# Is $Y_n$ a Markov chain?

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?

• In your example are $X_n$ two state $\pm 1$ variables? It looks like what you meant, but didn't say it. Nov 13, 2021 at 19:24
• Yes, it is. Sorry if I was not clear Nov 13, 2021 at 19:25
• You tried with $p=0.5$. Try a different value. Nov 13, 2021 at 19:29
• @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. Nov 13, 2021 at 19:37

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 $$p\ne 0.5$$.