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?
