ECMT6006/ECON4949 Mid-Semester Test
Suggested Solutions
9 April 2020
Note: This is an open-book exam. Please answer all questions. The total mark
is 20 and the breakdown is shown in square brackets. The duration of the exam
is 1 hour and 10 minutes.
Problems 1 (v) and (vi) have separate sub-questions for ECMT6006 and
ECON4949 students respectively, please answer the corresponding questions
based on your course code.
Problem 1. [13 marks] Consider a simple three time period model for returns
Rt, t = 1, 2, 3 of an asset. Let R0 = 10%, and the returns in later periods follow
an AR(1) process
Rt = φRt−1 + εt
where φ = 0.2 and ε1, ε2, ε3 are i.i.d. shocks with probability distribution
εt =
{
6%, with probability 0.4
−4%, with probability 0.6
for t = 1, 2, 3. Let Ft be the information set available at time t. Please answer
the following questions.
(i) What is the probability distribution of R1? [1]
Solution. Given R0 = 10% and φ = 0.2, it follows that
R1 = 0.2× 10% + ε1 = 2% + ε1.
Therefore, the probability distribution of R1 is
R1 =
{
8%, with probability 0.4
−2%, with probability 0.6
or, equivalently,
P (R1 = 8%) = 0.4 and P (R1 = −2%) = 0.6.
(ii) What is the probability distribution of R2? [1]
1
Solution. Note that
R2 = 0.2×R1 + ε2
=
0.2× 8% + 6% = 7.6% if ε1 = 6%, ε2 = 6%
0.2× 8%− 4% = −2.4% if ε1 = 6%, ε2 = −4%
0.2× (−2%) + 6 = 5.6% if ε1 = −4%, ε2 = 6%
0.2× (−2%)− 4% = −4.4% if ε1 = −4%, ε2 = −4%
Since ε1 and ε2 are independent, we have
P (ε1 = 6%, ε2 = 6%) = 0.4× 0.4 = 0.16,
P (ε1 = 6%, ε2 = −4%) = 0.4× 0.6 = 0.24,
P (ε1 = −4%, ε2 = 6%) = 0.6× 0.4 = 0.24,
P (ε1 = −4%, ε2 = −4%) = 0.6× 0.6 = 0.36.
Therefore, the probability distribution of R2 is
P (R2 = 7.6%) = 0.16,
P (R2 = −2.4%) = 0.24,
P (R2 = 5.6%) = 0.24,
P (R2 = −4.4%) = 0.36.
(iii) Compute E(R2),Var(R2),Skew(R2) and Kurt(R2), the unconditional mean,
variance, skewness and kurtosis of R2. [2]
Solution. We have the mean and variance as
E(R2) = 7.6%× 0.16− 2.4%× 0.24 + 5.6%× 0.24− 4.4%× 0.36
= 0.4%, [0.5]
Var(R2) = (7.6%− 0.4%)2 × 0.16 + (−2.4%− 0.4%)2 × 0.24
+ (5.6%− 0.4%)2 × 0.24 + (−4.4%− 0.4%)2 × 0.36
= 0.002496. [0.5]
The standard deviation is
Std(R2) =
√
Var(R2) = 5%.
So the skewness and kertosis are
Skew(R2) =
[
(7.6%− 0.4%)3 × 0.16 + (−2.4%− 0.4%)3 × 0.24
+(5.6%− 0.4%)3 × 0.24 + (−4.4%− 0.4%)3 × 0.36] /(5%)3
= 0.387. [0.5]
Kurt(R2) =
[
(7.6%− 0.4%)4 × 0.16 + (−2.4%− 0.4%)4 × 0.24
+(5.6%− 0.4%)4 × 0.24 + (−4.4%− 0.4%)4 × 0.36] /(5%)4
= 1.298. [0.5]
You will get the marks if the formulae are correct.
2
(iv) Explain what mean, variane, skewness and kurtosis tell about the distri-
butional properties of a random variable. [2]
Solution. The mean of the random variable tells the average value that
the random variable takes. [0.5] The variance tells the dispersion of the
distribution, and it measures how far the values of the random variable
can take are spread out from the average value. [0.5] The skewness of a
distribution tells whether the distribution of symmetric or not. The dis-
tribution with positive skewness has long right tail and that with negative
sknewness has long left tail. [0.5] The kurtosis measures the fatness of
the tails in the distribution. If a distribution has positive excess kurtosis
(kurtosis greater than 3) then the tails are fatter than normal tails. [0.5]
(v) (Answer if your course code is ECMT6006) Compute the conditional mean
E1(R2) := E(R2|F1) and conditional variance Var(R2|F1). [2]
Solution. If R1 = 8%, then by R2 = φR1 +ε2 = 0.2×8%+ε2 = 1.6%+ε2,
we know that R2 = 1.6% + 6% = 7.6% with probability 0.4 and R2 =
1.6%− 4% = −2.4% with probability 0.6. So the probability distribution
of R2 given R1 = 8% is
P (R2 = 7.6%|R1 = 8%) = 0.4,
P (R2 = −2.4%|R1 = 8%) = 0.6.
