If $R \perp A$ and $R \perp A | Y$ then $A \perp Y$ or $R \perp Y$ Let $R$, $Y$ and $A$ random variables. I need to prove the following statement:
$$\text{If } R \perp A \text{ and } R \perp A | Y \text{ then } A \perp Y \text{ or } R \perp Y$$
My initial hunch was going in the direction of proving the following two statements without much progress so far:
$$R \perp A \text{ and } R \perp A | Y \text{ and } A \not\!\perp\!\!\!\perp Y \implies R \perp Y$$
$$R \perp A \text{ and } R \perp A | Y \text{ and } R\not\!\perp\!\!\!\perp Y \implies A \perp Y$$
 A: This is not true in general. Let $Y$ take on values in $\{0,1,2\}$, each equally likely, and $R$ and $A$ both take on values on $\{0,1\}$. Suppose $R$ and $A$ are conditionally independent of $Y$, with
$$
\begin{array}{lcr}
a_0=P(A=1|Y=0)=1/2 &\qquad& r_0=P(R=1|Y=0)=1/2\\ 
a_1=P(A=1|Y=1)=1/4 && r_1=P(R=1|Y=1)=1/3 \\ 
a_2=P(A=1|Y=2)=1/2 &&  r_2=P(R=1|Y=2)=1/6
\end{array}
$$
Note that neither $A$ nor $R$ is independent of $Y$. However, $A$ and $R$ are independent, as
\begin{align}
P(A=R=1)
  &=\frac13\Big(P(A=R=1|Y=0)+P(A=R=1|Y=1)
\\&\qquad+P(A=R=1|Y=2)\Big)
\\&=\frac13\Big((1/2)(1/2)+ (1/4)(1/3)+(1/2)(1/6)\Big)
\\&=5/36
\\\\P(A=1)P(R=1)
  &=\frac13(a_0+a_1+a_2)\cdot \frac13(r_0+r_1+r_2)
\\&=\frac13(1/2+1/4+1/2)\cdot \frac13(1/2+1/3+1/6)
\\&=5/36
\end{align}

However, if $A,R,Y$ are events instead of random variables, the result is true.

Proposition: If $A, B$ and $C$ are events with $A\perp B$ and $(A\perp B)|C$, then either $A\perp C$ or $B\perp C$. 

Proof: Supposing $(A\perp B)|C$, let $a=P(A|C)$, $a'=P(A|C')$, similarly for $b,b'$, and let $c=P(C)$. You can show that
\begin{align}
P(AB)-P(A)P(B)
&=abc+a'b'(1-c)-\big(ac+a'(1-c)\big)\big(bc+b'(1-c)\big)
\\&=(a-a')(b-b')c(1-c).
\end{align}
Therefore, assuming $A\perp B$, the above quantity is zero, implying either $a=a'$, in which case $A\perp C$, or $b=b'$, in which case $B\perp C$, or $c\in\{0,1\}$, in which case both occur. $\square$
