Why a random walk with drift is nonstationary

Question

Consider the random walk with drift Xt = \delta + X{t-1} + \varepsilont, started at X0 = 0, where \varepsilont is white noise with mean 0 and variance \sigma^2. Compute \mathbb{E}[Xt] and \operatorname{Var}(Xt) as functions of t, argue the process is nonstationary, and find a transform that makes it stationary.

Accepted Answer

Unrolling the recursion from X0 = 0: Xt = \delta t + \sum{s=1}^{t} \varepsilons. Mean: \mathbb{E}[Xt] = \delta t, which grows linearly with t, so the mean is not constant. Variance: since the innovations are independent, \operatorname{Var}(Xt) = \sum{s=1}^t \operatorname{Var}(\varepsilons) = t\sigma^2, which grows without bound. Autocovariance: \operatorname{Cov}(Xt, X{t+k}) = \operatorname{Var}(Xt) = t\sigma^2, depending on t rather than only on the lag k. All three violate weak stationarity: the mean drifts, the variance explodes, and the autocovariance is time-dependent. Shocks are permane…

Why a random walk with drift is nonstationary

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