Hitting time $h_i(k)\geqslant h_i(j)\cdot h_j(k)$ Let $(X_n)_{n\in\mathbb{N}_0}$ be a Markov chain with state space $E$. The hitting time of a set $A\subseteq E$ is a RV
    $$
H(A)\colon\Omega\to\mathbb{N}_0\cup\left\{\infty\right\},~~\omega\mapsto\inf\left\{n\in\mathbb{N}_0 | X_n(\omega)\in A\right\}.
$$
    The probability starting from $i\in E$ that $(X_n)_{n\in\mathbb{N}_0}$ ever hits $A$  is
    $$
h_i(A):=\mathbb{P}_i(H(A)&lt;\infty).
$$
    For $i,j\in E$ we write $h_i(j):=h_i(\left\{j\right\})$.
    Prove, for all $i,j,k\in E$ that
    $$
h_i(k)\geqslant h_i(j)\cdot h_j(k).
$$
  


I tried to prove that using
$$
\begin{cases}h_i(A)=1, &amp; i\in A\\h_i(A)=\sum_{j\in E}p_{ij}h_j(A), &amp; i\notin A\end{cases}
$$

Although I do not know if I can use that here and how to continue I at least have
$$
h_i(k)=\sum_{s\in E}p_{is}h_s(k)
$$
and
$$
h_i(j)\cdot h_j(k)=\sum_{s\in E}\sum_{t\in E}p_{is}h_s(j)p_{jt}h_t(k)
$$
This follows from the (strong) Markov property of the Markov chain at the first hitting time of $j$.


Note that $h_i(k)\geqslant P_i(B)$, where $B$ denotes the event that the Markov chain hits $j$ and that, after it hits $j$, it hits $k$. Conditionally on the path of the Markov chain up to the hitting time $\tau$ of $j$, the probability to hit $k$ after $\tau$ is the probability that the Markov chain hits $k$ starting from $j$, that is, $h_j(k)$. Thus, $P_i(B)=h_i(j)\cdot h_j(k)$.
Edit 2 And again a new try...

Let $\tau:=H(j)&lt;\infty$. Otherwise the equation is fullfilled anyway, because then $h_i(k)\geq h_i(j)h_j(k))=0$.

It is
$$
P_i(B)=P(\exists s&gt;t&gt;0, X_s=k,X_t=j)
$$
Furthermore
$$
h_i(j)=P_i(\exists s&gt;0, X_s=j)=\frac{P(\exists s&gt;0, X_s=j,X_0=i)}{P(X_0=i)}
$$
and (using the strong Markov property)
$$
h_j(k)=P_j(\exists u&gt;0, X_u=k)\\=P(\exists u&gt;0,X_u=k|X_0=j)\\=P(\exists u&gt;0, X_u)k|X_{\tau}=j)\\
=\frac{P(\exists u&gt;0, X_u=k,X_{\tau}=j,\ldots,X_0=i)}{P(X_{\tau}=j,\ldots,X_0=i)}.
$$
Thus
$$
h_i(j)h_j(k)=\frac{P(\exists s&gt;0, X_s=j,X_0=i)}{P(X_0=i)}\cdot\frac{P(\exists u&gt;0, X_u=k,X_{\tau}=j,\ldots,X_0=i)}{P(X_{\tau}=j,\ldots,X_0=i)}.
$$
Because 
$$
P(\exists s&gt;0, X_s=j,X_0=i)=P(X_{\tau}=j,\ldots,X_0=i)
$$
it remains
$$
h_i(j)h_j(k)=\frac{P(\exists u&gt;0, X_u=k,X_{\tau}=j,\ldots,X_0=i)}{P(X_0=i)}\\
=P_i(\exists u&gt;0,X_u=k,X_{\tau}=j,\ldots,X_1=a_1)=P_i(B).
$$

On the other hand it is
$$h_i(k)\geqslant P_i(B),$$
so 
$$
h_i(k)\geqslant h_i(j)h_j(k).
$$