 < 1.99 (see below)
Thus it is not possible to reject the null hypothesis that the coefficient of r is zero
(b)
Derive an unbiased estimate of the variance of the residuals.
Estimate of variance = RSS/n-k = 1.419 / 92 = 0.0154
(c)
Test the null hypothesis that the coefficient of p in (i) is one and also that
the coefficient of gdp in (i) is one.
H 0 coefficient of p = 1
H1 coefficient of p  1
t ratio = (0.883 - 1)/ 0.05 =  2.34
This distrbiuted as a t distribution with 92 degrees of freedom if H 0 is
true.
The critical value for a two tailed test with 95 % confidence is 1.99
(approx). Since  2.34 > 1.99 reject H 0 .
H 0 coefficient of gdp = 1
H1 coefficient of gdp > 1
t ratio = (2.598 - 1)/0.236 = 6.771. This is distributed as a t distribution
with 92 degrees of freedom if H 0 is true..
The critical value of a one tailed test with 95 % confidence is 1.65
(approx). Since 6.77 > 1.65, reject H 0 .
(d)
In the light of your results in (b) and (c) comment further on the estimates
above.
Thus on these estimates money is clearly a "luxury" good and the demand
for money does not increase proportionately with inflation - in fact people
economise on money as inflation rises.
(e)
State the assumptions which your answers to (b) and (c) above require.
Are they likely to be true? Give details.
1. Model correct
2. E(uiuj) =  2
i=j
3.
= 0
ij
4. Large sample or normal disturbances.
This model is a standard demand for money function. In time series
models assumption (3) usually needs to be tested. Sample probably just
large enough to appeal to a central limit theorem.