Why is a deterministic trend process not stationary? I am confused on why a simple trend process is not stationary. Consider the following process:
$Y_t = a + bt + \epsilon_t$
The variance is clearly constant. However, the mean $bt$ is dependent on $t$. When shifted in time, the mean only depends on the time interval and is independent of history. For example, $Y_{0,t}$ and $Y_{t,2t}$ would have the same mean and variance. So why is this process not stationary?
Secondly, if we now consider the following process:
$Y_t = a + \sqrt{t}\epsilon_t$
Assume $\epsilon_t$ is standard normal.
In this case, the mean is constant, however, the variance is dependent on $t$. However, in this case the variance is proportional to the time interval, which means $Y_{0,t}$ and $Y_{t,2t}$ would have the same mean and variance. So why is this process not stationary?
If you could explain it intuitively rather than definition/proof that would be helpful. My understanding of a stationary process is that the first two moments of the process (mean and variance) remain the same when shifted in time or space. I don't think I have the right understanding of stationary processes.
 A: I think I nice way to get the intuition is to simulate 3 series for $t=0,...,500$ and plot them:


*

*Autoregressive Stationary Series: $A_{t}=0.05+0.95A_{t-1}+u_{t}$

*Random Walk with Drift: $R_{t}=0.05+1R_{t-1}+u_{t}$

*Explosive Series: $E_{t}=0.05+1.05E_{t-1}+u_{t}$


where $u_{t}$ is just some white noise, like iid $N(0,1)$. 
Look at $A$ and $R$:
 
The theoretical mean of $A$ is $1$ (red horizontal line) and its standard deviation is $3.2$. The graph will deviate from that mean over time, but not too far. $R$ will look qualitatively similar to $A$ early on, but begins to drift apart in the middle, but converges towards the end. In theory, the unconditional mean and variance of $R$ do not exist, and you can see that in the graph.
Now plot all 3 series on a graph with the same scale. 

Can you see how $E$ just makes the other two look like a straight line? The slope parameter in $E$ exceeds $1$ by 0.05, the same amount that it falls short for $A$, but what a difference it makes! Here, the average makes no sense at all.
The other point is that $A$ and $R$ look like the sorts of things we see every day, but we are bad at guessing which ones are stationary, especially with fewer data.       
This is shamelessly plagiarized from Econometric Methods by Jack Johnston and John DiNardo, which is sadly out of print.
A: You answer your question yourself: Because stationarity implies both a constant variance, and a constant mean.
If either term is dependent on time, the process is not stationary. In your first example, the mean is dependent on time, and in the second, variance is.
A: A process $Y_t$ is stationary when for any vector of times $(t_1,...,t_n)$ and for every time interval $\tau$ the joint distribution of the vector
$$
(Y_{t_1},...,Y_{t_n})
$$
coincides with the joint distribution of the vector
$$
(Y_{t_1+\tau},...,Y_{t_n+\tau})
$$
In your example, the distribution of the "vector"
$$
(Y_1)
$$
is normal (I'm assuming that the shocks $\varepsilon_t$ are normal in your example) with mean $a+b$ and variance equal to the variance of $\varepsilon_t$. On the other hand the distribution of the "vector" 
$$
(Y_2)=(Y_{1+\tau})
$$ 
where $\tau=1$ is normal with mean $a+2b$ and same variance. Therefore the process cannot be stationary. In the same way you prove that the second example you show is stationary (variance grows). For what we have said above you should see that a stationary process always has constant mean and variance.
I think you are confusing stationary processes, e.g. AR(1), with a process with stationary increments, i.e. random walks.
