ECO4185 - Financial Econometrics
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Additional practice questions - ile #1
ECO4185 - Financial Econometrics
1. Consider the linear regression model:
yt = β 1 + β2 x2t + ... + βkxkt + ut (1)
where our period of interest is t = 1, ...,T . We know that this model can equivalently be written as:
yt = xβ + ut (2)
or in matrix form as
y = Xβ + u (3)
In (2), xtis the (k × 1) vector xt = [1 x2t ... xkt]\ and β is the (k × 1) vector β = [β1 β2 ... βk]\ . In (3), β is the same as in (2), and y , X, and u are as deined in class. Remember that the dimension of y and u is (T × 1) and the dimension of X is (T × k). It will be useful to notice that
l x(x)」
X = (4)
「x
where x1 , x2 , ...,xT are the (k × 1) vectors xt , t = 1, ...,T, used in (2).
Assume that we estimate this model by OLS and obtain the vector and the residuals t , t = 1, ...,T .
a. We know that the OLS estimator of β has the following expression: = (X\X)−1 (X\ y). Show that, using the notation of (2), this expression is equivalent to = ( xtx)− 1 ( xtyt). Hint: use (4).
T
b. Show that, by construction, 对 xtt = 0, where 0 is a (k × 1) vector of zeros. Hint: use the fact t=1
that, as discussed in part a., = (xtx)− 1 ( xtyt).
T
c. Show that, by construction, 对 t = 0. Hint: use the result from part b. of the question. t=1
d. Our friend Ellen estimated (1) using a diferent sample of data that she collected. She computed
T T
that in her estimated OLS regression 对 t = 0 and s2 = 对 (t)2 = 2.43. She argues that this
implies that E(ut) = 0 and var(ut) = a2 < ∞ (constant) in the underlying true model (1). Do you agree with Ellen? Motivate your answer.
2. We are interested in the linear regression model:
yt = β 1 + β2 x2t + αzt + ut (5)
where t = 1, ...,T . We know that E(ut) = 0, var(ut) = a2 < ∞ , cov(ut,ut−j) = 0 for all j 0. In
Unfortunately the variable zt is unobservable and we cannot include it in the model that we are going to estimate. Therefore, we decide to drop this variable and focus on the alternative model
yt = β 1 + β2 x2t + vt (6)
The variable zt will now be captured in the innovation term, so vt = αzt + ut . Assume that α 0, i.e. we know that the variable zt has an impact on yt .
a. Write the expression for E(vt). Under which conditions is E(vt) = 0? Assume that we estimate the parameters of (6) by OLS. Explain how these estimates are going to be afected if E(vt) 0 but all the other assumptions of the classical linear regression model (CLRM) hold.
b. Write the expression for var(vt). Under which conditions is var(vt) inite and constant? Explain your answer.
c. Assume that var(zt) = 4 for 1 ≤ t ≤ τ and var(zt) = 1.5 for τ < t ≤ T, where τ is a time period within our interval of interest. Explain whether in this case the innovation vt is homoskedastic or heteroskedastic.
d. Write the expression for E(x2tvt). Under which conditions is E(x2tvt) = 0? Explain your answer.
3. In this question we will look into the intuition behind the GLS estimator. We are interested in the linear regression model:
yt = β 1 + β2 x2t+ β3 x3t + ut (7)
where t = 1, ...,T . We know that E(ut) = 0, cov(ut,ut−j) = 0 for all j 0, E(xtut) = 0, where
a. Divide all variables in (7) by ^rt and rewrite the model as:
qt = β 1 z1t+ β2 z2t+ β3 z3t + vt (8)
Deine all the variables in (8) in term of the variables in (7) and rt . Show your work.
b. Show that E(ztvt) = 0, where zt = [z1t z2t z3t]\ .
c. Show that the innovation vt in (8) is homoskedastic.
4. We are interested in estimating the following linear regression model using OLS:
yt = β 1 + β2 x2t + ut (9)
where t = 1, ...,T . In addition, we want to study whether the innovation ut could be heteroskedastic in our model and sample period of interest.
a. Our friend Mario suggests the following strategy. First, estimate (9) using homoskedasticity-only standard errors, and compute the residual sum of squares from this regression, let’s call it RSS .
