A first order autoregressive process with lag-1 auto-correlation 0.5 I was reading my professor's notes by myself, it says that:
''A first-order autoregressive process with lag-1 auto-correlation 0.5 can be generated from $\theta^{(t+1)}|\theta^{(t)} \sim N(0.5\theta^{(t)},1)$''
My question is:


*

*If it is a first-order autoregressive process, do we still need to state it is a process with lag-1? Do they mean the same thing? I looked that up on the website, it says that both of them means that the current value is based on the immediately preceding value. 

*I searched online it says that the auto-correlation is defined as a function as shown on wiki here: https://en.wikipedia.org/wiki/Autoregressive_model So what does our 0.5 mean here and why it follows a normal distribution with $0.5\theta^{(t)}$ mean and 1 as variance?
 A: The "lag-1" qualified here belongs to the autocorrelation coefficient, not the process, which is already specified as AR(1). He's saying "lag-1 autocorrelation coefficient is 0.5". The reason he has to qualify the autocorrelation is that AR(1) process's autocorrelation isn't chopped off at lag 1, it gradually declines as the lags increase, see an example of autocorrelation function (ACF) for AR(1) process:

(from https://onlinecourses.science.psu.edu/stat510/node/60/)
Your process is $\theta_{t+1}=0.5\theta_t+\varepsilon_{t+1}$, where the errors are standard normal $\varepsilon_{t+1}\sim\mathcal N(0,1)$
It's easy to see why autocorrelation is 0.5:
$$Var[\theta_{t+1},\theta_t]=0.5\sigma^2_\theta$$
Why AR(1) is called is "first order"? The reason is the lag operator notation. Define an operator: $Ly_t=y_{t-1}$. If you feed the series into this operator, it'll return a series that have previous period values of the input. So, you can write AR(p) process as follows: $(1-L)^p\theta_y=\varepsilon_t$. This is written as if you were raising the expression in the brackets into power p as if it were a polynomial, that's why it's called p-order process
For $p=1$ we have: $(1-L)\theta_t=\theta_t-\phi_1\theta_{t-1}=\varepsilon_t$
