$\sum X_k $ is a Markov Process but not Markov Chain

Question

If $(X_n)_{n\ge1}$ are independent r.v.s and $X_n\sim Uniform(1,\dots,n)$ then the sum $S_n:=\sum\limits_{k=0}^nX_k$ is a Markov process but not a Markov chain

Since $S_{n+1}=S_n+X_{n+1}$ the other r.v.s $S_k$ for $k<n$ are irrelevant. and it is not a Markov chain because the homogeneity fails for example

$P(S_n+1=n+1+k|S_n=k)=\frac{1}{n+1}\neq\frac{1}{j+1}=P(S_j+1=j+1+k|S_j=k)$ for an appropriate value $k$

or one can even choose $k$ s.t. $j<k<n$ hence one of the probabilites is zero, the other is not.

am I correct ?

Answer 1

It is still called a Markov chain but it is not time-homogeneous. See the formal definition of discrete time Markov chain in https://en.wikipedia.org/wiki/Markov_chain