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If $X_0,X_1,X_2,...$ is a Markov chain then why is $Z_n=X_0X_1...X_n$ also a Markov chain?

Could you explain briefly?

Many thanks**

RGS
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hatban
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  • Thanks for edit. I appologise that I do not know how to type like you did. – hatban Jan 25 '17 at 23:53
  • I think you could try and check the definition of markov chain against $Z_n $. My guess would be that it should be simple to show that $Z_n $ complies with the definition. – RGS Jan 26 '17 at 00:20

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This isn't true.

For isntance: let $(X_n)$ be the Markov chain defined as follows: Let $P(X_0=1)=P(X_0=2)=\frac{1}{2}$, and determine $X_{n+1}$ from $X_n$ using the transition probabilities $$ p_{11}=\frac{1}{4},\quad p_{12}=\frac{3}{4},\quad p_{21}=\frac{1}{2},\quad p_{22}=\frac{1}{2}. $$ You can show that for $Z_i:=\prod_{j\leq i}X_i$, $$ P(Z_2=4\mid Z_0=1,Z_1=2)=\frac{1}{2} $$ while $$ P(Z_2=4\mid Z_0=2, Z_1=2)=\frac{3}{4}. $$ If $(Z_n)$ were a Markov chain, then we would need to have $$ P(Z_2=4\mid Z_0=1,Z_1=2)=P(Z_2=4\mid Z_1=2)=P(Z_2=4\mid Z_0=2, Z_1=2). $$

Fundamentally, the problem with your assertion is this: knowledge of $Z_n$ is not enough to produce good knowledge of $X_n$ (it is only information about $X_n/X_{n-1}$ essentially), but knowledge of $Z_0,\ldots, Z_n$ IS enough to produce knowledge about $X_0,\ldots,X_n$.

Nick Peterson
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  • What if the question was If $X_0,X_1,X_2,...$ is a Markov Chain and $Z_n=X_0X_1...X_n$, explain why $Z_0,Z_1,Z_2,...$ is a Markov Chain? – ʎpoqou Jan 13 '19 at 16:40
  • @ʎpoqou How is that different than the question that was initially asked? – Nick Peterson Jan 14 '19 at 02:13
  • In the question I ask, you consider a sequence of $Z_0,Z_1,Z_2,...$. Unless you also assumed the op meant that this is the sequence to be considered. In any case. What if I added that each $X_i$ has a state space {0,1}? – ʎpoqou Jan 14 '19 at 09:04
  • @ʎpoqou That is precisely the sequence considered in the question above, yes. However, in the very special case you've just mentioned, it does turn out to be a Markov chain. Note that $Z_n\in{0,1}$ for all $n$, and think about the probability of transition between these states. If you have further questions, though, please don't follow up here; that's not what comments are for. Instead, ask an actual question! – Nick Peterson Jan 14 '19 at 16:06