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A Markov Chain $X_n$ with state space $S=(0,1,2)$ and transition probabilities $p_{00}=p_{01}=p_{11}=p_{12}=p_{20}=p_{22}=\frac{1}{2}$. Also, $P(X_0=0)=P(X_0=1)=P(X_0=2)=\frac{1}{3}$. I want to determine $P(X_2=0, X_4=0)$, but I don't know how. Is $$P(X_2=0, X_4=0)=P(X_2=0)P(X_4=0|X_2=0)?$$

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closed as off-topic by Did, max_zorn, saz, ancientmathematician, Cesareo Feb 1 at 7:57

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  • $\begingroup$ Yes $P(A\cap B)=P(A)P(B\mid A)$. Is this your question? $\endgroup$ – Did Jan 31 at 7:38
  • $\begingroup$ Yes, but now I am confused. Isn't $P(X_2=0|X_0=0)$ the same as the $p_{00}$ element in the transition matrix $P^2$ which is $\frac{1}{4}$? $\endgroup$ – Natalie_94 Jan 31 at 10:26
  • $\begingroup$ Yes $P(X_2=0\mid X_0=0)=\frac14$. To prove this, one can either compute $P^2$ (as you did), or identify the paths going from $0$ to $0$ in two steps: there is only one such path, which is $0\to0\to0$, and its probability is $\frac12\cdot\frac12=\frac14$, qed. Why the confusion? $\endgroup$ – Did Jan 31 at 10:55
  • $\begingroup$ Thanks! I was confused because gt6989b below said something else. $\endgroup$ – Natalie_94 Jan 31 at 13:13
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HINT

You are correct. Note also that $$ P(X_4=0|X_2=0) = P(X_2=0|X_0=0) = 1/4 $$ since there are only 4 paths of length 2 from 0 (all of equal weight), and 1 of them end up back at 0.

Can you compute $P(X_2=0)$? This should depend on the initial condition first, and then you can average out what the related weight would be for the overall probability...

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  • $\begingroup$ Since the probability distribution $p^{0}=(\frac{1}{3},\frac{1}{3},\frac{1}{3})$ I found that $p^{2}=(\frac{1}{3},\frac{1}{3},\frac{1}{3})$ so $P(X_2=0)=\frac{1}{3}$ or am I wrong? $\endgroup$ – Natalie_94 Jan 29 at 18:30
  • $\begingroup$ And I thought that $P(X_2=0|X_0=0)=\frac{1}{4}$ since I can only find 1 path of length 2 from 0 to 0? $\endgroup$ – Natalie_94 Jan 29 at 20:31
  • $\begingroup$ @Amanda there are 2 such paths: $0 \to 1 \to 0$ and $0 \to 2 \to 0$. $\endgroup$ – gt6989b Jan 29 at 23:18
  • $\begingroup$ @Amanda you are correct on $p^2$ and $P(X_2=0)$ $\endgroup$ – gt6989b Jan 29 at 23:19
  • $\begingroup$ I thought the only path was $0\to 0 \to 0$ since $p_{10}=p_{02}=0$, shouldn't $0 \to 1 \to 0$ and $0 \to 2 \to 0$ be impossible? $\endgroup$ – Natalie_94 Jan 30 at 8:05

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