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a) $P_0(T_0=n)$

b) $P_0(T_1=n)$

In class my we were learning about hitting times. $$T_a; \text{stands for the first time we hit state}\ \textit{"a"} \\\\ T_a=min\{n\ |n>0\ \text{and} \ X_n=a\}$$ My professor defined $$P_0(T_0=n)=P(T_0=n|X_0=0)$$ where we as hitting state zero in n-steps, $n\geq 0$. For a 2-state Markov chain we know, $$P(0,0)=1-p\ > \ \ P(0,1)=p\ \ \ P(1,1)=1-q\ \ \ P(1,0)=q$$ so I let n=4 and found each probability to see if there was a pattern that I could show for the general case, but I think this approach is wrong. I would like help and/or hints with both parts, if that's not to much trouble.

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1 Answer 1

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there are two states, 0 and 1. you start at 0 and don't end up there again until move $n$. From markovianity, the answer is $$ \pi(0) \cdot P(0,1) \cdot \underbrace{ P(1,1) \cdots P(1,1)}_{n-2} \cdot P(1,0) $$

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