Quote:

$\displaystyle (a_n)_{n\in N}$ a sequence of integers.

$\displaystyle P(Z_{n+1}=a_{n+1}\mid Z_0=a_0,\dots,Z_n=a_n)$$\displaystyle =P\big(\sum_{i=1}^{a_n} X_{i,n+1}=a_{n+1}\mid Z_0=a_0,\dots,Z_n=a_n\big)$

Whatever values $\displaystyle a_0,\dots,a_{n-1}$ take, the probability won't be changed as none of them intervene in $\displaystyle \sum_{i=1}^{a_n} X_{i,n+1}$.

So $\displaystyle P(Z_{n+1}=a_{n+1}\mid Z_0,\dots,Z_n)=P(Z_{n+1}=a_{n+1}\mid Z_n)$

Hence (Zn) is a Markov chain.

As to show that it's homogeneous, that's no big deal.