Given a joint probability function over the random vector $X = (X_1,X_2,X_3,X_4)$ that factorizes as
$p(x_1,x_2,x_3,x_4) = p(x_1,x_4 | x_2)p(x_2,x_3|x_1)$
show that $X_1 \perp X_2$.
After several hours I can't seem to find a solution for this. I know that $X_1$ and $X_2$ are considered independent from each other if the following statement holds:
$p(x_1,x_2) = p(x_1)p(x_1)$
Based on the symmetry property of $p(x_1,x_2) = p(x_2,x_1)$ I can also rewrite it as follows:
- $p(x_1|x_2)p(x_2) = p(x_1)p(x_2)$
- $p(x_2|x_1)p(x_1) = p(x_1)p(x_2)$
I tried to reformulate the factorization by applying the sum and product rule in order to show independence but I'm stuck. Somehow I have to isolate $p(x_1), p(x_2)$ in order to check the independence statements mentioned above. That's the most promising approach I tried so far:
$p(x_1,x_4|x_2)p(x_2,x_3|x_1) = \frac{p(x_2|x_1,x_4)p(x_1,x_4)}{p(x_2)} \frac{p(x_1|x_2,x_3)p(x_2,x_3)}{p(x_1)}$
$p(x_1,x_4|x_2)p(x_2,x_3|x_1) = \frac{p(x_2|x_1,x_4)p(x_4|x_1)p(x_1)}{p(x_2)} \frac{p(x_1|x_2,x_3)p(x_3|x_2)p(x_2)}{p(x_1)}$
$p(x_1,x_4|x_2)p(x_2,x_3|x_1) = p(x_2|x_1,x_4)p(x_4|x_1)p(x_1|x_2,x_3)p(x_3|x_2)$
I also considered marginalizing $p(x_1)$ and $p(x_2)$ but the following approach does not seem to help either and I'm also not sure if that's the way to go.
$p(x_1) = \sum_{x_2,x_3} p(x_1|x_2,x_3)p(x_2,x_3)$
$p(x_2) = \sum_{x_1,x_4} p(x_2|x_1,x_4)p(x_1,x_4)$
A simple hint to this problem would be of great help and would be much appreciated.
Since $p(x_1,x_2,x_3,x_4)=p(x_1,x_4|x_2){p(x_2,x_3|x_1)}$,
then integrate out $x_3$: $\int_{x_3}p(x_1,x_2,x_3,x_4)=p(x_1,x_2,x_4)=\int_{x_3}p(x_1,x_4|x_2)p(x_2,x_3|x_1)=p(x_1,x_4|x_2)p(x_2|x_1)$.
Multiply $p(x_2)$ to both sides, $p(x_1,x_2,x_4)p(x_2)={p(x_1,x_4|x_2)}p(x_2)p(x_2|x_1)$, that is exactly what we want: $p(x_2)=p(x_2|x_1)$.