Is a random walk weakly stationary?

Question

Define Xt = X{t-1} + \varepsilont with X0 = 0, where \varepsilont is white noise with mean 0 and variance \sigma^2. Is \{Xt\} weakly stationary? Answer yes or no, and justify.

Accepted Answer

Unrolling the recursion, Xt = \varepsilon1 + \varepsilon2 + \dots + \varepsilont, a running sum of the shocks. Mean. E[Xt] = 0 for all t. Fine so far. Variance. Because the shocks are uncorrelated, \operatorname{Var}(Xt) = \operatorname{Var}(\varepsilon1) + \dots + \operatorname{Var}(\varepsilont) = t\,\sigma^2. This grows without bound as t increases: the variance at t=100 is a hundred times the variance at t=1. Condition 2 (constant, finite variance) fails. So the random walk is not weakly stationary. Its spread fans out over time, which is why random-walk paths wander ever farther from the…

Is a random walk weakly stationary?

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