A Random Walk Model. This is based on Fuller (1976) and Hamilton (1994). Consider the following random walk model yt = yt−1 + ut t = 0, 1, . . . , T where ut ∼ IIN(0, σ2) and yo = 0.

(a) Show that yt can be written as yt = u1 + u2 + .. + ut with E(yt) = 0 and var(yt) = tσ2 so that yt ∼ N(0, tσ2).

(b) Square the random walk equation y2 t = (yt−1 + ut)2 and solve for yt−1ut. Sum this over t = 1, 2, . . ., T and show that

T t=1 yt−1ut = (y2T

/2) −

T t=1 u2t

/2 Divide by Tσ2 and show that

T t=1 yt−1ut/T σ2 is asymptotically distributed as (χ21

− 1)/2.

Hint: Use the fact that yT ∼ N(0, Tσ2).

(c) Using the fact that yt−1 ∼ N(0, (t−1)σ2) show that E

T t=1 y2 t−1



= σ2T (T −1)/2. Hint:

Use the expression for

T t=1 t in problem 6.

(d) Suppose we had estimated an AR(1) model rather than a random walk, i.e., yt = ρyt−1 +ut when the true ρ = 1. The OLS estimate is

ρ =

T t=1 yt−1yt/

T t=1 y2 t−1 = ρ +

T t=1 yt−1ut/

T t=1 y2 t−1 Show that plim T (ρ − ρ) = plim

T t=1 yt−1ut/T σ2

T t=1 y2 t−1/T 2σ2

= 0 Note that the numerator was considered in part (b), while the denominator was considered in part (c). One can see that the asymptotic distribution of ρ when ρ = 1 is a ratio of (χ21 − 1)/2 random variable to a non-standard distribution in the denominator which is beyond the scope of this book, see Hamilton (1994) or Fuller (1976) for further details. The object of this exercise is to show that if ρ = 1, √
T(ρ−ρ) is no longer normal as in the standard stationary least squares regression with |ρ| < 1. Also, to show that for the nonstationary (random walk) model, ρ converges at a faster rate (T ) than for the stationary case (
√
T). From part

(c) it is clear that one has to divide the denominator of ρ by T 2 rather than T to get a convergent distribution.