Independence in joint conditional probability density functions

Question

I have a random vector: $\textbf{X} = \{Y_1, Y_2, Y_3, Y_4\}$

The corresponding probability density function factorizes as follows: $p(y_1,~y_2,~y_3,~y_4) = p(y_1,~y_4|y_2) p(y_2,~y_3|y_1)$

Now I have to show that $Y1⊥Y2$ holds. So far I think the following holds:
$p(y_1,~y_4|y_2) ⊥ p(y_2,~y_3|y_1)$

thus $p(y_1|y_2) ⊥ p(y_2|y_1)$

Now I'm stuck, it seems intuitive that these are unconditionally independent since they are independent given each other. How can I continue from here? Am I thinking in the wrong way? any help is greatly appreciated!

Answer 1

Hint: The factorization $$ p(y_1,y_2,y_3,y_4)=p(y_1,y_4\mid y_2)p(y_2,y_3\mid y_1)\tag1 $$ is valid for all $y_1, y_2, y_3, y_4$, so if you sum (1) over all values of $y_3$ you get: $$ p(y_1, y_2, y_4)=p(y_1, y_4\mid y_2)p(y_2\mid y_1).\tag2 $$ Now use the definition of conditional probability to write $$ p(y_1,y_4\mid y_2)=\frac{p(y_1,y_2,y_4)}{p(y_2)} $$ and $$ p(y_2\mid y_1)=\frac{p(y_1,y_2)}{p(y_1)}. $$