Markov and independent random variables This is a part of an exercise in Durrett's probability book.

Consider the Markov chain on $\{1,2,\cdots,N\}$ with $p_{ij}=1/(i-1)$ when $j<i, p_{11}=1$ and $p_{ij}=0$ otherwise. Suppose that we start at point $k$. We let $I_j=1$ if $X_n$ visits $j$. Then $I_1,I_2,\cdots,I_{k-1}$ are independent.

I don't find it obvious that $I_1,\cdots,I_{k-1}$ are independent. It is possible to prove the independence if we calculate all $P(\cap_{j\in J\subset\{1,\cdots,k-1\}}I_j)$, but this work is long and tedious. Since the independence was written as an obvious thing in this exercise, I assume that there is an easier way.

For any $j$, observe that $X_{3}|X_{2}=j-1,X_{1}=j$ has the same distribution as $X_{2}|X_{2} \neq j-1, X_{1}=j$. Since $X_{2}=j-1$ iif $I_{j-1}=1$, by Markovianity conclude that $I_{j-1}$ is independent of $(I_{j-2},\ldots,I_{1})$ given that $X_{1}=j$.
Let's prove by induction that $I_{j-1}$ independent of $(I_{j-2},\ldots,I_{1})$ given that $X_{1}=k$.
I) $j=k$ follows straight from the first paragraph.
II) Now assume $I_{a-1}$ independent of $(I_{a-2},\ldots,I_{1})$ for all $a \geq j+1$. Thus, $(I_{k-1},\ldots,I_{j})$ is independent of $(I_{j-1},\ldots,I_{1})$. Hence, in order to prove that $I_{j-1}$ is independent of $(I_{j-2},\ldots,I_{1})$ we can condition on $(I_{k-1}=1,\ldots,I_{j}=1)$. This is the same as conditioning on $(X_{2}=k-1,\ldots,X_{k-j+1}=j)$. By markovianity and temporal homogeneity, $(X_{k-j+2}^{\infty}|X_{k-j+1}=j,\ldots,X_{1}=k)$ is identically distributed to $(X_{2}^{\infty}|X_{1}=j)$. Using the first paragraph, we know that $I_{j-1}$ is independent of $(I_{j-1},\ldots,I_{1})$ given that $X_{1}=j$. Hence, by the equality of distributions, $I_{j-1}$ is independent of $(I_{j-2},\ldots,I_{1})$ given that $X_{1}=k$.
