Are the Markov Chains associated with the below equations invariant and irreducible?
- PMF $f = (1/3, 1/3, 1/3)$ and kernel $$K = \begin{bmatrix} 1/2 & 1/2 & 0 \\ 1/2 & 1/2 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
- PDF $g(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}$ and $K(x,y) = g(y)$
- PDF $h(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}$ and $\int_A K(x,y)dy=1_A(x)$, for $A \subset \mathbb{R}$ and initial value $x$.
$f*K = K$ hence invariant. No other state can be reached from {3}, therefore not irreducible.
$g*K = \frac{1}{2\pi}e^{-1/2(x^2+y^2)} \neq K,$ so not invariant. For any $t>0$, then $K^t > 0$, so the integral is non-negative and hence irreducible.
If $\int K = 1_A$, then $K=0$? So, $h*K=0 \neq K$, so not invariant. Since $\int K = 1_A$, then $\int K^t >0$ , hence irreducible.
SOLUTION:
1: invariant, not irreducible. 2: invariant, irreducible. 3: invariant, not irreducible.
Where did i go wrong for 2 and 3?
Irreducibility for continuous MCs are defined as:
Given a distribution, $\mu$, over $E$, a MC is irreducible if for all points $x \in E$ and all measurable sets $A$ such that $\mu(A) > 0$ there exists some $t$ such that:
$$ \int_A K^t(x,y) dy > 0$$