Let $X,Y$ be an independent random variables. $R(\pi,X)$ be a posteriori distribution for a priori distribution $\pi$ and sample $X$. Show that $$R(\pi_0,(X,Y))=R(R(\pi_0,X),Y)$$
My work so far
Firsly let's write :
$R(\pi_0,(X,Y))=\frac{P((X,Y)|\pi_0)\cdot P(\pi_0)}{P(X)P(Y)}=\frac{P(X\cap\pi_0)P(Y\cap\pi_0)P(\pi_0)}{P(X)P(Y)}$. I used fact that $X,Y$ are idenpendent so $P(X,Y)=P(X)P(Y)$
Now the second expression :
$R(\pi_0,X)=\frac{P(X|\pi_0)P(\pi_0)}{P(X)}$
$$R(R(\pi_0,X),Y)=\frac{P(Y|\frac{P(X|\pi_0\cdot P(\pi_0)}{P(X)})\cdot (\frac{P(X|\pi_0)\cdot P(\pi_0)}{P(X)})}{P(Y)}$$.
So if the equalitty is true i have to show that $$\frac{P(Y|\frac{P(X|\pi_0\cdot P(\pi_0)}{P(X)})\cdot (\frac{P(X|\pi_0)\cdot P(\pi_0)}{P(X)})}{P(Y)}=\frac{P(X\cap\pi_0)P(Y\cap\pi_0)P(\pi_0)}{P(X)P(Y)} \Rightarrow$$
$$P(Y|\frac{P(X\cap\pi_0)}{P(X)})\cdot P(X| \pi_0)=P(X\cap \pi_0)P(Y\cap \pi_0)$$.
So the left side of equality is : $$P(Y|P(\pi_0 | X))=P(Y)P(\pi_0|X)P(X|\pi_0)=P(Y)P(\pi_0 \cap X) $$ But $P(Y) \neq P(Y\cap \pi_0)$.
Can you please show me mistake in my justification ?