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Consider vector $y_t$: $$y_t = \mu_0 + \delta d_t + x_t,$$ where $x_t$ is a zero mean stationary process and $d_t = I(t \geq T_B)$ with $1 < T_B < T$ and $I(\cdot)$ being an indicator function. I have to derive mean of $y_t$ and argue whether it is weakly stationary or not.

Since mean of $x_t$ is zero, we have that $$E(y_t) = \mu_0 + \delta E(d_t) = \mu_0 + \delta\frac{T - T_B}{T},$$ is it right? From the above formula I concider $y_t$ as a stationary process since it's mean (and variance) does not depend on $t$ but only on $T_B$ and $T$ which are fixed values.

On the other hand, for $t < T_B$ we have that $$E(y_t) = \mu_0$$ and for $t \geq T_B$ $$E(y_t) = \mu_0 + \delta\frac{T - T_B}{T}.$$

That is, mean of the process changes with $t$ which implies non stationarity. Which argumetation is correct?

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I do not think you are interpreting things correctly. If $t$ is greater than $T_B$, $\delta$ is added to the mean. So the mean is $\mu_0$ for $t < T_B$ and $\mu_0 + \delta$ for $t \ge T_B$: clearly time-dependent.

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  • $\begingroup$ Thanks for your answer. You are right, for $t \geq T_B$, $E(y_t) = \mu_0 + \delta$. However, my question is not only whether $y_t$ is stationary or not, but why different ways of figuring it out give different results. That is, form the second part of the question it is clear that mean is time variant. But from the first part, $E(y_t) = \mu_0 + \delta\frac{T - T_B}{T}$ which is time invariant. So, I would like to know what is wrong with the first part. $\endgroup$
    – tosik
    Jun 6, 2017 at 16:09
  • $\begingroup$ Whe $t > T_B$ the indicator function is 1, not $(T-T_B)/T$. So before $T_B$ you have a mean of $\mu_0$, after a mean of $\mu_= + \delta$. $\endgroup$
    – F. Tusell
    Jun 6, 2017 at 16:58

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