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A process $X$ is called a Markov chain if $P(X_n=i_1 | X_{n-1}=i_{n-1},... , X_1=i_1) = $$P(X_n=i_1 | X_{n-1}=i_{n-1})$.

Now, given $X_n, n\ge1$, iid, with $P(X_n = 1) = p$, $P(X_{n-1} = -1)= 1-p$, define $Y_n = X_nX_{n+1}$. Is $(Y_n)$ a Markov chain?

My intuition tells me it is not, so I have been looking for a counter example.

I tried calculating $P(Y_3 = 1 | Y_2 = 1, Y_1= -1)$ and $P(Y_3 = 1| Y_2 = 1)$ to see if they are equal, but I am having dificulties. My thoughts to calculate those probabilities were $Y_2 = 1, Y_1= -1$ imply that either $X_1 = 1, X_2 = -1, X_3 = -1$ or $X_1 = -1 X_2 = 1, X_3 = 1$. So $P(Y_3 = 1 | Y_2 = 1, Y_1= -1) = 0.5 * 0.5 + 0.5 * 0.5 = 0.5$. But also by this same reasoning, $P(Y_3 = 1| Y_2 = 1)$ would be 0.5. I feel I am making a mistake somewhere when calculating this. Can someone tell me how to calculate those probabilities correctly? Or am I wrong and this is a Markov chain?

Edit: I assumed $p = 1/2$ by mistake. But then how do I calculate those probabilities for a general p?

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  • $\begingroup$ In your example are $X_n$ two state $\pm 1$ variables? It looks like what you meant, but didn't say it. $\endgroup$ Nov 13, 2021 at 19:24
  • $\begingroup$ Yes, it is. Sorry if I was not clear $\endgroup$
    – You1234
    Nov 13, 2021 at 19:25
  • $\begingroup$ You tried with $p=0.5$. Try a different value. $\endgroup$ Nov 13, 2021 at 19:29
  • $\begingroup$ @herbsteinberg Oh, yes, you are correct, I should have used a general p... But I am not sure how to calculate those probabilities for a general p. $\endgroup$
    – You1234
    Nov 13, 2021 at 19:37

1 Answer 1

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For calculation (general form) $P(A|B)=\frac{PA\cap B)}{P(B)}$. Need $P(Y_3=1\cap Y_2=1)=P(X_4=1\cap X_3=1\cap X_2=1)$$+P(X_4=-1\cap X_3=-1\cap X_2=-1)=p^3+(1-p)^3$ $P(Y_2=1)=p^2+(1-p)^2$ $P(Y_3=1\cap Y_2=1\cap Y_1=-1)=p^3(1-p)+p(1-p)^3$ $P(Y_2=1\cap Y_1=-1)=p^2(1-p)+p(1-p)^2$.

Net result: $P(Y_3=1|Y_2=1)=\frac{p^3+(1-p)^3}{p^2+(1-p)^2}$ and $P(Y_3=1|Y_2=1\cap Y_1=-1)$$=\frac{p^3(1-p)+p(1-p)^3}{p^2(1-p)+p(1-p)^2}=p^2+(1-p)^2$.

Not equal for $p\ne 0.5$.

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