Consider a communication system which transmits the two digits $~0~$and$~1~$ through several stages. Let $~X_0~$ be the digit transmitted initially (leaving) $~0^{\text{th}}~$ stage and $~X_n~,~n\ge 1~$ be the digit leaving the $~n^{\text{th}}~$ stage. At every stage there is a constant probability$~q~$that the digit which enters is transmitted unchanged and the probability$~p~$other wise with $~p+q=1~$. Show that $~X_n~:~n\ge 0~$ is a Markov chain.
As I am a newcomer in this session, I have no idea how to proceed. Any help in this matter is really appreciable. Thank You.
Note: Please give a complete solution, so that it will help me to solve the similar problems in future.
I'm not entirely sure I've understood the question correctly, but I take it that $$ X_n=\cases{X_{n-1}& with probability $\ q\ $, or\\ 1-X_{n-1} & with probability $\ p=1-q\ $.} $$ If that's the case, then \begin{align} P\left(\left.X_n=b_n\right|X_{n-1}=b_{n-1},\right.&\left. X_{n-2}=b_{n-2},\dots,X_0=b_0\right)\\ &=\cases{q&if $\ b_n=b_{n-1}\ $, or\\ 1-q& if $\ b_n\ne b_{n-1}\ $.}\\ &= P\left(\left.X_n=b_n\right|X_{n-1}=b_{n-1}\right)\ , \end{align} which is the condition needed for $\ X_n\ $ to be a Markov chain. The state of the chain at stage $\ n\ $ being $\ X_n\ $, with two possible values $0$ and $1$, it's a two-state chain with transition matrix $$ \pmatrix{q&1-q\\ 1-q& q}\ . $$