Question about poles and zeros in AR filter
An AR process is defined in a way that it is inherently causal. It depends on past and present values, but not on the future. Thus, its ROC must be:


In the continuous case, $s$ such that $|s|&gt;\sigma$ for some $\sigma\in\mathbb{R}$.
In the discrete case, $z$ such that $|z|&gt;r$ for some $r\in\mathbb{R}$.


For an AR filter, all the zeros are at the origin. This could make the 3rd option you wrote correct, as the origin is indeed on the real line. The first one is definitely incorrect.

Regarding the poles, it depends on the coefficient of the process, $a$. If $|a|&lt;1$, the filter will be stable and all the poles will be on the left half plane of the $s$-plane, or inside the unit circle in the $z$-domain. If $|a|&gt;1$, the contrary will happen.

So the answer would be the 4th, if the filter is required to be stable (which makes sense). If the filter is unstable, then the 2nd one would be the one to chose.