Markov Chain initial state conditional probability

Question

Suppose ${X_0, X_1, . . . , }$ forms a Markov chain with state space S. For any n ≥ 1 and $i_0, i_1, . . . , ∈ S$, which conditional probability, $P(X_0 = i_0|X_1 = i_1)$ or $P(X_0 = i_0|X_n = i_n)$, is equal to $P(X_0 = i_0|X_1 = i_1, . . . , X_n = i_n)$?

I think it is the second one?? I do know the Markov property but I am not sure on how it applies to the initial state?

score 1 · Answer 1 · answered Mar 09 '20 at 06:22

1

If you are working with a stationary ergodic Markov chain, then what you mention is a time-reversible markov chain, which must satisfy certain criterions to be classified as one, make sure those apply to the one you are working with. But, the reversed process satisfies $$\mathbb P\left(X_n=i_{n}\vert X_{n+1}=i_{n+1},X_{n+2},X_{n+3},...\right)=\mathbb P\left(X_n=i_{n}\vert X_{n+1}=i_{n+1}\right)$$ in your case, just let $i_n=0$.

The other option mentioned would be the $n$-th step transition probability of going from $i_n$ to $i_0$.

answered Mar 09 '20 at 06:22

kiyomi

1,424

thanks so much for replying this is quite useful. But I think it doesn't quite answer my question probably due to my bad Latex. I have edited it again. I am not working with a time reversible chain. – confused doughnut Mar 09 '20 at 07:00
it is conditioned on all the events from 1-n. – confused doughnut Mar 09 '20 at 07:23

score 0 · Answer 2 · answered Mar 04 '23 at 05:24

\begin{aligned} P(X_0 = i_0|X_1 = i_1, . . . , X_n = i_n) &= \frac{P(X_0 = i_0, X_1 = i_1, . . . , X_n = i_n)}{P(X_1 = i_1, . . . , X_n = i_n)} & \text{(conditional probability)}\\ &= \frac{P(X_0 = i_0)P( X_1 = i_1 |X_0 = i_0)\cdots P(X_n = i_n | X_{n-1} = i_{n-1})}{P(X_1 = i_1)P(X_2 = i_2 | X_1 = i_1) \cdots P(X_n = i_n | X_{n-1} = i_{n-1})} & \text{(Markov chain joint distribution)}\\ &= \frac{P(X_0 = i_0)P( X_1 = i_1 |X_0 = i_0)}{P(X_1 = i_1)} &\\ &= \frac{P(X_0 = i_0, X_1 = i_1)}{P(X_1 = i_1)} &\\ &= P(X_0 = i_0 | X_1 = i_1) \end{aligned}

Markov Chain initial state conditional probability

2 Answers2