$ P(X_n = i_n \forall 0\leq n \leq k+1 ) = \prod_{s=0}^k p_{i_s,i_{s+1}}(1) $

Question

If $X$ is a random variable, more precisely a markov chain, then one should get that expression :

$$ P(X_n = i_n \ \ \forall 0\leq n \leq k+1 ) = \prod_{s=0}^k p_{i_s,i_{s+1}}(1) $$

where $ p_{i_s,i_{s+1}}(1) = P(X_{m+1} = i_{s+1} | X_{m} = i_{s} )$

with $k, m \in \mathbb N$

Any idea how come ?

I tried with the case $ k = 2 $ But even in that case I have no idea. For me it seems wrong.

It appears here : Grimmette page 273 http://home.ustc.edu.cn/~zt001062/PTmaterials/Grimmett&Stirzaker-1000%20Probability.pdf bottom of the page.

Answer 1

Markov property tells us that $P(X_2=i_2|X_1=i_1 \text{ and }X_0=i_0)=P(X_2=i_2|X_1=i_1)$ and conditional probability tells us $$P(X_2=i_2| X_1=i_1 \text{ and }X_0=i_0)=\frac{P(X_2=i_2\text{ and } X_1=i_1 \text{ and }X_0=i_0)}{P( X_1=i_1 \text{ and }X_0=i_0)},$$ so $$P(X_2=i_2\text{ and } X_1=i_1 \text{ and }X_0=i_0)=P( X_1=i_1 \text{ and }X_0=i_0)P(X_2=i_2|X_1=i_1),$$ again, conditional probability tells us that $P( X_1=i_1 \text{ and }X_0=i_0)=P(X_0=i_0)P(X_1=i_1|X_0=i_0),$ substituting we get $$P(X_2=i_2\text{ and } X_1=i_1 \text{ and }X_0=i_0)=P(X_0=i_0)P(X_1=i_1|X_0=i_0)P(X_2=i_2|X_1=i_1)$$