If, in the random telegraph signal process, we drop the requirement that P{X0 = 1} = P{X0 = −1} = 12 and only require that E[X0] = 0, then the process {X(t), t 0} need no longer be stationary. (It will remain stationary if X0 has a symmetric distribution in the sense that −X0 has the same distribution as X0.) However, the process will be weakly stationary since

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E[X(t)] = E[X0]E[(1)N()] = 0, Cov[X(t), X(ts)] = E[X(t)X(t + s)] =E[X]E[(-1))+N(t+s)] =E[X]e-2s from (10.19)