Random Walk Model

Question 1

(a)

Using Eviews to generate y=log(dix). The graph of y is showed below.

(b)

The unit root test is showed below. Specifically, “Intercept” and “Modified Akaike” are chosen to do the test.

For Y, ADF= -2.22 and the p-value=0.1992 > α=0.05, so we fail to reject the null hypothesis which is Y has a unit roo. Therefore, Y is non-stationary as well as has a unit root.

(c)

The first difference of y is generated using dy=d(y). The graph of dy is showed below.

(d)

The unit root test is showed below. Specifically, “Intercept” and “Modified Akaike” are chosen to do the test.

Since ADF= -47.9971 and the p-value=0.0001< α=0.05, so we reject the null hypothesis which is dy has a unit root. So dy dose not have a unit root.

(e)

The correlogram of dy is showed below.

There are some significant individual autocorrelations as the bars are outside the dotted line. For example, lag-1, lag-2, lag-6, etc.

(f)

The hypotheses of the joint test are:

H0: ρ1=ρ2=ρ3=⋯=ρ10=0

H1: At least on of ρ1⋯ρ10≠0

For lag-10, it can be seen from (e) that Q-statistic=66.509 and p-value=0 < 0.05, so we reject H0. To conclude, there is at least one lag of acf does not equal 0.

(g)

By estimating all the 10 AR models, the values of AIC and SC are showed below.

Therefore, if we adopt SC criterion, MA(2) model is preferred as it has the smallest SC value.

(h)

Using Eviews to estimate the MA(2) model in part (g).

So θ0=-0.00000956, θ1=-0.136581, θ2=-0.106777. So the fitted model is

ΔYt=-0.00000956+et-0.136581et-1-0.106777et-2

Then the correlogram of residuals of this MA(2) model is showed below.

There are some significant autocorrelations in the residuals as some bars are outside the dotted line, especially at lag-6 and lag-10.

(i)

Using Eviews to estimate the MA(10) model

Then the correlogram of residuals of this MA(10) model is showed below.

As can be seen there is no...