Then, estimate (9) using heteroskedasticity-robust standard errors, and compute the residual sum
∼ ∼
of squares from this regression, let’s call it RSS . According to Mario, inding that RSS ≈ RSS would be evidence in favor of ut being homoskedastic. Do you agree with Mario? Why or why not?
b. Assume that we decide to run the Goldfeld-Quandt (GQ) test to examine the possibility that the var(ut) is not constant over our period of interest. We split our sample T into two sub-samples, T1 and T2, with T1 = 122 and T2 = 62. Let a 1(2) denote the variance of ut in T1 and a2(2) the variance of ut in T2 .
i. Write the statement of the test. What is the null hypothesis that we are testing?
ii. We estimate (9) by OLS using data from the irst sub-sample (T1) and compute the residual sum of squares from this regression, RSS1 = 245. Then we repeat the same procedure using data from the second sub-sample (T2) and compute the residual sum of squares from this regression, RSS2 = 117. Use this information to compute s1(2) and s2(2), which are the standard estimators of a 1(2) and a2(2) . Then use these values to compute the GQ-statistic. Show your work.
iii. The F(120, 60) distribution tells us that the critical value corresponding to a 5% signiicance level is 1.6373. Given this information, can we reject our H0 at the 5% signiicance level? Explain your answer.
c. Suppose that I now tell you that I have actually generated the data used for this exercise.
In particular, I have used the following distribution to obtain the time series for the innovations:
ut N(0,at(2)), where at(2) = 1.5 for t = 1, 3, 5, ..., (T − 1) and at(2) = 2.5 for t = 2, 4, 6, ...,T . This
consistent with the results of the GQ test that you run in part b. of the question? Explain your
answer.
5. We want to estimate the following linear regression model:
st = β 1 + β2 yt+ β3 st− 1 + β4yt− 1 + ut (10)
where st is a stock market index and yt is the growth rate of real GDP. All the assumptions of the CLRM hold in (10).
a. Consider the alternative model:
st = V1 + V2yt + V3st− 1 + ut (11)
Show that this model is equivalent to (10) but with an additional restriction on the parameters β2 and β4 . Explain how you would test this restriction using the t-statistic approach. Explain how you would test this restriction using an F-statistic and the approach of the “restricted vs. unrestricted model” .
b. We realize that expected real GDP growth, in addition to actual real GDP growth, could also afect our stock market index of interest. For this reason, we think it would be better to estimate
the following model:
st = β 1 + β2 yt+ β3 st− 1 + β4 yt− 1 + β5 yt(e)+1|t + ut
(12)
where yt(e)+1|t is the time-t expected growth rate for time t + 1. Unfortunately, we do not have data about this variable.
i. Our friend Mariam argues that we should just drop the variable yt(e)+1|t and estimate (10) instead. Explain why if yt(e)+1|t does have an impact on st (i.e. β5 0) this would not be a good idea.
ii. Our other friend Kent suggest approximating yt(e)+1|t using the data that we do have. More speciically, he says that we could compute yt(e)+1|t = 0.5yt + 0.5yt− 1 and use these data, in addition to the actual data for the other regressors, to estimate (12). Do you think that Kent’s idea could work? Explain your answer.
iii. Our other friend Jin agrees about Kent’s approach, but proposes to approximate yt(e)+1|t as: yt(e)+1|t = (yt+ yt− 1 + yt−2 )/3. Do you think that Jin’s idea could work? Explain your answer.
Answers
1. a. From (4), we have that X\ = [x1 x2 ... xT ], which is a k × T matrix. Thus we can write:
T
(X\X) = x1x + x2 x + ... + xT x = 工xtx
t=1
In this expression, each xtx is a product between two vectors and gives a k × k matrix. Thus, (X\X) can be written as the sum of T matrices of dimension k × k .
Similarly, we have:
T
(X\ y) = x1y1 + x2 y2 + ... + xT yT = 工ytxt
t=1
where each xtyt is a product between a vector and a scalar and gives a k × 1 vector. Thus, (X\ y) can be written as the sum of T vectors of dimension k × 1.
Putting the results together:
= (X\X)− 1 (X\ y) = ( xtx)− 1 ( xt yt)
b. We irst substitute the expression for t to get:
T T T T
工xtt = 工xt(yt − x) =工xtyt −工xtx
Now substituting the expression for obtained in part a. we have:
xtt = xtyt − xt x ( xtx)− 1 ( xtyt) = xtyt − xtyt = 0
T
c. We found in part b. that 对 xtt = 0, where 0 is a (k × 1) vector of zeros. As long as (1) includes t=1
T
a constant, then this also implies that 对 t = 0. In fact, we can write:
t=1
l对(T) 1 · t」
T x2tt l 」0(0)
工xtt = t=1 =
t=1 T 「0
「xkt t
T T
so 对 t is just the irst element of the vector given by 对 xtt .
t=1 t=1
d. No, Ellen is not correct. We have just shown that if the parameters of the model are estimated T
using the OLS formulas, then it is always the case that 对 t = 0, regardless of whether E(ut) = 0 in
t=1
the underlying true model. With respect to var(ut), the formula for s2 will always give a constant number, regardless of whether var(ut) is actually constant in the underlying true model.
