Homework 1 Solutions

Abstract

1. To check that {X t } is white noise, we need to compute its means and covariances. For the means, EX t = EW t (1 − W t−1)Z t = (EW t)(1 − EW t−1)(EZ t) = 0. For the covariances, γ(s, t) = E If s = t then the last term is EZ s Z t = EZ s · EZ t = 0. Therefore {X t } is uncorrelated. If s = t then EZ s Z t = EZ 2 t = 1 and so γ(t, t) = EW 2 t (1 − W t−1) 2 = 1 4. Thus, {X t } has constant variance. Hence it is white noise. To show that {X t } is not i.i.d, note that X t−1 = 1 implies that W t−1 = 1, which implies that X t = 0. Therefore P (X t−1 = 1, X t = 1) = 0. Since this is not equal to P (X t−1 = 1)P (X t = 1) = 1/64, X t and X t−1 are not independent. 2. (a) X t = W t − W t−3 is a stationary process: EX t = EW t − EW t−3 = 0 and γ(s, t) = EX s X t = EW s W t + EW s W t−3 + EW s−3 W t + EW s−3 W t−3 = ½ {s=t} + ·½ {s=t−3} + ·½ {s−3=t} + ·½ {s−3=t−3} = 2 · ½ {|s−t|=0} + ½ {|s−t|=3} , which is a function of |s − t|. (b) X t = W 3 is a stationary process because EX t = EW 3 = 0 and EX s X t = EW 2 3 = 1. (c) X t = W 3 + t is not a stationary process because its mean is not constant: EX t = t.