1 ECMT2130 – 2021 semester 1 final exam solutions
1. (15 points) Matilde’s ARIMA EMH testing
Matilde has a 5 year history of daily data on the total return index for an asset denoted, vt, in period t.
She first wants to confirm her view that the natural log of the raw data is likely to have at least one unit root. She tests for at least one unit root in the natural log of the total return index using an Augmented Dickey Fuller test, allowing for both drift and a deterministic trend. Her test results are reported below, based on estimation of the regression:
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wherext =100 ×ln(vt).
∆xt =α+ βxt−1 +γt+δ∆xt−1 +et Standard error
2.98e-02 2.17e-03 9.68e-05 2.84e-02
Coefficient
Point estimate 1.19e-03 -2.09e-03 1.42e-04 -1.53e-02
(a) (1 point) What feature of the total value index justified using the natural log of the total value index in the Augmented Dickey Fuller test?
(b) (5 points) Write up the formal Augmented Dickey-Fuller test, allowing for both drift and a deter- ministic trend. Perform the test at the 5% level of significance, using the corresponding critical value, -3.41. Make sure you include all steps from the statement of the null and alternative hy- potheses to the final conclusion.
She then uses her data to test the weak form of the Efficient Markets Hypothesis.
She does so by computing the the daily rate of return on the asset, expressed as a percentage:
rt =100 ×(ln(vt)−ln(vt−1))
She estimates two models:
• an ARMA(1,1) model, rt = μ1 + φrt−1 + u1t + θu1t−1 with i.i.d. error term u1t ∼ N 0, σ12
• an ARMA(0,0) modelmodel (constant mean plus white noise), rt = μ2 + u2t with i.i.d. error
term u2t ∼ N 0, σ2
(c) (1 point) Was she using continuously-compounded rates of return or simple rates of return in the two models?
Her estimation results are shown in the tables below. ARMA(1,1) model estimates
μ1 φ θ σ12
Point estimate
-0.35 0.50 0.33 0.51 0.25
Standard error
ARMA(0,0) model estimates
Point estimate
0.05 0.01 0.25
Standard error
Maximum log-likelihood function values ARMA(1,1) maximum log-likelihood value = -897.29 ARMA(0,0) maximum log-likelihood value = -897.64
(d) (2 points) Briefly explain the difference between the strong and weak forms of the efficient markets hypothesis. [2 points]
(e) (5 points) Use her results to perform a Likelihood Ratio test of the joint exclusion restrictions on the lagged dependent variable and the lagged error in the ARMA(1,1) model. Perform the test at the 5% level of significance, being clear about the distribution that the decision rule is based on.
(f) (1 point) Briefly explain the implications of her research for the weak form of the efficient markets hypothesis.
(a) Assuming that expected returns are not trending, the growth rate of the total return index should be stable over time. That suggests that the total return index would exhibit exponential growth over the long-run. Taking the natural log of the total return index, would then transform the total return index data to a series with a linear trend, and more importantly, to a series where the changes from period to period can be expected to have the same variance. This makes the natural log of the total return index more suitable for use in testing for one or more unit roots.
(b) ADF test:
1. Test the hypothesis: H0: β = 0 against H1: β < 0 at the 5% level of significance.
2. The test statistic, DF, is computed as the t-ratio for β: DF = βˆ/SE(βˆ) = (−2.09e− 03)/(2.17e − 03) = −0.97.
3. Under the null hypothesis, and given that the sample is large, the test statistic is asymp- totically distributed according to the Augmented Dickey-Fuller distribution.
4. The test is a one-sided lower-tail test. The critical value is −3.41. The decision rule is reject the null hypothesis if the test statistic lies below -3.41. Otherwise, fail to reject the null hypothesis.
5. The computed test statistic lies outside of the rejection region. Thus, we fail to reject the null hypothesis at the 5% level of significance. There is insufficient evidence to warrant concluding that there are no unit roots in the natural log of the total return index.
(c) EMH versions:
The strong form of the EMH states that everything that is knowable - even unpublished infor- mation - has already been reflected in present prices. Even if you have some inside information and could legally trade based upon it, you would gain nothing by doing so.
The weak form of the EMH states that today’s prices reflect all the information in past prices and no form of technical analysis of past prices can be effectively utilized to aid investors in making trading decisions.
The weak form is harder to reject empirically because it limits itself with regard to the types of data that should not be useful to making trading decisions.
(d) She was using continuously-compounded rates of return in the two model because using the continuously compounding interest rate, the relationship between the total value at time T and total value at time 0 is given by:
Taking natural logs: Thus:
and setting T = 1:
(e) Likelihood ratio test:
VT =V0erT ln(VT ) = ln(V0) + rT
r = ln(VT ) − ln(V0)/T
r = ln(VT ) − ln(V0)
1. Testthehypothesis: H0: φ=θ=0againstH1: φ̸=0and/orθ̸=0atthe5%levelof significance.
2. The test statistic, LR, is computed as -2 times the difference between the restricted and unrestricted maximised log likelihood function values.