Similarly, if R1 = −2%, then
P (R2 = 5.6%|R1 = −2%) = 0.4,
P (R2 = −4.4%|R1 = −2%) = 0.6.
So the conditional mean of R2 given F1 is [1]
E(R2|F1) =
{
7.6%× 0.4− 2.4%× 0.6 = 1.6% with probability 0.4 = P (R1 = 8%)
5.6%× 0.4− 4.4%× 0.6 = −0.4% with probability 0.6 = P (R1 = −2%)
The conditional variance is [1]
Var(R2|F1) =
{
(7.6%− 1.6%)2 × 0.4 + (−2.4%− 1.6%)2 × 0.6 with probability 0.4
(5.6%− (−0.4%))2 × 0.4 + (−4.4%− 0.4%)2 × 0.6 with probability 0.6
You get marks if the formulae are correct.
(Answer if your course code is ECON4949) Compute the conditional mean
E1(R3) := E(R3|F1) and conditional variance Var(R3|F1). [2]
Solution. If R1 = 8%, then by R2 = φR1 +ε2 = 0.2×8%+ε2 = 1.6%+ε2,
we know that R2 = 1.6% + 6% = 7.6% with probability 0.4 and R2 =
1.6% − 4% = −2.4% with probability 0.6. Moreover, R3 = φR2 + ε3 =
0.2R2 +ε3. We know that R3 = 0.2×7.6%+6% = 7.52% with probability
0.4 × 0.4 = 0.16, R3 = 0.2 × 7.6% − 4% = −2.48% with probability
3
0.4 × 0.6 = 0.24, R3 = 0.2 × (−2.4%) + 6% = 5.52% with probability
0.6×0.4 = 0.24, and R3 = 0.2× (−2.4%)−4% = −4.48% with probability
0.6× 0.6 = 0.36
So the probability distribution of R2 given R1 = 8% is
P (R3 = 7.52%|R1 = 8%) = 0.16,
P (R3 = −2.48%|R1 = 8%) = 0.24,
P (R3 = 5.52%|R1 = 8%) = 0.24,
P (R3 = −4.48%|R1 = 8%) = 0.36.
Similarly, if R1 = −2%, then
P (R3 = 7.12%|R1 = −2%) = 0.16,
P (R3 = −2.88%|R1 = −2%) = 0.24,
P (R3 = 5.12%|R1 = −2%) = 0.24,
P (R3 = −4.88%|R1 = −2%) = 0.36.
So the conditional mean of R3 given F1 is [1]
E(R3|F1) =
{
7.52× 0.16− 2.48× 0.24 + 5.52× 0.24− 4.48× 0.36 = 0.32(%) with probability 0.4
7.12× 0.16− 2.88× 0.24 + 5.12× 0.24− 4.88× 0.36 = −0.08(%) with probability 0.6.
The conditional variance is [1]
Var(R3|F1) =
(7.52− 0.32)2 × 0.16 + (−2.48− 0.32)2 × 0.24
+(5.52− 0.32)2 × 0.24 + (−4.48− 0.32)2 × 0.36 with probability 0.4
(7.12 + 0.08)2 × 0.16 + (−2.88 + 0.08)2 × 0.24
+(5.12 + 0.08)2 × 0.24 + (−4.88 + 0.08)2 × 0.36 with probability 0.6.
You get marks if the formulae are correct.
(vi) (Answer if your course code is ECMT6006) Verify the law of iterated ex-
pectation E(R2) = E [E1(R2)] using numbers given in this problem. [2]
Solution. From the conditional expectation of E1(R2) = E(R2|F1) derived
in (v), you can see that
E [E1(R2)] = 1.6%× 0.4− 0.4%× 0.6 = 0.4%
This is the same as the unconditional expectation of R2 derived in question
(iii).
(Answer if your course code is ECON4949) Verify the law of iterated ex-
pectation E(R3) = E [E1(R3)] using numbers given in this problem. [2]
Solution. From the conditional expectation of E1(R3) = E(R3|F1) derived
in (v), you can see that
E [E1(R3)] = 0.32%× 0.4− 0.08%× 0.6 = 0.08%
4
Note that from the results in (v) we know the unconditional distribution
of R3 is
P (R3 = 7.52%) = 0.16× 0.4 = 0.064,
P (R3 = −2.48%) = 0.24× 0.4 = 0.096,
P (R3 = 5.52%) = 0.24× 0.4 = 0.096,
P (R3 = −4.48%) = 0.36× 0.4 = 0.144,
P (R3 = 7.12%) = 0.16× 0.6 = 0.096,
P (R3 = −2.88%) = 0.24× 0.6 = 0.144,
P (R3 = 5.12%) = 0.24× 0.6 = 0.144,
P (R3 = −4.88%) = 0.36× 0.6 = 0.216.
from which we can compute the unconditional expectation E(R3) = 0.08%.
You get the marks if the formulae and logic is correct.
