In my lecture notes, are results with respect to essential states in a markov chain are uniquely in finite dimension, in other words, I ve been given another proof for properties I already knew for reccurent states...
That's why now, I don't see why we need essential states.
I understand that beeing essential is less powerful (example of the biased random walk).
But, is it an important notion in general? What can one do with essential states that one can't with reccurent states?
When there are finitely many states, the notions of essential state and recurrent state coincide. Clearly a recurrent state is an essential state (in any Markov chain). If $i$ is an essential state which is not recurrent, then $$\mathbb P\left(\bigcap_{n=1}^\infty \{X_n\ne i\}\mid X_0=i \right)>0. $$ Thus there exists a sequence of states $j_n$ with $j_0=i$, $j_n\ne i$, $n\geqslant 1$ such that $$ \mathbb P\left(\bigcap_{n=1}^\infty \{X_n=j_n\}\mid X_0=j_0 \right) = \prod_{n=1}^\infty P_{j_{n-1}j_n}>0. $$ Taking negative logs on both yields $$ -\sum_{n=1}^\infty \log(P_{j_{n-1}j_n})<\infty, $$ and thus $\lim_{n\to\infty} P_{j_{n-1}j_n}=1$. Since there are finitely many states, $P_{j_{n-1}j_n}$ can only take finitely many values, and hence $$ \liminf_{n\to\infty} P_{j_{n-1}j_n}=1. $$ Therefore there exists $N$ such that for every $n\geqslant N$, $P_{j_{n-1}j_n}=1$. It follows that $j_N$ is accessible from $i$ but $i$ is not accessible from $j_N$, contradicting the assumption that $i$ was essential.