Suppose $\left\{ X_{n}\right\} _{n=1}^{\infty}$ is a Markov Chain taking real values. Are the following Markov Chains? $$Y_{n}=\sum_{i=1}^{n}X_{i} , Z_{n}=\left(X_{n},X_{n-1}\right)$$ Edit1 I removed some of the nonsense I wrote previously as it was quite wrong as pointed out by Lost1's answer.

Edit2 I have a followup question. Consider we now know that $Y_{n}$ is not necessarily a Markov Chain, what about about $Z_{n}=(Y_{n-1},Y_{n})$? My hopefully new found intuition tells me this should be a Markov chain since asking a question about pairs of $Y$ values is equivalent to a question about $X$ values. I tried to formalize it but I'm not sure whether my reasoning is correct: $$\mathbb{P}\left(Z_{n}=\left(y_{n-1},y_{n}\right)\,|\, Z_{n-1}=\left(y_{n-2},y_{n-1}\right),Z_{n-2}=\left(y_{n-3},y_{n-2}\right),...,Z_{2}=\left(y_{1},y_{2}\right)\right) =\mathbb{P}\left(Y_{n-1}=y_{n-1},Y_{n}=y_{n}\,|\, Y_{n-1}=y_{n-1},Y_{n-2}=y_{n-2},...,Y_{1}=y_{1}\right) =\mathbb{P}\left(X_{n}=y_{n}-y_{n-1}|\, X_{n-1}=y_{n-1}-y_{n-2},...,X_{1}=y_{2}-y_{1}\right)=\mathbb{P}\left(X_{n}=y_{n}-y_{n-1}|\, X_{n-1}=y_{n-1}-y_{n-2}\right) \overbrace{=}^{\dagger}\mathbb{P}\left(Y_{n-1}=y_{n-1},Y_{n}=y_{n}\,|\, Y_{n-1}=y_{n-1},Y_{n-2}=y_{n-2}\right)=\mathbb{P}\left(Z_{n}=\left(y_{n-1},y_{n}\right)\,|\, Z_{n-1}=\left(y_{n-1},y_{n-2}\right)\right) $$ Where the first two transitions are a result of the events in question being equivalent and the same for the last two transitions (this is the part I'm unsure of, in particular I'm not convinced of the marked equality).

I'd appreciate it if someone could tell me whether what I wrote is correct. appreciated.

  • $\begingroup$ Indeed, it is a markov chain, conditioning on previous two Y values is the same as conditioning Xn and Yn, which is enough to tell you where you currently are in the X chain, so you know the transition prob. $\endgroup$ – Lost1 Jul 4 '14 at 12:17
  • $\begingroup$ On the other hand, i dont think the step you highlighted is corrected. I will edit my answer later $\endgroup$ – Lost1 Jul 4 '14 at 12:19
  • $\begingroup$ i made an edit. $\endgroup$ – Lost1 Jul 4 '14 at 14:57
  • $\begingroup$ It just occurred to me $(Y_n,X_n)$ and $(Y_n,Y_{n-1})$ are bijective can be mapped one-to-one. For your claim $(Y_n,Y_{n-1})$ is Markov, it is probability easier to make this observation and just show $(Y_n,X_n)$ is Markov, which is, imo, a little easier. $\endgroup$ – Lost1 Jul 4 '14 at 16:44

$Y_n$ is not a Markov Chain, $Z_n$ is a Markov Chain.

Your proof is clearly wrong. You can easily find an example where the Markov Chain is not a random walk

Even take the Markov Chain with two states: 1<->2 where $p_{12}=0.3=p_{21}$, $0.7=p_{11}=p_{22}$. Basically if you only know the sum up to time $n$, assuming that $Y_n\geq 2$, you don't know which state you are in, so you have no idea what the transition probability is.

$Z_n$ is clearly a Markov Chain. What $Z_n$ is doing is remembering an extra state. All the exciting things are actually happening in $X_n$. When conditioning $(X_n,X_{n-1})$ on $(X_{n-1},X_{n-2})$, it is the same as conditioning on $X_{n-1}$ by Markov property of $X$. the second component of $Z_n$ is a deterministic function of this! To make this waffle precise.

$P(Z_n=(a,b)|Z_{n-1}=(c,d)) = P(X_n=a, X_{n-1}=b|X_{n-1}=c,X_{n-2}=d)=P(X_n=a,X_{n-1}=b|X_{n-1}=b,X_{n-2}=d,... X_0=...)=P(X_n=a,X_{n-1}=b|Z_{n-1}=(b,d),Z_{n-2}=(...),...,Z_1=(...))= $

Note I used the Markov property of $X$ and the observation that conditioning on $X_0,...X_{n-1}$ is the same as conditioning on $Z_1$,...$Z_{n-1}$.

This is what I would do after the last line before the dagger.

$$P(X_n=y_n-y_{n-1}|X_{n-1}=y_{n-1}-y_{n-2}) =P(X_n=y_n-y_{n-1}|X_{n-1}=y_{n-1}-y_{n-2},Y_{n-1}=y_{n-1}) =P(X_n=y_n-y_{n-1}|Y_{n-1}=y_{n-1},Y_{n-2}=y_{n-2}) =P(X_n=y_n-y_{n-1},Y_{n-1}=y_{n-1}|Y_{n-1}=y_{n-1},Y_{n-2}=y_{n-2}) =P(Y_n=y_n,Y_{n-1}=y_{n-1}|Y_{n-1}=y_{n-1},Y_{n-2}=y_{n-2}) $$

Note where the first equality holds because $Y_{n-1}$ is a function of $X_1,...X_{n-1}$. I am using a consequence of the Markov property: Conditioning on functions of $X_1,...X_{n-1}$ and $X_{n-1}$ is the same as conditioning just on $X_{n-1}$. This is a standard argument if you are familiar with measure theory.

  • $\begingroup$ In the counterexample you suggested with the two state space is it enough to notice that $\mathbb{P}\left(Y_{3}=1\,|\, Y_{2}=1\right)>0=\mathbb{P}\left(Y_{3}=1\,|\, Y_{2}=1,Y_{1}=1\right)$ in order to deduce the sum is not a markov chain? $\endgroup$ – LlamaMan Jul 3 '14 at 23:43
  • $\begingroup$ @NewGuy How can $Y_2$ be $1$ when $X_i\geq 1$? but this is the right sort of line. Here is a simple one: $P(Y_3=4|Y_2=3,Y_1=2)=p_{11}$ and $P(Y=4|Y_2=3,Y_1=1)=p_{21}$ are different, so they cannot both equal to $P(Y_3=4|Y_2=3)$... $\endgroup$ – Lost1 Jul 3 '14 at 23:47
  • $\begingroup$ It seems to be the only case if $Y$ is also Markov is that $X_1$,... $X_n$ are i.i.d random variables... $\endgroup$ – Lost1 Jul 3 '14 at 23:49
  • $\begingroup$ Oops, I wrote it down with ${0,1}$ as the states instead of ${1,2}$ in which case $Y_{2}$ can be $1$ but the probability I wrote no longer equals zero. But obviously I see what you mean now, thanks! Oh and as as side note I liked the phrase "to make this waffle precise" :) $\endgroup$ – LlamaMan Jul 4 '14 at 0:13
  • 1
    $\begingroup$ Thanks for updating your reply! $\endgroup$ – LlamaMan Jul 4 '14 at 16:30

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