Verifying that a branching process is a Markov chain I would like to verify that the following model of a branching process creates a Markov chain - a Markov chain here defined as having the property that $\mathrm{P}\left\{\xi_{k+1}=a_{k+1} | \xi_{0}, \ldots, \xi_{k}\right\}=\mathrm{P}\left\{\xi_{k+1}=a_{k+1} | \xi_{k}\right\}$ where we take $\mathrm{P}\left\{\xi_{k+1}=a_{k+1} | \xi_{k}\right\}$ to be the random variable $\sum_{j=1 }^ m \mathrm{P}\left\{\xi_{k+1}=a_{k+1} | \xi_{k}=b_j\right\}\mathrm {1}_{\{\xi_k=b_j\} } $ (assuming $\xi_k $ has range $\{b_1,...,b_m \} $), and where $\mathrm{P}\left\{\xi_{k+1}=a_{k+1} | \xi_{k}=b_j\right\} $ is defined as the conditional probability $
\mathrm {P }\left\{\xi_{k+1}=a_{k+1} , \xi_{k}=b_j\right\} /\mathrm {P }\{ \xi_{k}=b_j \}, \ \mathrm {P }\{ \xi_{k}=b_j \} >0.$

Let $\xi_0=1 $ and for $k \ge 0 $ let $\xi_{k+1 }  = \eta_1^{(k) }+...+\eta_{\xi_k } ^{(k) } $ where the random variables $\eta _j^{(k) } $ are  mutually independent for any $j $ and any $k \ge 0 $. Verify that $\{\xi_k \} $ is a Markov chain.

 A: I believe it is sufficient to check that for any sequence $i_1,...,i_{k+1 } $, $\mathrm{P}\left\{\xi_{k+1}\right.=i_{k+1} | \xi_{k}=i_{k}, \xi_{k-1}=i_{k-1}, \ldots \}=\mathrm{P}\left\{\xi_{k+1}=i_{k+1} | \xi_{k}=i_{k}\right\}$, considering that for a partition $\{A_m \}$  of  $\{\xi_k = b_j \}$,  $1_{\{\xi_k = b_j \}} =\sum 1_{A_m } $.

Thus I would need to verify that $\mathrm{P}\left\{\xi_{k+1}\right.=i_{k+1} | \xi_{k}=i_{k}, \xi_{k-1}=i_{k-1}, \ldots \}=\mathrm{P}\left\{\xi_{k+1}=i_{k+1} | \xi_{k}=i_{k}\right\}$

Considering that the sets $\{\xi_{k+1}=i_{k+1} , \xi_{k}=i_{k}\}$ and $\{\eta_{1}^{(k)}+\cdots+\eta_{i_{k}}^{(k)}=i_{k+1},\xi_{k}=i_{k}\} $ are equal. And by independence that $\mathrm {P } \{\eta_{1}^{(k)}+\cdots+\eta_{i_{k}}^{(k)}=i_{k+1},\xi_{k}=i_{k}\}= \mathrm {P } \{\eta_{1}^{(k)}+\cdots+\eta_{i_{k}}^{(k)}=i_{k+1}\} \mathrm {P }\{\xi_{k}=i_{k}\}  $ 
We get that $\mathrm{P}\left\{\xi_{k+1}=i_{k+1} | \xi_{k}=i_{k}\right\}= \mathrm {P } \{\eta_{1}^{(k)}+\cdots+\eta_{i_{k}}^{(k)}=i_{k+1}\}$ by cancelling the denominator in the former. 
An identical argument yields $\mathrm{P}\left\{\xi_{k+1}\right.=i_{k+1} | \xi_{k}=i_{k}, \xi_{k-1}=i_{k-1}, \ldots \}= \mathrm {P } \{\eta_{1}^{(k)}+\cdots+\eta_{i_{k}}^{(k)}=i_{k+1}\}$ and thus the desired equality.

In the above we use the property that $\eta _j ^ {(k) } $ are mutually independent means that $\eta_{1}^{(k)}+\cdots+\eta_{i_{k}}^{(k)}$ is independent of $\xi_{k}$.