(vii) Derive the one-period ahead return point forecast Rˆt for t = 1, 2, 3. [1]
Solution. The one-period ahead returen point forecasts are
Rˆt = Et−1(Rt) = E(φRt−1 + εt|Ft−1)
= E(φRt−1|Ft−1) + E(εt|Ft−1)︸ ︷︷ ︸
=0
= φRt−1 = 0.2Rt−1
for t = 1, 2, 3.
(viii) Derive the two standard deviation one-period ahead return interval fore-
cast for t = 1, 2, 3. [2]
Solution. The two standard deviation one-period ahead return interval
forecasts are
Et−1(Rt)± 2×
√
Vart−1(Rt)
where Vart−1(Rt) = Var(Rt|Ft−1) is the conditional variance of Rt given
Ft−1, and Et−1(Rt) = φRt−1 is derived in (vii). [1.5] Further, we may
derive
Vart−1(Rt) = Vart−1(φRt−1 + εt)
= Vart−1(εt) = Var(εt) = Var(εt)
by the conditional constancy and the fact that εt is independent of Ft−1.
Moreover Var(εt) = (6%)
2 × 0.4 + (−4%)2 × 0.6 = 0.0024. [0.5]
Problem 2. [7 marks] Figure 1 presents the sample autocorrelation plot for
1-minute returns on IBM and General Electric for the last 10 trading days of
2012.
5
(i) Please interpret Figure 1. [1]
Solution. We can see that although the sample ACF of IBM returns looks
more significant than that of General Electric returns from the absolute
values, none of the lags in the sample ACF is statistically significant ac-
cording to the robust 95% confidence interval for both stock returns. This
means there are no significant serial correlation exibited in both 1-minute
stock returns.
(ii) Describe how to conduct a robust test jointly for autocorrelation in returns
up to lag L using a regression-based approach. Be explicit about the
regression model, the null and alternative hypotheses, how to construct
the test statistic, and how to make testing decisions. [3]
Solution. Let (Yt) be the return series. We can test this by running the
following regression [0.5]
Yt = β0 + β1Yt−1 + · · ·+ βLYt−L + et,
and obtain robust standard errors for the parameter estimates. Then we
test the joint hypothesis [1]
H0 : β1 = β2 = · · · = βL = 0 v.s. H1 : βj 6= 0 for some j = 1, . . . , L
through a chi square test.
Since we may write the null hypothesis as
H0 : Rβ = 0
where β = (β0, . . . , βL)
′ and
R =
0 1 0 . . . 0
0 0 1 . . . 0
...
...
0 0 0 . . . 1
is an L× (L+ 1) restriction matrix. The test statistic is given by
W = (Rβˆ − 0)′
(
RVˆ R′
)−1
(Rβˆ − 0).
where βˆ is the OLS estimate of β and Vˆ is the estimated (robust) variance-
covariance matrix of βˆ. [1]
Under the null hypothesis, we have
W →d χ2L
because there are L linear restrictions in the test. We reject the null
hypothesis at 5% significance level (or other specified level) if the test
statistic W is greater than the critical value of the χ2L distribution (or
p-value is less than the significance level). [0.5]
6
(iii) The table below presents the results from such a test for three choices of
L. Interpret these results. [1]
Solution. We see from the table that IBM test statistics are greater than
the GE test statistics which shows that the serial correlation in the IBM
stock returns is relatively more significant than that in the GE stock re-
turns.
Only the joint test for IBM stock returns with L = 20 rejects the null
hypothesis of no serial correlation in the return series. In general, we can
conclude that the serial correlation in both of the 1-minute stock returns
are not very significant.
(iv) Figure 2 presents the sample “cross correlation” plot for 1-minute returns
on IBM and General Electric. It shows Corr
(
RIBMt , R
GE
t+j
)
for j ranging
from −10 to +10. Interpret this figure [1], and propose a regression-based
approach to test jointly for the predictability in IBM returns using (lagged)
General Electric returns. [1]
Solution. The sample cross-correlation plot clearly shows that the IBM
and GE 1-minute stock returns are contemporaneously correlated, and
this correlation is extremely significant. However, none of the nonzero
lags from −10 to +10 is significant in their cross-correlation. This implies
that it would be barely possible to predict the future returns of one stock
using the past returns of the other stock, at least at the mean level. [1]
Let Yt and Xt denote the IBM and GE stock returns, respectively, at time
t. Then we can easily test the predictability in IBM returns using (lagged)
GE returns by running the regression [0.5]
Yt = β0 + β1Xt−1 + · · ·+ βLXt−L + et
and obtain robust standard errors for the parameter estimates. Then we
test the joint hypothesis [0.5]
H0 : β1 = β2 = · · · = βL = 0 v.s. H1 : βj 6= 0 for some j = 1, . . . , L
through a chi square test. Since the null hypothesis imposes L linear
restrictions, the test statistic will follow a chi-squared distribution with
degree of freedom L under the null.
7
Figure 1: Sample autocorrelation for 1-minute returns on IBM and General
Electric.
8
Figure 2: Sample cross-correlation for 1-minute returns on IBM and General
Electric.
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