For the simple Deterministic Time Trend Model yt = + t + ut t = 1, .., T where ut IIN(0, 2). (a)

For the simple Deterministic Time Trend Model yt = α + βt + ut t = 1, .., T where ut ∼ IIN(0, σ2).

(a) Show that



αOLS − α

βOLS

− β



= (X



X)X



u =



T

T

 t=1 t T t=1 t

T t=1 t2

−1  T t=1  ut T t=1 tut



where the t-th observation of X, the matrix of regressors, is [1, t].

(b) Use the results that

T t=1 t = T (T + 1)/2 and

T t=1 t2 = T (T + 1)(2T + 1)/6 to show that plim (XX/T) as T →∞ is not a positive definite matrix.

(c) Use the fact that

 √

T(αOLS − α)

T

√

T(βOLS

− β)



= A(X



X)−1AA

−1(X



u) = (A

−1(X



X)A

−1)−1A

−1(X



u)

where A =

 √

T 0 0 T

√

T



is the 2 × 2 nonsingular matrix, to show that plim (A−1(XX)A−1) is the finite positive definite matrix Q =

⎛

⎜⎜⎝

1 1

2 1

2 1

3

⎞

⎟⎟⎠

and A

−1(X



u) =

⎛

⎝

T t=1 ut/

√

T

T t=1 tut/T

√

T

⎞

⎠

(d) Show that z1 =

T t=1 ut/

√

T is N(0, σ2) and z2 =

T t=1 tut/T

√

T is N(0, σ2(T + 1)(2T +

1)/6T 2) with cov(z1, z2) = (T + 1)σ2/2T, so that



z1 z2



∼ N

⎛

⎜⎜⎜⎝

0, σ2

⎛

⎜⎜⎜⎝

1 T + 1 2T T + 1 2T

(T + 1)(2T + 1)

6T 2

⎞

⎟⎟⎟⎠

⎞

⎟⎟⎟⎠

.

Conclude that as T →∞, the asymptotic distribution of



z1 z2



is N(0, σ2Q).

(e) Usin√g the results in parts

(c) and (d), conclude that the asymptotic distribution of T(αOLS − α)

T

√

T(βOLS

− β)



is N(0, σ2Q−1). Since βOLS has the factor T

√

T rather than the usual

√

T, it is said to be superconsistent. This means that not only does (βOLS

− β) converge to zero in probability limits, but so does T (βOLS

− β). Note that the normality assumption is not needed for this result. Using the central limit theorem, all that is needed is that ut is White noise with finite fourth moments, see Sims, Stock and Watson (1990) or Hamilton

(1994).