Prove d-separation path is blocked as long as NOT conditioning on the collider

Question

Getting into the causality tools. Suppose I have a causal graph $X\to R\to T\leftarrow U.$

I can work out that $R$ and $U$ are independent; i.e., $P(r, u) = P(r)\,P(u).$

Also $X$ and $T$ are conditionally independent given $R;$ i.e., $P(x|t,r) = P(x|r).$

I think there must be a way to prove:

1. independence of $X$ and $U;$ i.e., $P(x|u) = P(x)$
2. independence of $X$ and $U$ conditional on $R;$ i.e., $P(x|u,r) = P(x|r).$

Appreciate any help!

Answer 1

From your diagram, the collider blocks the causal pathway from $X$ to $U,$ and there is no back-door path. Hence they are independent. If you condition on $R,$ the path is still blocked by the collider at $T.$ The only situation in which you would get dependence is if you conditioned on the collider $T.$