2. a. E(vt) = E(αzt + ut) = αE(zt). Since α 0, this expression will be zero only if E(zt) = 0. If E(zt) 0 but all the other assumptions of the CLRM hold in (6), then our OLS estimate of β1 will be biased (we would obtain an estimate of β 1 + αE(zt) rather than just β 1 ) but our estimate of β2 will not be afected.
b. var(vt) = var(αzt + ut) = α2 var(zt) + a2 + 2αcov(zt,ut) = α2 var(zt) + a2 It follows that var(vt) is inite and constant if var(zt) is inite and constant.
c. In this case var(vt) = 4α2 + a2 for 1 ≤ t ≤ τ and var(vt) = 1.5α2 + a2 for τ < t ≤ T .
Since var(vt) is not constant across our period of interest, the innovation vt is heteroskedastic.
d. E(x2tvt) = E(x2t(αzt + ut)) = αE(x2tzt) + E(x2tut) = αE(x2tzt).
Thus E(x2tvt) = 0 if E(x2tzt) = 0.
3. a. yt^rt = β 1 1^rt + β2 x2t^rt + β3 x3t^rt + ut^rt
so: qt = yt^rt ; z1t = 1^rt ; z2t = x2t^rt ; z3t = x3t^rt ; vt = ut^rt .
b. E(ztvt) = [E(z1tvt) E(z2tvt) E(z3tvt)]′ . We have: E(z1tvt) = E(1^rt ut^rt) = 1rtE(ut) = 0 E(z2tvt) = E(x2t^rt ut^rt) = 1rtE(x2tut) = 0 E(z3tvt) = E(x3t^rt ut^rt) = 1rtE(x3tut) = 0
c. var(vt) = var(ut^rt) = 1rtvar(ut) = 1rta2 rt = a2
4. a. No, Mario’s strategy will not work. “Homoskedasticity-only” and “heteroskedasticity-robust” are
alternative ways of estimating the covariance matrix V, and therefore the standard errors for the
elements of . These methods do not afect the estimated that we obtain from OLS. This implies
that t = yt − 1 − 2 x2t is the same if we use “homoskedasticity-only” or “heteroskedasticity-robust”
∼
ind that RSS = RSS, even if ut is heteroskedastic.
b.i. H0: a 1(2) = a2(2) vs. H1: a 1(2) a2(2)
We test the null hypothesis that the variance of ut is the same in T1 and T2 .
ii. s1(2) = = 245120 = 2.04 s2(2) = = 11760 = 1.95 GQ-statistic= 2s1s22 = 1.046
iii. 1.046 < 1.6373, so we fail to reject the H0 at the 5% signiicance level. This means that we do not have strong enough evidence against the null hypothesis that a 1(2) = a2(2) .
c. In the GQ test, we failed to reject the H0 that the variance of ut is the same in the two sub- samples T1 and T2 in which we split our sample. As we said, this means that we do not have strong enough evidence against this H0 . But this test only focuses on diferences in var(ut) between two sub-samples, so it will not necessarily catch other forms of heteroskedasticity that may be present in our data, as for instance at(2) = 1.5 for t = 1, 3, 5, ..., (T − 1) and at(2) = 2.5 for t = 2, 4, 6, ...,T .
It is also worth mentioning that failing to reject the H0 in a GQ test does not mean that var(ut) is constant for the following additional reasons.
1. The GQ test might fail to catch discrete, one-time changes in var(ut) if the two sub-samples T1 and T2 that we choose are not correctly relecting this change (i.e. the point where we split the sample does not correspond to the point where the change in var(ut) actually happened).
2. In all tests there is always the possibility of a type II error, i.e. failing to reject the null when the alternative hypothesis is actually correct.
5. a. st − st− 1 = V1 + V2(yt − yt− 1 ) + V3st− 1 + ut
st = V1 + V2yt+ (1 + V3)st− 1 − V2yt− 1 + ut
The restriction is β4 = −β2 , or β4 + β2 = 0. We can test this restriction using a t-statistic. To
(β华4+ β华2), which we can obtain from our estimated .
b.i. If we drop yt(e)+1|t from the regression, the impact of this variable on st, which is measured by βyt(e)+1|t , will end up being captured by the innovation term. We can write this innovation as vt = βyt(e)+1|t + ut . Now, if yt(e)+1|t is correlated with one or more of the other regressors in (12),
then the orthogonality assumption E(xtvt) will not hold (here, xt = [1 yt st− 1 yt− 1]\ ). As a
ii. No, this idea will not work. If we compute yt(e)+1|t = 0.5yt + 0.5yt− 1 , then this variable will be a
linear function of two other regressors in the model, yt and yt− 1 . This implies that the matrix X of
to compute the inverse of this matrix, which we need to obtain .
be large. But we will still be able to obtain (unlike in part b.ii.) and our OLS estimator will still
2022-08-05