If $X_1 \rightarrow X_2 \rightarrow X_3 \rightarrow X_4$ is a Markov chain, is $X_2 \rightarrow X_3 \rightarrow X_4$ also one?

Question

If $X_1 \rightarrow X_2 \rightarrow X_3 \rightarrow X_4$ is a Markov chain, is $X_2 \rightarrow X_3 \rightarrow X_4$ also one?

The actual problem I want to solve is: given the term $E(log \frac{p(x_3 | x_1, x_2)p(x_4|x_1,x_2)}{p(x_3 | x_1)p(x_4|x_1)})$, how can I conclude $E(log \frac{p(x_2)}{p(x_3 | x_1)p(x_4|x_1)})$?

Answer 1

By assumption, $$ p(x_1, x_2, x_3, x_4) = p(x_1) p(x_2|x_1)p(x_3|x_2)p(x_4|x_3), $$ which implies $$ p(x_2, x_3, x_4|x_1) = p(x_2|x_1)p(x_3|x_2)p(x_4|x_3). $$

Now, note that $$ \begin{align} p(x_2, x_3, x_4) &= \int_{x_1} p(x_1,x_2,x_3,x_4)dx_1 \\ &= \int_{x_1}p(x_1)p(x_2,x_3,x_4| x_1)dx_1 \\ &=p(x_3|x_2)p(x_4|x_3)\int_{x_1}p(x_1)p(x_2|x_1)dx_1\\ &=p(x_2)p(x_3|x_2)p(x_4|x_3), \end{align} $$ which means that $X_2\rightarrow X_3 \rightarrow X_4$.

I am not sure what your actual problem is and how the above is relevant to its solution.