3. Under the null hypothesis, and assuming that the shocks are normally distributed, the sample is large enough to use the χ2 distribution (with 2 degrees of freedom because we are testing 2 restrictions jointly) for the LR test statistic.
4. The test is a one-sided upper-tail test. The critical value is χ2(0.95,2) = 5.9915. The decision rule is reject the null hypothesis if the test statistic lies above 5.9915. Otherwise, fail to reject the null hypothesis.
5. The LR test statistic is computed as LR = −2(Lr −Lu) = −2(−897.64−−897.29) = 0.70.
6. The computed test statistic lies outside of the rejection region (0.7 < 5.9915). Thus, we fail to reject the null hypothesis at the 5% level of significance. There is insufficient evidence to warrant including the first lag of the dependent variable and the first lag of the shock in the ARMA model.
(f) She has found no evidence to support ARMA structure that would enable the use of past price information to improve forecasts of future returns. Thus, she has found no evidence to support a violation of the weak form of the efficient markets hypothesis.
2. (15 points) APM testing
Aaron wants to estimate the CAPM Security Market Line (SML). He has monthly data for 20 years on:
• the average simple monthly returns for 500 randomly selected assets, rit;
• the average simple monthly rate of return on a short-term government bond, rft; and • the average simple monthly rate of return on a proxy of the market portfolio, rmt
The SML model specification is:
r ̄ i = γ 0 + γ 1 β i + e i
Aaron’s supervisor suggests that he make it possible to estimate the SML regression by first running a time-series regression for each asset to get estimates of βi for each asset. These estimates could then be used as proxies for the true value of βi.
Aaron follows this advice and runs the time-series regression for each asset, using all 20 years of data. This produces a single estimate of βi for each asset, βˆi.
Aaron also averages the simple monthly returns for each asset over the full 20 years to obtain the dependent variable in the SML model specification.
(a) (3 points) Explain the consequences for Aaron’s OLS estimates of the SML model coefficients if the measurement error in βˆi is positively correlated with βˆi.
(b) (3 points) Outline and justify the steps in a portfolio formation strategy that Aaron could use to mitigate the problem identified in part A.
Assume that Aaron has managed to mitigate the problem identified in part C and has used OLS to estimate the coefficients of the following augmented version of the SML regression model:
r ̄ i = γ 0 + γ 1 βˆ i + γ 2 βˆ i2 + e i
as well as the coefficients on the standard SML regression model:
r ̄ i = γ 0 + γ 1 β i + e i
The cross-section regressions used 15 observations. The OLS results are reported in the table below.
Coefficient
Augmented model Point estimate 0.18
Augmented model Standard error
Standard SML model Point estimate
Standard SML model Standard error 0.05
0.12 0.33 0.15 0.49 0.04
Over the sample period, the average risk-free monthly simple rate of return was 0.26%.
Over the sample period, the average monthly simple rate of return on the proxy for the market portfolio was 0.77%.
(c) (5 points) Report the results, at the 5% level of significance, of an exclusion restriction test for the βˆi2 regressor in the augmented regression, being clear about your rationale for the distribution of the test statistic.
(d) (4 points) Discuss the consistency of the reported estimates with the implications of the risk-free asset CAPM.
(a) If the measurement error in βˆi is correlated with βˆi then the cross-section regressions estimating the SML will be affected by the problems associated with classic errors in variables problem. Specifically, the coefficient estimates will be biased and inconsistent.
This is because, the measurement error will become part of the regression error term and that part of the error term will be correlated with a regressor, thus violating the zero conditional mean assumption underpinning the usage of OLS estimation.
(b) To address the classic errors in variables problem, Aaron could use returns on portfolios of assets to reduce the size of the measurement error in the βˆi. He would then be working with portfolios of assets to estimate his SML regression.
To work with portfolios, Aaron would need to split the sample, using the earlier period to estimate a βˆi for each asset that could be used to rank assets so that portfolios could be formed with significant variation in the regressorβˆ.
This would lead to selection bias issues aggregating large positive measurement errors in the high β portfolios and large negative measurement errors in the low β portfolios. This selection bias could be mitigated by using the portfolios formed using data for the first part of the available time to then estimate portfolio β’s using the subsequent returns data.
(c) Exclusion test:
1. Test the hypothesis: H0: γ2 = 0 against H1: γ2 ̸= 0 at the 5% level of significance.
2. The test statistic, t, is computed as the t-ratio for γ : t∗ = (γˆ −0)/SE(γˆ ) = −0.06/0.04 = 222
3. Under the null hypothesis, and given that the sample is small, the test statistic has a Student’s t distribution if the errors in the equation are normally distributed. This dis- tribution of the errors can be justified on various grounds by the student (e.g. the errors in the portfolio regression are aggregations of errors for assets in the portfolios and so a central limit theorem could be invoked to argue that the aggregate errors are normally distributed).
With 15 observations and 2 regressors, the test statistic then has a Student’s t distribution with 15 − 3 = 12 degrees of freedom.
