Let Xo = 0 and Xt => Z; where Z; independently takes values +1,-1 with i=1 probability 1/2 each. Then (t) = E(Z;) = 0 but o(t) = var(Z;) = t, which is not constant so {X} is neither weakly nor strongly stationary. Exercise: Show that (s, t) = s and p(s, t) = /s/t.