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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.

nmasanta
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  • Since the transition probabilities are constant, the distribution of $X_n$ is independent of the past states $X_{n-1},...,X_{0}$, thus it is a markov chain – fGDu94 Dec 23 '19 at 04:28
  • In the definition of $X_n$, is all we care about the digit ($0$ or $1$), or also the $n^{\mathrm{th}}$ "stage"? – Math1000 Dec 23 '19 at 06:56

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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 $$ \pmatrix{q&1-q\\ 1-q& q}\ . $$

lonza leggiera
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  • Sorry for the late response @lonzaleggiera . The fact is that form another source I found that the transition matrix for this case is $$\pmatrix{1-q&q\ q& 1-q}\ .$$ – nmasanta Dec 29 '19 at 16:25
  • I can't see how that would be the transition matrix for the process as you've described it. In your question, you define the probability $\ q\ $ as the probability "that the digit which enters is transmitted unchanged"—i.e. the probability the next state (the bit transmitted) is $\ 0\ $ given that the current state (the bit that had entered) was $\ 0\ $, namely $\ p_{00}\ $, is $\ q\ $, and the probability the next state is $\ 1\ $ given that the current state was $\ \ 1$, namely $\ p_{11}\ $, is also $\ q\ $. – lonza leggiera Dec 29 '19 at 17:53
  • Are you sure that your "other source" is using precisely the same definitions as the ones you've given here? – lonza leggiera Dec 29 '19 at 17:55
  • Yes, it is. @lonzaleggiera – nmasanta Dec 29 '19 at 18:48