4. The test is a two-sided test. At the 5% level of significance, with 12 degrees of freedom for a Student’s t distribution, the upper critical value is 2.179 and the lower critical value is -2.179. The decision rule is reject the null hypothesis if the test statistic lies below -2.179 or above 2.179. Otherwise, fail to reject the null hypothesis. More succinctly, reject the null hypothesis at the 5% level of significance iff |t∗| > 2.179.
5. The computed test statistic does not lie in the rejection region. Thus, we fail to reject the null hypothesis at the 5% level of significance. There is insufficient evidence to warrant concluding that β2 has a role to play in explaining variation in asset returns.
(d) The lack of evidence to support inclusion of β2 in the SML regression is consistent with the CAPM which implies that only sensitivity to excess returns on the market portfolio should have explanatory power.
The risk-free asset version of the CAPM also implies that the intercept of the model should approximately equal the risk-free rate of return. The estimated intercept in the standard model is not significantly different from the average risk-free rate of return over the sample period, at least at the 5% level of significance. So the intercept is also consistent with the version of the CAPM based on access to a risk-free asset.
The risk-free asset version of the CAPM also implies that the slope coefficient on the βˆ in the SML regression should approximately equal the excess return on the market portfolio. The estimated slope coefficient in the standard model is not significantly lower than the average excess return on the market over the sample period at the 5% of significance. This is also consistent with the version of the CAPM based on access to a risk-free asset.
It would be acceptable to use the base more or the augmented model for these tests relating to the slope coefficient on the beta and to the intercept.
Using the base model, and justifying that choice because the coefficients are more precisely estimated and linearity failed to be rejected in the previous question parts, our t-test statistic
for the intercept is:
t∗0 = 0.33 − 0.26 = 1.4 0.05
Likewise, our t-test statistic for the slope coefficient on the beta estimate regressor is:
t∗0 = 0.49−(0.77−0.26) =−2 =−0.67 0.03 3
3. (15 points) Bilbo’s GARCH model
Bilbo models the daily continuously-compounded rate of return on an asset, rt, as:
• ut = σtet
• the shocks, et, are i.i.d. et ∼ N (0, 1)
• intheunrestrictedmodel,σ2 =α +α u2
• andintherestrictedmodel,σ2 =γ +γ u2
Estimates for the unrestricted model and the restricted model, using 530 observations, are shown in the following tables.
Unrestricted model: Coefficient Point estimate
α0 0.01 α1 0.05 α2 0.12 α3 0.07 β1 0.64
Standard error 0.003
The maximised log-likelihood value for the unrestricted model is 85.0.
Restricted model:
Coefficient
Point estimate 0.01
Standard error 0.003
The maximised log-likelihood value for the restricted model is 80.0.
(a) (2 points) What restrictions are required to simplify the unrestricted model to the restricted model?
(b) (7 points) Perform a likelihood ratio test of the restrictions needed to simplify the GARCH(1,3) model to a GARCH(1,1) model. Write up the hypothesis test in full using a 5% level of significance.
(c) (2 points) What typical characteristic of high-frequency financial returns is the variance equation in the GARCH model intended to capture?
(d) (2 points) What is the estimated unconditional variance of the the error term, ut, in the GARCH(1,1) model? Show your working.
(e) (2 points) What is the estimated unconditional variance of the the error term, ut, in the GARCH(1,3) model? Show your working.
(a) The required restrictions are α2 = α3 = 0.
(b) Likelihood ratio test:
1. Test the hypothesis: H0: α2 = α3 = 0 against H1: α2 ̸= 0 and/or α3 ̸= 0 at the 5% level of significance.
2. The test statistic, LR, is computed as -2 times the difference between the restricted and unrestricted maximised log likelihood function values.
3. Under the null hypothesis, and assuming that the shocks are normally distributed, the sample is large enough to use the χ2 distribution (with 2 degrees of freedom because we are testing 2 restrictions jointly) for the LR test statistic.
4. The test is a one-sided upper-tail test. The critical value is χ2(0.95,2) = 5.9915. The decision rule is: reject the null hypothesis if the test statistic lies above 5.9915. Otherwise, fail to reject the null hypothesis.
5. The LR test statistic is computed as LR = −2(Lr − Lu) = −2(80.0 − 85.0) = 10.0.
6. The computed test statistic lies in the rejection region (10 > 5.9915). Thus, we reject the null hypothesis at the 5% level of significance. There is sufficient evidence to warrant including the second and third lags of the squared residuals in the equation describing the evolution of return variance.
(c) The variance equation in the GARCH model is intended to capture clustered volatility, a typical feature of high-frequency financial returns.
(d) The estimated unconditional variance of the the error term, ut, in the GARCH(1,1) model is: E(u2t ) = γˆ0 = 0.01 = 0.067
1−(γˆ1 +δˆ1) 1−(0.33+0.52)
(e) The estimated unconditional variance of the the error term, ut, in the GARCH(1,3) model is:
E(u2t ) = αˆ0 = 0.01 = 0.083 1−(αˆ1 +αˆ2 +αˆ3 +βˆ1) 1−(0.05+0.12+0.07+0.64)
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